CONTENTS

VECTOR & THREE DIMENSIONAL GEOMETRY

Topic Page No.

Theory 01 34

Exercise # 1 Part - I : Subjective Question 35 45

Part - II : Only one option correct type
Part III : Match the column

Exercise - 2 Part - I : Only one option correct type 45 55

Part - II : Numerical value questions
Part - III : One or More than one options correct type
Part - IV : Comprehension

Exercise - 3 56 68
Part - I : JEE(Advanced) / IIT-JEE Problems (Previous Years)
Part - II : JEE(Main) / AIEEE Problems (Previous Years)

Answer Key 69 71

High Level Problems (HLP) 72 75

Answer Key (HLP) 75 75

JEE (ADVANCED) SYLLABUS

Vector : Addition of vectors, scalar multiplication, dot and cross products, scalar triple products and their
geometrical interpretations.
Three dimensions : Direction cosines and direction ratios, equation of a straight line in space, equation of a
plane, distance of a point from a plane.

JEE (MAIN) SYLLABUS

Vector : Vectors and scalars, addition of vectors, components of a vector in two dimensions and three
dimensional space, scalar and vector products, scalar and vector triple product.
3-D : Coordinates of a point in space, distance between two points, section formula, direction ratios and
direction cosines angle between two intersecting lines. Skew lines, the shortest distance between them and its
equation. Equations of a line and a plane in different forms, intersection of a line and a plane, coplanar lines.

© Copyright reserved.
All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the
enrolled student of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.
Vector & Three Dimensional Geometry

What if angry vectors veer Round your sleeping head, and from. There's never need to fear Violence of the poor world's abstract storm. ........ Warren,
Robert PennNature is an infinite sphere of which the centre is everywhere and the circumference nowhere ........ Pascal, Blaise

Vector quantities are specified by definite magnitude and definite direction. A vector is generally
represented by a directed line segment, say AB . A is called the initial point and B is called the
terminal point. The magnitude of vector AB is expressed by AB .
:
A vector of zero magnitude i.e. which has the same initial and terminal point, is called a zero vector. It
is denoted by O. The direction of zero vector is indeterminate.
:
A vector of unit magnitude in the direction of a vector a is called unit vector along a and is denoted by
a
a , symbolically a .
|a|

Two vectors are said to be equal if they have the same magnitude, direction and represent the same
physical quantity.
:

Two vectors are said to be collinear if their directed line segments are parallel irrespective of their
directions. Collinear vectors are also called parallel vectors. If they have the same direction( )

they are named as like vectors but if they have opposite direction ( ) then they are named as
unlike vectors.
Symbolically, two non-zero vectors a and b are collinear if and only if, a b , where R
a1 a a
a b a1i a2 j a3k = b1i b2 j b3k a1 = b1, a2 = b2, a3 = b3 = 2 = 3 (= )
b1 b2 b3
a1 a 2 a 3
Vectors a = a1 i + a2 j + a3k and b = b1 i + b2 j + b3 k are collinear if = =
b1 b 2 b3

Note : If a,b are non zero, non collinear vectors, such that xa yb x 'a y 'b x x' , y y' ,
(where

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 1
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 1 : Find unit vector of i 2j 3k
Solution : a = i 2j 3k
2 2 2
if a = ax i + ay j + azk then | a | = ax ay az
a 1 2 3
|a| = 14 a = = i j + k
|a| 14 14 14

Example # 2 : a = (x + 1) i (2x + y) j + 3 k and b = (2x 1) i + (2 + 3y) j + k

find x and y for which a and b are parallel.
x 1 3
Solution : a and b are parallel = = x = 4/5 , y = 19/25
2 3y 1

:
A given number vectors are called coplanar if their line segments are all parallel to the same plane.
Note that two vectors are always coplanar .

If a is a vector and m is a scalar, then m is a vector parallel to a whose magnitude is m times that
of a . This multiplication is called scalar multiplication. If a and a are vectors and m, n are scalars,
then :
(i) m (a) (a) m ma (ii) m (na) n(ma) (mn)a
(iii) (m n) a ma na (iv) m (a b) ma mb

Self Practice Problems :

(1) Given a regular hexagon ABCDEF with centre O, show that
(i) OB OA = OD OE (ii) EA = 2 OB + OF (iii) AD + EB + FC = 4 AB
(2) Let ABCDEF be a regular hexagon. If AD = x BC and CF = y AB then find xy.
(3) The sum of the two unit vectors is a unit vector. Show that the magnitude of the their difference
is 3 .
Answers : (2) 4

(i) If two vectors a and b are represented by OA and OB , then their sum a b is a
vector represented by OC , where OC is the diagonal of the parallelogram OACB.
(ii) a b b a (commutative) (iii) (a b) c a (b c) (associative)
(iv) a 0 a 0 a (v) a ( a) 0 ( a) a
(vi) |a b| |a| |b| (vii) | a b | || a | | b ||

Example # 3 : The two sides of ABC are given by AB = 2 i + 4 j + 4 k , AC = 2 i + 2 j + k . Then find the
length of median through A.
Solution : Let D be mid point of BC
1 A
In ABC, AB + BD = AD AB + BC = AD
2
AB (AB BC)
= AD
2
AB AC 4i 6 j 5k 77
= AD | AD | = = B D C
2 2 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 2
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Example # 4 : In a triangle ABC, D, E, F are the mid-points of the sides BC, CA and AB respectively then
A
prove that, AD = ( BE + CF ).
1
Solution : AD = 3 GD = 3. (GB GC) where D is mid-point of BC F 2
G
E
2
3 2 2 1
= EB FC = ( BE + CF )
2 3 3 B D C

:
Let O be a fixed origin, then the position vector of a point P is the vector OP . If a and b are position
vectors of two points A and B, then
AB = b a = position vector (p.v.) of B position vector (p.v.) of A.

Distance between the two points A (a) and B (b) is AB = a b

If a and b are the position vectors of two points A (x1, y1, z1) and B(x2, y2, z2),
na mb
then the p.v. of a point R which divides AB in the ratio m: n is given by r
m n
nx1 mx2 ny1 my2 nz1 mz2
Here R = , ,
n m n m n m

a b x1 x 2 y1 y 2 z1 z2
Note : Position vector of mid point M of AB is . Here M = , ,
2 2 2 2
Example # 5 : Let O be the centre of a regular pentagon ABCDE and OA = a .
Then AB 2BC 3CD 4DE 5EA =
Solution : OA a,OB b,OC c,OD d,OE e
AB 2BC 3CD 4DE 5EA = b a +2 c b +3 d c +4 e d +5 a e

= 5a a b c d e = 5a , (since a b c d e = 0)

Example # 6 : In a triangle ABC, D and E are points on BC and AC respectively, such that BD = 2DC and
BP
AE = 3EC. Let P be the point of intersection of AD and BE. Find using vector method.
PE
Solution : Let the position vectors of points B and C be respectively b and c referred to A as origin of
reference.
BP PD A(0)
Let = and =
PE AP 3
3 c 2c b 1 E
b
2c b 3 4 3 1
AD = , AE = c AP = = P
3 4 1 1 2 µ 1
C(c)
B(b) D
comparing the coefficient of b & c
1 1 3 2
= and =
1 3( 1) 4( 1) 3( 1)
solving above equations we get = 8/3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 3
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Self Practice Problems

(4) Express vectors BC , CA and AB in terms of the vectors OA , OB and OC
(5) If a, b are position vectors of the points(1, 1),( 2, m), find the value of m for which a and
b are collinear.
(6) The vertices P, Q and S of a PQS have position vectors p, q and s respectively.
(i) Find the position vector of t of point T in terms of p, q and s , such that
ST : TM = 2 : 1 and M is mid-point of PQ.
(ii) If the parallelogram PQRS is now completed. Express , the position vector
of the point R in terms of p, q and s

(7) In a quadrilateral ABCD, AB = p , BC = q , DA = p q . If E is the mid point of BC and F is

4
the point on DE such that DF = DE. Show that the points. A,F,C are collinear.
5

(8) Point L, M, N divide the sides BC, CA, AB of ABC in the ratios 1 : 4, 3 : 2, 3 : 7 respectively.
Prove that AL + BM + CN is a vector parallel to CK , when K divides AB in the ratio 1 : 3.

Answers : (4) BC OC OB , CA OA OC , AB OB OA (5) m=2

1
(6) (i) t = (p q s) (ii) r = (q p s)
3

Distance between any two points (x1, y1, z1) and (x 2, y2, z2) is given as (x1 x2 )2 (y1 y 2 )2 (z1 z2 )2

Let PA, PB and PC are distances of the point P(x, y, z) from the coordinate axes OX, OY and OZ
respectively then PA = y2 z2 , PB = z2 x2 , PC = x 2 y2

Example # 7 : Find the locus of a point which is equidistance from A (0,2,3) and B (2, 2, 1).
Solution : let P (x, y, z) be any point which is equidistance from A (0,2,3) and B (2, 2, 1)
PA = PB
2 2 2 2 2 2
= x 2y z+1=0

Example # 8 : Find the locus of a point which moves such that the sum of its distances from points A(0, 0, )
and B(0, 0, ) is constant.
Solution : Let the variable point whose locus is required be P(x, y, z)
Given PA + PB = constant = 2a (say)
(x 0)2 (y 0)2 (z )2 + (x 0)2 (y 0)2 (z )2 = 2a
x2 y2 (z )2 = 2a x2 y2 (z )2
x2 + y2 + z2 + 2
+ 2z = 4a2 + x2 + y2 + z2 + 2
2z 4a x2 y2 (z )2
z2 2
4z 4a2 = 4a x2 y2 (z )2 + a2 2z = x2 + y2 + z2 + 2
2z
a2
2
x2 y2 z2
or, x2 + y2 + z2 1 = a2 2
+ + =1
a2 a2 2
a2 2
a2
This is the required locus.
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 4
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Self practice problems :
(9) One of the vertices of a cuboid is (0, 2, 1) and the edges from this vertex are along the
positive x-axis, positive y-axis and positive z-axis respectively and are of lengths 2, 2, 3
respectively find out the vertices.

(10) Show that the points (0, 4, 1), (2, 3, 1), (4, 5, 0) and (2, 6, 2) are the vertices of a square.

(11) Find the locus of point P if AP2 BP2=20, where A (2, 1, 3) and B ( 1, 2, 1).

Answers : (9) (2,2, 1), (2, 4, 1), (2, 4, 2), (2, 2, 2), (0, 2, 2), (0, 4, 2), (0, 4, 1).
(11) x + y + 2z = 6

x1 x2 x 3 y1 y2 y 3 z1 z2 z3
G , ,
3 3 3

ax1 bx2 cx3 ay1 by2 cy3 az1 bz2 cz3

I , , Where AB = c, BC = a, CA = b
a b c a b c a b c

Example # 9 : Show that the points A(2, 3, 4), B( 1, 2, 3) and C( 4, 1, 10) are collinear. Also find the
ratio in which C divides AB.
Solution : Given A (2, 3, 4), B ( 1, 2, 3), C ( 4, 1, 10).

Let C divide AB internally in the ratio k : 1, then

k 2 2k 3 3k 4 k 2
C , , = 4 3k = 6 k= 2
k 1 k 1 k 1 k 1
2k 3 3k 4
For this value of k, = 1, and = 10
k 1 k 1
Since k < 0, therefore C divides AB externally in the ratio 2 : 1 and points A, B, C are collinear.

Example # 10 :The vertices of a triangle are A(5, 4, 6), B(1, 1, 3) and C(4, 3, 2). The internal bisector of BAC
meets BC in D. Find AD.
Solution : AB = 42 52 32 5 2
AC = 12 12 42 3 2
Since AD is the internal bisector of BAC
BD AB 5
D divides BC internally in the ratio 5 : 3
DC AC 3
5 4 3 1 5 3 3( 1) 5 2 3 3 23 12 19
D , , or, D= , ,
5 3 5 3 5 3 8 8 8
2 2 2
23 12 19 1530
AD = 5 4 6 = unit
8 8 8 8

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 5
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 11 : If the points P, Q, R, S are (4, 7, 8), ( 1, 2, 1), (2, 3, 4) and (1,2,5) respectively, show that PQ
and RS intersect. Also find the point of intersection.
Solution : Let the lines PQ and RS intersect at point A.
4 2 7 8
Let A divide PQ in the ratio : 1, ( 1) then A , , . .... (1)
1 1 1
k 2 2k 3 5k 4
Let A divide RS in the ratio k : 1, then A , , ..... (2)
k 1 k 1 k 1

From (1) and (2), we have,

4 k 2
k + 4k + 4 = k + 2 + k + 2 2 k+3 3k 2=0 .....(3)
1 k 1
2 7 2k 3
2 k 2 + 7k + 7 = 2 k + 3 + 2k + 3 4 k+5 5k 4 = 0 .....(4)
1 k 1
8 5k 4
..... (5)
1 k 1
Multiplying equation (3) by 2, and subtracting from equation (4), we get + k = 0 or, =k
Putting = k in equation (3), we get 2 2 + 3 3 2=0 =1=k
Clearly = k = 1 satisfies eqn. (5), hence our assumption is correct.
1 4 2 7 1 8 3 5 9
A , , or, A , , .
2 2 2 2 2 2

Self practice problems :

(12) Find the ratio in which yz plane divides the line joining the points A (4, 3, 5) and B (7, 4, 5).

(13) Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 2, 1) to the line
joining the point B(1, 4, 6) and C(5, 4, 4).

8
(14) Two vertices of a triangle are (4, 6, 3) and (2, 2, 1) and its centroid is , 1, 2 . Find the
3
third vertex.

3 7 1
(15) Show that , , is the circumcentre of the triangle whose vertices are
2 2 2
A (2, 3, 2), B (0, 4, 1) and C (3, 3, 0) and hence find its orthocentre.

Answers : (12) 4 : 7 Externally (13) (3, 4, 5) (14) (2, 5, 2) (15) (2, 3, 2)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 6
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(i) : Let be the angles which a directed line makes with the positive
directions of the axes of x, y and z respectively, then cos , cos cos are called the
direction cosines of the line. The direction cosines are usually denoted by , m, n.

Thus = cos , m = cos , n = cos .

(ii) If , m, n be the direction cosines of a line, then 2
+ m2 + n2 = 1

(iii) : Let a, b, c be proportional to the direction cosines , m, n then a, b, c are

called the direction ratios.
If a, b, c, are the direction ratios of any line L, then ai bj ck will be a vector parallel to the
line L.
If , m, n are direction cosines of line L, then i + m j + n k is a unit vector parallel to the line
L.

(iv) If , m, n be the direction cosines and a, b, c be the direction ratios of a vector, then ( , m, n)
a b c a b c
= , , or , ,
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
a b c a b c a b c a b c a b c a b2 c2

(v) If the coordinates P and Q are (x 1, y1, z1) and (x 2, y2, z2), then the direction ratios of line PQ are,
x 2 x1
a = x2 x1, b = y2 y1 & c = z2 z1 and the direction cosines of line PQ are = ,
| PQ |
y 2 y1 z z1
m= and n = 2 .
| PQ | | PQ |

cos3
Example # 12 : If a line makes angle , , with the co-ordinate axes. Then find the value of .
cos
cos3 4cos3 3cos
Solution : = = (4cos2 3)
cos cos
= 4(cos2 + cos2 + cos2 ) 3 3 3=4 9= 5 Ans. 5

Example # 13 : If the direction ratios of two lines are given by mn 4n + 3 m = 0 and + 2m + 3n = 0 then
find the direction ratios of the lines.
m n m n
Solution : Eliminating we have m = ± 2 n = = & = =
2 1 1
Ans. ( ), , 2 where R {0}

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 7
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Self practice problems:

(16) Find the direction cosines of a line lying in the xy plane and making angle 30° with x-axis.
(17) A line makes an angle of 60° with each of x and y axes, find the angle which this line makes
with z-axis.
(18) A plane intersects the co-ordinates axes at point A(2, 0, 0), B(0, 4, 0), C(0, 0, 6) ; O is origin.
Find the direction ratio of the line joining the vertex B to the centroid of face ABC.
3 1 2 8
Answers : (16) = ,m=± , n=0 (17) 45° (18) , ,2
2 2 3 3

It is the smaller angle formed when the initial points or the terminal points of the two vectors are brought
together. Note that 0º 180º .

a . b a b cos , (0 )

Note: (a) If is acute, then a . b > 0 and if is obtuse, then a . b < 0.

(b) a . b 0 a b (a 0, b 0)

(c) Maximum value of a . b is | a | | b | (d) Minimum value of a . b is |a | |b |

B(b)
As shown in Figure, projection of vector OB (or b ) along vector OA ( or a )
b.a
is OL = | b | cos b.a
a
L
O
Properties of Dot Product | b | cos b.a
A(a)

a . b
(i) Projection of a on b
|b|
(ii) a. b b . a (commutative)
(iii) a . (b c) a . b a . c (distributive)
(iv) (ma) . b = a . (mb) = m (a . b) , where m is a scalar.
(v) i . i = j . j = k . k = 1; i . j = j . k = k . i = 0
2
(vi) a . a a a2

(vii) If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k , then a . b = a1b1 + a2b2 + a3b3

2 2 2 2 2 2
a a1 a2 a3 , b b1 b2 b3

(viii) a b = | a |2 | b |2 2 | a || b | cos , where is the angle between the vectors

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 8
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Example # 14 : Find the value of p for which the vectors a 3i 2j 9k and b i pj 3k are
(i) perpendicular (ii) parallel
Solution : (i) a b a . b =0 3i 2 j 9k . i pj 3k = 0

3 + 2p + 27 = 0 p= 15
(ii) vectors a = 3i 2j 9k and b = i pj 3k are parallel iff
3 2 9 2 2
= = 3= p=
1 p 3 p 3
Example # 15 : If a , b , c are three vectors such that each is inclined at an angle /3 with the other two and
| a | = 1, | b | = 2, | c | = 3, then find the scalar product of the vectors 2 a + 3 b 5 c and
4a 6 b + 10 c .
Solution : Dot products is 82 a 182 b 502 c + a . b ( 12 + 12) + b . c (30 + 30) + c . a (20 20)

=8 18 (4) 50(9) + 60 2.3cos = 188 522 = 334

3

Example # 16 : Find the values of x for which the angle between the vectors
a = 2x2 i + 4x j + k and b = 7 i 2 j + x k is obtuse.

a . b
Solution : The angle between vectors a and b is given by cos =
| a || b |

a . b
Now, is obtuse cos <0 <0 a . b <0 [ | a |, | b | 0 ]
| a || b |
1
14x2 8x + x < 0 7x (2x 1) < 0 x(2x 1) < 0 0<x <
2
Hence, the angle between the given vectors is obtuse if x (0, 1/2)

Example # 17 : If a = i + j + k and = 2 a j + 3 k , then find

(i) Component of b along a . (ii) Component of b in plane of a & b but to a .

a . b
Solution : (i) Component of b along a is a ; Here a . b = 2 1 + 3 = 4 and | a |2 = 3
| a |2

a . b 4 4
Hence a = a = (i+ j + k)
| a |2 3 3

a . b 1
(ii) Component of b in plane of a & b but to a is b a.= 2i 7j 5k
| a |2 3

Example # 18 : Find the projection of the line joining A(1, 2, 3) and B( 1, 4, 2) on the line having direction ratios
2, 3, 6.
Solution : AB 2i 2j k B
A
4 6 6 8
Projection of AB 2i 2j k on 2i 3 j 6k is
4 9 36 7 90° 90°
P L M Q
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 9
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Self Practice Problems :

|a b|
(19) If a and b are unit vectors and is angle between them, prove that tan = .
2 |a b|

(20) Find the values of x for which the angle between the vectors a = 2x i 4j 3k and
b = is 90º

(21) if a, b, c are the pth, qth, rth terms of a HP then find the angle between the vectors
1 1 1
= (q r) i + (r q) j + (p q) k and v = i + j + k.
a b c

(22) The points O, A, B, C, D are such that OA a , OB b , OC 2a 3b , OD a 2b

Given that the length of OA is three times the length of OB . Show that BD and AC are
perpendicular.

(23) ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that
AB2 + AC2 + AD2 = GB2 + GC2 + GD2 + 3GA2

(24) A (2, 3, 2), B (1, 5, 4,), C(0, 1, 2) D (4, 0, 3). Find the projection of line segment AB on CD
line.

(25) The projections of a directed line segment on co-ordinate axes are 3, 4, 12. Find its length and
direction cosines.
2 2
Answers : (20) x = 12/7 (21) /2 (24) (25) 13, , ,
3 13 13 13

(i) If a , b are two vectors and is the angle between them, then a x b a b sin n , where n is
the unit vector perpendicular to both a and b such that a , b and n forms a right handed screw
system.
(ii) Geometrically a x b = area of the parallelogram whose two adjacent sides are represented by

a and b .

(iii) axb bxa (not commutative)

(iv) (m a) b =a (m b) = m (a b) , where m is a scalar.
(v) a x (b c) (a xb) (a xc) (distributive)
(vi) a b 0 a and b are parallel (collinear) (a 0,b 0) i.e. a K b , where K is a scalar.

(vii) i i j j k k 0; i j k, j k i, k i j
i j k
(viii) If a = a1 i +a2 j + a3 k and b = b1 i + b2 j + b3 k , then a b a1 a2 a3
b1 b2 b3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 10
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
r(a b)
(x) a and b is
|a b|

(xii) If a, b and c are the position vectors of 3 points A, B and C respectively, then the vector area of
1
ABC = (a b b c c a) . The points A, B and C are collinear if
2
a b b c c a 0
1
(xiii) Area of any quadrilateral whose diagonal vectors are d1 and d2 is given by d1 d2
2

Example # 19 : Given a = i + j k, b = i + 2 j + k and c = i +2j k , then find a unit vector

perpendicular to both a + b and b + c .
Solution : × is to both and
a +b = 3 j , b +c = 2 i + 4 j
3 j × ( 2 i + 4 j ) is to both or 6 j × i = 6k hence a unit vector is k .

Example # 20 : If =2i +3j k , = i+2j 4k , = i + j + k , then find value ( × ).( × .

Solution : × = 10 i + 9 j + 7 k and × =4 i 3 j k their dot product = 40 27 7 = 74

Example # 21 : Let OA = a + 3 b , OB = 5 a + 4b and OC = 2 a b where O is origin. Let p denote the area

of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as
p
adjacent sides. Find .
q
Solution : We have, p = Area of the quadrilateral OABC
1 1 1
p = | OB AC | = | OB (OC OA) | p = | (5a 4b) (4b a) |
2 2 2
1
p= | 20(a b) 4(b a) | = 12 | a b | .....(i)
2
and q = Area of the parallelogram with OA and OC as adjacent sides
q = | OA OC | = = 7| a b | .....(ii)
p 12
From (i) and (ii), we get =
q 7
Self Practice Problems :

(26) If p and q are unit vectors forming an angle of 30º. Find the area of the parallelogram
having a p 2q and b 2p q as its diagonals.

(27) ABC is a triangle and EF is any straight line parallel to BC meeting AC, AB in E, F
respectively. If BR and CQ be drawn parallel to AC, AB respectively to meet EF in R and Q
respectively, prove that ARB = ACQ.

Answers : (26) 3/4 sq. units

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 11
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(i) Vector equation: Vector equation of a straight line passing through a fixed point with position
vector a and parallel to a given vector b is r = a + b where is a scalar.

(ii) Vector equation of a straight line passing through two points with position vectors a & b is
r= a + (b a ).

(iii) The equation of a line passing through the point (x 1, y1, z1) and having direction ratios a, b, c
x x 1 y y1 z z1
is = = = r. This form is called symmetric form. A general point on the line
a b c
is given by (x + ar, y + br, z + cr).

(iv) The equation of the line passing through the points (x1, y1, z1) and (x2, y2, z2) is
x x1 y y1 z z1
= =
x 2 x1 y 2 y1 z2 z1

(v) Reduction of cartesian form of equation of a line to vector form & vice versa
x x1 y y1 z z1
= = = r (x1 i + y1 j + z1 k ) + (a i + b j + c k ).
a b c

(vi) The equations of the bisectors of the angles between the lines r = a + b and
r =a + c are : r = a + t b c and r = a + p c b .

Example # 22 : Find the equation of the line through the points (4, 5, 8) and ( 1, 2, 7) in vector form as well
as in cartesian form.
Solution : Let A (4, 5, 8) , B ( 1, 2, 7)

Now a = OA = 4i 5 j + 8k and b = OB = i + 2 j + 7k

Equation of the line through A( a ) and B( b ) is r = a + t ( b a )

or r =4i 5 j + 8k + t ( 5 i + 7 j k) ..... (1)

Equation of AB in cartesian form is

Example # 23 : Find the equation of the line passing through point (1, 0, 2) having direction ratio 3, 1, 5.
Prove that this line passes through (4, 1, 7).

Solution. equation of line is ,

Now 1 , so line passes through point (4, 1,7)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 12
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 24 : Find the equation of the line drawn through point ( 1, 7, 0) to meet at right angles the

line
2 1 2

Solution : Given line is ..... (1)

2 1 2
Let P ( 1, 7, 0)
Co-ordinates of any point on line (1) may be taken as Q (2r + 2, r 3, 2r + 1)
Direction ratios of PQ are 2r + 3, r 10, 2r + 1
Direction ratios of line AB are 2, 1, 2
Since PQ AB
2(2r + 3) + ( r 10) ( 1) + 2 (2r + 1) = 0 r= 2
Therefore, direction ratios of PQ are 1 , 8, 3

Equation of line PQ is
1 8 3
Example # 25 : A line passes through the point 3i and is parallel to the vector i j k and another line
passes through the point i j and is parallel to the vector i k , then find the point of
intersection of lines.
Solution : A point on the first line is 3i s( i j k) ....(i)
A point on the second line is i j t(i k) ....(ii)
At the point of intersection (i) and (ii) are same.
3 s = 1 + t, s = 1, s = t s =t=1
hence the point is 3i ( i j k) = 2i j k Ans. (2,1,1)
Self practice problems:
x 3 y 1 z 7
(28) Find the equation of the line parallel to line and passing through the point

(2, 3, 2).

Answers : (28)
4 1 5

P(x1,y1,z1)

Q(x2,y2,z2) F(x ,y ,z ) x x2 y y2 z z2
a b c

R(x ,y ,z )

x x2 y y2 z z2
Let L = = is a given line and P(x1, y1, z1) is given point as shown in figure.
a b c
Let F(x , y , z ) = (ar+x2, br+ y2, cr+ z2) be the foot of the point P (x1, y1, z1) with respect to the line
L. Apply PF.(ai bj ck) 0 .N in (1) we get F

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 13
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Now for calculating the reflection R(x , y , z ) of the point P (x1, y1, z1) with respect to the line L, apply
midpoint formula (midpoint of P & R is F)
PF.(ai bj ck)
PF = PQ sin =
a2 b2 c2

x 1 y 3 z 2
Example # 26 : Find the length of the perpendicular from P (2, 3, 1) to the line .
2 3 1
Solution : Co-ordinates of any point on given line may be taken as Q (2r 1, 3r + 3, r 2)
Direction ratios of PQ are 2r 3, 3r + 6, r 3
Direction ratios of AB are 2, 3, 1
Since PQ AB
15
2 (2r 3) + 3 (3r + 6) 1( r 3) = 0 14r + 15 = 0 r=
14
22 3 13 531
Q , , PQ = units.
7 14 14 14

Self practice problems :

(29) Find the length and foot of perpendicular drawn from point (2,3,4) to the line
. Also find the image of the point in the line.

Answers : (29) 3 5,N (2, 6, 2), (2, 9, 8)

If two lines have direction ratios a 1, b1, c1 and a 2, b2, c2 respectively, then we can consider two vectors
parallel to the lines as a1 i + b1 j + c1 k and a2 i + b2 j + c2 k and angle between them can be given as.
a1a2 b1b2 c1c 2
cos = .
2 2 2 2
a
1 b1 c
1 a 2 b22 c 22
(i) The lines will be perpendicular if a1a2 + b1b2 + c1c2 = 0
a b c
(ii) The lines will be parallel if 1 = 1 = 1
a2 b2 c2

Example # 27 : What is the angle between the lines whose direction cosines are
3 1 3 3 1 3
, , and , ,
4 4 2 4 4 2
Solution : Let be the required angle, then cos = 1 2
+ m1m2 + n1n2
3 3 1 1 3 3 3 1 3 1
= . = = 120°,
4 4 4 4 2 2 16 16 4 2
Example # 28 : P is a point on line r = 5i 7 j 2k s(3i j k) and Q is a point on the line
r = +t 3i 2j 4k . If PQ is parallel to the vector, 2i 7j 5k , find P and Q
Solution : is parallel to 2i 7j 5k
8i 4j 8k t( 3i 2j 4k) s 3i j k =2 = 8 3t 3s
7 = 4 + 2t + s 5 = 8 + 4t s
solving, =t=s= 1
P = (5, 7, 2) (3, 1, 1) = (2, 8, 3) Q = ( 3, 3, 6) ( 3, 2, 4) = (0, 1, 2)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 14
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Self practice problems :
(30) Find the angle between the lines whose direction cosines are given by + m + n = 0 and
2
+m 2
n =0
2

(31) Let P (6, 3, 2), Q (5, 1, 4), R (3, 3, 5) are vertices of a find Q.

(32) Show that the direction cosines of a line which is perpendicular to the lines having
directions cosines 1
m1 n1 and 2
m2 n2 respectively are proportional to
m 1 n2 m2n1 , n1 2 n2 1, 1m2 2m1

Answers : (30) 60° (31) 90°

:
Lines in space which do not intersect and are also not parallel are called skew line.

If lines r = a + p & r =b + q are skew lines then (b a) . (p x q) 0

If lines are not skew lines then they are coplanar which means if (b a) . (p x q) 0 , then lines are coplanar.

(i) Shortest distance (d) between lines r = a + p & r = b + q

(b a) . (p x q)
is d
px q

x y z x ' y ' z '

(ii) Shortest distance (d) between two skew lines = = and
m n ' m' n'

' ' '

is d = m n (mn m n)2
' m' n'

For Skew lines the direction of the shortest distance would be perpendicular to both the lines.
If d = 0, the lines are coplanar
(iii) Shortest distance between two parallel lines r1 a1 Kb and r2 a2 Kb , is given by

b x (a 2 a1 )
d
b

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 15
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 29 : Find the shortest distance and the vector equation of the line of shortest distance between the
lines given by r 3i 8j 3k 3i j k and r 3i 7j 6k 3i 2j 4k
x 3 y 8 z 3
Solution : Equation of given lines in cartesian form is = (say L 1)
3 1 1
x 3 y 7 z 6
and = (say L 2)
3 2 4
Let L on L1 is (3 + 3, + 8, + 3) and M on L2 is ( 3 3, 2 7, 4 + 6)
Direction ratios of LM are 3 + 3 + 6, 2 + 15, 4 3.
Since LM L1
3 (3 + 3 + 6) 1 ( 2 + 15) + 1 ( 4 3) = 0 or, 11 + 7 = 0 ..... (1)
Again LM CD
3 (3 + 3 + 6) + 2 ( 2 + 15) + 4 ( 4 3) = 0 or, 7 29 = 0 ..... (2)

A L B
90°

90°
C M D

Solving (1) and (2), we get = 0, =0 L (3, 8, 3), M ( 3, 7, 6)

2 2 2
Hence shortest distance LM = (3 3) (8 7) (3 6) = 270 = 3 30 units

Vector equation of LM is r 3i 8j 3k t 6i 15j 3k

x 3 y 8 z 3
Note : Cartesian equation of LM is .
6 15 3
Self practice problems:
x 1 y 2 z 3 x 2 y 4 z 5
(33) Find the shortest distance between the lines and .
2 3 4 3 4 5
Find also its equation.
1
Answers : (33) , 6x y = 10 3y = 6z 25
6

(i) The scalar triple product of three vectors a , b and c is defined as: a x b . c a b c ,

sin . cos where is the angle between a , b (i.e. a ^b = ) and is the angle between
a x b and c (a × b) ^ c = ) . It is (i.e. a b . c ) also written as a b c and spelled as box
product.
(ii) Scalar triple product geometrically represents the volume of the parallelopiped whose three
coterminous edges are

represented by a, b and c i.e. V | [a b c ] |

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 16
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
(iii) In a scalar triple product the position of dot and cross can be interchanged i.e.
a . (b x c) (a x b) . c [ a b c] [b c a ] [c a b ]
(iv) a . (b x c) a . (c x b) i.e. [ a b c ] [ a c b ]
a1 a2 a3
(v) If a = a1 i + a2 j + a3 k ; b = b1 i +b2 j +b3 k and c = c1 i + c2 j + c3 k , then [ a b c ] b1 b2 b3
c1 c2 c3
.
In general, if a a1 a2m a3n ; b b1 b2m b3n and c c1 c 2m c 3 n
a1 a2 a3
then a b c b1 b2 b3 m n , where , m and n are non-coplanar vectors.
c1 c 2 c3

(vi) If a , b , c are coplanar, then [a b c] 0.

(vii) If a , b , c are non-coplanar, then [a b c] 0 for right handed system and [a b c] 0 for left
handed system.

(viii) [i j k] = 1
(ix) [K a b c] K [a b c]
(x) [(a b)c d] [a c d] [b c d]
(xi) a b b c c a = 0 and a b b c c a =2 a b c

a.a a.b a.c

2
(xii) abc = b.a b.b b.c
c.a c.b c.c

(a) The volume of the parallelopiped whose three coterminous edges are a , b and c is V [a b c ]

(b) The volume of the tetrahedron OABC with O as origin and the position vectors of A, B and C being
1
a, b and c respectively is given by V abc
6

(c) If the position vectors of the vertices of tetrahedron are a , b , c and d , then the position vector of its
1
centroid is given by (a b c d) .
4
Note : that this is also the point of concurrency of the lines joining the vertices to the centroids of the
opposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it is
equidistant from the vertices and the four faces of the tetrahedron.

Example # 30 : The volume of the paralleopiped whose edges are represented by 12 i + k , 3 j k,

2i + j 15 k is 546, then find .
12 0
Solution : V= | 0 3 1 | 546 = |12 × 44 6 | = 3 , 179
2 1 15

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 17
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 31 : Find the volume of the tetrahedron whose four vertices have position vectors a , b , c and d .
Solution : Let four vertices be A, B, C, D with position vectors a , b , c and d respectively.
DA = ( a d ) DB = ( b d ) DC = ( c d )
1
Hence volume V = [a d b d c d]
6
1 1
= (a d ). [( b d ) × ( c d )] = (a d ) . [b × c b × b + c ×d ]
6 6
1 1
= {[ a b c ] [a b d] + [a c d ] [ d b c ]} = {[ a b c ] [a b d] + [a c d ] [ b c d ]}
6 6

Example # 32 : Prove that vectors r1 = (sec2 A, 1, 1) ; r2 = (1, sec2B, 1) ; r3 = (1, 1, sec2 C) are always non-
coplanar vectors if A, B, C (0, ).
sec 2 A 1 1
Solution : Condition of coplanarity gives D = 0 1 sec 2 B 1 =0
2
1 1 sec C
sec2A [sec2Bsec2C 1] 1(sec2c 1) + 1(1 sec2B) = 0
(1 + tan2 A)(tan2 B + tan2 C + tan2 B tan2 C) tan2 C tan2 B = 0
tan2 B tan2 C + tan2 A tan2 B + tan2 C tan2A + tan2 A tan2 B tan2 C = 0
divide by tan 2 A tan2 B tan2 C
cot2A + cot2B + cot2C = 1 it is a not possible

Example # 33 : If two pairs of opposite edges of a tetrahedron are mutually perpendicular, show that the third
pair will also be mutually perpendicular.
Solution : Let OABC be the tetrahedron, where O is the origin and co-ordinates of A, B, C are
(x1, y1, z1), (x2, y2, z2), (x3, y3, x3) respectively.
Let OA BC and OB CA .
We have to prove that OC BA .
Now, direction ratios of OA are x 1, y1, z1 and of BC are (x3 x2), (y3 y2), (z3 z2).
OA BC and OB CA
x1(x3 x2) + y1(y3 y2) + z1(z3 z2) = 0 and x2(x1 x3) + y2(y1 y3) + z2(z1 z3) = 0
A (x1, y1, z 1)

O (0, 0, 0)

B C
(x2, y2, z 2) (x3, y3, z3)

Adding above two equations we get x3(x1 x2) + y3(y1 y2) + z3(z1 z2) = 0
OC BA ( direction ratios of OC are x3, y3, z3 and that of BA are (x1 x2), (y1 y2), (z1 z2))

Self practice problems :

(34) Show that a . (b c) (a b c) 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 18
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
(35) One vertex of a parallelopiped is at the point A (1, 1, 2) in the rectangular cartesian co- ordinate. If
three adjacent vertices are at B( 1, 0, 2), C(2, 2, 3) and D(4, 2, 1), then find the volume of the
parallelopiped.

(36) Show that the vector a, b, c are coplanar if and only if b c , c a , a b are coplanar.

(37) Show that {( a + b + c ) × ( c b )} . a = 2 a b c .

(38) Find the value of m such that the vectors 2i j k , i 2j 3k and 3i mj 5k are coplanar.

(39) Find the value of for which the four points with position vectors j k , 4i 5j k , 3i 9 j 4k , and
4i 4 j 4k are coplanar.
Answer : (35) 72 (38) 4 (39) =1

Let a , b and c be any three vectors, then the expression a x (b x c ) is a vector & is called a
vector triple product. This vector is perpendicular to a and lies in plane containing vectors b and c

a x ( b x c ) = (a . c)b (a . b)c
(a x b) x c = (a . c)b (b . c)a
In general (a x b) x c a x (b x c)

Example # 34 : [ a ×(3 b +2 c ) b ×( c 2a) 2 c ×( a 3 b )] =

Solution : Let b × c = p , c × a = q , a × b = r
[ p q r ] = [ a b c ]2 ....(i)
a × (3 b + 2 c ) = 3 r 2q etc.
E = [3 r 2 q p + 2 r , 2q + 6p ]
0 2 3
= [0 p 2 q + 3 r , p + 0 q + 2 r , 6p + 2q + 0 r ] = 1 0 2 [ a b c ]2 = 18 [ a b c ]2
6 2 0

3b a
Example # 35 : If a , b , c are non-coplanar unit vectors such that ( a × b ) × c = . Then find
2

angles which makes c with a & b ( a and b are non-collinear)

3b a 3b a
Solution : (a ×b ) × c = ( a . c .), b (b .c )a =
2 2

3 1
a .c = , and b .c = .
2 2
3 1
cos = and cos = .
2 2
2
= and =
6 3
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 19
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 36 : Prove that a {b (c d)} = (b . d)(a c) (b. c) (a d)
Solution : We have, a {b (c d)} = a {(b . d) c (b . c) d}
= a {(b . d) c} a {(b . c) d} [by dist. law]
= (b . d) (a c) (b . c) (a d) .

Self Practice Problems :

(40) Prove that a {a (a b)} (a . a) (b a) .
(41) Let b and c be noncollinear vectors. If a is a vector such that a . (b c) = 4 and
a (b c) = (x2 2x + 6) b + c siny, then find x and y.
(42) Find a unit vector coplanar with i j 2k and i 2 j k and perpendicular to i j k is

Answer : (41) x = 1 & y = (4n + 1) /2 , n (42) ±

2

Given a finite set of vectors a,b,c,...... , then the vector r xa yb zc ........ is called a linear
combination of a,b,c,...... for any x, y, z..... R. We have the following results :

(a) If a,b are non zero, non collinear vectors, then xa yb x 'a y 'b x x' , y y'

(b) Fundamental Theorem in plane : Let a,b be non zero, non collinear vectors, then any vector r
coplanar with a,b can be expressed uniquely as a linear combination of a and b
i.e. there exist some unique x, y R such that xa yb r .

(c) If a,b,c are non zero, non coplanar vectors, then

xa yb zc x'a y'b z'c x x' , y y' , z z'

(d) Fundamental theorem in space: Let a,b,c be non zero, non coplanar vectors in space. Then any
vector r can be uniquely expressed as a linear combination of a,b,c i.e. there exist
some unique x,y, z R such that xa yb zc r.

(e) If x1 ,x2 ,......, xn are n non zero vectors and k 1,k2,.....,kn are n scalars and if the linear
combination k1x1 k2 x2 ....... k n xn 0 k1 0, k2 0 , ....., kn 0 , then we say that vectors
x1 , x2 , ......,xn are linearly independent vectors.

(f) If k1x1 k2 x2 k 3 x3 ...... kr xr ...... kn xn 0 and if there exists at least one k r 0, then
x1 , x2 , ......, xn are said to be .
If kr 0 then xr is expressed as a linear combination of vectors x1, x2 ,........, xr 1, xr 1, .........., xn
Note :
In general, in 3 dimensional space every set of four vectors is a linearly dependent system.

i , j , k are Linearly Independent set of vectors. For K1 i + K2 j + K3 k = 0 K1= K2= K3 = 0

Two vectors a and b are linearly dependent a is parallel to b i.e. a b 0 linear dependence
of a and b . Conversely if a b 0 then a and b are linearly independent.
If three vectors a, b, c are linearly dependent, then they are coplanar i.e. [a b c] = 0. Conversely if
[a b c] 0 then the vectors are linearly independent.
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 20
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 37 : If a , b , c are three non-coplanar vectors, solve the vector equation r . a = r .b = r .c = 1
Solution : since a , b , c are three non-coplanar vectors therefore a × b , b × c & c × a are also non-
coplanar vectors
Let r = x ( a × b ) + y ( b × c ) + z ( c × a ).
Then, r . a = 1 1 = y [( b × c ) a .]
1 1 1
y= , similarly x = z = r = (( a × b ) + ( b × c ) + ( c × a ))
[a b c] [a b c] [a b c]

Example # 38 : Given that position vectors of points A, B, C are respectively

a 2 b + 3 c , 2 a + 3 b 4 c , 7 b + 10 c then prove that vectors AB and AC are linearly
dependent.
Solution : Let A, B, C be the given points and O be the point of reference then
OA = a 2 b + 3 c , OB = 2 a + 3 b 4c and OC = 7 b + 10 c
Now AB = p.v. of B p.v. of A
= OB OA = ( a + 5 b 7 c ) and AC = p.v. of C p.v of A
= OC OA = (a 5b 7c) = AB
AC = AB
where = 1. Hence AB and AC are linearly dependent.

Example # 39 : Prove that the vectors 5 a + 6 b + 7 c , 7 a 8 b + 9 c and 3 a + 20 b + 5 c are linearly

dependent, where a , b , c being linearly independent vectors.
Solution : We know that if these vectors are linearly dependent , then we can express one of them as a
linear combination of the other two.
Now let us assume that the given vector are coplanar, then we can write
5 a + 6b + 7c = ( 7a 8 b + 9 c ) + m (3 a + 20 b + 5 c ) where , m are scalars
Comparing the coefficients of a , b and c on both sides of the equation
5 = 7 + 3m, 6= 8 + 20 m, 7 = 9 + 5m
1
= = m. Hence the given vectors are linearly dependent.
2

Self Practice Problems :

1 1
(43) Given that x (p . x) p q , show that p . x p . q and find x in terms of p and q .
p2 2
(44) If x . a = 0, x . b = 0 and x . c = 0 for some non-zero vector x , then show that
[a b c] = 0
(r . a) (b c) (r . b) (c a) (r . c) (a b)
(45) Prove that r = + +
[ a b c ] [ a b c ] [ a b c ]
where a, b, c are three non-coplanar vectors

(46) Does there exist scalars u, v, w such that ue1 ve2 we3 i where e1 k,
e2 j k , e3 j 2k ?
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 21
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
(47) If a and b are non-collinear vectors and A =(x + 4y) a + (2x + y + 1) b and
B = (y 2x + 2) a + (2x 3y 1) b , find x and y such that 3A 2B .

(48) If vectors a, b,c be linearly independent, then show that

(i) a 2b 3c , 2a 3b 4c , b 2c are linearly dependent
(ii) a 3b 2c , 2a 4b c , 3a 2b c are linearly independent.

(49) Prove that a vector r in space can be expressed linearly in terms of three non-coplanar, non-
zero vectors a, b, c in the form

[r b c] a [ r c a ] b [ r a b ] c
r
[ a b c]
p . q
Answers : (43) x = q p (46) No (47) x = 2, y = 1
2 | p |2

Three points A,B,C with position vectors a, b, c respectively are collinear, if & only if there exist scalars
x, y, z not all zero simultaneously such that xa yb zc 0 = 0 , where x + y + z = 0.

Four points A, B, C, D with position vectors a, b, c, d respectively are coplanar if and only if there exist
scalars x, y, z, w not all zero simultaneously such that xa+yb+zc + wd = 0 , where x + y + z + w = 0.

Example # 40 : Prove that four points 2a 3b c , a 2b 3c , 3a 4b 2c and a 6b 6c are coplanar.

Solution
PS : Let the given four points be P, Q, R and S respectively. These points are coplanar if the
vectors PQ , PR and PS are coplanar. These vectors are coplanar iff one of them can be
expressed as a linear combination of other two. So let PQ = x PR + y PS
a 5b 4c = x a b c +y a 9b 7c

a 5b 4c = (x y) a + (x 9y) b + ( x + 7y) c
x y = 1, x 9y = 5, x + 7y = 4 [Equating coeff. of a, b, c on both sides]
1 1
Solving the first two equations of these three equations, we get x = ,y= .
2 2
These values also satisfy the third equation. Hence the given four points are coplanar.

Self Practice Problems :

(50) If a, b, c, d are any four vectors in 3-dimensional space with the same initial point and such
that 3a 2b c 2d 0 , show that the terminal A, B, C, D of these vectors are coplanar. Find
the point (P) at which AC and BD meet. Also find the ratio in which P divides AC and BD.

3a c
Answers : (50) p divides AC in 1 : 3 and BD in 1 : 1 ratio
4

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 22
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

If line joining any two points on a surface lies completely on it then the surface is a plane.
OR
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this surface
is called a plane. This fixed line is called the normal to the plane.

(i) Vector form : The equation (r r0 ) . n 0 represents a plane containing the point with position
vector is a vector normal to the plane.
The above equation can also be written as r . n d , where d = r0 . n

(ii) Cartesian form : The equation of a plane passing through the point (x1, y1, z1) is given by
a (x x1) + b( y y1) + c (z z1) = 0 where a, b, c are the direction ratios of the normal
to the plane.

(iii) Normal form : Vector equation of a plane normal to unit vector and at a distance d from the
origin is r . n = d. Normal form of the equation of a plane is x + my + nz = p, where, ,m, n are
the direction cosines of the normal to the plane and p is the distance of the plane from the
origin.

(iv) General form : ax + by + cz + d = 0 is the equation of a plane, where a, b, c are the

direction ratios of the normal to the plane.

(v) Plane through three points : The equation of the plane through three non collinear points
x x3 y y3 z z3
(x1, y1, z1), (x2, y2, z2), (x3, y3, z3) is x1 x3 y1 y3 z1 z3 = 0
x2 x3 y2 y3 z2 z3
x y z
(vi) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is 1
a b c
Note :
Equation of yz plane, xz plane and xy plane is x = 0, y = 0 and z = 0
: To reduce any equation
ax + by + cz d = 0 to the normal form, first write the constant term on the right hand side
2 2 2
and make it positive, then divide each term by a b c , where a, b, c are coefficients of x,
y and z respectively e.g.
ax by cz d
+ + =
2 2 2 2 2 2 2 2 2 2
a b c a b c a b c a b2 c2
Where (+) sign is to be taken if d > 0 and ( ) sign is to be taken if d < 0.
A plane ax + by + cz + d = 0 divides the line segment joining (x1, y1, z1) and (x 2, y2, z2). in the
ax1 by1 cz1 d
ratio
ax2 by 2 cz2 d
Coplanarity of four points

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 23
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
The points A(x1 y1 z1), B(x2 y2 z2) C(x3 y3 z3) and D(x 4 y4 z4) are coplanar then
x2 x1 y2 y1 z2 z1
x3 x1 y3 y1 z3 z1 =0
x4 x1 y4 y1 z 4 z1
Example # 41 : Find the equation of the plane upon which the length of normal from origin is 10 and direction
ratios of this normal are 3, 2, 6.
Solution : If p be the length of perpendicular from origin to the plane and , m, n be the direction
cosines of this normal, then its equation is
x + my + nz = 10 ..... (1)
Direction ratios of normal to the plane are 3, 2, 6
3 2 6
Direction cosines of normal to the required plane are = ,m= ,n=
7 7 7
3 2 6
Equation of required plane is x + y + z = 10 or, 3x + 2y + 6z = 70
7 7 7

Example # 42 :Find the plane through the points (2, 3,3), ( 5, 2, 0), (1, 7, 1)

Solution : =0 or =0 2x + y 3z + 8 = 0

Example # 43 : If P be any point on the plane x + my + nz = p and Q be a point on the line OP such that
OP . OQ = p2, show that the locus of the point Q is p( x + my + nz) = x 2 + y2 + z2.
Solution : Let P ( , , ), Q (x1, y1, z1)
Direction ratios of OP are , , and direction ratios of OQ are x1, y1, z1.

Since O, Q, P are collinear, we have = k (say) ..... (1)

x1 y1 z1
As P ( , , ) lies on the plane x + my + nz = p,
+m +n =p or k( x1 + my1 + nz1) = p ..... (2)
2 2 2
Given OP . OQ = p2 x12 y12 z12 = p2

or, k 2 (x12 y12 z12 ) x12 y12 z12 = p2 or, k (x12 y12 z12 ) = p2 ....(3)

x1 my1 nz1 1
On dividing (2) by (3), we get or, p ( x1 + my1 + nz1) = x12 y12 z12
x12 y12 z12 p
Hence the locus of point Q is p ( x + my + nz) = x2 + y2 + z2.

Example # 44 : A moving plane passes through a fixed point ( , ) and cuts the coordinate axes A, B, C . Find
the locus of the centroid of the tetrahedron OABC.
x y z
Solution : Let the plane be = 1, 0 (0,0,0) , A (a, 0,0), B (0, b, 0)
a b c
a b c
C (0,0,c) . Centroid of OABC is , ,
4 4 4

The plane passes through ( , , ) 1 ......(i)

a b c

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 24
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
a b c
Centroid , x = ,y = ,z= or a = 4x , b = 4y , c = 4z
4 4 4
Now (1) gives the locus of G as =4
x y z

Self practice problems :

(51) Check whether given points are coplanar if yes find the equation of plane containing them
A (0, 1, 1), B (4, 5, 1), C (3, 9, 4), D ( 4, 4, 4)
(52) Find the plane passing through point (3, 2, 1) and perpendicular to the line joining the
points (2, 4, 3) and (3, 1, 5).

(53) Find the equation of plane passing through the point (2, 4, 6) and making equal intercepts on
the coordinate axes.
(54) Find the equation of plane passing through (1, 2, 3) and (2, 3, 3) and perpendicular to the
plane 2x + y 3z + 4 = 0.

(55) Find the equation of the plane parallel to and and passing through (2, 1, 3).

(56) Find the equation of the plane passing through the point (1, 1, 1) and perpendicular to the
planes x + 2y + 3z 7 = 0 and 2x 3y + 4z = 0.
Answers : (51) yes, 5x 7y + 11z + 4 = 0 (52) x 5y + 2z + 5 = 0
(53) x + y + z =12 (54) 9x 15y + z + 24 = 0
(55) x y+z=4 (56) 17x + 2y 7z = 26

A plane divides the three dimensional space in two equal parts. Two points A (x 1 y1 z1)
and B (x2 y2 z2) are on the same side of the plane ax + by + cz + d = 0 if ax1 + by1 + cz1 + d and
ax2 + by2 + cz2 + d are both positive or both negative and are opposite side of plane if both of these
values are in opposite sign.

Example # 45 : Show that the points (1, 2, 3) and (2, 1, 4) lie on opposite sides of the plane
x + 4y + z 3 = 0.
Solution : Since the numbers 1+ 4 × 2 + 3 3 = 9 and 2 4+4 3= 1 are of opposite sign, then points
are on opposite sides of the plane.

P(x 1,y 1, z1)

F(x , y , z )

R(x ,y , z )

Let P ax + by + cz + d = 0 is a given plane and P(x1, y1, z1) is given point as shown in figure.
Let F(x , y , z ) be the foot of the point P (x 1, y1, z1) with respect to the plane P.
And R(x , y , z ) be the reflection of point P (x 1, y1, z1) with respect to the plane P.
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 25
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
(i) Distance of the point (x , y , z ) from the plane ax + by + cz+ d = 0 is given by
ax1 by1 cz1 d
.
2 2 2
a b c
(ii) The length of the perpendicular from a point having position vector a to plane r . n = d is
|a . n d|
p= .
|n|
(iii) The coordinates of the foot (F) of perpendicular from the point (x1, y1, z1) to the plane
x x1 y y1 z z1 (ax1 by1 cz1 d)
ax + by + cz + d = 0 are =
a b c a 2 b2 c 2
(iv) The coordinates of the Image (R) of point (x1, y1, z1) to the plane
x x1 y y1 z z1 2(ax1 by1 cz1 d)
ax + by + cz + d = 0 are =
a b c a2 b2 c 2

Example # 46 : Find the image of the point P (3, 5, 7) in the plane 2x + y + z = 0.

Solution : Given plane is 2x + y + z = 0 ..... (1)
Direction ratios of normal to plane (1) are 2, 1, 1
Let Q be the image of point P in plane (1). Let PQ meet plane (1) in R then PQ plane (1)
Let R (2r + 3, r + 5, r + 7)
Since R lies on plane (1)
2(2r + 3) + r + 5 + r + 7 = 0 or, 6r + 18 = 0 r= 3
R ( 3, 2, 4)
Let Q ( , , )
Since R is the middle point of PQ
3 5 7
3= = 9 and 2 = = 1 and 4 = =1
2 2 2
Q = ( 9, 1, 1).

Example # 47 : A plane passes through a fixed point (a, b, c). Show that the locus of the foot of perpendicular
to it from the origin is the sphere x 2 + y2 + z2 ax by cz = 0
Solution : Let the equation of the variable plane be x + my + nz + d = 0 ..... (1)
Plane passes through the fixed point (a, b, c) a + mb + nc + d = 0 ..... (2)
Let P ( , , ) be the foot of perpendicular from origin to plane (1).
Direction ratios of OP are
O(0, 0, 0)

P( , , )

0, 0, 0 i.e. , ,
From equation (1), it is clear that the direction ratios of normal to the plane i.e. OP are , m, n;
, , and , m, n are the direction ratios of the same line OP
1
= = = (say) =k ,m=k ,n=k ..... (3)
m n k
Putting the values of , m, n in equation (2), we get ka + kb + kc + d = 0 ..... (4)
Since , , lies in plane (1) +m +n +d=0 ..... (5)
Putting the values of , m, n from (3) in (5), we get k 2 + k 2 + k 2 + d = 0 ..... (6)
or k 2 + k 2 + k 2 ka kb kc = 0 [putting the value of d from (4) in (6)]
or 2
+ 2+ 2 a b c =0
Therefore, locus of foot of perpendicular P ( , , ) is x2 + y2 + z2 ax by cz = 0 ..... (7)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 26
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Self practice problems :

(57) Find the intercepts of the plane 3x + 4y 7z = 84 on the axes. Also find the length of
perpendicular from origin to this plane and direction cosines of this normal.

(58) Find : (i) perpendicular distance (ii) foot of perpendicular

(iii) image of (1, 1, 1) in the plane 3x + 4y 12z + 13 = 0
84 3 4 7
Answers : (57) a = 28, b = 21, c = 12, p = ; , ,
74 74 74 74
17
(58) (i) (ii) ( 1, 1/2, 1) (iii) ( 3, 0,1)
2

(i) Consider two planes ax + by + cz + d = 0 and a x + b y + c z + d = 0. Angle between these

planes is the angle between their normals. Since direction ratios of their normals are (a, b, c)
and (a , b , c ) respectively, hence , the angle between them, is given by
aa' bb' cc '
cos =
2 2
a b c2 a'2 b' 2 c'2
a b c
Planes are perpendicular if aa + bb + cc = 0 and planes are parallel if = =
a' b' c '
n1 . n2
(ii) The angle between the planes r . n1 = d1 and r . n2 = d2 is given by, cos =
| n1 | | n2 |

Planes are perpendicular if n1 . n2 = 0 & planes are parallel if n1 = n2 .

| d1 d2 |
Distance between two parallel planes ax + by + cz + d 1 = 0 and ax + by + cz + d2 = 0 is
2 2
a b c2

Example # 48 : Find the distance between the parallel planes 2x y + 2z + 3 = 0 and 4x 2y + 4z + 5 = 0

Solution : Given planes are 2x y + 2z + 3 = 0 and 2x y + 2z + 5/2 = 0

Required distance between planes =

2 2 2 6

(i) The equations of the planes bisecting the angle between two given planes
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are
a1x b1y c1z d1 a2 x b2 y c 2 z d2
=
2 2 2
a1 b 1 c 1 a22 b22 c 22
a1x b1y c1z d1 a2 x b2 y c 2 z d2
=
2 2 2
a1 b 1 c 1 a22 b22 c 22
(ii) If a1 + b1 + c1 + d1 and a2 + b2 + c2 + d2 are of same/opposite sign then (1)/(2)
gives equation of angle bisector of region containing point , ,

(iii) If a1a2 + b1b2 + c1c2 > 0, then equation (1)/(2) gives obtuse/acute angle bisector
and if a1a2 + b1b2 + c1c2 < 0, then equation (1)/(2) gives acute/obtuse angle bisector.
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 27
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(i) Any plane passing through the line of intersection of non parallel planes or equation of
the plane through the given line in non symmetric form.
a1x + b1y + c1z + d1 = 0 & a2x + b2y + c2z + d2 = 0 is
a1x + b1y + c1z + d1 + (a2x + b2y + c2z + d2) = 0, where R
(ii) The equation of plane passing through the intersection of the planes r . n1 = d1 & r . n2 = d2
is. r (n1 + n2 ) = d 1 d2 where is arbitrary scalar

Example # 49 : Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 2 = 0,
2x + 3y z + 3 = 0 and perpendicular to the plane x + y + z = 0
Solution : The plane is x + 2y + 3z + 2 + (2x + 3y z + 3) = 0
or (1 + 2 ) x + (2 + 3 ) y + (3 )z+2+3 =0
It is perpendicular to x + y + z = 0
3
1+2 +2+3 +3 = 0 or 2 + 3 = 0 =
2
Substituting we get 4x + 5y 9z+5=0

Example # 50 : Find the equation of the plane through the point (1, 1, 1) which passes through the line of
intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0.
Solution : Given planes are x + y + z 6 = 0 ..... (1)
and 2x + 3y + 4z + 5 = 0 ..... (2)
Given point is P (1, 1, 1).
Equation of any plane through the line of intersection of planes (1) and (2) is
x + y + z 6 + k (2x + 3y + 4z + 5) = 0 ..... (3)
If plane (3) passes through point P, then
3
1+1+1 6 + k (2 + 3 + 4 + 5) = 0 or, k=
14
From (3) required plane is 20x + 23y + 26z 69 = 0

Example # 51 Let planes are 2x + y + 2z = 9 and 3x 4y + 12z + 13 = 0. Which of these bisector planes
bisects the acute angle between the given planes. Does origin lie in the acute angle or obtuse
angle between the given planes ?

Solution : Given planes are 2x y 2z + 9 = 0 ..... (1)

and 3x 4y + 12z + 13 = 0 ..... (2)
2x y 2z 9 3x 4y 12z 13
Equations of bisecting planes are
2 2 2 2
( 2) ( 1) ( 2) 3 ( 4)2 (12)2
or, 13 [ 2x y 2z + 9] = ± 3 (3x 4y + 12z + 13)
or, 35x + y + 62z = 78, ..... (3) [Taking +ve sign]
and 17x + 25y 10z = 156 ..... (4) [Taking ve sign]
Now a1a2 + b1b2 + c1c2 = ( 2) (3) + ( 1) ( 4) + ( 2) (12)
= 6 + 4 24 = 26 < 0
Bisector of acute angle is given by 35x + y + 62z = 78
a1a2 + b1b2 + c1c2 < 0, origin lies in the acute angle between the planes.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 28
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Example # 52 : If the planes x cy bz = 0, cx y + az = 0 and bx + ay z = 0 pass through a straight line,
then find the value of a2 + b2 + c2 + 2abc.
Solution : Given planes are x cy bz = 0 ..... (1)
cx y + az = 0 ..... (2)
bx + ay z = 0 ..... (3)
Equation of any plane passing through the line of intersection of planes (1) and (2) may be
taken as x cy bz + (cx y + az) = 0
or, x (1 + c) y (c + ) + z ( b + a ) = 0 ..... (4)
If planes (3) and (4) are the same, then equations (3) and (4) will be identical.
1 c (c ) b a
b a 1
(i) (ii) (iii)
From (i) and (ii), a + ac = bc b
(a bc)
or, = ..... (5)
(ac b)

From (ii) and (iii),

c+ = ab + a2 or = ..... (6)
1 a2
(a bc) (ab c)
From (5) and (6), we have .
ac b (1 a2 )
or, a a3 + bc a2bc = a2bc + ac2 + ab2 + bc
or, a2bc + ac2 + ab2 + a3 + a2bc a = 0 or, a2 + b2 + c2 + 2abc = 1.

Self practice problems:

(59) Find the equation of plane passing through the line of intersection of the planes
2x 7y + 4z = 3 and 3x 5y + 4z = 11 and the point ( 2, 1, 3).

(60) Find the equations of the planes bisecting the angles between the planes x + 2y + 2z 3 = 0,
3x + 4y + 12z + 1 = 0 and sepecify the plane which bisects the acute angle between them.

(61) Show that the origin lies in the acute angle between the planes
x + 2y + 2z 9 = 0 and 4x 3y + 12z + 13 = 0

(62) Prove that the planes 12x 15y + 16z 28 = 0, 6x + 6y 7z 8 = 0 and

2x + 35y 39z + 12 = 0 have a common line of intersection.
Answers : (59) 15x 47y + 28z = 7
(60) 2x + 7y 5z = 21, 11x + 19y + 31z = 18; 2x + 7y 5z = 21

x x1 y y1 z z1
(i) If is the angle between line = = and the plane ax + by + cz + d = 0, then
m n

a b m c n
sin = .
(a 2
b 2
c 2
) 2
m2 n2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 29
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
b . n
(ii) Vector form: If is the angle between a line r = ( a + b ) and r . n = d then sin = .
|b| |n|
m n
(iii) Condition for perpendicularity = = b xn = 0
a b c
(iv) Condition for parallel a + bm + cn = 0 b .n = 0

x x1 y y1 z z1
(i) Cartesian form: Line = = would lie in a plane
m n
ax + by + cz + d = 0, if ax 1 + by1 + cz1 + d = 0 & a + bm + cn = 0.

(ii) Vector form: Line r = a + b would lie in the plane r . n = d if b . n = 0 & a . n = d

Example # 53 : Find the distance of the point (1, 0, 3) from the plane x y z = 9 measured parallel to the
x 2 y 2 z 6
line .
2 3 6
Solution : Given plane is x y z = 9 ..... (1)
x 2 y 2 z 6
Given line AB is ..... (2)
2 3 6
Equation of a line passing through the point Q(1, 0, 3) and parallel to line (2) is
x 1 y z 3
= r. ..... (3)
2 3 6
Co-ordinates of any point on line (3) may be taken as
P (2r + 1, 3r, 6r 3)
If P is the point of intersection of line (3) and plane (1), then P lies on plane (1),

(2r + 1) (3r) ( 6r 3) = 9
r=1
or, P (3, 3, 9)
Distance between points Q (1, 0, 3) and P (3, 3, 9)
B

2 2 2
PQ = = 4 9 36 = 7.

Example # 54 : Find the equation of the plane passing through (1, 2, 0) which contains the line
x 3 y 1 z 2
.
3 4 2
Solution : Equation of any plane passing through (1, 2, 0) may be taken as
a (x 1) + b (y 2) + c (z 0) = 0 ..... (1)
where a, b, c are the direction ratios of the normal to the plane. Given line is
x 3 y 1 z 2
..... (2)
3 4 2
If plane (1) contains the given line, then
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 30
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
3a + 4b 2c = 0 ..... (3)
Also point ( 3, 1, 2) on line (2) lies in plane (1)
a ( 3 1) + b (1 2) + c (2 0) = 0
or, 4a b + 2c = 0 ..... (4)
a b c
Solving equations (3) and (4), we get
8 2 8 6 3 16
a b c
or, = k (say). ..... (5)
6 2 13
Substituting the values of a, b and c in equation (1), we get
6 (x 1) + 2 (y 2) + 13 (z 0) = 0.
or, 6x + 2y + 13z 10 = 0. This is the required equation.

x 1 y 1 z 3
Example # 55 : Find the equation of the projection of the line on the plane x + 2y + z = 9.
2 1 4
x 1 y 1 z 3
Solution : Let the given line AB be ..... (1)
2 1 4
Given plane is x + 2y + z = 9 ..... (2)
Let DC be the projection of AB on plane (2)
Clearly plane ABCD is perpendicular to plane (2).
Equation of any plane through AB may be taken as (this plane passes through the point
(1, 1, 3) on line AB)
a (x 1) + b (y + 1) + c (z 3) = 0 ..... (3)
where 2a b + 4c = 0 ..... (4)
[ normal to plane (3) is perpendicular to line (1)]
Since plane (3) is perpendicular to plane (2),
a + 2b + c = 0 ..... (5)
a b c
Solving equations (4) & (5), we get .
9 2 5
Substituting these values of a, b and c in equation (3), we get
9 (x 1) 2 (y + 1) 5 (z 3) = 0
or, 9x 2y 5z + 4 = 0 ...... (6)
Since projection DC of AB on plane (2) is the line of intersection of plane ABCD and plane (2),
therefore equation of DC will be
9x 2y 5z 4 0 .....(i)
and ..... (7)
x 2y z 9 0 .....(ii)
Let , m, n be the direction ratios of the line of intersection of planes (i) and (ii)
9 2m 5n = 0 ..... (8) and + 2m + n = 0 ..... (9)
m n m n
2 10 5 9 18 2

Let any point on line (7) is ( , , 0) 9 2 +4=0

1 17
+2 9=0 = , =
2 4
1 17
So equation of line is 2 4

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 31
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
Self practice problems :
x 2 y 3 z 6
(63) Find the values of a and b for which the line is perpendicular to the plane
a 4 2
3x 2y + bz + 10 = 0.
x 1 y 2 z 3
(64) Find the equation of the plane containing the lines and
2 3 3
x 2 y 3 z 4
.
3 4 5
x 2 y 3 z 4
(65) Find the plane containing the line = = and parallel to the line
2 3 5
x 1 y 1 z 1
= =
1 2 1
x 1 y 2 z 3 x 4 y 1
(66) Show that the line = = & = = z are intersecting each other.
2 3 4 5 2
Find their intersection point and the plane containing the line.
Answers : (63) a = 6, b = 1 (64) 3x y z + 2 = 0
(65) 13x + 3y 7z 7 = 0 (66) ( 1, 1, 1) & 5x 18y + 11z 2 = 0

A straight line in space is characterised by the intersection of two planes which are not parallel and
therefore, the equation of a straight line is a solution of the system constituted by the equations of the
two planes, P1 a1x + b1y + c 1z + d1 = 0 and P2 a2x + b2y + c2z + d2 =0. This form is also known as
non-symmetrical form.

To find the equation of the line in symmetrical form, we must know (i) its direction ratios (ii) coordinate
of any point on it.

(i) Direction ratios: Let , m, n be the direction ratios of the line. Since the line lies in both the
planes, it must be perpendicular to normals of both planes.
So a1 + b1m + c1n = 0, a2 + b2m + c2n = 0. From these equations, proportional values of
m n
, m, n can be found by cross-multiplication as = =
b1c 2 b2c1 c1a 2 c 2a1 a1b2 a2b1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 32
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
(ii) Point on the line Note that as , m, n cannot be zero simultaneously, so at least one must
be non-zero. Let a1b2 a2b1 0, then the line cannot be parallel to xy plane, so it intersect it.
Let it intersect xy-plane in (x1, y1, 0). Then a1x1 + b1y1 + d1 = 0 and a2x1 + b2y1+ d2= 0.Solving
these, we get a point on the line.

Note : If 0, then we can take a point on yz-plane as (0, y1, z1) and if m 0, then we can take a point on xz-
plane as (x1, 0, z1).

Condition of coplanarity if both the lines are in general form Let the lines be
ax + by + cz + d = 0 = a x + b y + c z + d & x+ y+ z+ =0= x+ y+ z+
a b c d
a' b' c ' d'
They are coplanar if =0

' ' ' '

Example # 56 : Find the equation of the line of intersection of planes 4x + 4y 5z = 12, 8x + 12y 13z = 32
in the symmetric form.
Solution : Given planes are 4x + 4y 5z 12 = 0 ..... (1)
and 2x 3y + 4z = 5 ..... (2)
Let , m, n be the direction ratios of the line of intersection :
then 4 4m 3n = 0 ..... (3)
m n m n
and 42 12m + 13n = 0 or,
1 2 1
Hence direction ratios of line of intersection are 1, 2, 1.
Let the line of intersection meet the xy-plane at P ( , , 0).
Then P lies on planes (1) and (2)
=4 or, 2 3 =5 ..... (5)
= 2, = 3
x 2 y 3 z
Hence equation of line of intersection in symmetrical form is .
1 2 1

Example # 57 : Find the angle between the lines x 3y 4 = 0, 4y z + 5 = 0 and

x + 3y 11 = 0, 2y z + 6 = 0.
x 3y 4 0
Solution : Given lines are ..... (1)
4y z 5 0
x 3y 11 0
and ..... (2)
2y z 6 0
Let 1
, m1, n1 and 2
, m2, n2 be the direction cosines of lines (1) and (2) respectively
line (1) is perpendicular to the normals of each of the planes
x 3y 4 = 0 and 4y z + 5 = 0
1
3m1 + 0.n1 = 0 ..... (3) and 0 1
+ 4m1 n1 = 0 ..... (4)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 33
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
1 m1 n1
Solving equations (3) and (4), we get =
3 0 0 ( 1) 4 0
1 m1 n1
or, = k (let).
3 1 4
Since line (2) is perpendicular to the normals of each of the planes
x + 3y 11 = 0 and 2y z + 6 = 0,
2
+ 3m2 = 0 ..... (5) and 2m2 n2 = 0 ..... (6)

2 n2
2
= 3m2 or, = m2 and n2 = 2m 2 or, = m 2.
3 2
2 m2 n2
= t (let).
3 1 2
If be the angle between lines (1) and (2), then cos = 1 2
+ m1m2 + n1n2
= (3k) ( 3t) + (k) (t) + (4k) (2t) = 9kt + kt + 8kt = 0 = 90°.

x 3 y 1 z 2
Example # 58 : Show that the lines and are coplanar. Also find the
2 3 1
equation of the plane containing them.
x 3 y 1 z 2 x 7 y z 7
Solution : Given lines are = r (say)..... (1) and = R (say) ..... (2)
2 3 1 3 1 2
If possible, let lines (1) and (2) intersect at P.
Any point on line (1) may be taken as (2r + 3, 3r 1, r 2) = P (let).
Any point on line (2) may be taken as ( 3R + 7, R, 2R 7) = P (let).
2r + 3 = 3R + 7 or, 2r + 3R = 4 ..... (3)
Also 3r 1 = R or, 3r R = 1 ..... (4)
and r 2 = 2R 7 or, r 2R = 5. ..... (5)
Solving equations (3) and (4), we get, r = 1, R = 2
Clearly r = 1, R = 2 satisfies equation (5).
Hence lines (1) and (2) intersect.
lines (1) and (2) are coplanar.
x 3 y 1 z 2
Equation of the plane containing lines (1) and (2) is 2 3 1 =0
3 1 2
or, (x 3) ( 6 1) (y + 1) (4 + 3) + (z + 2) (2 9) = 0
or, 7 (x 3) 7 (y + 1) 7 (z + 2) = 0 or, x 3 + y + 1 + z + 2 = 0 or, x + y + z = 0.

Self practice problems:

(67) Find the equation of the line of intersection of the plane

x y + 2z = 5, 3x + y + z = 6.

(68) Prove that the three planes 2x + y 4z 17 = 0, 3x + 2y 2z 25 = 0, 2x 4y + 3z + 25 = 0

intersect at a point and find its co-ordinates.

4y 9 z
Answers : (67) = = (68) (3, 7, 1)
5 1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 34
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Marked questions are recommended for Revision.

PART - I : SUBJECTIVE QUESTIONS

Section (A) : Position vector, Direction Ratios & Direction cosines

A-1. (i) Let position vectors of points A,B and C are a , b and c respectively. Point D divides line
segment BC internally in the ratio 2 : 1. Find vector AD .

(ii) Let ABCD is parallelogram. Position vector of points A,C and D are a , c and d respectively .
If E divides line segement AB internally in the ratio 3 : 2 then find vector DE .

(iii) Let ABCD is trapezium such that AB = 3 DC . E divides line segement AB internally in the ratio
2 : 1 and F is mid point of DC. If position vector of A,B and C are a,b and c respectively then
find vector FE .

A-2. In a ABC, AB = 6i 3j 3k ; AC = 3i 3j 6k
D and D are points trisections of side BC
Find AD and AD .

A-3. If ABCD is a quadrilateral, E and F are the mid-points of AC and BD respectively, then prove that
AB AD CB CD 4EF

A-4. Let ABCD is parallelogram where A = (1,2,4) , B = (8,7,9) and D = (6,1,5) . Find direction cosines of line
AC

A-5_. Let one vertex of cuboid whose faces are parallel to coordinate planes is (1, 2, 3). If mid-point of one of
the edge in (0, 0,0) then find possible
(i) total surface area of cuboid. (ii) length of body diagonal.
A-6. Find the direction cosines , m, n of line which are connected by the relations + m + n = 0,
2mn + 2m n = 0.
Section (B) : Dot Product, Projection and Cross Product
B-1. Show that the points A, B, C with position vectors 2i j k , i 3 j 5k and 3i 4j 4k respectively are
the vertices of a right angled triangle. Also find the remaining angles of the triangle.

B-2. If a, b, c are three mutually perpendicular vectors of equal magnitude, prove that a b c is equally
inclined with vectors a, b and c .

B-3. (i) Find the projection of b c on a where a = i 2j k , b = i 3j k and c = i k .

(ii) Find the projection of the line segment joining (2, 1, 3) and (4, 2, 5) on a line which makes
equal acute angles with co-ordinate axes.

(iii) P and Q are the points ( 1, 2, 1) and (4, 3, 5) respectively. Find the projection of PQ on a line
which makes angles of 120º and 135º with y and z axes respectively and an acute angle with x-
axis.
2 2
a b a b
B-4. Prove that =
a2 b2 |a| |b|

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 35
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

B-5. If a, b are two unit vectors and is the angle between them, then show that:
1 1
(a) sin a b (b) cos a b
2 2 2 2

B-6. If two vectors a and b are such that | a | = 2, | b | = 1 and a·b = 1, then find the value of
· (2a 7b) .

B-7. For an y two vector s u & v , prove that

(a) (u.v)2 | u v |2 | u |2 | v | 2 & (b) (1 | u |2 )(1 | v |2 ) (1 u.v)2 | u v (u v) |2

B-8. If the three successive vertices of a parallelogram have the position vectors as,
A ( 3, 2, 0); B (3, 3, 1) and C ( 5, 0, 2). Th en find
(a) position vector of the fourth vertex D
(b) a vector having the sam e direction as that of AB but magnitude equal to AC
(c) the angle between AC and BD .

B-9. If a,b,c are three vectors such that | a | = 5, | b | = 12 and | c | = 13, and a b c o , find the value of
a.b b.c c.a .

B-10. ABCD is a parallelogram in which AB = 3i 6 j 3k and AD = 6i 6 j 3k . P is a point of AB such

that AP : PB = 1 : 2 and Q is a point on BC such that BQ : QC = 2 : 1. Find angle between DQ and PC.

B-11. Prove using vectors : If two medians of a triangle are equal, then it is isosceles.

B-12. (i) Find the angle between the lines whose direction cosines are given by the equations :
3 + m + 5n = 0 and 6 mn 2n + 5 m = 0
(ii) Find the angle between the lines whose direction cosines are given by + m + n = 0 and 2
+
m2 = n2.

B-13. Position vectors of A, B, C are given by a, b, c where a b b c c a = 0. If AC =

then find BC if BC = 14.

B-14. A vector c is perpendicular to the vectors 2 i 3j k, i 2j 3k and satisfies the

condition c . 2 i j k + 6 = 0. Find the vector c

B-15. (a) Show that the perpendicular distance of the point c from the line joining a and b is
b c c a a b
.
b a
(b) Given a parallelogram ABCD with area 12 sq. units. A straight line is drawn through the mid
point M of the side BC and the vertex A which cuts the diagonal BD at a point ' O '. Use vectors
to determine the area of the quadrilateral OMCD.

B-16. P, Q are the mid-points of the non-parallel sides BC and AD of a trapezium ABCD. Show that
APD = CQB.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 36
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Section (C) : Line

C-1. Find the coordinates of the point when the line through (3, 2, 5) and ( 2, 3, 5) crosses the xy plane.

x y 1 z 2
C-2. Find the foot of the perpendicular from (1, 6, 3) on the line = = .
1 2 3

C-3. (i) Find the cartesian form of the equation of a line whose vector form is given by .
r 2i j 4k (i j 2k)
(ii) Find the vector form of the equation of a line whose cartesian form is given by
2x 4 3y 6
= = .
1 2 1

C-4. Find the distance between points of intersection of

x 1 y 2 z 3 x 4 y 1
Lines = = & = =z and
2 3 4 5 2
Lines r = ( i j k)+ (3 i j)& r = (4 i k)+ (2 i 3k )

C-5. Show that the foot of the perpendicular from the origin to the join of A( 9, 4, 5) and B (11, 0, 1) is the
mid point of AB. Also find distance of point (2, 4, 4) from the line AB.

x 3 y 3 z
C-6. Find the equation of the two lines through the origin which intersect the line = = at an
2 1 1
angle of /3.

C-7. The foot of the perpendicular from (a, b, c) on the line x = y = z is the point (r, r, r), then find the value of
r.

C-8. Find the shortest distance between the lines :

r = (4i j) (i 2j 3k) and r = (i j 2k) (2i 4j 5k)

x 3 y 8 z 3 x 3 y 7 z 6
C-9. Let L1 and L2 be the lines whose equation are and = =
3 1 1 3 2 4
respectively. A and B are two points on L1 and L2 respectively such that AB is perpendicular both the
lines L1 and L2.Find points A, B and hence find shortest distance between lines L 1 and L2

C-10. If r = (i 2j 3k) (i j k) and r = (i 2j 3k) (i j k) are two lines, then find the equation
of acute angle bisector of two lines.

C-11. The edges of a rectangular parallelopiped are a, b, c; show that the angles between two of the four
2 2 2 2 2 2 2 2 2
diagonals are given by cos 1
2 2 2
or cos 1 2 2 2
or cos 1 2 2 2
.
a b c a b c a b c

C-12. Show that equation of angle bisectors of line r a b and r = b + µ a are r = (a b) + ba ab

C-13. Prove that the shortest distance between the diagonals of a rectangular parallelepiped whose
bc ca ab
coterminous sides are a, b, c and the edges not meeting it are , ,
2 2 2 2 2
b c c a a b2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 37
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Section (D) : STP, VTP, Vector equation, LI/LD

D-1. Show that {( a + b + c ) × ( c b )} . a = 2 a b c .

D-2. Given unit vectors m , n and p such that m n = p mx n = then find value of n p m in

terms of .

D-3. Let a=a1 i+a2 j+a3k , b=b1i+b 2 j+b3k and c=c1i+c 2 j+c 3 k be three non-zero vectors such that c is a unit
2
a1 a2 a3
vector perpendicular to both a and b . If the angle between aandb is , then b1 b2 b3 is equal
6
c1 c 2 c3
to:

D-4. Examine for coplanarity of the following sets of points

(a) 4 i + 8 j + 12 k , 2 i + 4 j + 6 k , 3 i + 5 j + 4 k , 5 i + 8 j + 5 k .
(b) 3 a + 2b 5 c , 3a + 8b + 5c , 3 a + 2b + c , a + 4b 3 c . Where a , b , c are noncoplanar

D-5. The vertices of a tetrahedron are P(2, 3, 2), Q(1, 1, 1), R(3, 2, 1) and S (7, 1, 4).
(i) Find the volume of tetrahedron
(ii) Find the shortest distance between the lines PQ & RS.

D-6. Are the following set of vectors linearly independent?

(i) a = i 2 j + 3k , b = 3 i 6 j + 9k
(ii) a = 2i 4k , b = i 2j k, c = i 4 j + 3k

D-7. Find value of x R for which the vectors a = (1, 2, 3), b = ( 2, 3, 4), c = (1, 1, x) form a linearly
dependent system.

D-8. If a , b , c are non-coplanar vectors and v . a v .b v . c = 0, then find value of v .

D-9. Let a i 2j 3k , b 2i j k, c 3i 2j k and d = 3 i j 2k , then

(i) if a (b c) = pa qb rc , then find value of p, q and r.
(ii) find the value of ( a × b ) × ( a × c ). d

1 1
D-10. Given that x (p . x) p q , then show that p.x (p.q) and hence find x in terms of p and q .
p2 2

D-11. Let there exist a vector x satisfying the conditions x × a = c d and x + 2 d = v d . Find x in

terms of a , c and d

Section (E) : Plane

E-1. Find equation of plane
(i) Which passes through (0, 1, 0), (0, 0, 1), (1, 2, 3)
(ii) Which passes through (0, 1, 0) and contains two vectors i j k & 2i j.
(iii) Whose normal is i j k & which passes through (1, 2, 1).
(iv) Which makes equal intercepts on co-ordinate axis and passes through (1, 2, 3)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 38
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

E-2. Find the ratio in which the line joining the points (3, 5, 7) and ( 2, 1, 8) is divided by the yz-plane. Find
also the point of intersection of the plane and the line

E-3. Find the locus of the point whose sum of the square of distances from the planes x + y + z = 0, x z=0
and x 2y + z = 0 is 9

E-4. The foot of the perpendicular drawn from the origin to the plane is (4, 2, 5), then find the vector
equation of plane.

E-5. Let P (1, 3, 5) and Q( 2, 1, 4) be two points from which perpendiculars PM and QN are drawn to the x-z
plane. Find the angle that the line MN makes with the plane x + y + z = 5.

E-6. The reflection of line = = about the plane x 2y + z 6 = 0 is

3 4 5

x 1 y 2 z 3
E-7. Find the equation of image of the line = = in the plane 3x 3y + 10z = 26.
9 1 3

E-8. Find the angle between the plane passing through points (1, 1, 1), (1, 1, 1), ( 7, 3, 5) & x z plane.

3 y z 2
E-9. Find the equation of the plane containing parallel lines (x 4) = = and
4 5
(x 3) = (y + 2) = z

E-10. Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line
x 3 y 2 z
= =
3 6 2

E-11. If the acute angle that the vector, i j k makes with the plane of the two vectors
2i 3j k and i j 2k is cot 1
2 then find the value of ( + )

E-12. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line

.
2 7 5

E-13. Find the vector equation of a line passing through the point with position vector and
perpendicular to the plane + 2 = 0.
Also, find the point of intersection of this line and the plane.

E-14. Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining
the points (1, 4, 2) and (2, 3, 5). Also find the coordinates of the foot of the perpendicular and the
perpendicular distance of the point (4, 0, 3) from the above found plane.

E-15. Find the equation of the planes passing through points (1, 0, 0) and (0, 1, 0) and making an angle of
0.25 radians with plane x + y 3 = 0

E-16. Find the distance between the parallel planes r . (2i 3j 6k) = 5 and r . (6i 9 j 18k) + 20 = 0.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 39
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

E-17. If the planes x cy bz = 0, cx y + az = 0 and bx + ay z = 0 pass through a straight line, then find
the value of a2 + b2 + c2 + 2abc is :

E-18. (i) If n is the unit vector normal to a plane and p be the length of the perpendicular from the origin
to the plane, find the vector equation of the plane.
(ii) Find the equation of the plane which contains the origin and the line of intersection of the
planes r . a = p and r . b = q

PART - II : ONLY ONE OPTION CORRECT TYPE

Section (A) : Position vector, Direction Ratios & Direction cosines

A-1. If the vector b is collinear with the vector a 2 2, 1, 4 and b = 10, then:

(A) a b =0 (B) a 2b = 0 (C) 2a b =0 (D) 3a b =0

A-2. OABCDE is a regular hexagon of side 2 units in the XY plane as shown in figure . O being the origin
and OA taken along the X axis. A point P is taken on a line parallel to Z axis through the centre of the
hexagon at a distance of 3 units from O in the positive Z direction. Then vector AP is:
y

D C

E B

x
O A

z
(A) i 3j 5k (B) i 3j 5k (C) i 3j 5k (D) i 3j 5k

A-3. If the sum of the squares of the distances of a point from the three coordinate axes be 36, then its
distance from the origin is
(A) 6 (B) 3 2 (C) 2 3 (D) 6 2

A-4. A line makes angles with the coordinate axes. If + = 90º, then =
(A) 0 (B) 90º (C) 180º (D) 45°

Section (B) : Dot Product, Projection and Cross Product

1
B-1. The vector 2i 2 j k is:
2
(A) a unit vector
(B) makes an angle with the vector 2 i 4j 3k
3
1
(C) parallel to the vector i j k
2
(D) perpendicular to the vector 3 i 2j 2k

B-2. If | a | = 5, | a b | = 8 and | a + b | = 10, then | b | is equal to :

(A) 1 (B) 57 (C) 3 (D) 57
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 40
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

B-3. If a b = c , b c = a , then find value of | 3a 4b 12c | if a, b, c are vectors of same magnitude.

(A) 11 (B) 12 (C) 13 (D) 14

B-4. Angle between diagonals of a parallelogram whose side are represented by a 2i j k and
b i j k
1 3 8 4
(A) cos 1
(B) cos 1 (C) sin 1
(D) tan 1
2 5 9 3

B-5. A, B, C & D are four points in a plane with position vectors a , b , c & d respectively such that

a d . b c b d . c a = 0. Then for the triangle ABC, D is its :

(A) incentre (B) circumcentre (C) orthocentre (D) centroid

B-6. Given a = 10, b = 2 and a.b = 12, then find a b.

(A) 12 (B) 16 (C) 8 (D) 32

B-7. Unit vector perpendicular to the plane of the triangle ABC with position vectors of the vertices A, B, C,
is (where is the area of the triangle ABC).
axb bxc cxa axb b xc c xa
(A) (B)
2

a xb b xc c xa axb c b c xa
(C) (D)
4 2

B-8. ABC is a triangle where A = (2, 3, 5), B = ( 1, 2, 2) and C( , 5, ), if the median through A is equally
inclined to the positive axes, then + is
(A) 7 (B) 6 (C) 15 (D) 9

Section (C) : Line

C-1. If a line has a vector equation r 2i 6j i 3 j , then which of the following statement(s) is/are

NOT correct?
(A) the line is parallel to 2 i 6j (B) the line passes through the point 3 i 3j
(C) the line passes through the point i 9j (D) the line is parallel to XY-plane

C-2. Let a = i + j and b = 2 i k . The point of intersection of the lines r × a = b × a and

r × b = a × b is :
(A) i + j + 2k (B) 3 i j + k (C) 3 i + j k (D) i j k

x 1 y 2 z x 7 y z
C-3. T & are coplaner
4 1 1 6 2
(A) 2, 8 (B) 2, 8 (C) 3, 5 (D) 1, 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 41
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x 1 y 2 z 3
C-4. Equation of the angle bisector of the angle between the lines = = &
1 1 1
x 1 y 2 z 3
= = is :
1 1 1
x 1 y 2 x 1 y 2 z 3
(A) = ;z 3=0 (B) = =
2 2 1 2 3
y 2 z 3 x 1 y 2
(C) x 1 = 0 ; = (D) = ;z 3=0
1 1 2 3

x y z x 1 y 2 z 3 x k y 1 z 2
C-5. If the lines = = , = = and = = are concurrent then
1 2 3 3 1 4 3 2 h
1 1
(A) h = 2, k = 6 (B) h = , k = 2 (C) h = 6, k = 2 (D) h = 2, k =
2 2

C-6. Points X and Y are taken on the sides QR and RS, respectively of a parallelogram PQRS, so that
QX = 4XR and RY = 4YS. The line XY cuts the line PR at Z. Find the ratio PZ : ZR.
(A) 4 : 21 (B) 3 : 4 (C) 21 : 4 (D) 4 : 3

Section (D) : STP, VTP, Vector equation, LI/LD

D-1. The value of a 2b c a b a b c is equal to the box product :

(A) a b c (B) 2 a b c (C) 3 a b c (D) 4 a b c

D-2. For a non zero vector A if the equations A . B = A . C and A B = A C hold simultaneously,
then :
(A) A is perpendicular to B C (B) A B
(C) B C (D) C A

D-3. Let a = xi 12j k , b = 2i 2xj k and c = i k . If the ordered set bca is left handed, then :
(A) x (2, ) (B) x ( , 3) (C) x ( 3, 2) (D) x { 3, 2}

D-4. If a = i + j k, b = i j + k, c is a unit vector such that c.a = 0, [c ab] = 0 then a unit vector d both a
and c is perpendicular to
1 1 1 1
(A) (2i j + k) (B) (j + k) (C) (i + j) (D) (i + k)
6 2 2 2

D-5. If a = i + j + k and b = 2i + k, then the vector c satisfying the conditions.

(i) that it is coplanar with a and b
(ii) that its projection on b is 0
(A) 3i + 5j + 6k (B) 3i 5j + 6k (C) 6i + 5k (D) i + 2j + 2k

D-6. If a x b = c x d and a x c = b x d , then the vectors a d and b c are :

(A) non-collinear (B) linearly independent
(C) perpendicular (D) parallel

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 42
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

D-7. Vector of length 3 unit which is perpendicular to i j k and lies in the plane of i j k and 2i 3 j , is
3 3 3 3
(A) (i 2j k ) (B) ( 2i j k) (C) ( 7i 8 j k ) (D) ( 7i 8j k )
6 6 114 114

D-8. If a , b , c be the unit vectors such that b is not parallel to c and a x 2b x c b , then the angle
that a makes with b and c are respectively:
2 2
(A) & (B) & (C) & (D) &
3 4 3 3 2 3 2 3

D-9. If a , b , c are linearly independent vectors, then which one of the following set of vectors is linearly
dependent?
(A) a b , b c , c a (B) a b , b c , c a (C) a xb , b x c , c xa (D) a 2b 3c, b c a, a c

D-10. Let a, b, c are three non-coplanar vectors such that r1 a b c , r2 b c a , r3 c a b,

r 2a 3b 4c . If r r
1 1 2 r2 3 r3 , then the values of 1 , 2 and 3
respectively are
(A) 7, 1, 4 (B) 7 / 2, 1, 1 / 2 (C) 5 / 2, 1, 1/2 (D) 1 / 2, 1, 7 / 2

D-11. Vector x satisfying the relation A.x c and A x B is

cA (A B) cA (A B) cA (A B) cA 2(A B)
(A) (B) 2
(C) 2
(D)
|A| |A| |A| | A |2

D-12. The value of r if exist where r = a b and r c d is

a.d a.d a.d a.d
(A) a + b (B) a b (C) a b (D) a + b
b.d b.d b.d b.d

Section (E) : Plane

E-1. The equation of a plane which passes through (2, 3, 1) & is perpendicular to the line joining the points
(3, 4, 1) & (2, 1, 5) is given by:
(A) x + 5y 6z + 19 = 0 (B) x 5y + 6z 19 = 0
(C) x + 5y + 6z + 19 = 0 (D) x 5y 6z 19 = 0

E-2. The reflection of the point (2, 1, 3) in the plane 3x 2y z = 9 is :

26 15 17 26 15 17 15 26 17 26 17 15
(A) , , (B) , , (C) , , (D) , ,
7 7 7 7 7 7 7 7 7 7 7 7

x 2 y 1 z 2
E-3. The distance of the point ( 1, 5, 10) from the point of intersection of the line,
3 4 12
and the plane, x y + z = 5, is :
(A) 10 (B) 11 (C) 12 (D) 13

E-4. The distance of the point (1, 2, 3) from the plane x y + z = 5 measured parallel to the line,
x y z
, is :
2 3 6
(A) 1 (B) 6/7 (C) 7/6 (D) 1/6

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 43
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x 1 y 3 z 2
E-5. The distance of the point P(3, 8, 2) from the line = = measured parallel to the plane
2 4 3
3x + 2y 2z + 17 = 0 is
(A) 2 (B) 3 (C) 5 (D) 7

E-6. If line r = ( i 2 j k ) + (2 i j 2 k ) is parallel to the plane r .(3 i 2 j m k )=14, then the value of m is
(A) 2 (B) 2
(C) 0 (D) can not be predicted with these informations

E-7. The locus represented by xy + yz = 0 is

(A) A pair of perpendicular lines (B) A pair of parallel lines
(C) A pair of parallel planes (D) A pair of perpendicular planes

E-8. The equation of the plane passing through the point ( 1, 3, 2) and perpendicular to planes
x + 2y + 2z = 5 and 3x + 3y + 2z = 8, is
(A) 2x 4y + 3z 8 = 0 (B) 2x 4y 3z + 8 = 0
(C) 2x + 4y + 3z + 8 = 0 (D) 2x 4y + 3z 8 = 0

E-9. If a plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes (where 'O' is the origin),
then the area of the triangle ABC is equal to
1 1
(A) b2c 2 c 2a2 a2b2 (B) (bc + ca + ab)
2 2
1 1
(C) abc (D) (b c)2 (c a)2 (a b) 2
2 2

E-10. Given the vertices A (2, 3, 1), B (4, 1, 2), C (6, 3, 7) & D ( 5, 4, 8) of a tetrahedron. The length of the
altitude drawn from the vertex D is:
(A) 7 (B) 9 (C) 11 (D) 13

PART - III : MATCH THE COLUMN

1. Column Column
(A) If the vectors a 3i j 2k , b i 3 j 4k and (p) 2

c 4i 2j 6k constitute the sides of a ABC

and length of the median bisecting the vector c is , then 2

(B) Let p is the position vector of the orthocentre and g is the (q) 3
position vector of the centroid of the triangle ABC, where
circumcentre is the origin. If p = K g , then K is equal to :

(C) Twice of the area of the parallelogram constructed on (r) 6

the vectors a p 2q and b 2 p q , where p and
q are unit vectors containing an angle of 30º, is :

(D) Let u , v and w are vector such that u v w 0. (s) 5

If | u | 3, | v | 4, |w| 5 then u.v v.w w.u is

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 44
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2. Match the following set of lines to the corresponding type :

Column Column
x 1 y 2 z 3 x 1 y 2 z 3
(A) & (p) parallel but not coincident
1 2 3 2 2 2
x 1 y 2 z 3 x 5 y 2 z 3
(B) & (q) intersecting
1 2 3 1 2 3
x 1 y 2 z 3 x 3 y 4 z 1
(C) & (r) skew lines
2 2 2 1 1 1
x 1 y 2 z 3 x y 1 z
(D) & (s) Coincident
2 2 3 2 3 1

3. Column Column
(A) The volume of the parallelopiped constructed on
the diagonals of the faces of the given rectangular (p) 3
parallelopiped is m times the volume of the given parallelopiped.
Then m is equal to
2
12 b a b
(B) If x satisfying the conditions b.x & b x a is x 2
(q) 2
|b| | b |2
then can be
(C) The points (0, 1, 1), (4, 5, 1), (3, 9, 4) and (r) 4
( 4, 4, k) are coplanar, then k =

(D) In ABC the mid points of the sides AB, BC and CA are (s) 8
respectively ( , 0, 0) (0, m ,0) and (0, 0, n).
AB2 BC2 CA 2
Then 2
is equal to
m2 n2

Marked questions are recommended for Revision.

PART - I : ONLY ONE OPTION CORRECT TYPE

1. Let a, b, c be vectors of length 3, 4, 5 respectively. Let a be perpendicular to b c , b to c a and
c to a b . Then a b c is equal to :
(A) 2 5 (B) 2 2 (C) 10 5 (D) 5 2

2. Four coplanar forces are applied at a point O. Each of them is equal to k and the angle between two
consecutive forces equals 45º as shown in the figure. Then the resultant has the magnitude equal to :

(A) k 2 2 2 (B) k 3 2 2 (C) k 4 2 2 (D) k 4 2 2

1
3. Taken on side AC of a triangle ABC, a point M such that AM = AC . A point N is taken on the side
3
CB such that BN = CB , then for the point of intersection X of AB and MN which of the following
holds good?
1 1 3
(A) XB = AB (B) AX = AB (C) XN = MN (D) XM = 3 XN
3 3 4
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 45
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

4. If 3 non zero vectors a, b, c are such that a b = 2(a c) , | a | = | c | = 1; | b | = 4 the angle between
1
b and c is cos 1 then b = c µa where | | + | µ | is -
4
(A) 6 (B) 5 (C) 4 (D) 0

5. If is the angle between the vectors p = a i + b j + c k and vector q = b i + c j + a k then range of is

(A) [0, /3] (B) [ /3, 2 /3] (C) [0, 2 /3] (D) [0, 5 /6]

6. If the unit vectors e1 and e 2 are inclined at an angle 2 and | e1 e2 | < 1, then for [0, ], may lie
in the interval :
5 5
(A) 0, (B) , (C) , (D) ,
6 6 2 6 2 6

7. A line makes angles , , , with the four diagonals of a cube, then cos 2 + cos 2 + cos2 + cos2 is
equal to
(A) 1/3 (B) 2/3 (C) 4/3 (D) 5/3

9. Given a xi yj 2k , b i j k, c i 2j ; (a ^ b) = , a.c 4 , then

2
(A) [a b c]2 | a | (B) [a b c] | a | (C) [a b c] 0 (D) [a b c] | a |2

10. The vectors i 2 j 3k , 2 i j k and 3 i j 4k are so placed that the end point of one vector is the
starting point of the next vector. Then the vectors are :
(A) Not coplaner (B) Coplaner but cannot form a triangle
(C) Coplaner but can form a triangle (D) Coplaner & can form a right angled triangle

11. Let a, b and c be non-coplanar unit vectors equally inclined to one another at an acute angle . Then
[a b c] in terms of is equal to :

(A) (1 + cos ) cos 2 (B) (1 + cos ) 1 2cos2

(C) (1 cos ) 1 2 cos (D) (1 sin ) 1 2 cos

12. Consider a tetrahedron with faces f1, f 2, f3, f4. Let a1 , a2 ,a3 , a4 be the vectors whose magnitudes are
respectively equal to the areas of f 1, f2, f3, f4 and whose directions are perpendicular to these faces in
the outward direction. Then,
(A) a1 a2 a3 a4 = 0 (B) a1 a3 a2 a4
(C) a1 a2 a3 a4 (D) a1 a2 a3 a4 = 0

13. Let r be a vector perpendicular to a + b + c , where [ a b c ] = 2. If

r = ( b × c ) + m( c × a ) + n( a × b ), then ( + m + n) is equal to
(A) 2 (B) 1 (C) 0 (D) 1

14. If a , b , c are three non-coplanar non-zero vectors and r is any vector in space, then
(a b) × (r c) + (b c) × (r a) + (c a) × (r b) is equal to
(A) 2[ a b c ] r (B) 3[ a b c ] r (C) [ a b c ] r (D) 4[ a b c ] r

15. If b and c are two non-collinear vectors such that a || ( b × c ), then ( a × b ) . ( a × c ) is equal to
(A) a2 (b.c) (B) b2 (a.c) (C) c 2 (a.b) (D) a2 (b.c)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 46
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

16. Let 2c b = 4a 3d . If d c a and abd are natural numbers with H.C.F. equal to 1 then how
many statement are true among below six statement.
(i) abd 2 (ii) abc 3

(iii) dc b 4 (iv) dc a 1

(v) a b × c d = 2c 3d (vi) a b × c d =
(A) 2 (B) 4 (C) 6 (D) 0
17. Let u and v are unit vectors and w is a vector such that (u v) u = w and w u v then the
value of u v w is equal to
(A) 1 (B) 2 (C) 0 (D) 1

18 . Find the shortest distance between any two opposite edges of a tetrahedron formed by the planes
y + z = 0, x + z = 0, x + y = 0, x + y + z = 3 a .
(A) a (B) 2a (C) a / 2 (D) 2 a.

19. A plane meets the coordinate axes in A, B, C and ( ) is the centroid of the triangle ABC, then the
equation of the plane is
x y z x y z 3x 3y 3z
(A) 3 (B) 1 (C) 1 (D) x + y + z =1

x 1 y 2 z 3
20. Equation of plane which passes through the point of intersection of lines = = and
3 1 2
x 3 y 1 z 2
= = and at greatest distance from the point (0, 0, 0) is:
1 2 3
(A) 4x + 3y + 5z = 25 (B) 4x + 3y + 5z = 50 (C) 3x + 4y + 5z = 49 (D) x + 7y 5z = 2

21. y + 3z + 4 = 0 = ax + y z + 2 and
x 3y + z = 0 = x + 2y + z + 1 are co-planar is :
(A) 2 (B) 4 (C) 6 (D) 0

22. A line having direction ratios 3, 4, 5 cuts 2 planes 2x 3y + 6z 12 = 0 and 2x 3y + 6z + 2 = 0 at

point P & Q, then find length of PQ
35 2 35 2 35 2 35 2
(A) (B) (C) (D)
12 24 6 8

23. A line L1 having dierection ratios 1, 0, 1 lies on xz plane. Now this xz plane is rotated about z-axis by an
angle of 90°. Now the new position of L 1 is L2. The angle between L1 & L2 is :
(A) 30° (B) 60° (C) 90° (D) 45°

24._ Focii of ellipse lie on plane 2x + 3y + 6z = 0 are (54, 18, 27) and ( 54, 18, 27). If eccentricity of
3
ellipse is then find equation of directrices .
5
x 150 y 50 z 75 x 150 y 50 z 75
(A) = = , = =
3 6 2 3 6 2
x 150 y 50 z 75 x 150 y 50 z 75
(B) = = , = =
3 6 2 3 6 2
x 150 y 50 z 75 x 150 y 50 z 75
(C) = = , = =
3 6 2 3 6 2
x 150 y 50 z 75 x 150 y 50 z 75
(D) = = , = =
3 6 2 3 6 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 47
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

PART-II: NUMERICAL VALUE QUESTIONS

INSTRUCTION :
The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto two digit.
If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal
placed.

1. Given f2(x) + g2(x) + h2(x) 9 and U(x) = 3f(x) + 4g(x) + 10h(x), where f(x), g(x) and h(x) are continuous
x R then maximum value of U(x) is.

2. If in a plane A1, A2, A3,......, A20 are the vertices of a regular polygon having 20 sides and O is its centre
19
and (OA i OA i 1 ) = (OA 2 OA1 ) then | | is
i 1

| PA |2 | PB |2 | PC |2
3. In an equilateral ABC find the value of where P is any arbitrary point lying on
R2
its circumcircle, is

4. Let A = 2 i + k , B = i + j + k and C = 4 i 3 j + 7 k . If a vector satisfies

R x B C x B and R . A 0 then 2 2 2 is

x 2 y 3 z k
5. A line = = cuts the y-z plane and the x-y plane at A and B respectively. If
1 2 3

AOB = , then k, where O is the origin, is

2

6. Given four non zero vectors a, b, c and d . The vectors a, b and c are coplanar but not collinear pair

by pair and vector d is not coplanar with vectors a, b and c and ( a b ) ( b c) , (d a)

3

and (d b) , if (d c) cos 1(mcos ncos ) then m n is :

a2 b x c b2 c x a c 2 a xb
7. If the circumcentre of the tetrahedron OABC is given by , where

a, b & c are the position vectors of the points A, B, C respectively relative to the origin 'O' such that
[ a b c ] = 36 then is

8. Given three point on x y plane as O(0, 0), A(1, 0) & B( 1, 0). Point P moving on the given plane
satisfying the condition (PA . PB) + 3 (OA . OB) = 0
If the maximum & minimum values of | PA | | PB | is M & m respectively then the value of M2 + m2 is

9. If the volume of tetrahedron formed by planes whose equations are y + z = 0, z + x = 0, x + y = 0 and

x + y + z = 1 is t cubic unit, then the value of t is

10. If r represents the position vector of point R in which the line AB cuts the plane CDE, where position
vectors of points A, B, C, D, E are respectively a i 2j k , b 2i j 2k , c 4 j 4k ,
d 2i 2j 2k and e 4i j 2k , then r is :

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 48
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

11. Line L1 is parallel to vector 3i 2j 4k and passes through a point A(7, 6, 2) and line L 2 is parallel
to a vector 2i j 3k and passes through a point B(5, 3, 4). Now a line L 3 parallel to a vector
r 2i 2j k intersects the lines L1 and L2 at points C and D respectively, then | 4CD | is equal to :

12. L is the equation of the straight line which passes through the point (2, 1, 1); is parallel to the plane
4x + y + z + 2 = 0 and is perpendicular to the line of intersection of the planes 2x + y = 0 = x y + z. If
the point (3, , ) lies on line L, then absolute value of is

x 4 y 6 z 1
13. The lines = = and 3x 2y + z + 5 = 0 = 4x + 3y 4z 3k are coplanar, then the
3 5 2
value of k is

x 1 y 2 z 3
14. About the line = = the plane 3x + 4y + 6z + 7 = 0 is rotated till the plane passes
2 3 1
through the origin. Now 4x + y + z = 0 is the equation of plane in new position. The value of 2 + 2
is

15. The value of tan 3 , where is the acute angle between the plane faces of a regular tetrahedron, is

16. R and r are the circum radius and in radius of a regular tetrahedron respectively in terms of the length
k of each edge. If R2 + r2 = k2 then value of is

17. A line L on the plane 2x + y 3z + 5 = 0 is at a distance 3 unit from the point P(1, 2, 3). A spider starts
x 1 y 2 z 3
moving from point A and after moving 4 units along the line it reaches to point P.
2 1 3
and from P it jumps to line L along the shortest distance and then moves 12 units along the line L to
reach at point B. The distance between points A and B is

18. The length of edge of a regular tetrahedron D-ABC is 'a'. Point E & F are taken on the edges AD and
BD respectively. Such that E divide DA and F divide BD in the ratio 2 : 1 each. The area of CEF is
3 2
equal to a , then value of k is :
k

y z x z
19. If 'd' be the shortest distance between the lines + = 1; x = 0 and = 1; y = 0 and if
b c a c
1 1 1
2
= + 2 + 2 then is
d a2 b c

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

1. A vector a has components 2p and 1 with respect to a rectangular cartesian system. The system is
rotated through a certain angle about the origin in the counterclockwise sense. If with respect to the
new system, a has components p + 1 and 1, then
1 1
(A) p = (B) p = 1 (C) p = 1 (D) p =
3 3

2. If z1 ai b j and z2 ci d j are two vectors in i and j system, where z1 z2 = r and

z1 . z2 = 0, then w1 ai c j and w 2 b i d j satisfy :
(A) w1 = r (B) w 2 = r (C) w1 .w 2 = 0 (D) | w1 | |w2 |

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 49
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

3. If a, b, c, x, y, z R such that ax + by + cz = 2, then which of the following is always true

(A) (a2 + b2 + c2)(x2 + y2 + z2) 4 (B) (x2 + b2 + z2)(a2 + y2 + c2) 4
(C) (a2 + y2 + z2)(x2 + b2 + c2) 4 (D) (a2 + b2 + z2)(x2 + y2 + c2) 4

4. The direction cosines of the lines bisecting the angle between the lines whose direction cosines are
1
, m 1, n1 and 2, m2, n2 and the angle between these lines is , are
m m2 n n2
(A) 1 2
, 1 2
, 1 2
(B) 1 2
, 1 , 1
cos cos cos 2cos 2cos 2cos
2 2 2 2 2 2
m m n n m m n n
(C) 1 2
, 1 2
, 1 2
(D) 1 2
, 1 2
, 1 2

2 sin 2 sin 2 sin 2 sin 2 sin 2 sin

2 2 2 2 2 2

5. The value(s) of [0, 2 ] for which vector a i 3 j (sin2 ) k makes an obtuse angle with the

z-axis and the vectors b (tan ) i j 2 sin k and c = (tan ) i + (tan ) j 3 cosec k are
2 2
orthogonal, is/are
(A) tan 1 3 (B) tan 1
2 (C) + tan 1
3 (D) 2 tan 1
2

11
6. The vector i xj 3k is rotated through an angle of cos 1
and doubled in magnitude, then it
14
becomes 4i (4x 2)j 2k . The value of ' x ' CANNOT be :
2 2 20
(A) (B) (C) (D) 2
3 3 17

7. The vertices of a triangle are A (1, 1, 2), B(4, 3, 1) and C(2, 3, 5). A vector representing the bisector of
the angle A is :
(A) (B) (C) (D) 2i 2j k

8. The vector c , parallel to the internal bisector of the angle between the vectors a 7i 4j 4k and
b 2i j 2k with c = 5 6 , is :
5 5 5 5
(A) i 7j 2k (B) i 7j 2k (C) i 7j 2k (D) i 7j 2k
3 3 3 3

9. A line passes through a point A with position vector 3 i j k and is parallel to the vector
2i j 2k . If P is a point on this line such that AP = 15 units, then the position vector of the point P
is/are
(A) 13 i 4j 9k (B) 13 i 4j 9k (C) 7 i 6j 11k (D) 7 i 6j 11k
x 1
y 1 z x 1 y 3 z 1
10. Acute angle between the lines = = and = = where > m > n, and
m n m n
, m, n are the roots of the cubic equation x3 + x2 4x = 4 is equal to :
3 65 13 2
(A) cos 1 (B) sin 1 (C) 2cos 1 (D) tan 1
13 9 18 3

x 2 y 1 z 1
11. The line = = intersects the curve xy = c2, z = 0 if c is equal to :
3 2 1
(A) 1 (B) 5 (C) 5 (D) 1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 50
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2
x 1 y 2 z 3 x 1 2y 4 3z 9 x y 2 z 3
12. Three distinct lines , , are
3 2 1 5 3 1 3 2
concurrent the value of may be :
(A) 1 (B) 1 (C) 2 (D) 2

x 6 y 10 z 14
13. The line = = is the hypotenuse of an isosceles right angle triangle whose opposite
5 3 8
vertex is (7, 2, 4) Then the equation of remaining sides is/are -
x 7 y 2 z 4 x 7 y 2 z 4
(A) = = (B) = =
3 6 2 2 3 6
x 7 y 2 z 4 x 7 y 2 z 4
(C) = = (D) = =
3 6 2 2 3 6

14. Two lines are

x 1 y 1 z 1 x 1 y 1 z 1
L1 : ; L2 :
1 2 2 2 1 2
Equation of line passing through (2, 1, 3) and equally inclined to L 1 & L2 is/are
x 2 y 1 z 3 x 3 y 2 z 5
(A) (B)
2 2 3 1 1 2
x 2 y 1 z 3 x y 1 z 6
(C) (D)
1 1 3 2 2 3

15. Which of the followings is/are correct :

(A) The angle between the two straight lines r = 3 i 2 j + 4k + ( 2 i + j + 2 k ) and

1 4
r = i+3j 2k + (3 i 2 j + 6 k ) is cos
21
(B) (r.i) (i r ) (r.j) ( j r ) (r.k) (k r ) = 0 .
(C) The force determined by the vector r = (1, 8, 7) is resolved along three mutually perpendicular
directions, one of which is in the direction of the vector a 2i 2j k . Then the vector component of
7
the force r in the direction of the vector a is (2i 2j k)
3
1
(D) The cosine of the acute angle between any two diagonals of a cube is .
3

16. If the distance between points ( , 5 , 10 ) from the point of intersection of the line.
r = (2 i j+ 2k ) + (2 i 4 j + 12 k ) and plane r . ( i j + k ) = 5 is 13 units, then value of may be
80
(A) 1 (B) 1 (C) 4 (D)
63

17. A vector v = aj bk is coplanar with the vectors i j 2 k and i 2j k and is orthogonal to

the vector 2 i j k . It is given that the length of projection of v along the vector i j k is equal to
6 3 . Then the value of ab may be
2

(A) 81 (B) 9 (C) 9 (D) 81

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 51
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

18. a and b are two given unit vectors at right angle. The unit vector equally inclined with a , b and
a × b will be:
1 1 1 1
(A) a b a b (B) a b a b (C) a b a b (D) a b a b
3 3 3 3

19. Let a = , b= and c = be three vectors. A vector in the plane of b and c

2
whose length of projection on a is of , is
3
(A) 2i + 3j 3k (B) 2i + 3j + 3k (C) 2i j + 5k (D) i 5j + 3k

20. The position vectors of the angular points of a tetrahedron are A(3i 2j k) , B(3i j 5k) , C(4i 3k)
and D(i) . Then the acute angle between the lateral face ADC and the base face ABC is :
5 2 5 2
(A) tan 1
(B) tan 1
(C) cot 1 (D) cot 1
2 5 2 5

3
21. If a, b, c and d are unit vectors such that (a b) . (c d) = and a . c = , then
2
(A) a , b, c are coplanar if =1 (B) Angle between b and d is 30° if = 1
(C) angle between b and d is 150° if = 1 (D) If = 1 then angle between b and c is 60°

22. The volume of a right triangular prism ABCA1B1C1 is equal to 3. If the position vectors of the vertices of
the base ABC are A(1, 0, 1), B(2,0, 0) and C(0, 1, 0), then position vectors of the vertex A 1 can be:
(A) (2, 2, 2) (B) (0, 2, 0) (C) (0, 2, 2) (D) (0, 2, 0)

23. The coplanar points A , B , C , D are (2 x , 2 , 2) , (2 , 2 y , 2) , (2 , 2 , 2 z) and (1 , 1 , 1)

respectively, then
1 1 1 x 1 y 1 z 1
(A) =1 (B) =2
x y z x y z
1 1 1 x y z
(C) + + =1 (D) + + +2=0
1 x 1 y 1 z 1 x 1 y 1 z

24. Which of the following statement(s) is/are correct :

(A) If a,b,c are non coplanar and d is any vector, then
[d b c] a [d c a] b [d a b] c [a b c] d 0
(B) If is incentre of ABC then | BC | A | CA | B | AB | C= 0
(C) Any vector in three dimension can be written as linear combination of three non coplanar vectors.
a b c
(D) In a triangle, if position vector of vertices are a,b,c , then position vector of incentre is .
3
25. Let V = 2i j k& W=i 3k . If U is a unit vector, then the value of the scalar triple product
U V W may be :

(A) 59 (B) 10 6 (C) 59 (D) 60

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 52
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

26. If A B = a , A . a = 1 and A B = b , then

a b a | a |2 1
(A) A = a b a (B) B =
| a |2 | a |2
b a a | a |2 1
(C) A = b a a (D) B =
| a |2 | a |2

x 4 y 2 z k2
27. The line = = lies in the plane 2 x 4 y + z = 1 . Then the value of k
k 1 2
cannot be :
(A) 1 (B) 1 (C) 2 (D) 2
x x2 y y2 z z2
28. Equation of the plane passing through A(x1, y1, z1) and containing the line = = is
d1 d2 d3
x x1 y y1 z z1 x x2 y y2 z z2
(A) x 2 x1 y2 y1 z 2 z1 = 0 (B) x1 x2 y1 y 2 z1 z 2 = 0
d1 d2 d3 d1 d2 d3
x d1 y d2 z d3 x y z
(C) x1 y1 z1 = 0 (D) x1 x2 y1 y 2 z1 z 2 = 0
x2 y2 z2 d1 d2 d3

29. A line = = intersects the plane x y + 2z + 2 = 0 at point A. The equation of the

2 3 4
straight line passing through A lying in the given plane and at minimum inclination with the given line
is/are
x 1 y 1 z 1
(A) = = (B) 5x y + 4 = 0 = 2y 5z 3
1 5 2
x 2 y 6 z 3
(C) 5x + y 5z + 1 = 0 = 2y 5z 3 (D) = =
1 5 2
30. If the -plane 7x + ( + 4)y + 4z r = 0 passing through the points of intersection of the planes
2x 3y z +1= 0 and x y 2z + 3 = 0 and is perpendicular to the plane 3x y 2z = 4 and
12 78 57
, , is image of point (1, 1, 1) in plane, then

(A) =9 (B) = 117 (C) = 9 (D) = 117

31. The planes 2x 3y 7z = 0, 3x 14y 13z = 0 and 8x 31y 33z = 0
(A) pass through origin (B) intersect in a common line
(C) form a triangular prism (D) pass through infinite the many points

32. If a,b,c and d are the position vectors of the points A, B, C and D respectively in three dimensional
space no three of A, B, C, D are collinear and satisfy the relation 3a 2b c 2d = 0 , then :
(A) A, B, C and D are coplanar
(B) The line joining the points B and D divides the line joining the point A and C in the ratio 2 : 1.
(C) The line joining the points A and C divides the line joining the points B and D in the ratio 1 : 1.
(D) The four vectors a,b,c and d are linearly dependents.
PART - IV : COMPREHENSION
Comprehension # 1
In a parallelogram OABC, vectors a, b, c are respectively the position vectors of vertices A, B, C with
reference to O as origin. A point E is taken on the side BC which divides it in the ratio of 2 : 1 internally.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 53
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Also, the line segment AE intersect the line bisecting the angle O internally in point P. If CP, when
extended meets AB in point F. Then
1. The position vector of point P, is
3a c a c a c a c
(A) (B)
3c 2a a c 3c 2a a c

2a c a c 3a c a c
(C) (D)
3c 2a a c 3c 2a a c

2. The position vector of point F, is

1 a a 2a a
(A) a c (B) a c (C) a c (D) a c
3 c c c c

3. The vector AF , is given by

a a 2a 1 a
(A) c (B) c (C) c (D) c
c c c 3 c

Comprehension # 2
Let a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 be two planes, where d1, d2 > 0. Then origin lies
in acute angle if a1a2 + b1b2 + c1c2 < 0 and origin lies in obtuse angle if a1a2 + b1b2 + c1c2 > 0.
Further point (x 1, y1, z1) and origin both lie either in acute angle or in obtuse angle ,
if (a 1x1 + b1y1 + c 1z1 + d1) (a2x1 + b2y1 + c 2z1 + d2) > 0, one of (x1, y1, z1) and origin lie in acute angle and
the other in obtuse angle, if (a1x1 + b1y1 + c1z1 + d1) (a2x1 + b2y1 + c2z1 + d2) < 0
4. Given the planes 2x + 3y 4z + 7 = 0 and x 2y + 3z 5 = 0, if a point P is (1, 2, 3) and O is origin,
then
(A) O and P both lie in acute angle between the planes
(B) O and P both lie in obtuse angle between the planes
(C) O lies in acute angle, P lies in obtuse angle.
(D) O lies in obtue angle, P lies in acute angle.
5. Given the planes x + 2y 3z + 5 = 0 and 2x + y + 3z + 1 = 0. If a point P is (2, 1, 2) and O is origin,
then
(A) O and P both lie in acute angle between the planes
(B) O and P both lie in obtuse angle between the planes
(C) O lies in acute angle, P lies in obtuse angle. (D) O lies in obtue angle, P lies in acute angle.

6. Given the planes x + 2y 3z + 2 = 0 and x 2y + 3z + 7 = 0, if the point P is (1, 2, 2) and O is origin,

then
(A) O and P both lie in acute angle between the planes
(B) O and P both lie in obtuse angle between the planes
(C) O lies in acute angle, P lies in obtuse angle. (D) O lies in obtue angle, P lies in acute angle.

Comprehension # 3

If a, b, c & a',b',c' are two sets of non-coplanar vectors such that a.a' =b.b'=c.c' = 1 , then the two
b x c c x a a x b
systems are called Reciprocal System of vectors and a = ,b and c .
[a b c ] [a b c] [ a b c]

w7. Find the value of a a b b c c .

(A) 0 (B) a b c (C) a b c (D) a b c

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 54
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

a b c
8. Find value of such that a b + b c + c a = .
[abc]
(A) 1 (B) 1 (C) 2 (D) 2

9. If [(a × b ) × (b × c ) (b × c ) × (c × a ) (c × a ) × (a × b )] = [abc] n , then find n.

(A) n = 4 (B) n = 4 (C) n = 3 (D) n = 3

Comprehension # 4
The vertices of square pyramid are A(0, 0, 0), B(4, 0, 0), C(4, 0, 4), D(0, 0, 4) and E(2, 6, 6)
10. Volume of the pyramid is :
(A) 32 (B) 16 (C) 8 (D) 4

11. Centroids of triangular faces of square pyramid are

(A) Non-coplanar (B) Coplanar but the plane is not parallel to base plane
(C) Coplanar & plane is parallel to base plane (D) Co-linear

12. The distance of the plane EBC from ortho-centre of ABD is :

12
(A) 2 (B) 5 (C) (D) 10
10
Comprehension # 5

General equation of a sphere is given by x 2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, where ( u, v, w) is

the centre and u2 v2 w2 d is the radius of the sphere.

Let P be a any plane and F is the foot of perpendicular from centre(C) of the sphere to this plane.
If CF > u2 v2 w2 d then plane P neither touches nor cuts the sphere.
2 2 2
If CF = u v w d then plane P touches the sphere.
2 2 2
If CF < u v w d then intersection of plane P and sphere is a circle with
2 2 2 2
radius =

13. Find the equation of the sphere having centre at (1, 2, 3) and touching the plane
x + 2y + 3z = 0.
(A) x2 + y2 + z2 2x 4y 6z = 0 (B) x2 + y2 + z2 2x + 4y 6z = 0
(C) x2 + y2 + z2 2x 4y + 6z = 0 (D) x2 + y2 + z2 + 2x 4y 6z = 0
14. A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn
from origin to this plane is:
(A) x2 + y2 + z2 x 2y 3z = 0 (B) x2 + 2y2 + 3z2 x 2y 3z = 0
(C) x + 4y + 9z + x + 2y + 3 = 0
2 2 2
(D) x2 + y2 + z2 + x + 2y + 3z = 0

x 1 y 2 z 3
15. Find the length of the chord intercepted on the line by the sphere
1 2 3
22
x2 + y2 + z2 2x 2y z = 0.
3
(A) 56 (B) 54 (C) 9 (D) 6

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 55
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

* Marked Question s may hav e more th an on e correct option.

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

1. Let P, Q, R and S be the points on the plane with position vectors 2i j , 4 i , 3 i + 3 j and
3 i + 2 j respectively. The quadrilateral PQRS must be a IIT-JEE-2010, Paper-1, (3, 1), 84]
(A) parallelogram, which is neither a rhombus nor a rectangle
(B) square
(C) rectangle, but not a square
(D) rhombus, but not a square

2i j 3k
2. If a and b are vectors in space given by a = and b = , then the value of
5 14
2a b . is [IIT-JEE-2010, Paper-1, (3, 0), 84]

3. Two adjacent sides of a parallelogram ABCD are given by [IIT-JEE-2010, Paper-2, (5, 2), 79]
AB = 2i 10j 11k and AD = i 2j 2k . The side AD is rotated by an acute angle in the plane
of the parallelogram so that AD becomes AD . If AD makes a right angle with the side AB, then the
cosine of the angle is given by
8 17 1 4 5
(A) (B) (C) (D)
9 9 9 9

x y z
4. Equation of the plane containing the straight line = = and perpendicular to the plane
2 3 4
x y z x y z
containing the straight lines = = and = = is [IIT-JEE-2010, Paper-1, (3, 1), 84]
3 4 2 4 2 3
(A) x + 2y 2z = 0 (B) 3x + 2y 2z = 0 (C) x 2y + z = 0 (D) 5x + 2y 4z = 0

x 1
5. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A y = 0
z 0
has exactly two distinct solutions, is [IIT-JEE-2010, Paper-1, (3, 1), 84]

(A) 0 (B) 29 1 (C) 168 (D) 2

6. If the distance between the plane Ax 2y + z = d and the plane containing the lines
= = and = = is 6 , then |d| is
2 3 4 3 4 5
[IIT-JEE-2010, Paper-1, (3, 0), 84]

7. If the distance of the point P(1, 2, 1) from the plane x + 2y 2z = , where > 0, is 5, then the foot of
the perpendicular from P to the plane is [IIT-JEE-2010, Paper-2, (5, 2), 79]
8 4 7 4 4 1 1 2 10 2 1 5
(A) , , (B) , , (C) , , (D) , ,
3 3 3 3 3 3 3 3 3 3 3 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 56
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

8. Match the statements in Column-I with those in Column-II. [IIT-JEE-2010, Paper-2, (8, 0), 79]
Column-I Column-II
(A) A line from the origin meets the lines (p) 4
8
z 1 3 = y 3 =
= = and
1 1 2 1
at P and Q respectively. If length PQ = d, then d 2 is
(B) The values of x satisfying (q) 0
3
tan 1(x + 3) tan 1(x 3) = sin 1 are
5
(C) Non-zero vectors a , b and c satisfy a . b = 0, (r) 4
and 2 . If a µb 4c
then possible value of µ are
(D) Let f be the function on [ , ] given by (s) 5
9x
sin
2
f(0) = 9 and f(x) = for x 0. The value (t) 6
x
sin
2
2
of f(x) dx is

9. Let a i j k, b i j k and c i j k be three vectors. A vector in the plane of a and b ,

1
whose projection on c is , is given by [IIT-JEE 2011, Paper-1, (3, 1), 80]
3
(A) i 3 j 3k (B) 3i 3j k (C) 3i j 3k (D) i 3j 3k

10*. The vector(s) which is/are coplanar with vectors i j 2k and i 2j k , and perpendicular to the
vector i j k is/are [IIT-JEE 2011, Paper-1, (4, 0), 80]
(A) j k (B) i j (C) i j (D) j k

11. Let a i k, b i j and c i 2j 3k be three given vectors. If r is a vector such that

r b c b and r . a 0 , then the value of r . b is [IIT-JEE 2011, Paper-2, (4, 0), 80]

12. Match the statements given in Column-I with the values given in Column-II
[IIT-JEE 2011, Paper-2, (8, 0), 80]
Column-I Column-II
(A) If a j 3 k, b j 3 k and c 2 3 k form a triangle, (p)
6
then the internal angle of the triangle between a and b is
b
2
(B) If (f(x) 3x) dx = a2 b2, then the value of f is (q)
a
6 3
2 5/6
(C) The value of sec ( x) dx is (r)
ln3 7 / 6 3
1
(D) The maximum value of Arg for |z| = 1, z 1 is given by (s)
1 z

(t)
2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 57
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

13. The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1, 1, 4) with the
plane 5x 4y z = 1. If S is the foot of the perpendicular drawn from the point T(2, 1,4) to QR, then the
length of the line segment PS is [IIT-JEE 2012, Paper-1, (3, 1), 70]
1
(A) (B) 2 (C) 2 (D) 2 2
2

14. If a, b and c are unit vectors satisfying | a b |2 + | b c |2 + | c a |2 = 9, then |2 a + 5 b + 5 c |

is
[IIT-JEE 2012, Paper-1, (4, 0), 70]

15. The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and
2
x y + z = 3 and at a distance from the point (3, 1, 1) is [IIT-JEE 2012, Paper-2, (3, 1), 66]
3
(A) 5x 11y + z = 17 (B) (C) x + y + z = 3 (D) x 2y = 1 2

16. If a and b are vectors such that a b 29 and a (2i 3j 4k) = (2i 3j 4k) × b , then a
possible value of (a b) . is [IIT-JEE 2012, Paper-2, (3, 1), 66]
(A) 0 (B) 3 (C) 4 (D) 8

x 1 y 1 z x 1 y 1 z
17*. If the straight lines = = and = = are coplanar, then the plane(s)
2 k 2 5 2 k
containing these two lines is(are) [IIT-JEE 2012, Paper-2, (4, 0), 66]
(A) y + 2z = 1 (B) y + z = 1 (C) y z = 1 (D) y 2z = 1

18. Let and determine diagonals of a parallelogram PQRS and

PT i 2j 3k be another vector. Then the volume of the parallelepiped determined by the vectors
PT, PQ and PS is [JEE (Advanced) 2013, Paper-1, (2, 0)/60]
(A) 5 (B) 20 (C) 10 (D) 30

x 2 y 1 z
19. Perpendicular are drawn from points on the line to the plane x + y + z = 3. The feet of

perpendiculars lie on the line [JEE (Advanced) 2013, Paper-1, (2, 0)/60]
(A) (B) (C) (D)

20.* A line l passing through the origin is perpendicular to the lines

l1 : (3 + t) i + ( 1 + 2t) j + (4 + 2t) k , <t<
l2 : (3 + 2s) i + (3 + 2s) j + (2 + s) k , <s<
Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and
l1 is(are) [JEE (Advanced) 2013, Paper-1, (4, 1)/60]
7 7 5 7 7 8
(A) , , (B) ( 1, , 1, 0) (C) (1, 1, 1) (D) , ,
3 3 3 9 9 9

21. Consider the set of eight vectors V = . Three non-coplanar vectors can be

chosen from V in 2p ways. Then p is [JEE (Advanced) 2013, Paper-1, (4, 1)/60]
y z y z
22.* Two lines L1 : x = 5, = and L2 : x = , = are coplanar. Then can take value(s)

[JEE (Advanced) 2013, Paper-2, (3, 1)/60]

(A) 1 (B) 2 (C) 3 (D) 4
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 58
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

23. Match List I with List II and select the correct answer using the code given below the lists :
[JEE (Advanced) 2013, Paper-2, (3, 1)/60]
List - I List - II
(P) Volume of parallelopiped determined by vectors (1) 100
a,b and c is 2. Then the volume of the parallelepiped
determined by vectors 2(a b),3(b c) and (c a) is
(Q) Volume of parallelepiped determined by vectors a,b (2) 30
and c is 5. Then the volume of the parallelepiped
determined by vectors 3(a b),(b c) and 2 (c a) is
(R) Area of a triangle with adjacent sides determined by (3) 24
vectors a and b is 20. Then the area of the triangle
with adjacent sides determined by vectors 2a 3b
and is
(S) Area of a paralelogram with adjacent sides determined by (4) 60
vectors a and b is 30. Then the area of the parallelogram
with adjacent sides determined by vectors (a b) and a is
Codes :
P Q R S
(A) 4 2 3 1
(B) 2 3 1 4
(C) 3 4 1 2
(D) 1 4 3 2

24. Consider the lines L1 : , L2 : and the planes P 1 : 7x + y + 2z = 3,

1 1 2
P2 : 3x + 5y 6z = 4. Let ax + by + cz = d the equation of the plane passing through the point of
intersection of lines L1 and L2, and perpendicular to planes P1 and P2.
Match List - I with List- II and select the correct answer using the code given below the lists :
[JEE (Advanced) 2013, Paper-2, (3, 1)/60]
List- I List- II
P. a= 1. 13
Q. b= 2. 3
R. c= 3. 1
S. d= 4. 2
Codes :
P Q R S
(A) 3 2 4 1
(B) 1 3 4 2
(C) 3 2 1 4
(D) 2 4 1 3

25*. Let x,y and z be three vectors each of magnitude 2 and the angle between each pair of them is .
3
If a is a nonzero vector perpendicular to x and y z and b is a nonzero vector perpendicular to y
and z x , then [JEE (Advanced) 2014, Paper-1, (3, 0)/60]
(A) (B) (C) (D)

26. From a point P( , , ), perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and
y = x, z = 1. If P is such that QPR is a right angle, then the possible value(s) of is(are)
[JEE (Advanced) 2014, Paper-1, (3, 0)/60]
(A) 2 (B) 1 (C) 1 (D) 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 59
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

27. Let a, b and c be three non-coplanar unit vectors such that the angle between every pair of them is
p2
2q2 r 2
. If a b b c pa qb rc , where p,q and r are scalars, then the value of is
3 q2
[JEE (Advanced) 2014, Paper-1, (3, 0)/60]

28. [JEE (Advanced) 2014, Paper-2, (3, 1)/60]

List I List II
3
P. Let y(x) = cos(3 cos 1 x), x [ 1, 1], x ± . Then 1. 1
2
1 2 d2 y(x) dy(x)
equals
y(x) dx2 dx
Q. Let A1, A2,......, An (n > 2) be the vertices of a regular polygon of n 2. 2
sides with its centre at the origin. Let ak be the position vector of

the point Ak, k = 1, 2,...., n. If (ak ak 1 ) (ak .ak 1 ) , then

k 1 k 1

the minimum value of n is

x2 y2
R. If the normal from the point P(h, 1) on the ellipse 1 is 3. 8
6 3
perpendicular to the line x + y = 8, then the value of h is
S. Number of positive solutions satisfying the equation 4. 9
1 1 2
is tan tan tan
2x 1 4x 1 x2
P Q R S
(A) 4 3 2 1
(B) 2 4 3 1
(C) 4 3 1 2
(D) 2 4 1 3

29* . In R3, consider the planes P 1 : y = 0 and P 2 : x + z = 1. Let P3 be a plane, different from P 1 and P2,
which passes through the intersection of P1 and P2 . If the distance of the point (0, 1, 0) from P3 is 1
and the distance of a point ( , , ) from P3 is 2, then which of the following relation is (are) true ?
[JEE(Advanced)2015,P-1 (4, 2)/ 88]
(A) 2 + + 2 + 2 = 0 (B) 2 +2 +4=0
(C) 2 + 2 10 = 0 (D) 2 +2 8=0

30* . In R3, let L be a straight line passing through the origin. Suppose that all the points on L are at a
constant distance from the two planes P 1 : x + 2y z + 1 = 0 and P2 : 2x y + z 1 = 0. Let M be the
locus of the feet of the perpendiculars drawn from the points on L to the plane P 1. Which of the
following points lie(s) on M? [JEE(Advanced)2015,P-1 (4, 2)/ 88]
5 2 1 1 1 5 1 1 2
(A) (B) (C) (D)
6 3 6 3 6 6 6 3 3

31*. Let PQR be a triangle. Let a QR, b RP and c PQ . If | a | = 12, | b | 4 3 and b.c 24 , then
which of the following is(are) true? [JEE(Advanced)2015, P-1 (4, 2)/88]
| c |2 | c |2
(A) | a | 12 (B) | a | 30 (C) | a b c a | 48 3 (D) a. b 72
2 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 60
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

32. Column- Column-

[JEE (Advanced) 2015, P-1 (2, 1)/ 88]
(A) In R2, if the magnitude of the projection vector of the vector (P) 1
i j on 3i j is 3 and if = 2 + 3 ,
then possible value(s) of | | is (are)

(B) Let a and b be real numbers such that the function (Q) 2
2
, x 1
f(x) is differentiable for all x R.
bx a2 , x 1
Then possible value(s) of a is (are)

(C) Let 1 be a complex cube root of unity. (R) 3

If (3 3 + 2 2)4n + 3 + (2 + 3 3 2)4n + 3 + ( 3 + 2 + 3 )
2 4n + 3
= 0,
then possible value(s) of n is (are)

(D) Let the harmonic mean of two positive real numbers a and b be 4. (S) 4
If q is a positive real number such that a, 5,q, b is an arithmetic
progression, then the value(s) of |q a| is (are)
(T) 5

33. Column- Column-

(A) In a triangle XYZ, let a, b and c be the lengths of the sides (P) 1
opposite to the angles X, Y and Z, respectively. If 2(a2 b2) = c2
and = , then possible values of n for which
sinZ
cos(n ) = 0 is (are)
(B) In a triangle XYZ, let a, b and c be the lengths of the sides (Q) 2
opposite to the angles X, Y and Z, respectively. If
a
1 + cos2X 2cos2Y = 2sinXsinY, then possible value(s) of is (are)
b

(C) In R2, let 3 i j, i 3 j and be the position vectors (R) 3

of X, Y and Z with respect to the origin O, respectively. If the
distance of Z from the bisector of the acute angle of OX and OY

3
is , then possible value(s) of | | is (are)
2

(D) Suppose that F( ) denotes the area of the region bounded by (S) 5
x = 0, x = 2, y2 = 4x and y = | x 1| + | x 2| + x, where
8
{0, 1}. Then the value(s) of F( ) + 2 , when = 0 and
3
= 1, is (are)
(T) 6
[JEE (Advanced) 2015, P-1 (2, 1)/ 88]

34. Suppose that p,q and r are three non-coplanar vectors in R3. Let the components of a vector s
along p,q and r be 4, 3 and 5, respectively. If the components of this vector s
along and are x, y and z, respectively, then the value of 2x + y + z is
[JEE (Advanced) 2015, P-2 (4, 0) / 80]

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 61
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

35*. Consider a pyramid OPQRS located in the first octant (x 0, y 0, z 0) with O as origin, and OP and
OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with
OP = 3. The point S is directly above the mid point T of diagonal OQ such that TS = 3. Then

(A) the acute angle between OQ and OS is [JEE (Advanced) 2016, Paper-1, (4, 2)/62]
3
(B) the equation of the plane containing the triangle OQS is x y = 0
3
(C) the length of the perpendicular from P to the plane containing the triangle OQS is
2
15
(D) the perpendicular distance from O to the straight line containing RS is
2

36. Let P be the image of the point (3, 1, 7) with respect to the plane x y + z = 3. Then the equation of the
x y z
plane passing through P and containing the straight line is
1 2 1
[JEE (Advanced) 2016, Paper-2, (3, 1)/62]
(A) x + y 3z = 0 (B) 3x + z = 0 (C) x 4y + 7z = 0 (D) 2x y = 0

1 r
37*. Let u = u1i + u2 j + u3k be a unit vector in R3 and w =
6
(i + j + 2k) . Given that there exists a vector u
r r
in R3 such that u 1 and w.(u ) 1. Which of the following statements(s) is (are) correct?
r
(A) There is exactly one choice for u
r
(B) There are infinitely many choices for such u [JEE (Advanced) 2016, Paper-2, (4, 2)/62]
(C) If u lies in the xy-plane then u1 = u2
(D) If u lies in the xz-plane then 2 u1 = u3

38. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that
OP . OQ + OR . OS = OR . OP + OQ . OS = OQ . OR + OP . OS
Then the triangle PQR has S as its [JEE(Advanced) 2017, Paper-2,(3, 1)/61]
(A) centroid (B) orthocenter (C) incentre (D) circumcenter

39. The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes
2x + y 2z = 5 and 3x 6y 2z = 7, is [JEE(Advanced) 2017, Paper-2,(3, 1)/61]
(A) 14x + 2y 15z = 1 (B) 14x + 2y + 15z = 3
(C) 14x 2y + 15z = 27 (D) 14x + 2y + 15z = 31

Comprehension (Q.40 & 41)

Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides QR, RP, PQ ,
respectively, of a triangle PQR.

40. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is
[JEE(Advanced) 2017, Paper-2,(3, 0)/61]
3 3 5 5
(A) (B) (C) (D)
2 2 3 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 62
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

41. | OX OY| = [JEE(Advanced) 2017, Paper-2,(3, 0)/61]

(A) sin(P + Q) (B) sin(P + R) (C) sin(Q + R) (D) sin2R

42*. Let P1 : 2x + y z = 3 and P2 : x + 2y + z = 2 be two planes. Then, which of the following statement(s) is
(are) TRUE? [JEE(Advanced) 2018, Paper-1,(4, 2),60]
(A) The line of intersection of P1 and P2 has direction ratios 1, 2, 1
3x 4 1 3y z
(B) The line = = is perpendicular to the line of intersection of P1 and P2
9 9 3
(C) The acute angle between P1 and P2 is 60º
(D) If P3 is the plane passing through the point (4, 2, 2) and perpendicular to the line of intersection of
2
P1 and P2, then the distance of the point (2, 1, 1) from the plane P 3 is
3

43. Let a and b be two unit vectors such that a . b 0 . For some x, y R, let c xa yb (a b) . If
| c | = 2 and the vector c is inclined at the same angle to both a and b , then the value of 8 cos2 is
_______. [JEE(Advanced) 2018, Paper-1, (3, 0), 60]

44. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is
perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the
z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length
of PR is _______ . [JEE(Advanced) 2018, Paper-2,(3, 0)/60]

45. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis and
1 1 1
z-axis, respectively, where O(0, 0, 0) is the origin. Let S , , be the centre of the cube and T be
2 2 2
the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p = SP , q =
SQ , r = SR and t = ST , then the value of |( p × q ) × ( r × t )| is ______ .
[JEE(Advanced) 2018, Paper-2,(3, 0), 60]

46*. Let L1 and L2 denote the lines r = i + ( i + 2 j + 2 k ), R and r = (2 i j + 2 k ), R

respectively. If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of
the following options describe(s) L 3 ? [JEE(Advanced) 2019, Paper-1,(4, 1)/62]
1
(A) r = (2 i + k ) + t(2 i + 2 j k ), t R
3
2
(B) r = (4 i + j + k ) + t(2 i + 2 j k ), t R
9
2
(C) r = (2 i j + 2 k ) + t(2 i + 2 j k ), t R
9

(D) r = t(2 i + 2 j k ), t R

47. Three lines are given by

r i, R
r (i j) , n R and r (i j k ), R
Let the lines cut the plane x + y + z = 1 at he points A, B and C respectively. If the area of the triangle
ABC is then the value of (6 )2 [JEE(Advanced) 2019, Paper-1,(4, 1)/62]

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 63
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

48*. Three lines

L1 : r = i , R
L2 : r = k + µ j , µ R and
L3 : r = i j k, R.
are given. For which point(s) Q on L 2 can we find a point P on L1 and a point R on L3 so that P,Q and R
are collinear. [JEE(Advanced) 2019, Paper-2 ,(4, 1)/62]
1 1
(A) k j (B) k j (C) k (D) k j
2 2

49. Let a = 2 i j k and b i 2 j k be two vectors. Consider a vector c a b, , R. If the

projection of c on the vector (a b) is 3 2 , then the minimum value of (c (a b)) .c equal to
[JEE(Advanced) 2019, Paper-2 ,(4, 1)/62]

PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)

1. Let and . Then the vector b satisfying a b c 0 and a.b 3 is

[AIEEE 2010 (4, 1), 144]
(1) (2) (3) (4)

2. If the vectors , b 2 i 4j k and c i j µk are mutually orthogonal, then

( , µ) = [AIEEE 2010 (4, 1), 144]
(1) (2, 3) (2) ( 2, 3) (3) (3, 2) (4) ( 3, 2)

3. Statement -1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x y + z = 5.
Statement -2 : The plane x y + z = 5 bisects the line segment joining A(3, 1,6) and B(1, 3, 4).
(1) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement -1.
(2) Statement-1 is true, Statement-2 is false. [AIEEE 2009 (4, 1), 144]
(3) Statement -1 is false, Statement -2 is true.
(4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.

4. A line AB in three-dimensional space makes angles 45º and 120º with the positive x-axis and the
positive
y-axis respectively. If AB makes an acute angle with the positive z-axis, then equal
[AIEEE 2010 (4, 1), 144]
(1) 45º (2) 60º (3) 75º (4) 30º
1 1
5. If a (3i k) and b (2i 3 j 6k) , then the value of (2a b) . [(a b) (a 2b)] is:
10 7
[AIEEE 2011, I, (4, 1), 120]
(1) 5 (2) 3 (3) 5 (4) 3

6. The vectors a and b are not perpendicular and c and d are two vectors satisfying : b c b d and
a.d 0 . Then the vector d is equal to : [AIEEE 2011, I, (4, 1), 120]
b.c a.c b.c a.c
(1) b c (2) c b (3) b c (4) c b
a.d a.b a.b a.b

7. If the vector p i + j + k , i + q j + k and i + j + r k (p q r 1) are coplanar, then the value of

pqr (p+q+r) is-
[AIEEE 2011, II, (4, 1), 120]
(1) 2 (2) 0 (3) 1 (4) 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 64
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

8 . Let a , b , c be three non-zero vectors which are pairwise non-collinear. If a + 3 b is collinear with c
and b + 2 c is collinear with a , then a + 3 b + 6 c is : [AIEEE 2011, II, (4, 1), 120]
(1) a (2) c (3) 0 (4) a + c

y 1 z 3 5
9. If the angle between the line x = = and the plane x + 2y + 3z = 4 is cos 1
, then
2 14
equals :
[AIEEE 2011, I, (4, 1), 120]
2 3 2 5
(1) (2) (3) (4)
3 2 5 3

10. Statement-1 : The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line :
x y 1 z 2
[AIEEE 2011, I, (4, 1), 120]
1 2 3
x y 1 z 2
Statement-2 : The line : bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).
1 2 3
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true.

11. The distance of the point (1, 5, 9) from the plane x y + z = 5 measured along a straight line
x = y = z is : [AIEEE 2011, II, (4, 1), 120]
(1) 10 3 (2) 5 3 (3) 3 10 (4) 3 5

x y 2 z 3
12. The length of the perpendicular drawn from the point (3, 1, 11) to the line = = is :
2 3 4
[AIEEE 2011, II, (4, 1), 120]
(1) 29 (2) 33 (3) 53 (4) 66

13. Let a and b be two unit vectors. If the vectors c a 2b and d 5a 4b are perpendicular to each
other, then the angle between a and b is : [AIEEE-2012, (4, 1)/120]
(1) (2) (3) (4)
6 2 3 4

14. A equation of a plane parallel to the plane x 2y + 2z 5 = 0 and at a unit distance from the origin is :
[AIEEE 2012, (4, 1), 120]
(1) x 2y + 2z 3=0 (2) x 2y + 2z + 1 = 0 (3) x 2y + 2z 1 = 0 (4) x 2y + 2z + 5 = 0

x 1 y 1 z 1 x 3 y k z
15. If the line = = and = = intersect, then k is equal to :
2 3 4 1 2 1
[AIEEE 2012, (4, 1), 120]
2 9
(1) 1 (2) (3) (4) 0
9 2

16. Let ABCD be a parallelogram such that AB q , AD p and BAD be an acute angle. If r is the
vector that coincides with the altitude directed from the vertex B to the side AD, then is given by :
[AIEEE-2012, (4, 1)/120]
3(p . q) p . q p . q 3(p . q)
(1) r 3q p (2) r q p (3) r q p (4) r 3q p
(p . p) p . p p . p (p . p)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 65
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

17. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

[AIEEE - 2013, (4, 1),360]
3 5 7 9
(1) (2) (3) (4)
2 2 2 2

18. If the lines = = and = = are coplanar, then k can have

1 1 k 2 1
[AIEEE - 2013, (4, 1),360]
(1) any value (2) exactly one value (3) exactly two values (4) exactly three values

19. If the vectors AB 3i 4k and are the sides of a triangle ABC, then the length of
the median through A is [AIEEE - 2013, (4, 1),360]
(1) 18 (2) 72 (3) 33 (4) 45

x 1 y 3 z 4
20. The image of the line in the plane 2x y + z + 3 = 0 is the line :
3 1 5
[JEE(Main)2014,(4, 1),120]
x 3 y 5 z 2 x 3 y 5 z 2
(1) (2)
3 1 5 3 1 5
x 3 y 5 z 2 x 3 y 5 z 2
(3) (4)
3 1 5 3 1 5

21 . The angle between the lines whose direction cosines satisfy the equations + m + n = 0 and
2
= m2 + n2 is [JEE(Main)2014,(4, 1), 120]
(1) (2) (3) (4)
6 2 3 4
2
22. If a b b c c a a b c then is equal to [JEE(Main)2014,(4, 1), 120]
(1) 0 (2) 1 (3) 2 (4) 3
y 1
23. The distance of the point (1,0,2) from the point of intersection of the line = = and the
3 4 12
plane x y + z = 16, is [JEE(Main)2015,(4, 1), 120]
(1) 2 14 (2) 8 (3) 3 21 (4) 13

24. The equation of the plane containing the line 2x 5y + z = 3, x + y + 4z = 5 and parallel to the plane
x + 3y + 6z = 1, is [JEE(Main)2015,(4, 1), 120]
(1) 2x + 6y + 12z = 13 (2) x + 3y + 6z = 7
(3) x + 3y + 6z = 7 (4) 2x + 6y + 12z = 13

25. Let a, b and c be three non-zero vectors such that no two of them are collinear and
1
(a b) c | b || c | a . If is the angle between vectors b and c , then a value of sin is
3
[JEE(Main)2015,(4, 1), 120]
2 2 2
(1) (2) (3) (4)
3 3 3 3

26. If the line, lies in the plane, lx + my z = 9, then l2 + m2 is equal to

[JEE(Main)2016,(4, 1),120]
(1) 18 (2) 5 (3) 2 (4) 26

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 66
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

3
27. Let a, b and c be three unit vectors such that a b c = (b c) . If b is not parallel to c , then
2
the angle between a and b is [JEE(Main)2016,(4, 1), 120]
2 5 3
(1) (2) (3) (4)
2 3 6 4
28. The distance of the point (1, 5, 9) from the plane x y + z = 5 measured along the line x = y = z is
[JEE(Main)2016,(4, 1), 120]
10 20
(1) 10 3 (2) (3) (4) 3 10
3 3

29. If the image of the point P(1, 2, 3) in the plane, 2x + 3y 4z + 22 = 0 measured parallel to the line,
x y z
= = is Q, then PQ is equal to : [JEE(Main)2017,(4, 1), 120]
1 4 5
(1) 3 5 (2) 2 42 (3) 42 (4) 6 5
30. The distance of the point (1, 3, 7) from the plane passing through the point (1, 1, 1), having normal
x 1 x 2 x 4 x 2 y 1 z 7
perpendicular to both the lines = = and , is
1 2 3 2 1 1
[JEE(Main)2017,(4, 1), 120]
20 10 5 10
(1) (2) (3) (4)
74 83 83 74

31. Let and b i j . Let c be a vector such that = 3, a b c = 3 and the angle

between c and a b be 30º. Then a.c is equal to [JEE(Main)2017,(4, 1), 120]

25 1
(1) (2) 2 (3) 5 (4)
8 8
32. If L1 is the line of intersection of the planes 2x 2y + 3z 2 = 0, x y + z + 1 = 0 and L 2 is the line of
intersection of the planes x + 2y z 3 = 0, 3x y + 2z 1 = 0, then the distance of the origin from the
plane, containing the lines L1 and L2 , is : [JEE(Main)2018, (4, 1), 120]
1 1 1 1
(1) (2) (3) (4)
2 2 2 4 2 3 2

33. Let u be a vector coplanar with the vectors a = 2 i + 3 j k and b = j + k . If u is perpendicular to a

2
and u . b = 24, then u is equal to : [JEE(Main)2018, (4, 1), 120]
(1) 256 (2) 84 (3) 336 (4) 315

34. The length of the projection of the line segment joining the points (5, 1, 4) and (4, 1,3) on the plane ,
x + y + z = 7 is : [JEE(Main)2018, (4, 1), 120]
1 2 2 2
(1) (2) (3) (4)
3 3 3 3
35. If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
[JEE(Main) 2019, Online (09-01-19),P-2 (4, 1), 120]
(1) ab' + bc' + 1 = 0 (2) bb' + cc' + 1 = 0 (3) cc' + a + a' = 0 (4) aa' + c + c' = 0

36. Let a i j 2k , b b1i b2 j 2 k and c 5i j 2 k be three vectors such that the projection
vector of b on a is a . If a + b is perpendicular to c , then | b | is equal to :
[JEE(Main) 2019, Online (09-01-19),P-2 (4, 1), 120]
(1) 22 (2) 4 (3) 32 (4) 6

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 67
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

37. A tetrahedron has vertices P(1,2,1), Q(2,1,3), R( 1,1,2) and O(0,0,0) the angle between the faces OPQ
and PQR is : [JEE(Main) 2019, Online (12-01-19),P-1 (4, 1), 120]
19 7 17 9
(1) cos 1 (2) cos 1 (3) cos 1 (4) cos 1
35 31 31 35

38. Let S be the set of all real values of such that a plane passing through the points ( 2 , 1, 1),
(1, 2, 1) and (1, 1, 2) also passes through the point ( 1, 1, 1). Then S is equal to -
[JEE(Main) 2019, Online (12-01-19),P-2 (4, 1), 120]
(1) {1, 1} (2) { 3 } (3) { 3 3} (4) {3, 3}

39. The magnitude of the projection of the vector 2 i 3j k on the vector perpendicular to the plane
containing the vectors i j k and i 2 j 3k , is :
3 3
(1) 3 6 (2) 6 (3) (4)
2 2
[JEE(Main) 2019, Online (08-04-19),P-1 (4, 1), 120]

40. If the volume of parallelepiped formed by the vectors i j k, j k and i + k is minimum , then is
equal to : [JEE(Main) 2019, Online (12-04-19),P-1 (4, 1), 120]
1 1
(1) 3 (2) 3 (3) (4)
3 3

x 2 y 1 2
41. The vertices B and C of a ABC lie on the line, such that BC = 5 units. Then the area
3 0 4
(in sq. units) of this triangle, given that the point A(1, 1, 2), is :
[JEE(Main) 2019, Online (09-04-19),P-2 (4, 1), 120]
(1) 5 17 (2) 6 (3) 34 (4) 2 34

42. Let A(3, 0, 1), B(2, 10, 6) and C(1,2,1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos( GOA) (O being the origin) is equal to :
[JEE(Main) 2019, Online (10-04-19),P-1 (4, 1), 120]
1 1 1 1
(1) (2) (3) (4)
6 10 2 15 15 30

x 1 y 1 z
43. A perpendicular is drawn from a point on the line to the plane x + y + z = 3 such that
2 1 1
the foot of the perpendicular Q also lies on the plane x y + z = 3. Then the co-ordinates of Q are :
(1) (4, 0, 1) (2) (2, 0, 1) (3) ( 1, 0, 4) (4) (1, 0, 2)
[JEE(Main) 2019, Online (10-04-19),P-2 (4, 1), 120]
44. A vector a i 2j k ( R) lies in the plane of the vectors, b i j and c i j 4k .
If a bisects the angle between b and c , then: [JEE(Main) 2020, Online (07-01-20),P-1 (4, 1), 120]
(1) a . k + 4 = 0 (2) a . k + 2 = 0 (3) a . i + 1 = 0 (4) a . i + 3 = 0

45. Let a, band c be three unit vectros such that a b c 0 . If a. b b. c c.a and
d a b b c c a , then the ordered pair, , d is equal to:
[JEE(Main) 2020, Online (07-01-20),P-2 (4, 1), 120]
3 3 3 3
(1) ,3a c (2) ,3c b (3) ,3b c (4) ,3a b
2 2 2 2
46. If the distance between the plane, 23x 10y 2z +48 = 0 and the plane containing the lines
x 1 y 3 z 1 x 3 y 2 z 1 k
and R is equal to , then k is equal to ___________
2 4 3 2 6 633
[JEE(Main) 2020, Online (09-01-20),P-2 (4, 0), 120]
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 68
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

EXERCISE - 1
PART I
Section (A) :

A-1. (i) (ii) (iii)

3 5 6

A-2. 5i j 4k , 4i j 5k A-4. , , , , ,
7 7 7 7 7 7
1 1 2 2 1 1
A-5. (i) 40, 38, 32 (ii) 41 , 26 , 17 A-6. , , or , ,
6 6 6 6 6 6
Section (B) :
1
35 1
6
B-1. A = cos , B = cos , C=
41 41 2
10 7
B-3. (i) (ii) (iii) 2 2 2 B-6. 0
6 3

6 2
B-8. (a) i + j + k (b) (6 i j + k) (c) B-9. 169
19 3
2 1
B-10. = cos 1
B-12. (i) cos 1
(ii) 60º B-13. ±( )
3 713 6

B-14. 3 i j k B-15. (b) 5 unit sq.

Section (C) :

1 5 x 2 y 1 z 4
C-1. , ,0 C-2. (1, 3, 5) C-3. (i) (ii) r (2i 2j k) (3i 4 j k)
2 2 1 1 2
x y z x y z
C-4. 26 C-5. 3 C-6. = = , = =
1 2 1 1 1 2
a b c 6
C-7. r= C-8. unit C-9. A = (3, 8, 3), B = ( 3, 7, 6), AB = 3 30
3 5

C-10. r = (i 2j 3k) t ( j k) , where t is parameter

Section (D) :

1 2
D-2. sin cos D-3. (a1 a22 a32 ) (b12 b22 b32 )
4
3
D-4. (a) Coplanar (b) Non-coplanar D-5. (i) 1/2 unit3 (ii) unit
35
D-6. (i) No (ii) Yes D-7. x=1 D-8. v =0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 69
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(p . q) p
D-9. (i) p = 0; q = 10; r = 3 (ii) 100 D-10. x q
2 | p |2
2
D-11. x =
d.a

Section (E) :
E-1. (i) 4x y z+1=0 (ii) x + 2y + 3z 2=0 (iii) x + y + z 4=0 (iv) x + y + z 6=0
E-2. 3 : 2, (0,13/5, 2) E-3. x +y +z =9
2 2 2
E-4. r . (4i 2j 5k) = 45
4 y 2 x 4 y 1 z 7
E-5. sin 1 E-6. = = E-7. = =
30 3 4 5 9 1 3

E-8. /2 E-9. 11x y 3z = 35 E-10. 7 E-11. 0

E-12. 8x 13y + 15z + 13 = 0

E-13. r = +t , , , E-14. x y + 3z 2 = 0 ; (3, 1, 0) ; 11

35 35 35
5
E-15. x+y± 2 z = 1 E-16. unit E-17. 1 E-18. (i) r . n = ±p (ii) r . ( a q pb ) = 0
3
PART II
Section (A) :
A-1. (C) A-2. (C) A-3. (B) A-4. (B)
Section (B) :
B-1. (D) B-2. (B) B-3. (C) B-4. (C) B-5. (C) B-6. (B) B-7. (B)
B-8. (C)
Section (C) :
C-1. (A) C-2. (C) C-3. (A) C-4. (A) C-5. (D) C-6. (C)
Section (D) :
D-1. (C) D-2. (C) D-3. (C) D-4. (B) D-5. (A) D-6. (D)
D-7. (D) D-8. (D) D-9. (B) D-10. (B) D-11. (B) D-12. (B)
Section (E) :
E-1. (A) E-2. (B) E-3. (D) E-4. (A) E-5. (D) E-6. (A)
E-7. (D) E-8. (A) E-9. (A) E-10. (C)
PART - III
1. (A) (r), (B) (q), (C) (q), (D) (s)
2. (A) (q), (B) (p), (C) (s), (D) (r)
3. (A) (q), (B) (p, r), (C) (r), (D) (s)

EXERCISE - 2
PART I
1. (D) 2. (C) 3. (C) 4. (A) 5. (C) 6. (A) 7. (C)
8. (C) 9. (D) 10. (B) 11. (C) 12. (A) 13. (C) 14. (A)
15. (A) 16. (C) 17. (A) 18 . (D) 19. (A) 20. (B) 21. (A)
22. (A) 23. (B) 24. (A)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 70
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

PART II
1. 33.54 2. 19.00 3. 06.00 4. 69.00 5. 04.50 6. 02.00
7. 72.00 8. 34.00 9. 00.66 or 00.67 10. 04.24 11. 36.00 12. 01.75
13. 10.66 or 10.67 14. 32.00 15. 22.62 or 22.63 16. 00.41 or 00.42
17. 13.00 18. 07.20 19. 04.00
PART III
1. (AB) 2. (ABC) 3. (ABCD) 4. (BCD) 5. (BD) 6. (ABC)
7. (ABCD) 8. (AC) 9. (BD) 10. (BC) 11. (BC)
12. (B) 13. (AB) 14. (ABCD) 15. (ABCD) 16. (BD) 17. (D)
18. (AB) 19. (AC) 20. (AD) 21. (AC) 22. (AD) 23. (AB)
24. (ABC) 25. (ABC) 26. (AD) 27. (BCD) 28. (AB) 29. (ABCD)
30. (AD) 31. (ABD) 32. (ACD)

PART IV
1. (A) 2. (A) 3. (D) 4. (B)
5. (C) 6. (A) 7. (A) 8. (B) 9. (A) 10. (A) 11. (C)
12. (C) 13. (A) 14. (A) 15. (A)

EXERCISE - 3
PART I
1. (A) 2. 5 3. (B) 4. (C) 5. (A)
6. 6 7. (A) 8. (A) (t), (B) (p, r), (C) (q) (JEE given q, s) (D) (r)
9. (C) 10*. (A, D) 11. 9 12. (A) (q), (B) (p), (C) (s), (D) (t)
13. (A) 14. 3 15. (A) 16. (C) 17*. (BC) 18. (C) 19. (D)
20.* (BD) 21. 8
C3 24 = 32 22.* (AD) 23. (C) 24. (A) 25*. (ABC)
26. (C) 27. (4) 28. (A) 29*. (B,D) 30*. (A,B) 31*. (ACD)
32. (A) P,Q ; (B) P, Q ; (C) P,Q,S,T ; (D) Q, T
33. (A) P,R,S ; (B) P ; (C) P,Q ; (D) S, T 34. BONUS 35*. (BCD)
36. (C) 37*. (BC) 38. (B) 39. (D) 40. (A) 41. (A) 42*. (CD)
43. (3) 44. (8) 45. (0.5) 46. (ABC) 47. (0.75) 48. (AD) 49. (18.00)

PART II

1. (4) 2. (4) 3. (1) 4. (2) 5. (1) 6. (4) 7. (4)

8. (3) 9. (1) 10. (2) 11. (1) 12. (3) 13. (3) 14. (1)
15. (3) 16. (2) 17. (3) 18. (3) 19. (3) 20. (3) 21. (3)
22. (2) 23. (4) 24. (3) 25. (1) 26. (3) 27. (3) 28. (1)
29. (2) 30. (2) 31. (2) 32. (4) 33. (3) 34. (2) 35. (4)
36. (4) 37. (1) 38. (3) 39. (3) 40. (3) 41. (3) 42. (3)
43. (2) 44. (2) 45. (4) 46. 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 71
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1. Using Vectors prove that

(i) cos(A B) = cosA cosB + sinA sinB (ii) sin(A + B) = sinA cosB + cosA sinB

2. Using vectors, prove that the altitudes of a triangle are concurrent.

3. Prove that the direction cosines of a line equally inclined to three mutually perpendicular lines having
m m2 m3 n1 n2 n3
1
, m1, n1 ; 2, m2, n2 ; 3, m 3, n3 are 1 2 3
, 1 , .
3 3 3

4. If A (a) , B (b) , C (c) , D (d) are the position vector of cyclic quadrilateral then find the value of
|a b b d d a| |b c c d d b|
+ . (It is given that no angle of cyclic quadrilateral ABCD is
[(b a).(d a)] [(b c).(d c)]
right angle)

5. Prove that the volume of tetrahedron bounded by the planes,

2p3
r.(mj nk) 0 , r .(nk i) 0 , r .( i m j) 0 , r.( i m j n k) p is .
3 mn

6. In a ABC, let M be the mid point of segment AB and let D be the foot of the bisector of C. Then
ar( CDM) 1 a b 1 A B A B
prove that = = tan cot .
ar( ABC) 2 a b 2 2 2

7. Let ABC be a triangle.Points M, N and P are taken on the sides AB, BC and CA respectively such that
AM BN CP
= = = . Prove that the vectors AN , BP and CM form a triangle. Also find for which
AB BC CA
the area of the triangle formed by these vectors is the least.

8. In any triangle, show that the perpendicular bisectors of the sides are concurrent.

9. Let ABC be an acute-angled triangle AD be the bisector of BAC with D on BC and BE be the altitude
from B on AC. Show that CED > 45º.

10. In a quadrilateral ABCD, it is given that AB || CD and the diagonals AC and BD are perpendicular to
each other. Show that
(a) AD. BC AB.CD (b) AD + BC AB + CD

11. A, B, C, D are four points in space. using vector methods, prove that
AC2 + BD2 + AD2 + BC2 AB2 + CD2 what is the implication of the sign of equality.

12. The direction cosines of a variable line in two near by positions are l, m, n; l + l, m + m, n + n. Show
that the small angle between the two position is given by ( )2 = ( l)2 + ( m)2 + ( n)2.

a2 b2 c2
13. In a ABC, prove that distance between centroid and circumcentre is R2
9
where R is the circumradius and a, b, c denotes the sides of ABC.

14. Prove that the square of the perpendicular distance of a point P (p, q, r) from a line through A(a, b, c)
and whose direction cosines are , m, n is {(q b) n (r c) m}2.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 72
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

15. (i) Let 1

& 2
be two skew lines. If P, Q are two distinct points on 1
and R, S are two distinct points
on 2 , then prove that PR can not be parallel to QS.
(ii) A line with direction cosines proportional to (2, 7 5) is drawn to intersect the lines
and . Find the coordinate of the points of

intersection and the length intercepted on it. Also find the equation of intersecting straight line.

16. The base of the pyramid AOBC is an equilateral triangle OBC with each side equal to 4 2 . ' O ' is the
origin of reference, AO is perpendicular to the plane of OBC and | AO | = 2 . Then find the cosine of
the angle between the skew straight lines one passing through A and the mid point of OB and the other
passing through 'O' and the mid point of BC.

17. If D, E, F be three point on BC, CA, AB respectively of a ABC. Such that the line AD, BE, CF are
BD CE AF
concurrent then find the value of . . .
CD AE BF

18. Without expanding the determinant, Prove that

na1 b1 na 2 b 2 na3 b 3 a1 a2 a3
nb1 c 1 nb 2 c 2 nb3 c3 = (n + 1)
3
b1 b2 b3
nc1 a1 nc 2 a 2 nc 3 a3 c1 c 2 c3

19. (i) OABC is a regular tetrahedron D is circumcentre of OAB and E is mid point of edge AC.
Prove that DE is equal to half the edge of tetrahedron.
(ii) If V be the volume of a tetrahedron and V be the volume of the tetrahedron formed by the
centroids and V = k V then find the value of k.

20. Given that u i 2j 3k v 2i j 4k , w 2i j 3k and (u.R 10) i + (v.R 20) j+ (w.R 20) k = 0
then find R

21. AB , AC and AD are three adjacent edges of a parallelopiped . The diagonal of the parallelopiped
passing through A and directed away from it is vector a . The vector area of the faces containing
vertices A , B , C and A , B , D are b and c respectively i.e. AB AC = b and AD AB c . If
|a|
projection of each edge AB and AC on diagonal vector a is , then find the vectors AB, AC and
3
AD in terms of a, b, c and | a | .

22. Prove that if the equation r a b and r c d are consistent a.d 0,b.c 0,d b 0 then

b.c a.d 0

23. If a & b are two non collinear vector a . b 0 a (a (a (a.............. (a (a b)) = (a b)

2018 times

24. Let P be an interior point of a triangle ABC and AP, BP, CP meet the sides BC, CA, AB is D, E, F
AP AF AE
respectively. Show that = + .
PD FB EC

25. ABCD is a quadrilateral and P, Q are mid-points of CD, AB respectively. Let AP, DQ meet at X and BP,
CQ meet at Y. Prove that Area of ADX + Area of BCY = Area of quadrilateral PXQY.

26. Find the locus of the centroid of the tetrahedron of constant volume 64K3 , formed by the three
co-ordinates planes and a variable plane.
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 73
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

27. Lengths of two opposite edges of a tetrahedron are a and b. Shortest distance between these edges is
d and the angle between them is . Prove that its volume is (1/6) abd sin .

28. If r.a = 0, r . b = 1 and [r a b] = 1 , ab 0 and | a |2 | b |2 (a . b)2 = 1 then find r in terms of a & b

29. Let u & v be unit vectors. If w is a vector such that w w xu v , then prove that
1
u v .w and the equality holds if and only if u is perpendicular to v .
2

30. Let A is set of all possible planes passing through four vertices of given cube. Find number of ways of
selecting four planes from set A, which have one common point.

31. Let OABC is a regular tetrahedron and P is any point in space. If edge length of tetrahedron is 1 unit,
find the least value of 2(PA 2 + PB2 + PC2 + PO2).

32. Find the minimum value of x2 + y2 + z2 when ax + by + cz = p.

x a d y a z a d x b c y b z b c
33. The lines = = , = = are coplanar and these

determine a single plane if . Find the equation of the plane in which they lie.

1 1 1
34. Consider the plane E : r = 1 + 2 + 0
1 0 1
F is a plane containing the point A ( 4, 2, 2) and parallel to E. Suppose the point B is on the plane E,
such that B has a minimum distance from point A. If C( 3, 0, 4) lies in the plane F. Then find the area of
ABC.

35. Through a point P (h, k, ) a plane is drawn at right angles to OP to meet the co-ordinate axes in A, B
p5
and C. If OP = p, show that the area of ABC is , where O is the origin.
2hk

36. If A (a) , B (b) and C (c) are three non collinear points and origin does not lie in the plane of the points
A, B and C, then for any point P (p) in the plane of the ABC, prove that ;
(i) abc p . a xb b xc cxa

(ii) A point v is on plane of ABC such that vector ov is to plane of ABC. Then show that
[a b c](a b b c c a)
v = , where is the vector area of the ABC.
4 2

37. Prove that the equation of the sphere which passes through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1)
and having radius as small as possible is 3 x2 2 x 1 = 0.

x y z 1
38. Prove that the line = = lies in the plane x + y + z = 1. Find the lines in the plane through the
1 1 2
1
point (0, 0, 1) which are inclined at an angle cos 1 with the line.
6

39. Find the equation of the sphere which has centre at the origin and touches the line
2(x + 1) = 2 y = z + 3.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 74
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

40. A mirror and a source of light are situated at the origin O and at a point on OX (x-axis) respectively. A
ray of light from the source strikes the mirror and is reflected. If the Drs of the normal to the plane are
1,

41. A variable plane x + my + nz = p (where , m, n are direction cosines) intersects with co-ordinate axes
at points A, B and C respectively show that the foot of normal on the plane from origin is the
orthocentre of triangle ABC and hence find the coordinates of circumcentre of triangle ABC.

42. A rectangle whose vertices are (5,3, 3), (5,9,9), (0,5,11) and (0, 1, 1) is rotated about its diagonal
1 2 2
(whose direction cosines are , , in such a way that new position of rectangle is perpendicular to
3 3 3
its old position find the coordinates of new position of the vertices whose position is changed.

43 -Y plane) whose equation is (x a)2 + (y b)2 + (z c)2 = r2

x y z
(i) A light source lies on the line = = above X-Y plane at infinite distance from X-Y plane.
a b c
If a = c and b = 0 then find the equation of -Y plane.

(ii) A light source is at (5,0,3), a = c = 2, b = 0, r = 1. What is the locus name of boundary of

-Y plane.

1
4. 0 7. = 15. (ii) (2, 8, 3) & (0, 1, 2) ; 78 ;
2

1
16. 17. 1 19. (ii) 27 20. R 10i
2

c a 1 a (b c) 3(b a) 1 a (b c) 3(c a)
21. AB AD 3 ; AC = a + + ; AD = a +
| a |2 3 | a |2 | a |2 3 | a |2 | a |2

2018
23. 26. xyz = 6k3 28. r = (a . b) a + | a |2 b + (a b) 30. 135

p2 9
31. 3 32. 33. x 2y + z = 0 34.
a2 2

x y z 1 x y z 1
38. = = and = = 39. 9(x2 + y2 + z2) = 5
1 15 1 15 2 1 15 1 15 2

2
1 2 2 p p p m2p p n2p
40. , , 41. , ,
3 3 3 2 2m 2n

42. {(5, 3,3), (5,11,5)} or {( 3,5, 1), (8,3,9)} 43 (i) x2 + 2y2 = 2r2 (ii) Parabola

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVVTTD- 75
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 Marked questions are recommended for Revision.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA

PART - I : SUBJECTIVE QUESTIONS

Hkkx - I : fo"k;kRed iz'u ¼SUBJECTIVE QUESTIONS½

Section (A) : Position vector, Direction Ratios & Direction cosines

[k.M (A) : ewyHkwr] foHkktu lw=k vkSj ljy js[kk dk lehdj.k
A-1. (i) Let position vectors of points A,B and C are a , b and c respectively. Point D divides line segment
BC
internally in the ratio 2 : 1. Find vector AD .
(ii) Let ABCD is parallelogram. Position vector of points A,C and D are a , c and d respectively .
If E divides
line segement AB internally in the ratio 3 : 2 then find vector DE .

(iii) Let ABCD is trapezium such that AB = 3 DC . E divides line segement AB internally in the ratio 2 : 1
and F is mid point of DC. If position vector of A,B and C are a,b and c respectively then find vector FE .
(i) ekuk fd fcUnq A,B vkSj C ds fLFkr lfn’'k Øe'k% a , b vkSj c gSA fcUnq D js[kk[k.M BC dks fcUnq D ij 2 : 1
esa
vUr% foHkkftr djrk gSA rc lfn'k AD Kkr dhft,A
(ii) ekuk ABCD ,d lekUrj prqHkZt gS fcUnq A,C vkSj D ds fy, fLFkr lfn'k Øe'k% a , c vkSj d gSA ;fn E
js[kk[k.M
AB dks 3 : 2 esa vkUrfjd foHkkftr djrk gSA rc lfn'k DE Kkr dhft,A
(iii) ekuk ABCD ,d leyEc prqHkZt bl izdkj gS fd AB = 3 DC gSA E js[kk[k.M AB dks 2 : 1 esa vUr%foHkkftr
djrk gSA
rFkk F, DC dk e/; fcUnq gSA ;fn A,B vkSj C ds fLFkfr lfn'k Øe'k% a, b vkSj c gS] rks lfn'k FE gS &

2c  b – 3a 3c  5a – 8d 5b  a – 6c
Ans. (i) (ii) (iii)
3 5 6

Aa

Sol. (i)
B b  C c 
D
2:1

2c  b
Position vector of D is
3
2c  b
D dk fLFkfr lfn'k gS
3
2c  b
vector lfn'k AD = –a
3
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 1
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

D C

(ii)

A B
3 E 2
Vector lfn'k DC = c – d
Vector lfn'k c – d = c – d
Vector lfn'k AE =
3
5
c–d  
3
Position vector of E is a  (c – d)
5
3
E dk fLFkfr lfn'k a  (c – d) gSA
5
5a  3c – 3d 5a  3c – 8d
vector lfn'k DE = – d =
5 5

(iii)

D 1:1 C
F

E
A B
2:1

Vector lfn'k AB = b – a

b–a
Vectorlfn'k DF = FC =
6

AF = AC + CF = (c – a) +
1
6
a–b 
vector lfn'k AE =
2
3
b–a  
FE = AE – AF =
2
3

 1 
b – a –  c – a  a – b 
 6 
  
=
5
6

b–a + a–c
5b  a – 6c
=
6

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 2
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

A-2. In a ABC es a , AB = 6iˆ  3jˆ  3kˆ ; AC = 3iˆ 3jˆ 6kˆ [16JM120001]

D and D are points trisections of side BC
Find AD and AD .

 ABC es a , AB = 6iˆ  3ˆj  3kˆ ; AC = 3iˆ 3jˆ 6kˆ
D vkS j D Hkq t k BC ds
f=kfoHkktu ds fcUnq gS A
AD rFkk AD Kkr dhft,A

Ans. 5iˆ ˆj 4kˆ , 4iˆ ˆj 5kˆ .

Sol.

2 · AB 1 · AC 12iˆ 6ˆj 6kˆ  3iˆ 3 ˆj 6kˆ

AD = =
3 3
ˆ ˆ
15i 3 j 12k ˆ
= = 5iˆ ˆj 4kˆ
3
1 · AB 2 · AC 6iˆ 3ˆj 3kˆ  6iˆ 6 ˆj 12kˆ
AD = = = 4iˆ ˆj 5kˆ .
3 3

A-3. If ABCD is a quadrilateral, E and F are the mid-points of AC and BD respectively, then prove that
AB  AD  CB  CD  4EF
;fn ABCD ,d prq H kq Z t gS rFkk E vkS j F Øe'k% Hkq t k AC rFkk BD ds e/; fcUnq gS ] rks iz n f'kZ r
dhft, fd AB  AD  CB  CD  4EF

ac
Sol. E 
2

bd
F 
2
AB  AD  CB  CD
 b  d (a  c) 
= b – a  d – a  b – c  d – c = 2[b  d – a – c] = 4  –  = 4EF
 2 2 

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 3
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

A-4. Let ABCD is parallelogram where A = (1,2,4) , B = (8,7,9) and D = (6,1,5) . Find direction cosines of line
AC
ekuk ABCD ,d lekUrj prqHkZt gS tgk¡ A = (1,2,4) , B = (8,7,9) vkSj D = (6,1,5) rc js[kk AC dh fnd~ dksT;k,sa
Kkr dhft,A
 6 2 3   –6 –2 –3 
Ans.  7 , 7 , 7 ,  7 , 7 , 7 
   
Sol. Mid point of BD is M (7,4,7)
(6,1,5)
D C

A B (8,7,9)
(1,2,4)

Direction ratio of AM is (6,2,3)

6 2 3  –6 –2 –3 
direction cosines of AC is  , ,  or  , , 
 7 7 7   7 7 7 

A-5._ Let one vertex of cuboid whose faces are parallel to coordinate planes is (1, 2, 3). If mid-point of one of
the edge in (0, 0,0) then find possible

(i) total surface area of cuboid. (ii) length of body diagonal.

ekuk fd ?kukHk ftlds Qyd funsZ'kkad leryksa ds lekUrj gS] dk ,d 'kh"kZ (1, 2, 3) gS ;fn fdukjs ds ,d dk e/;
fcUnq (0, 0, 0) gSA rc laHkkfor gSA
(i) ?kukHk dk dqy i`"Bh; {ks=kQy (ii) fod.kZ dh yEckbZ

Ans. (i) 40, 38, 32 (ii) 41 , 26 , 17

Sol.

B1

B2
B3

(1, 2, 3)

(0, 0, 0) can be either B1 or B2 or B3

Length of sides of cuboid are either 1, 2, 6 or 1,4,3 or 2, 2, 3

 Length of body diagonal of cuboid can be 41 , 26 , 17 B1

B2 (Raj.)-324005
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in B3
ADVQE- 4
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

(1, 2, 3)
Vector & Three Dimensional Geometry

 Surface area of cuboid can be 40, 38, 32

Hindi.

(0, 0, 0) B1 ;k B2 ;k B3 gks ldrk gSA

?kukHk dh Hkqtk dh yEckbZ 1, 2, 6 ;k 1,4,3 ;k 2, 2, 3 gSA

 /kukHk ds fod.kZ dh yEckbZ 41 , 26 , 17

 ?kukHk dk i`"Bh; {ks=kQy 40, 38, 32 gSA

A-6. Find the direction cosines , m, n of line which are connected by the relations  + m + n = 0, 2mn + 2m
– n = 0.
js[kk dh fnd~dksT;k,sa , m, n Kkr dhft, tcfd ;s  + m + n = 0 vkSj 2mn + 2m – n = 0 }kjk lEcfU/kr gSA
1 1 2 2 1 1
Ans.  ,  , or ;k , ,
6 6 6 6 6 6
Sol. n = – ( + m) ........(1) 2mn + 2ml – nl = 0 ............(2)
– 2k ( + m) + 2m + ( + m)  = 0

 – 2m – 2m2 + 2m + 2 + m = 0  2 + m – 2m2 = 0   1, –2

m
Case - I :
We have ;gk¡  1   =m So blfy, n = – 2 So vr%,  : m : n = 1 : 1 : – 2.
m
1 1 2 1 1 2
 direction cosines are fnd~dksT;k,¡ , , or ;k  ,  , gSA
6 6 6 6 6 6
Case - II :
We have ;gk¡  2   = – 2m  :m:n=–2:1:1
m

2 1 1 2 1 1
hencevr% direction cosines are fnd~dksT;k,¡ , , or ;k , , gksxhA
6 6 6 6 6 6

Section (B) : Dot Product, Projection and Cross Product

[k.M (B) : vfn'k xq.ku] iz{ksi vkSj lfn'k xq.ku

B-1. Show that the points A, B, C with position vectors 2iˆ  ˆj  kˆ , ˆi  3ˆj  5kˆ and
3iˆ  4ˆj  4kˆ respectively are the vertices of a right angled triangle. Also find the remaining
angles of the triangle.
n'kkZb;s fd fcUnq A, B, C gS ftuds fLFkfr lfn'k 2iˆ  ˆj  kˆ , ˆi  3ˆj  5kˆ vkSj 3iˆ  4ˆj  4kˆ ledks.k f=kHkqt ds 'kh"kZ
gS rFkk f=kHkqt ds 'ks"k dks.k Hkh Kkr dhft, &

35 6 
Ans. A = cos–1 ,B = cos–1 ,C =
41 41 2
Solution : We have,
AB = (iˆ  3jˆ  5k)
ˆ – (2iˆ  ˆj  k)
ˆ = ˆi  2jˆ  6kˆ

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 5
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

BC = (3iˆ  4jˆ  4k)ˆ – (iˆ  3ˆj  5k)

ˆ = 2iˆ  ˆj  kˆ
and, CA = (2iˆ  ˆj  k)
ˆ – (3iˆ  4jˆ  4k)ˆ = ˆi  3jˆ  5kˆ

Since AB + BC + CA = (ˆi  2jˆ  6k) ˆ + (2iˆ  ˆj  k)

ˆ + (ˆi  3jˆ  5k)
ˆ = 0
So A, B and C are the vertices of a triangle.
Now, BC . CA = (2iˆ  ˆj  k) ˆ . (ˆi  3jˆ  5k)
ˆ = –2 – 3 + 5 = 0

  BC  CA BCA =  ABC is a right angled triangle.
2
Since A is the angle between the vectors AB and AC . Therefore

AB . AC ( ˆi  2jˆ  6k)

ˆ . (iˆ  3 ˆj  5k)
ˆ
cos A = =
| AB || AC | ( 1)2  ( 2)2  ( 6)2 12  ( 3) 2  ( 5) 2
1  6  30 35 35 35
= = =  A = cos–1
1  4  36 1  9  25 41 35 41 41
BA . BC (iˆ  2jˆ  6k)
ˆ . (2iˆ  ˆj  k)
ˆ
cos B = =
| BA || BC | 12  22  62 22  ( 1)2  (1)2
226 6 6
 cos B = = B = cos–1
41 6 41 41
gy : ;gk¡, AB = (iˆ  3jˆ  5k)
ˆ – (2iˆ  ˆj  k)
ˆ = ˆi  2jˆ  6kˆ
BC = (3iˆ  4jˆ  4k) ˆ – (iˆ  3ˆj  5k)
ˆ = 2iˆ  ˆj  kˆ
rFkk, CA = (2iˆ  ˆj  k)ˆ – (3iˆ  4jˆ  4k)ˆ = ˆi  3jˆ  5kˆ

pwafd AB + BC + CA = (ˆi  2jˆ  6k) ˆ + (2iˆ  ˆj  k)

ˆ + (ˆi  3jˆ  5k)
ˆ = 0
blfy, A, B rFkk C f=kHkqt ds 'kh"kZ gSA
vc, BC . CA = (2iˆ  ˆj  k) ˆ . (ˆi  3jˆ  5k)
ˆ = –2 – 3 + 5 = 0

  BC  CA BCA =  ABC ledks.k f=kHkqt gSA
2
AB vkSj AC ds e/; dks.k A gS blfy,

AB . AC ( ˆi  2jˆ  6k)

ˆ . (iˆ  3 ˆj  5k)
ˆ
cos A = =
| AB || AC | ( 1)  ( 2)2  ( 6)2
2
12  ( 3) 2  ( 5) 2
1  6  30 35 35 35
= = =  A = cos–1
1  4  36 1  9  25 41 35 41 41
BA . BC (iˆ  2jˆ  6k)
ˆ . (2iˆ  ˆj  k)
ˆ
cos B = =
| BA || BC | 12  22  62 22  ( 1)2  (1)2
226 6 6
 cos B = = B = cos–1
41 6 41 41

B-2. If a, b, c are three mutually perpendicular vectors of equal magnitude, prove that a  b  c is
equally inclined with vectors a, b and c .
;fn a, b, c rhu ijLij yEcor leku ifjek.k ds gSA rc fl) dhft, fd a  b  c , lfn'k a, b rFkk
c ls cjkcj dks.k ij >qdh gqbZ gSA

Solution : Let | a | = | b | = | c | =  (say). Since a, b, c are mutually

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 6
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

perpendicular vectors, therefore a . b = b . c = c . a = 0 ..............(i)

2
Now, abc = a . a + b . b + c . c + 2a . b + 2b . c + 2c . a

= | a |2 | + | b |2 + | c |2 = 32 [ | a | = | b | = | c | = ]
 |abc | = 3 ..............(ii)

Suppose a  b  c makes angles 1, 2, 3 with a, b and c respectively. Then,

a . (a  b  c) a . a a . b a . c | a |2
cos1 = = =
| a || a  b  c | | a || a  b  c | | a || a  b  c |
|a|  1  1 
= = = [Using (ii)]  1 = cos–1  
|abc | 3 3  3
 1   1 
  Similarly, 2 = cos–1   and 3 = cos 
–1

 3  3
 Because 1 = 2 = 3 , so a  b  c is equally inclineded with a, b and c
gy : ekuk | a | = | b | = | c | =  (say). pwafd a, b, c yEcor lfn'k gS
a . b = b . c = c . a =0 ..............(i)
2
vc, abc = a . a + b . b + c . c + 2a . b + 2b . c + 2c . a

= | a |2 | + | b |2 + | c |2 = 32 [ | a | = | b | = | c | = ]
 |abc | = 3 ..............(ii)

ekukfd a  b  c , a, b , c ds lkFk 1, 2, 3 dks.k cukrs gS,

a . (a  b  c) a . a a . b a . c | a |2
cos1 = = =
| a || a  b  c | | a || a  b  c | | a || a  b  c |
|a|  1  1 
= = = [ (ii) ls ]  1 = cos–1  
|abc | 3 3  3
 1   1 
  blhizdkj, 2 = cos–1   vkS j 3 = cos–1  
 3  3
 ijUrq 1 = 2 = 3 , blfy, a  b  c , a, b vkSj c cjkcj >qdh gqbZ gSA

B-3. (i) Find the projection of b  c on a where a = ˆi  2jˆ  kˆ , b = ˆi  3ˆj  kˆ and c = ˆi  kˆ .

b  c dk a ij iz{ksi Kkr dhft, tgk¡ a = ˆi  2jˆ  kˆ , b = ˆi  3ˆj  kˆ rFkk c = ˆi  kˆ gSA

(ii) Find the projection of the line segment joining (2, – 1, 3) and (4, 2, 5) on a line which makes equal
acute angles with co-ordinate axes.
fcUnq (2, – 1, 3) vkSj (4, 2, 5) dks feykus okyh ljy js[kk dk] ml ljy js[kk ij iz{ksi Kkr djks tks v{kksa ds
lkFk leku U;wudks.k cukrh gSA

(iii) P and Q are the points (–1, 2, 1) and (4, 3, 5) respectively. Find the projection of PQ on a line which
makes angles of 120º and 135º with y and z axes respectively and an acute angle with x-axis.
fcUnq P rFkk Q Øe'k% (–1, 2, 1) rFkk (4, 3, 5) gSaA PQ dk ml ljy js[kk ij iz{ksi Kkr dhft, tks y rFkk z v{k
ls Øe'k% 120º rFkk 135º dk dks.k cukrh gS rFkk x-v{k ls U;wudks.k cukrh gSA [16JM120008]

10 7
Ans. (i) (ii) (iii) 2 2 –2
6 3
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 7
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. (i) b  c = ˆi  3jˆ  kˆ  ˆi  kˆ = 2iˆ  3jˆ  2kˆ

(b  c) ·a (2iˆ  3ˆj  2k)
ˆ · (iˆ  2jˆ  k)
ˆ 262 10
Projection of b  c on a = = = =
|a| (1)  (2)  (1)
2 2 2
1 4  1 6

Hindi (i) b  c = ˆi  3jˆ  kˆ  ˆi  kˆ = 2iˆ  3jˆ  2kˆ

(b  c) ·a (2iˆ  3ˆj  2k)
ˆ · (iˆ  2jˆ  k)
ˆ 262 10
b  c dk a ij iz{ksi = = = =
|a| (1)  (2)  (1)
2 2 2
1  4  1 6

–1 –1
(iii)  m = cos 120° = ; n  cos135    2 + m2 + n2 = 1
2 2
1 1 1 1
 2 +  1  2 =   = as acute angle with x-axis
4 2 4 2

1 1 1

, m , n
2 2 2
 required projection = |(x2 – x1)  + (y2 – y1) m + (z2 – z1)n|
5 1
= |5 × 1/2 + 1 × (– 1/2) + 4 × (– 1/ 2 ) = – –2 2  2 ( 2 – 1)
2 2
–1 –1
Hindi  m = cos 120° = ; n  cos135    2 + m2 + n2 = 1
2 2
1 1 1
 2 +  1  2 =  
4 2 4
1
  = tSlk fd js[kk x-v{k ds lkFk U;wu dks.k cukrh gSA
2

1 1 1

, m , n
2 2 2
 vHkh"V iz{ksi = |(x2 – x1)  + (y2 – y1) m + (z2 – z1)n|
5 1
= |5 × 1/2 + 1 × (– 1/2) + 4 × (– 1/ 2 ) = – –2 2  2 ( 2 – 1)
2 2

2 2
 a b   ab 
Prove that  2  2  = 
 | a | | b | 
B-4.
a 
 b   
2 2
 a b   ab 
fl) dhft, :  2  2 = 
a   | a | | b | 
 b   

| a |2 | b |2 2a . b 1 1 2a . b
Sol.. LHS + –   –
|a| 4
| b |4 | a |2 | b |2 | a |2 | b |2 | a |2 | b |2
| a |2  | b |2 –2a . b 1 1 2a . b
RHS   – .Hence Proved.
| a |2 | b |2 | a |2 | b |2 | a |2 | b |2
| a |2 | b |2 2a . b 1 1 2a . b
Hindi ck;k¡ i{k + –   –
|a| 4
| b |4 | a 2| | b 2| | a |2 | b |2 | a |2 | b |2
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 8
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

| a |2  | b |2 –2a . b 1 1 2a . b
nk;k¡ i{k   – . vr% fl) gqvkA
| a |2 | b |2 | a |2 | b |2 | a |2 | b |2

B-5. If a, b are two unit vectors and  is the angle between them, then show that:
;fn a, b nks bdkbZ lfn'k gS vkSj  muds e/; dks.k gS] rks iznf'kZr dhft, fd
 1  1
(a) sin  ab (b) cos  ab
2 2 2 2
Sol. (a) | a – b |2  | a |2  | b |2 – 2(a . b)  | a – b |2  1  1– 2(1)  (1)  cos 

 | a – b |2  1  1– 2(1)  (1)  cos   | a – b |2  2 [1 – cos]  | a – b |2  4 sin2
2
 1
 sin = |a–b|
2 2
(b) | a  b |2  | a |2  | b |2 2(a . b)  | a  b |2 = 1 + 1 + 2 × 1 × 1 cos
  1
 | a  b |2  2[1  cos ]  4cos2  cos  |ab|
2 2 2

B-6. If two vectors a and b are such that | a | = 2, | b | = 1 and a·b = 1, then find the value of
(3a – 5b) · (2a  7b) . [16JM120004]
;fn nks lfn'k a rFkk b bl izdkj gS fd | a | = 2, | b | = 1 rFkk a·b = 1 rks (3a – 5b) · (2a  7b) dk eku
Kkr dhft,A
Ans. 0

Sol. Here, two vectors a and b are such that | a | = 2, | b | = 1 and a · b = 1,

Now, (3a – 5b) · (2a  7b) = 6 | a |2 – 10 a·b + 21 a·b – 35 | b |2
= 6(2)2 – 10 × 1 + (21 × 1) – 35(1)2
= 24 – 10 + 21 – 35 = 45 – 45 = 0
Hindi. ;gk¡ , nks lfn'k a rFkk b bl izdkj gS fd | a | = 2, | b | = 1 rFkk a · b = 1,
vc, (3a – 5b) · (2a  7b) = 6 | a |2 – 10 a·b + 21 a·b – 35 | b |2
= 6(2)2 – 10 × 1 + (21 × 1) – 35(1)2 = 24 – 10 + 21 – 35 = 45 – 45 = 0

B-7. For any two vectors u & v , prove that [16JM120003]

(a) (u.v)  | u  v |  | u | | v |
2 2 2 2
& (b) (1 | u | )(1 | v | )  (1 u.v)2  | u  v  (u  v) |2
2 2

fdUgh nks lfn'kks a u vkS j v ds fy, fl) djks fd &

(a) (u.v)2  | u  v |2  | u |2 | v |2 & (b) (1 | u |2 )(1 | v |2 )  (1 u.v)2  | u  v  (u  v) |2
Sol. (a) | v | 2 | u | 2 cos 2  + | v | 2 | u | 2 sin 2  = | v | 2 | u | 2
(b) LHS 1 + | v | 2 +| u | 2 + | u | 2 | v | 2
RHS 1 + |u| 2 | v | 2 cos 2  – 2| u | | v | cos+ | u | 2 +| v | 2 +| u | 2 | v | 2 sin 2  + 2 u . v + 0 + 0
= 1 + | v | 2 +| u | 2 + | u | 2 | v | 2
LHS = RHS
Hence Proved. ( bfr fl}e )

B-8. If the three successive vertices of a parallelogram have the position vectors as,
A ( 3,  2, 0); B (3,  3, 1) and C ( 5, 0, 2). Then find
(a) position vector of the fourth vertex D

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 9
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(b) a vector having the same direction as that of AB but magnitude equal to AC
(c) the angle between AC and BD .
;fn ,d lekUrj prq H kq Z t ds rhu Øekxr 'kh"kks ± ds fLFkfr lfn'k A ( 3,  2, 0); B (3,  3, 1) vkS j
C ( 5, 0, 2) gS a ] rks Kkr dhft, &
(a) prq F kZ 'kh"kZ D dk fLFkfr lfn'k
(b) ,d lfn'k ftldh fn'kk AB ds leku gS ys f du ekika d AC ds cjkcj gS A
(c) AC vkS j BD ds chp dks . k
6 2
Ans. (a) – î + ĵ + k̂ (b) (6 î – ĵ + k̂ ) (c)
19 3
Sol. (a) (3iˆ  3ˆj  kˆ  d)  2iˆ  2jˆ  2kˆ  d  ˆi  ˆj  kˆ
(b) AB  6iˆ  ˆj  kˆ
AC  8iˆ  2jˆ  2kˆ  AC  64  4  4 = 72

72 6
Required vector is (6iˆ  ˆj  k)
ˆ = (6ˆi  ˆj ˆk )
38 19

AC. BD 32  8 24 1

(c) BD  4iˆ  4ˆj  cos   = = = 
AC BD 72 32 6 2 · 4 2 2

2
  =
3
Hindi (a) (3iˆ  3ˆj  kˆ  d)  2ˆi  2ˆj 2k
ˆ  d  ˆi  ˆj  kˆ
(b) AB  6iˆ  ˆj  kˆ
AC  8iˆ  2jˆ  2kˆ  AC  64  4  4 = 72

72 6
vHkh"B lfn'k (6iˆ  ˆj  k)
ˆ = (6ˆi  ˆj ˆk )
38 19

AC. BD 32  8 24 1

(c) BD  4iˆ  4ˆj  cos   = = = 
AC BD 72 32 6 2 · 4 2 2

2
  =
3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 10
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

B-9. If a,b,c are three vectors such that | a | = 5, | b | = 12 and | c | = 13, and a  b  c  o , find the value of
a.b  b.c  c.a . [16JM120005]
;fn a,b rFkk c rhu ,sls lfn'k gSa fd | a | = 5, | b | = 12 rFkk | c | =13, rFkk a  b  c  o gS] rks
a.b  b.c  c.a dk eku Kkr dhft,A
Ans. – 169
Sol. Consider abc = o  a  b  c  = o 2

 a2  b2  c 2 + 2a.b  2b.c  2c.a = 0  25 + 144 + 169 + 2(a.b  b.c  c.a) = 0

 2 (a.b  b.c  c.a) = – 338  a.b  b.c  c.a = – 169

B-10. ABCD is a parallelogram in which AB = 3iˆ 6jˆ 3kˆ and AD = 6iˆ 6jˆ 3kˆ . P is a point of AB such
that AP : PB = 1 : 2 and Q is a point on BC such that BQ : QC = 2 : 1. Find angle between DQ and PC.
ABCD ,d lekUrj prq H kq Z t gS ftles a AB = 3iˆ 6jˆ 3kˆ rFkk AD = 6iˆ 6jˆ 3kˆ gS A P, AB ij ,d
fcUnq bl iz d kj gS fd AP : PB = 1 : 2 rFkk BC ij fcUnq Q bl iz d kj gS fd BQ : QC = 2 : 1 rc DQ
vkS j PC ds e/; dks . k Kkr dhft,A
2
Ans.  = cos–1
3 713
Sol. Let A be the origin AP = ˆi 2jˆ kˆ
ekuk A ew y fcUnq gS rc AP = ˆi 2jˆ kˆ
18iˆ 12kˆ 3iˆ 6ˆj 3kˆ
AQ = = 7iˆ 2jˆ 5kˆ
3

DQ  7iˆ  2jˆ  5 kˆ  6iˆ  6jˆ  3 kˆ = ˆi  8ˆj  2 kˆ

PC  9iˆ  6 kˆ  (iˆ  2jˆ k)
ˆ  DQ . PC  | DQ | | PC | cos 
2
8 – 16 + 15 = 69. 93 cos  2 = 3 23. 31 cos   = cos–1
3 713
B-11. Prove using vectors : If two medians of a triangle are equal, then it is isosceles.
lfn'k fof/k ls n'kkZb;s fd ;fn f=kHkqt dh nks ekf/;dk,a cjkcj gS rc ;g lef}ckgq gSA

Sol. Let ABC be a triangle and let BE and CF be two equal medians. Taking A as the origin, let the
position vectors of B and C be b and c respectively. Then,
1 1
P.V. of E = c and P.V. of F = b
2 2
1 1
 BE = (c  2b)  CF = (b  2c)
2 2

Now, BE = CF  | BE | = | CF |
2 2
1 1
 | BE |2 = | CF |2  (c  2b) = (b  2c)
2 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 11
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 1
 | c  2b |2 = | b  2c |2  | c  2b |2 = | b  2c |2
4 4
 (c  2b) . (c  2b) = (b  2c) . (b  2c) 
  c . c – 4b . c + 4b . b = b . b – 4b . c + 4c . c
 | c |2 – 4b . c + 4 | b |2 = | b |2 – 4b . c + 4 | c |2 3 | b |2 = 3 | c |2  | b |2 = | c |2 
  AB = AC Hence triangle ABC is an isosceles triangle.
Hindi. ekuk ABC ,d f=kHkqt gS rFkk BE rFkk CF nks cjkcj ekf/;dk,a gSA A dks ewyfcUnw ysrs gq, ekuk B vkSj C ds
1 1
fLFkfr lfn'k Øe'k% b vkSj c gS rc E dk fLFkfr lfn'k = c rFkk F dk fLFkfr lfn'k = b
2 2
1 1
 BE = (c  2b)  CF = (b  2c)
2 2

vc, BE = CF  | BE | = | CF |
2 2
1 1
 | BE |2 = | CF |2  (c  2b) = (b  2c)
2 2

1 1
 | c  2b |2 = | b  2c |2  | c  2b |2 = | b  2c |2
4 4
 (c  2b) . (c  2b) = (b  2c) . (b  2c) 
  c . c – 4b . c + 4b . b = b . b – 4b . c + 4c . c
 | c |2 – 4b . c + 4 | b |2 = | b |2 – 4b . c + 4 | c |2 3 | b |2 = 3 | c |2  | b |2 = | c |2 
  AB = AC vr% ABC lef}ckgq f=kHkqt gSA

B-12. (i) Find the angle between the lines whose direction cosines are given by the equations :
3 + m + 5n = 0 and 6 mn – 2n + 5 m = 0
(ii) Find the angle between the lines whose direction cosines are given by  + m + n = 0 and 2 + m2 =
n2.
(i) mu ljy js[kkvksa ds e/; dks.k Kkr djks ftudh fnd~ dksT;k;s fuEu lehdj.kksa dks larq"V dj jgh gS&
3 + m + 5n = 0 vkSj 6 mn – 2n + 5 m = 0
(ii) mu js[kkvksa ds e/; dks.k Kkr dhft, ftudh fnd~dksT;k,¡ fuEu lEcU/kksa ls nh tkrh gS] 
  + m + n = 0 rFkk 2 + m2 = n2.
 1
Ans. (i) cos–1   (ii) 60º
6
Sol. (i) We have :
3l + m + 5n = 0 ...(1)
6mn – 2nl + 5lm = 0 ...(2)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 12
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

From (1), m = –(3l + 5n) ...(3)

Putting the value in (2), we get
–6(3 + 5n)n – 2n – 152 – 25n = 0  –18 n – 30 n2 – 2n – 152 – 25 n = 0
–30 n2 – 45n – 152 = 0  2n2 + 3n + 2 = 0 [Divide by (–15)]
(2n + ) (n + ) = 0 
Either 2n +  = 0 or n+=0
() when 2n +  = 0   = –2n
from (3), m = –(–6n + 5n) = n
(II) when n +  = 0   = –n
from (3), m = –(–3n + 5n) = –2n
 Direction ratios of two lines are
–2n, n , n and –n, –2n, n
–2, 1, 1 and 1, 2, –1
( 2)(1)  (1)(2)  (1)( 1)
Angle between two lines  cos  = .
( 2)2  12  12 12  22  ( 1)2
 1
Hence  = cos–1  
6

Hindi. fn;k gS
3l + m + 5n = 0 ...(1)
6mn – 2nl + 5lm = 0 ...(2)
lehdj.k (1) ls, m = –(3l + 5n) ...(3)
fuEu dk eku lehdj.k (2) esa j[kus ij
– 6(3 + 5n)n – 2n – 152 – 25n = 0  –18 n – 30 n2 – 2n – 152 – 25 n = 0
–30 n2 – 45n – 152 = 0   2n2 + 3n + 2 = 0 [(–15) ls Hkkx nsus ij]
(2n + ) (n + ) = 0
;k rks tc 2n +  = 0 ;k n +  = 0
() tc 2n +  = 0   = –2n
lehdj.k (3), ls m = –(–6n + 5n) = n
(II) tc n +  = 0   = –n
lehdj.k (3) ls, m = –(–3n + 5n) = –2n
vr% nksuks ljy js[kkvksa ds fnd~ vuqikr fuEu gksxs
–2n, n , n rFkk –n, –2n, n
–2, 1, 1 rFkk 1, 2, –1
( 2)(1)  (1)(2)  (1)( 1)
nksuks ljy js[kkvksa ds e/; dks.k cos  =
( 2)2  12  12 12  22  ( 1)2
 1
 vr%  = cos–1  
 
6

(ii) + m + n = 0 ................(1) 2 + m2 = n2 ................(2)

Put n = – ( + m) in (2)   2 + m2 = 2 + m2 + 2m  m = 0

Case-I if  = 0 ; m  0 then from (1) m = – n

m n 1 1
   direction cosine are : 0, ,
0 1 1 2 2
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 13
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Case-II if   0 ; m = 0, then from (1),  = – n

m n 1 1
     direction cosine are : , 0,
1 0 1 2 2
Let  be the angle between the lines
1 1 
cos = 0 + 0 +  cos = =
2 2 3
Hindi +m+n=0 ........(1) 2 + m2 = n2 ........(2)

(2) esa n = – ( + m) j[kus ij  2 + m2 = 2 + m2 + 2m m = 0

m n
Case-I ;fn  = 0 ; m  0 rks (1) ls m = –n   
0 1 1
1 1
fnd~ dksT;k,sa gS 0, ,
2 2
m n 1 1
Case-II ;fn   0 ; m = 0, rks (1) ls  = –n    fnd~ dksT;k,sa gS : , 0,
1 0 1 2 2
1
ekuk  js[kkvksa ds e/; dks.k gS cos = 0 + 0 +
2
1 
cos = =
2 3

B-13. Position vectors of A, B, C are given by a, b, c where a  b  b c  c  a = 0. If AC = 2iˆ – 3jˆ  6kˆ

then find BC if BC = 14. [DRN2037]
 
A, B, C dsfLFkfr lfn'k Øe'k% a, b, c gS tcfd a  b  b c  c  a = 0. ;fn AC = 2iˆ – 3jˆ  6kˆ rc BC
Kkr dhft;s ;fn BC = 14.
Ans. ± ( 4iˆ – 6ˆj  12kˆ )
Sol. a  b – c  b c a – c c  (a – c)  b  c  (a – c) = 0  (a – c)  (b – c) = 0
     2iˆ – 3ˆj  6kˆ 

CA CB = 0  BC is | | to AC 
BC = ± 14 
  = ± ( 4iˆ – 6ˆj  12kˆ )

 7 
ˆ ˆ ˆ ˆ ˆ
B-14. A vector c is perpendicular to the vectors 2 i  3 j  k , i  2 j  3k ˆ and satisfies the

 
condition c . 2 ˆi  ˆj  kˆ + 6 = 0. Find the vector c

,d lfn'k c lfn'kks a 2 ˆi  3 ˆj  kˆ ,oa ˆi  2 ˆj  3kˆ ds yEcor~ gS vkS j c .  2 ˆi  ˆj  kˆ  + 6 =

0 dks la r q " V djrk gS A lfn'k c dks Kkr dhft,A
Ans. 
3  ˆi  ˆj  kˆ 
Sol. Let c   ˆi  ˆj  kˆ
c  2iˆ  3ˆj – kˆ   2+ 3 –  = 0 ....(1)
c  ˆi  2jˆ  3kˆ    – 2 + 3 = 0 ....(2)
and given condition
c.(2iˆ – ˆj  k)
ˆ 6 0  2 –     6  0 ....(3)
Solving (1), (2) & (3)
=–3 =3  = 3. Hence c  3(–iˆ  ˆj  k)
ˆ

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 14
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Hindi. ekukfd c  ˆi  ˆj  kˆ

c  2iˆ  3ˆj – kˆ   2+ 3 –  = 0 ....(1)
c  ˆi  2jˆ  3kˆ    – 2 + 3 = 0 ....(2)
vkSj fn;k x;k izfrcU/k
c.(2iˆ – ˆj  k)
ˆ 6 0  2 –     6  0 ....(3)
(1), (2) vkSj (3) dks gy djus ij =–3 =3
 = 3. vr% c  3(–iˆ  ˆj  k)
ˆ

B-15. (a) Show that the perpendicular distance of the point c from the line joining a and b is
bc  c a  a  b
. [16JM120007]
ba
(b) Given a parallelogram ABCD with area 12 sq. units. A straight line is drawn through the
mid point M of the side BC and the vertex A which cuts the diagonal BD at a point ' O '. Use
vectors to determine the area of the quadrilateral OMCD.
(a) iznf'kZr dhft, fd a ,oa b dks feykus okyh js[kk ls fcUnq c dh yEcor~ nwjh
bc  c a  a  b
gSA
ba
(b) ,d lekUrj prqHkqZt ABCD dk {ks=kQy 12 oxZ bdkbZ fn;k x;k gSA Hkqtk BC ds e/; fcUnq M
vkSj 'kh"kZ A ls ,d ljy js[kk [kahpha xbZ gS tks fod.kZ BD dks fcUnq ' O ' ij dkVrh gSA prqHkqZt OMCD dk
{ks=kQy lfn'k fof/k dh lgk;rk ls Kkr dhft,A
Ans. (b) 5 unit sq. oxZ bdkbZ

Sol. (a)
1 1
Let  distance of c from line joining a and b is p. Now  = AB  AC  AB  p
2 2
| AB  AC | | (b  a)  (c  a) | | ab  bc  c a |
 p= = =
| AB | |ba| |b a|
(b) Equation of line AM is
 d
r   b  
 2 


Equation of line BD is
r  b   (d – b)
to obtain point of intersection

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 15
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 d   2
  b   = b  (d – b)  =1– &   =1– or =
 2  2 2 3

2  d
hence point O is  b   . Area OMCD = Area OMC + Area OCD
3  2

1 1  d   b 2d  1  b 2d   –2 2 
=  b       +      b  d
2 3  2   3 3  2  3 3   3 3 

1 1 d  1 1
=  b  2d   b)   (b  2d – 4d  b)
2 9 2  2 9
1 3 1 1 15
= bd  | 6b  d |   | bd|
18 2 18 18 2
15
=  12  5 sq. units
18  2
Hindi (a)

ekukfd c dh a vkSj b dks tksM+us okyh js[kk ls yEcor nwjh p gSA

1 1
vc = AB  AC  AB  p
2 2
| AB  AC | | (b  a)  (c  a) | | ab  bc  c a |
 p= = =
| AB | |ba| |b a|
(b) js[kk AM dk lehdj.k gS
 d
r   b  
 2 


js[kk BD dk lehdj.k gS r  b   (d – b)
izfrPNsn fcUnq izkIr djus ds fy,
 d   2
  b   = b  (d – b)  =1– &   =1– or =
 2  2 2 3

 d
2
vr% fcUnq O,  b   gSA
 2
3
OMCD dk {kS=kQy = OMC dk {kS=kQy + OCD dk {kS=kQy
1 1  d   b 2d  1  b 2d   –2 2 
=  b       +      b  d
2 3  2   3 3  2  3 3   3 3 

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 16
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 1 d  1 1
=  b  2d   b)   (b  2d – 4d  b)
2 9 2  2 9
1 3 1 1 15 15
= bd  | 6b  d |   | bd| =  12  5 oxZ bdkbZ
18 2 18 18 2 18  2

B-16. P, Q are the mid-points of the non-parallel sides BC and AD of a trapezium ABCD. Show that
APD = CQB.
P, Q leyEc prqHkqZt ABCD dh vlekUrj Hkqtkvksa BC vkSj AD ds e/; fcUnq gSA n'kkZb;s fd
APD = CQB.

Solution : Let AB = b and AD = d

Now DC is parallel to AB  there exists a scalar t such that DC = t AB = t b
 AC = AD + DC = d  t b
1 1
The position vectors of P and Q are (b  d  t b) and d respectively.
2 2
Now 2 APD = AP × AD
1 1
= (b  d  t b) × d = (1 + t) (b  d)
2 2
1 
Also 2 CQB = CQ × CB =  d  (d  tb) × [b  (d  t b)]
 2 

 1  1
=   d  t b    d  (1  t) b  =  (1  t) (d  b)  t (b  d)
 2  2
1 1
= (1  t  2t) (b  d) = (1  t) (b  d) = 2 APD Hence Prove.
2 2
Solution : ekuk AB = b rFkk AD = d
vc DC , AB ds lekUrj gS  vfn'k t fo|eku gS tcfd DC = t AB = t b
 AC = AD + DC = d  t b
1 1
P vkSj Q ds fLFkfr lfn'k (b  d  t b) rFkk d gSA
2 2
vc 2 APD = AP × AD
1 1
= (b  d  t b) × d = (1 + t) (b  d)
2 2
1 
rFkk 2 CQB = CQ × CB =  d  (d  tb) × [b  (d  t b)]
2 

 1  1
=   d  t b    d  (1  t) b  =  (1  t) (d  b)  t (b  d)
 2  2
1 1
= (1  t  2t) (b  d) = (1  t) (b  d) = 2 APD vr% fl) gqvkA
2 2

Section (C) : Line

[k.M (C) : js[kk

C-1. Find the coordinates of the point when the line through (3, 2, 5) and (–2, 3, –5) crosses the xy plane.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 17
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ml fcUnq ds funsZ'kkad Kkr dhft, tcfd js[kk] xy lery dks dkVrh gqbZ fcUnqvksa (3, 2, 5) vkSj (–2, 3, –5) ls
xqtjrh gSA

1 5 
Ans  2 , 2 ,0 
 
Sol. (x, y, 0) divides line segment joining (3, 2, 5) and (–2, 3, –5) in the ratio 1 : 1 internally.
32 1 23 5
x=  ,y= 
2 2 2 2
Sol. (x, y, 0) fcUnqvksa (3, 2, 5) vkSj (–2, 3, –5) dks feykus okyh js[kk dks 1 : 1 es vkUrfjd foHkkftr djrh gSA
32 1 23 5
x=  ,y= 
2 2 2 2

x y 1 z2
C-2. Find the foot of the perpendicular from (1, 6, 3) on the line = = .
1 2 3
x y 1 z2
fcUnq (1, 6, 3) ls ljy js[kk = = ij Mkys x;s yEc ikn ds funsZ'kkad Kkr dhft,A
1 2 3
Ans. (1, 3, 5)
x y 1 z2
Sol. Any point on the line = = k (say) ...(1)
1 2 3
is P(k, 2k + 1, 3k + 2)
Let it be the foot of  from Q(1, 6, 3) on the line (1)
Then d.r.s. of PQ are k – 1, 2k + 1 – 6, 3k + 2 – 3 i.e., k – 1, 2k – 5, 3k – 1
As PQ is  to the line (1), so
(k – 1) × 1 + (2k – 5) × 2 + (3k – 1) × 3 = 0  k – 1 + 4k – 10 + 9k – 3 = 0  14k – 14 = 0  k = 1.
 The foot of the  is given by P(1, 2 × 1 + 1, 3 × 1 + 2) = (1, 3, 5).

x y 1 z2
Hindi. ljy js[kk = = = k (ekuk) ...(1)
1 2 3
ij fLFkr P(k, 2k + 1, 3k + 2) fcUnq gksxk
ekuk ;gh fcUnq Q(1, 6, 3) ls ljy js[kk (1) ij Mkys x;s yEc ikn ds funsZ'kkad Kkr djksA
rks PQ ds fnd~ vuqikr gSA k – 1, 2k + 1 – 6, 3k + 2 – 3 vFkkZr~ k – 1, 2k – 5, 3k – 1
pw¡fd PQ ljy js[kk (1) ds yEcor~ gS] blfy;s (k – 1) × 1 + (2k – 5) × 2 + (3k – 1) × 3 = 0
 k – 1 + 4k – 10 + 9k – 3 = 0  14k – 14 = 0  k = 1.
vr% yEc ikn dk funsZ'kad P(1, 2 × 1 + 1, 3 × 1 + 2) = (1, 3, 5) gksxasA

C-3. (i) Find the cartesian form of the equation of a line whose vector form is given by .
r  2iˆ  ˆj  4kˆ  (iˆ  ˆj  2k)
ˆ

(i) ,d js[kk dk lfn'k lehdj.k r  2iˆ  ˆj  4kˆ  (iˆ  ˆj  2k)

ˆ gS] rks bldk dkrhZ; lehdj.k Kkr dhft,A
2x  4 3y  6
(ii) Find the vector form of the equation of a line whose cartesian form is given by =
1 2
–6z  6
= . [16JM120009]
1
2x  4 3y  6 –6z  6
(ii) js[kk ds lehdj.k dk lfn'k :i fyf[k, ftldk dkrhZ; :i = = gSA
1 2 1
x  2 y 1 z  4
Ans. (i)   (ii) r  (2iˆ  2jˆ  k)
ˆ  (3iˆ  4jˆ  k)
ˆ
1 1 2
Sol. (i)  r  2i – j  4k  (iˆ  ˆj – 2k)
ˆ ˆ ˆ ˆ

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 18
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Let (xiˆ  yjˆ  zk)

ˆ be a point lying on the line

ekukfd nh xbZ js[kk ij fLFkr dksbZ fcUnq (xiˆ  yjˆ  zk)

ˆ gSA

 xiˆ  yjˆ  zkˆ = î (2 + ) + ĵ ( – 1) + (4 – 2) x–2=;y+1=  ;

x  2 y 1 z – 4
 equation of line is  
1 1 –2
x  2 y 1 z – 4
 js[kk dk lehdj.k   gSA
1 1 –2
C-4. Find the distance between points of intersection of
x 1 y2 z3 x4 y 1
Lines = = & = =z and
2 3 4 5 2
Lines r = ( î  ĵ  k̂ ) +  (3 î  ĵ ) & r = (4 î  k̂ ) +  (2 î  3 k̂ )
fuEufyf[kr js[kkvksa ds izfrPNsn fcUnqvksa ds e/; nwjh Kkr dhft, &
x 1 y2 z3 x4 y 1
js[kkvksa = = ,oa = =z rFkk
2 3 4 5 2
js[kkvksa r = ( î  ĵ  k̂ ) +  (3 î  ĵ ) & r = (4 î  k̂ ) +  (2 î  3 k̂ )
Ans. 26
x 1 y  2 z  3
Sol.    r (Let) ... (1)
2 3 4
 (2r + 1, 3r + 2, 4r + 3) represents any point on (1)
x4 y 1 z  0
=  ... (2)
5 2 1
To find point of intersection of (1) and (2)
2r  1– 4 3r  2 – 1 4r  3 2r – 3 3r  1 4r  3
     4r – 6 = 15r + 5 11r = – 11 r = – 1
5 2 1 5 2 1
 point of intersection of (1) and (2) is (– 1, – 1, – 1)
r  (iˆ  ˆj – k)
ˆ  (3iˆ – ˆj) ... (1) r  (4iˆ – k)
ˆ  (2iˆ  3k)
ˆ ... (2)
For their point of intersection
3 + 1 = 4 + 2  3 – 2 – 3 = 0 ... (3)
1–=0  =1 ... (4)
and – 1 = – 1 + 3 =0
 point of intersection is (4, 0, – 1)
 required distance = (4  1)2  1  0 = 25  1 = 26
x 1 y  2 z  3
Hindi.    r (ekuk) ... (1)
2 3 4
 (2r + 1, 3r + 2, 4r + 3) js[kk (1) ij fLFkr fdlh fcUnq dks iznf'kZr djrk gSA
x4 y 1 z  0
=  ... (2)
5 2 1
(1) vkSj (2) dk izfrPNsn fcUnq izkIr djus ds fy, ––
2r  1– 4 3r  2 – 1 4r  3 2r – 3 3r  1 4r  3
     4r – 6 = 15r + 5 11r = – 11 r = – 1
5 2 1 5 2 1
 (1) vkSj (2) dk izfrPNsn fcUnq (– 1, – 1, – 1) gSA
r  (iˆ  ˆj – k)
ˆ  (3iˆ – ˆj) ... (1) r  (4iˆ – k)
ˆ  (2iˆ  3k)
ˆ ... (2)
buds izfrPNsn fcUnq ds fy,
3 + 1 = 4 + 2  3 – 2 – 3 = 0 ... (3)
1–=0  =1 ... (4)
vkSj – 1 = – 1 + 3  = 0
 izfrPNsn fcUnq (4, 0, – 1) gSA  vHkh"V nwjh = (4  1)2  1  0 = 25  1 = 26
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 19
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

C-5. Show that the foot of the perpendicular from the origin to the join of A(–9, 4, 5) and B (11, 0, –1) is the
mid point of AB. Also find distance of point (2, 4, 4) from the line AB.

iznf'kZr dhft, fd fcUnqvksa A(–9, 4, 5) vkSj B (11, 0, –1) dks feykus okyh js[kk ij ewyfcUnq ls Mkys x;s yEc dk
ikn AB dk e/; fcUnq gS rFkk js[kk AB ls fcUnq (2, 4, 4) dh nwjh Hkh Kkr dhft,A
Ans. 3

Sol.

 P (mid-point of AB) is (1, 2, 2)

 dr’s of OP are : 1, 2, 2 and dr’s of AB are : 20, – 4, – 6
 1 × 20 + 2 × (– 4) + 2 × (– 6) = 20 – 12 – 8 = 0
 OP  AB
as the midpoint of (0, 0, 0) and (2, 4, 4) is (1, 2, 2) so the distance = 1  22  22  3

Ans. 3

Hindi.

 P (AB dk e/;fcUnq), (1, 2, 2) gS  OP ds fnd~vuqikr gSa 1, 2, 2

vkSj AB ds fnd~vuqikr gSa 20, – 4, – 6
 1 × 20 + 2 × (– 4) + 2 × (– 6) = 20 – 12 – 8 = 0  OP  AB
pwafd (0, 0, 0) vkSj (2, 4, 4) dk e/; fcUnq (1, 2, 2) gS blfy, nwjh = 1  22  22  3

x3 y3 z
C-6. Find the equation of the two lines through the origin which intersect the line = = at an
2 1 1
angle of /3.
x3 y3 z
ewyfcUnq ls xqtjus okyh nks ljy js[kkvksa ds lehdj.k Kkr dhft, tks js[kk = = dks /3 dks.k ij
2 1 1
dkVrh gaSA
x y z x y z
Ans. = = , = =
1 2  1 1 1 2

Sol.

Any point on the given line is (2r + 3, r + 3, r)

Dr’s of new line are  2r + 3, r + 3, r

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 20
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 2(2r  3)  1(r  3)  1.r

cos 
3 6 (2r  3)2  (r  3)2  r 2
Solving above we get r = – 1, – 2
Dr’s are 1, 2, – 1 or – 1, 1, – 2
x y z x y z
equation of the line is   or  
1 2 1 1 1 2

Hindi.

nh xbZ js[kk ij dksbZ fcUnq gS (2r + 3, r + 3, r)

ubZ js[kk ds fnd~vuqikr gSa  2r + 3, r + 3, r
 2(2r  3)  1(r  3)  1.r
cos 
3 6 (2r  3)2  (r  3)2  r 2
mijksDr dks gy djus ij gesa izkIr gksrk gSA r = – 1, – 2
fnd~vuqikr 1, 2, – 1 ;k – 1, 1, – 2 gSaA
x y z x y z
js[kk dk lehdj.k   ;k   gSA
1 2 1 1 1 2

C-7. The foot of the perpendicular from (a, b, c) on the line x = y = z is the point (r, r, r), then find the value of
r.
fcUnq (a, b, c) ls js[kk x = y = z ij Mkys x;s yEc dk ikn fcUnq (r, r, r) gS] rks r dk eku Kkr dhft,A
abc
Ans. r =
3

Sol.

x=y=z=r
(r – a)r + (r – b)r + (r – c)r = 0
a + b + c = 3r

C-8. Find the shortest distance between the lines :

r = (4iˆ  ˆj)   (iˆ  2jˆ  3k)
ˆ and r = (iˆ  ˆj  2k)
ˆ   (2iˆ  4jˆ  5k)
ˆ

js[kkvksa r = (4iˆ  ˆj)   (iˆ  2jˆ  3k)

ˆ vkSj r = (iˆ  ˆj  2k)
ˆ   (2iˆ  4jˆ  5k)
ˆ ds chp U;wure nwjh Kkr dhft,A
6
Ans. unit bdkbZ
5
(iˆ – ˆj  2kˆ – 4iˆ  ˆj) · [(iˆ  2jˆ – 3k)
ˆ  (2iˆ  4ˆj – 5k)]
ˆ (– 3iˆ  2k)
ˆ . (2iˆ – j) 6
Sol. SD = = =
(iˆ  2jˆ – 3k)
ˆ  (2iˆ  4ˆj – 5k)
ˆ ˆ ˆ
| 2i – j | 5

x 3 y 8 z3 x3 y7 z6

C-9. Let L1 and L2 be the lines whose equation are   and = =
3 1 1 3 2 4
respectively. A and B are two points on L1 and L2 respectively such that AB is perpendicular both the
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 21
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

lines L1 and L2.Find points A, B and hence find shortest distance between lines L 1 and L2

x 3 y 8 z3 x3 y7 z6

ekukfd js[kk;sa L1 rFkk L2 ftuds lehdj.k Øe'k%   rFkk = = gaSA A rFkk
3 1 1 3 2 4
B nks fcUnq Øe'k% L1 vkSj L2 ij bl izdkj fLFkr gSa fd AB nksuksa js[kkvksa L1 rFkk L2 ds yEcor~ gSA nksuksa fcUnq A o B
Kkr dhft, rFkk L1 o L2 ds e/; U;wure nwjh Hkh Kkr dhft,A [16JM120010]
Ans. A = (3, 8, 3), B = (–3, – 7, 6), AB = 3 30
Sol. Let the co-ordinates of A be (3 + 3, 8 – ,  + 3) and the co-ordinates of B be (–3 – 3, 2 – 7, 4 +
6).
Then direction ratios of AB are 3 + 3 + 6, – – 2 + 15,  – 4 – 3
AB L1 so 3(3 + 3 + 6) – (– – 2 + 15) + (– 4 – 3) = 0 i.e.
11 + 7 = 0
and AB L2 so – 3(3 + 3 + 6) + 2(– – 2 + 15) + 4( – 4 – 3) = 0 i.e. –7 – 29 = 0
 ==0 so the point A is (3, 8, 3) and the point B is (–3, –7, 6)
 AB = 36  225  9 = 270 = 3 30
Hindi. ekukfd A ds funsZ'kkad A(3 + 3, 8 – ,  + 3) gaS rFkk B ds funsZ'kkad (–3 – 3, 2 – 7, 4 + 6) gaSA
rc AB ds fnd~vuqikr 3 + 3 + 6, – – 2 + 15,  – 4 – 3 gksaxs
AB L1 blfy, 3(3 + 3 + 6) – (– – 2 + 15) + (– 4 – 3) = 0 vFkkZr~ 11 + 7 = 0
rFkk AB L2 blfy, –3(3 + 3 + 6) + 2(– – 2 + 15) + 4( – 4 – 3) = 0 vFkkZr~ –7 – 29 = 0
 ==0
blfy, fcUnq A (3, 8, 3) gS rFkk fcUnq B (–3, –7, 6) gSA
 AB = 36  225  9 = 270 = 3 30

C-10. If r = (iˆ  2jˆ  3k)

ˆ   (iˆ  ˆj  k)
ˆ and r = (iˆ  2jˆ  3k)
ˆ   (iˆ  ˆj  k)
ˆ are two lines, then find the equation
of acute angle bisector of two lines.
;fn r = (iˆ  2jˆ  3k) ˆ   (iˆ  ˆj  k)
ˆ vkSj r = (iˆ  2jˆ  3k)
ˆ   (iˆ  ˆj  k)
ˆ nks js[kk,¡ gSa] rks nksuksa js[kkvksa ds
U;wudks.k lef}Hkktd dk lehdj.k Kkr dhft,A
Ans. r = (iˆ  2jˆ  3k)
ˆ  t ( ˆj  k)
ˆ , where t is parameter (tgk¡ t izkPky gSA)
1 1  (–1)(1)  (1)(–1) 1
Sol. Angle between two vectors =  – . Hence obtuse angle between them.
3 3 3
 ˆi – ˆj  kˆ ˆi  ˆj – kˆ  2
Vector along acute angle bisector =   –   ;  – ˆj  kˆ   t( ˆj – k)
ˆ
 3 3  3  

hence equation of acute angle bisector = (iˆ  2jˆ  3k) ˆ  t( ˆj – k)ˆ

1 1  (–1)(1)  (1)(–1) 1
Hindi nks lfn'kksa ds chp dks.k = – . vr% buds e/; vf/kd dks.k gS
3 3 3
 ˆi – ˆj  kˆ
ˆi  ˆj – kˆ  2
U;wu dks.k lef}Hkktd ds vuqfn'k lfn'k gS =   –
 ;  – ˆj  kˆ   t( ˆj – k)
ˆ
 3 3  3  

vr% U;wu dks.k lef}Hkktd dh lehdj.k gS = (iˆ  2jˆ  3k)

ˆ  t( ˆj – k)
ˆ

C-11. The edges of a rectangular parallelopiped are a, b, c; show that the angles between two of the four
  a2  b2 – c 2     a2 – b2  c 2     a2 – b2 – c 2  
diagonals are given by cos1    2 2  or cos 1

    
 or cos 1
   2 2  .
  a  b  c    a  b  c    a  b  c 
2 2 2 2 2

,d vk;rkdkj lekUrj"kV~Qyd ds fdukjs a, b, c gS] iznf'kZr dhft, fd pkjksa fod.kksZa ds e/; dks.k
  a2  b2 – c 2     a2 – b2  c 2     a2 – b2 – c 2  
cos1    2 
2 
or cos 1
   2 2 
 or cos
1
   2 2   . gksrk gSA
  a  b 2
 c    a  b 2
 c    a  b  c 
2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 22
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol.

Dr's of OP are a, b, c ; Dr's of AM are –a, b,c

a( a)  b(b)  c(c) –a2  b2  c 2
If  be the angle between OP and AM, we get cos   =
a2  b2  c 2 a 2  b 2  c 2 a2  b2  c 2

Similarly we can find the angle between other pairs and hence angles in general between six pairs are
given by
 a 2  b 2  c 2 
  cos1  2 2 
 a b c 
2

Note : All +ve or -ve signs can not be taken because angle between diagonals are 0 or  radian is
not possible.

Hindi

OP ds fnd~ vuqikr a, b,c gSaA AM ds fnd~ vuqikr – a, b,c gaSA

;fn  , OP rFkk AM ds e/; dks.k gS] rks
a( a)  b(b)  c(c) –a2  b2  c 2
cos   =
a2  b2  c 2 a 2  b 2  c 2 a2  b2  c 2
blh rjg ge nwljs ;qXeksa ds e/; dks.k Kkr dj ldrs gSaA
 a 2  b 2  c 2 
bl izdkj lkekU; :i esa N% ;qXeksa ds e/; dks.k   cos1  2 
}kjk fn;k tkrk gS
 a b c 
2 2

uksV : ,d lkFk lHkh /kukRed ;k lHkh _.kkRed fpUg ugha fy;s tk ldrs gS D;ksafd fod.kks± ds e/; dks.k 0 ;k 
jsfM;u laHko ugha gSA

C-12.Show that equation of angle bisectors of line r  a  b and r = b + µ a are r = (a  b) +  b a  a b  

n'kkZb;sa fd js[kkvksa r  a  b vkSj r = b + µ a ds dks.k v)Zdksa ds lehdj.k r = (a  b) +  b a  a b gSA 
Sol. Point of intersection of r  a  b and r = b + µ a is a + b
Now angle bisector of vectors a & b is a + b and angle bisector of vector a and – b is a – b so
 
equation of angle bisectors of given lines are r = a  b +  â  bˆ  
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 23
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Hindi : r  a  bvkSj r = b + µ a dk izfrPNsn fcUnq a + b gSA

vc o ds e/; dks.k v)Zd rFkk a o – b ds e/; dks.k v)Zd a – b gSA blfy, nh xbZ js[kkvksa ds dks.k v)Zdksa ds
lehdj.k r =  a  b  +   â  bˆ  gSA

C-13. Prove that the shortest distance between the diagonals of a rectangular parallelepiped whose
bc ca ab
coterminous sides are a, b, c and the edges not meeting it are , ,
b c
2 2
c a
2 2
a  b2
2

fl) dhft, vk;rkdkj lekUrj "kV~Qyd ds fod.kksa ds e/; y?kqÙke nwjh ftldh vklUu Hkqtk,a a, b, c gS rFkk ;g
bc ca ab
fdukjksa dks ugha fey jgh gS , , gSA
b c
2 2
c a
2 2
a  b2
2

Sol. Let coordinates of rectangular parallelepiped are A(0,0,0); B(a,0,0); C(a,b,0); D(0,b,0); E(0,0,c);
x y z
F(a,0,c); G(a,b,c); H(0,b,c). Then equation of diagonal AG is   and equation of BC is
a b c
x a y 0 z0
 
0 1 0
ˆ ˆj  (aiˆ  bjˆ  ck))
(ai).( ˆ ac
 shortest distance between AG and BC = =
ˆj  (aiˆ  bjˆ  ckˆ a  c2
2

bc
Similarly shortest distance between AG and DC =
b  c2
2

ba
Similarly shortest distance between AG and BF =
b2  a2
Sol. ekuk lekUrj "kV~Qyd ds funsZ'kkad A(0,0,0); B(a,0,0); C(a,b,0); D(0,b,0); E(0,0,c); F(a,0,c); G(a,b,c);
x y z x a y 0 z0
H(0,b,c) rc fod.kZ AG dk lehdj.k   AG vkSj BC ds e/; nqjh  
a b c 0 1 0
ˆ ˆj (aiˆ  bjˆ  ck))
(ai).( ˆ ac
 AG rFkk BC ds e/; nqjh = =
ˆj  (aiˆ  bjˆ  ckˆ a  c2
2

bc
blhizdkj AG rFkk DC ds e/; nqjh =
b  c2
2

ba
blhizdkj AG rFkk BF ds e/; y?kqÙke nqjh =
b  a2
2

Section (D) : STP, VTP, Vector equation, LI/LD

[k.M (D) : vfn'k f=kd~xq.ku]lfn'k f=kd~xq.ku] lfn'k lehdj.k] jsf[k; ijra=k@jsf[k; Lora=k

D-1. Show that n'kkZb;s fd {( a + b + c ) × ( c – b )} . a = 2 a b c  .

Sol. Required vHkh"V = ( a × c – a × b + b × c – c × b ) . a = b c a  –  c b a  a b c 

= 2 a b c 
   
D-2. Given unit vectors m̂ , n̂ and p̂ such that  m ˆ nˆ  = pˆ  mˆ x nˆ  =  then find value of
 
nˆ pˆ m
ˆ  in terms of .

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 24
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

   
bdkbZ lfn'k m̂ , n̂ ,oa p̂ bl izdkj gS fd  mˆ nˆ  = pˆ  mˆ x nˆ  =  rks nˆ pˆ mˆ  dk eku 
 
ds inksa esa Kkr dhft,A
Ans. sin  cos 
Sol. nˆ pˆ mˆ  = pˆ m
ˆ nˆ  = p.(m ˆ = pˆ . m
ˆ ˆ  n) ˆ  nˆ cosa = ˆ  nˆ cosa = sina cosa
m

D-3. Let a=a1ˆi+a2 ˆj+a3kˆ , b=b1ˆi+b2 ˆj+b3kˆ and c=c1ˆi+c 2 ˆj+c 3 kˆ be three non-zero vectors such that c

is a unit vector perpendicular to both a and b . If the angle between aandb is , then
6
2
a1 a2 a3
b1 b2 b3 is equal to:
c1 c 2 c3
ekuk a=a1ˆi+a2 ˆj+a3kˆ , b=b1ˆi+b2 ˆj+b3 kˆ vkSj c=c1 ˆi+c 2 ˆj+c 3 kˆ rhu v'kwU; lfn'k bl rjg gS fd

c ,d bdkbZ lfn'k gS tks a rFkk b nksuksa ds yEcor~ gSA ;fn a rFkk b ds e/; dks.k gS] rks
6
2
a1 a2 a3
b1 b2 b3 cjkcj gS&
c1 c 2 c3
1 2
Ans. (a1  a22  a32 ) (b12  b22  b32 )
4
2
a1 a2 a3
a 2b 2
Sol. b1 b2 b3 = [a b c]2 = ((a  b).c)2 = (absin c.c) 2 =
4
c1 c 2 c3
1 2
= (a1  a22  a32 ) (b12  b22  b32 )
4
D-4. Examine for coplanarity of the following sets of points
fuEu fcUnqvksa ds leqP;;ksa dh leryrk dh tk¡p dhft, &
(a) 4 î + 8 ĵ + 12 k̂ , 2 î + 4 ĵ + 6 k̂ , 3 î + 5 ĵ + 4 k̂ , 5 î + 8 ĵ + 5 k̂ .
(b) 3 a + 2 b – 5 c , 3 a + 8 b + 5 c , –3 a + 2 b + c , a + 4 b – 3 c . Where a , b , c are noncoplanar
3 a + 2 b – 5 c , 3 a + 8 b + 5 c , –3 a + 2 b + c , a + 4 b – 3 c tgk¡ a , b , c vleyrh; gSA
Ans. (a) Coplanar (b) Non-coplanar (a) leryh; (b) vleryh;

Sol. (a) Let A  4iˆ  8jˆ  12kˆ ; B  2iˆ  4jˆ  6kˆ ; C  3iˆ  5jˆ  4kˆ ; D  5iˆ  8jˆ  5kˆ
AB  – 2iˆ – 4jˆ – 6kˆ ; BC  ˆi – ˆj – 2kˆ ; CD  2iˆ  3jˆ  kˆ
–2 4 6
[AB BC CD] = 1 1 2 = [– 2 (1 + 6) + 4(1 + 4) – 6(3 – 2)] = – 14 + 20 – 6 = 0
2 3 1
Hence coplanar
(b) Using these 4 points let us create any 3 vectors v1  6b  10c, v 2  6a  6c
v3  2a  2b  2c check coplanarity of these vectors
0 6 10
[v1 v 2 v 3 ] = 6 0 6 [a b c]  0 hence non coplanar
2 2 2
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 25
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Hindi. (a) ekuk A  4iˆ  8jˆ  12kˆ ; B  2iˆ  4jˆ  6kˆ ; C  3iˆ  5jˆ  4kˆ ; D  5iˆ  8jˆ  5kˆ
AB  – 2iˆ – 4jˆ – 6kˆ ; BC  ˆi – ˆj – 2kˆ ; CD  2iˆ  3jˆ  kˆ
–2 4 6
[AB BC CD] = 1 1 2 = [– 2 (1 + 6) + 4(1 + 4) – 6(3 – 2)] = – 14 + 20 – 6 = 0
2 3 1
vr% leryh; gS
(b) bu 4 fcUnqvksa dk mi;ksx djds 3 lfn'k cukus ij v1  6b  10c, v 2  6a  6c
v3  2a  2b  2c budh leryrk dh tk¡p djus ij
0 6 10
[v1 v 2 v 3 ] = 6 0 6 [a b c]  0 vr% vleryh; gSA
2 2 2

D-5. The vertices of a tetrahedron are P(2, 3, 2), Q(1, 1, 1), R(3, –2, 1) and S (7, 1, 4).
(i) Find the volume of tetrahedron
(ii) Find the shortest distance between the lines PQ & RS.
prq"Qyd ds 'kh"kZ P(2, 3, 2), Q(1, 1, 1), R(3, –2, 1) vkSj S (7, 1, 4) gS rc Kkr dhft;s
(i) prq"Qyd dk vk;ru
(ii) js[kk PQ rFkk RS ds e/; y?kqÙke nwjh

3
Ans. (i) 1/2 unit3 (ii) unit
35

Sol.

1 2 1
1
(i) v 1 5 1
6
5 2 2
1
= 1 ( 10  2)  2(2  5)  1 ( 2  25)
6
1 1
= ( 12  14  23) =
6 2
ˆ ˆ ˆ
(ii) ˆ (3i  j  5k) =
(iˆ  5ˆj  k).
3
9  1  25 35

D-6. Are the following set of vectors linearly independent?

D;k fuEufyf[kr lfn'kksa ds leqPp; ,d?kkrr~% Lora=k gaS \
(i) a = î  2 ĵ + 3 k̂ , b = 3 î 6 ĵ + 9 k̂

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 26
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(ii) a = 2 î  4 k̂ , b = î  2 ĵ  k̂ , c = î  4 ĵ + 3 k̂
Ans. (i) No ugha (ii) Yes gk¡
Sol. (i) b  3a linearly dependent (,d ?kkrr% ijra=k)
(iii) [a b c]  0 linearly independent (,d ?kkrr% LorU=k)

D-7. Find value of xR for which the vectors a = (1, –2, 3), b = (–2, 3, – 4), c = (1, – 1, x) form a linearly
dependent system.
x ds ,sls okLrfod eku Kkr dhft, ftuds fy, lfn'k a = (1, –2, 3), b = (–2, 3, – 4), c = (1, – 1, x) ,d
?kkrr% ijrU=k fudk; cukrs gSA
Ans. x = 1
Sol. For linearly dependent vectors (,d ?kkrr% ijrU=k lfn'kksa ds fy;sa)
(i – 2j + 3k) + m(– 2i + 3j – 4k) + n(i – j + xk) = 0  – 2m + n = 0, – 2 + 3m – n = 0
1 –2 1
3 – 4m + nx = 0  –2 3 –1 = 0 is x = 1
3 –4 x

D-8. If a , b , c are non-coplanar vectors and v . a  v . b  v . c = 0, then find value of v .

;fn a , b , c vleryh; lfn'k gS vkSj v . a  v . b  v . c = 0 gks] rks v dk eku Kkr dhft,A
Ans. v =0
Sol. Let v  1a  2b  3c [ a, b, c are non coplanar]

v.a = v.b = v.c = 0    1a.a +  2a.b + 3 a.c = 0

1b.a +  2b.b + 3b.c = 0   1c.a +  2c.b + 3 c.c = 0

a.a a.b a.c

Only possible values of 1, 2, 3 = 0 as b.a b.b b.c 0
c.a c.b c.c

Hindi ekuk v  1a  2b  3c [ a, b, c vleryh; gS ]

v.a = v.b = v.c = 0    1a.a +  2a.b + 3 a.c = 0

1b.a +  2b.b + 3b.c = 0   1c.a +  2c.b + 3 c.c = 0

a.a a.b a.c

1, 2, 3 dk lEHko eku 'kwU; gksxk D;ksafd b.a b.b b.c 0
c.a c.b c.c

D-9. Let a  ˆi  2 ˆj  3kˆ , b  2 ˆi  ˆj  kˆ , c  3 ˆi  2 ˆj  kˆ and d = 3 ˆi  ˆj  2kˆ , then

(i) if a  (b  c) = pa  qb  rc , then find value of p, q and r.
(ii) find the value of ( a × b ) × ( a × c ). d
ekukfd a  ˆi  2 ˆj  3kˆ , b  2 ˆi  ˆj  kˆ , c  3 ˆi  2 ˆj  kˆ
vkSj d = 3 ˆi  ˆj  2kˆ gS] rks
(i) ;fn a  (b  c) = pa  qb  rc gks] rks p, q vkSj r ds eku Kkr dhft,A
(ii) ( a × b ) × ( a × c ). d dk eku Kkr dhft,A
Ans. (i) p = 0; q = 10; r =  3 (ii) – 100

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 27
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. (i) a  (b  c) = (a·c)b  (a.b)c  10 b  3 c = pa  qb  rc

p = 0, q = + 10, r = – 3 [ a,b,c are non coplanar]
(ii) (a  b) × (a  c) . d    
= { (a  b).c a – (a  b).a c } · d = [a b c] a.d – 0 = 20 × (– 5) = –100

Hindi (i) a  (b  c) = (a·c)b  (a.b)c  10b  3c = pa  qb  rc

p = 0, q = + 10, r = – 3 [ a,b,c vleryh; gaS]
(ii) (a  b) × (a  c) . d    
= { (a  b).c a – (a  b).a c } · d = [a b c] a.d – 0 = 20 × (– 5) = –100

1 1
D-10. Given that x  2
(p . x) p  q , then show that p.x  (p.q) and hence find x in terms of p and q .
p 2

1 1
fn;k x;k gS fd x  2
(p . x) p  q , rks iznf'kZr dhft, fd p.x  (p.q) vkSj x ds eku dks p vkSj q ds inksa
p 2
esa Kkr dhft,A
(p . q) p
Ans. x  q
2 | p |2
Sol. Dot with p & obtain answer. p ds lkFk vfn'k xq.kuQyu djus ij mÙkj izkIr gksrk gSA
1 1 (p . q) p
x (p.q) p  q  x  q–
| p |2 2 2 | p |2

D-11. Let there exist a vector x satisfying the conditions x × a = c  d and x + 2 d = v  d . Find x in  
terms of a , c and d
ekuk lfn'k x bl izdkj vfLrRo gS fd izfrcU/k x × a = c  d vkSj x + 2 d =  v  d dks larq"V djrk gSA
a , c vkSj d dks x ds inksa esa Kkr dhft,A
d  (c  d) – 2 | d |2 a
Ans. x =
d.a
Sol. d ×  x  a = d × c  d  
   
 d.a x – d.x a = d × c  d   ....(i)

Now d.x + 2 d.d = d . v  d = 0  

2
 d.x = – 2 d ....(ii)

2
d  (c  d) – 2 d a
From (i) and (ii) we get x = ((i) vkSj (ii) ls½
d.a

Section (E) : Plane

[k.M (E) : lery
E-1. Find equation of plane
(i) Which passes through (0, 1, 0), (0, 0, 1), (1, 2, 3)
(ii) Which passes through (0, 1, 0) and contains two vectors ˆi  ˆj  kˆ & 2iˆ  ˆj .
(iii) Whose normal is ˆi  ˆj  kˆ & which passes through (1, 2, 1).
(iv) Which makes equal intercepts on co-ordinate axis and passes through (1, 2, 3)
lery dk lehdj.k Kkr dhft,
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 28
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(i) tks fcUnqvksa (0, 1, 0), (0, 0, 1), (1, 2, 3) ls xqtjrk gSA
(ii) tks fcUnqvksa (0, 1, 0) ls xqtjrk gS rFkk ˆi  ˆj  kˆ rFkk 2iˆ  ˆj dks lekfgr djrk gS
(iii) ftldk vfHkyEc ˆi  ˆj  kˆ gS tks fcUnq (1, 2, 1) ls xqtjrk gSA
(iv) tks funsZ'kkad v{kksa ij cjkcj vUr[k.M dkVrh gS rFkk (1, 2, 3) ls xqtjrh gSA
Ans. (i) 4x – y – z + 1 = 0 (ii) x + 2y + 3z – 2 = 0 (iii) x+y+z–4=0
(iv) x+y+z–6=0
x y 1 z
Sol. (i) 0 1 1 = 0  x(–3 – 1) – (y – 1)(0 – 1) + z(1) = 0
1 1 3
–4x + y – 1 + z = 0  4x – y – z + 1 = 0
x y 1 z
(ii) 1 1 1 = 0  x(–1) – (y – 1)(2) + z(–1 – 2) = 0
2 1 0
–x – 2y + 2 – 3z = 0  x + 2y + 3z – 2 = 0
(iii) x+y+z=1+2+1  x+y+z–4=0
(iv) x+y+z=1+2+3  x+y+z–6=0

E-2. Find the ratio in which the line joining the points (3, 5,–7) and (–2, 1, 8) is divided by the yz-plane. Find
also the point of intersection of the plane and the line [16JM120011]
fcUnq (3, 5,–7) vkSj (–2, 1, 8) dks tksM+us okyh js[kk dks yz-lery fdl vuqikr esa foHkkftr djrk gS] Kkr dhft,
lkFk lkFk gh lery vkSj js[kk dk izfrPNsn fcUnq Hkh Kkr dhft,A
Ans. 3 : 2, (0,13/5, 2)

Sol.

 2  3   5 8  7 
 R , ,
  1  1   1 
R lies on y –z plane
2  3 3
 =0  
 1 2
 13 
the ratio is 3 : 2  R  0, , 2
 5 

Hindi

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 29
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 2  3   5 8  7 
 R , ,
  1  1   1 
fcUnq R lery yz ij fLFkr gS
2  3 3
 =0  
 1 2
 13 
vHkh"V vuqikr 3 : 2 gS  R  0, , 2
 5 

E-3. Find the locus of the point whose sum of the square of distances from the planes x + y + z = 0, x – z = 0
and x – 2y + z = 0 is 9
ml fcUnq dk fcUnqiFk Kkr dhft,] ftldh leryksa x + y + z = 0, x – z = 0 rFkk x – 2y + z = 0 ls nwfj;ksa ds oxkZs
dk ;ksx 9 gksA
Ans. x2 + y2 + z2 = 9
Sol. Let point is (, , ) ekuk dksbZ fcUnq (, , ) gS] rks
2 2 2
     2  
   9   2(     )2  3(   )2  (  2   )2  54
3 2 6
6  6  6  54 
2 2 2
 Locus is x2 + y2 + z2 = 9
 fcUnqiFk x2 + y2 + z2 = 9 gSA

E-4. The foot of the perpendicular drawn from the origin to the plane is (4, –2, –5), then find the vector
equation of plane.
ewyfcUnq ls ,d lery ij Mkys x;s yEc dk ikn (4, –2, –5) gS] rks lery dk lfn'k lehdj.k Kkr dhft,A
Ans. r . (4iˆ  2jˆ  5k)
ˆ = 45

Sol.
Equation of plane is lery dk lehdj.k gSA
r – (4iˆ – 2jˆ – 5k)ˆ  · 4iˆ – 2jˆ – 5kˆ  0   
r · 4iˆ – 2jˆ – 5kˆ  45

E-5. Let P (1, 3, 5) and Q(–2, 1, 4) be two points from which perpendiculars PM and QN are drawn to the x-z
plane. Find the angle that the line MN makes with the plane x + y + z = 5.
ekuk fd P (1, 3, 5) rFkk Q(–2, 1, 4) nks fcUnq gS ftuls x-z lery ij Øe'k% PM rFkk QN yEc Mkys tkrs gaSA
js[kk MN }kjk lery x + y + z = 5 ds lkFk cuk;k x;k dks.k Kkr dhft,A
4
Ans. sin–1
30
Sol. P(1, 3, 5) ; Q (–2, 1, 4) ; M(1, 0, 5) ; N(–2, 0, 4)
ˆ ˆ ˆ
ˆ sin  = MN . (i  j  k) = 3  1 = 4
NM = (3iˆ  k)
| MN | 3 10 3 30
4
 Angle b/w plane and line MN is sin–1
30
Hindi. P(1, 3, 5) ; Q (–2, 1, 4) ; M(1, 0, 5) ; N(–2, 0, 4)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 30
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

MN . (iˆ  ˆj  k)
ˆ 3 1 4
NM = (3iˆ  k)
ˆ sin  = = =
| MN | 3 10 3 30

4
 js[kk MN rFkk lery ds e/; dks.k sin–1 gSA
30
x –1 y–2 z–3
E-6. The reflection of line = = about the plane x – 2y + z – 6 = 0 is
3 4 5
x –1 y–2 z–3
js[kk = = dk lery x – 2y + z – 6 = 0 ds lkis{k izfrfcEc gksxkA
3 4 5
x–3 y2 z–5
Ans. = =
3 4 5
Sol.  3.1 – 2.4 + 5 ×1 = 0, line is parallel to the plane
 reflection of line will also have same direction ratios i.e. 3, 4, 5
Also mirror image of (1, 2, 3) will be on required line.
x –1 y–2 z–3  1– 4  3 – 6 
= = = – 2 2 2 2 
 (x, y, z) = (3, – 2, 5)
1 –2 1  1  1  (–2) 
x–3 y2 z–5
 equation of straight line = =
3 4 5
Hindi  3.1 – 2.4 + 5 ×1 = 0, js[kk lery ds lekUrj gSA
 vr% js[kk ds izfrfcEc ds fnd~ vuqikr ogh vFkkZr~ 3, 4, 5 gksxsaA
x –1 y–2 z–3  1– 4  3 – 6 
iqu% (1, 2, 3) dk niZ.k izfrfcEc vHkh"B js[kk ij gksxkA = = = – 2 2 2 2 
1 –2 1  1  1  (–2) 
 (x, y, z) = (3, – 2, 5)
x–3 y2 z–5
 ljy js[kk dk lehdj.k = =
3 4 5

x 1 y2 z3
E-7. Find the equation of image of the line = = in the plane 3x – 3y + 10z = 26.
9 1 3
x 1 y2 z3
js[kk = = dk lery 3x – 3y + 10z = 26 esa izfrfcEc dk lehdj.k Kkr
9 1 3
dhft,A[16JM120012]
x4 y 1 z7
Ans. = =
9 1 3

Sol.

 A is mirror image of A w.r.t the given plane

x –1 y 2 z 3  3 – 6 – 30 – 26 
    –2  
3 3 10  9  9  100 
x –1 y – 2 z 3  59 
     2  –   A(4, – 1, 7)
3 3 10  118 
x – 4 y 1 z  7
  Equation of required line  
9 1 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 31
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Hindi.

 A fn;s x;s lery ds lkis{k A dk izfrfcEc gSA

x –1 y 2 z 3  3 – 6 – 30 – 26 
    –2  
3 3 10  9  9  100 
x –1 y – 2 z 3  59 
     2  –   A(4, – 1, 7)
3 3 10  118 
x – 4 y 1 z  7
  vHkh"V js[kk dk lehdj.k  
9 1 3

E-8. Find the angle between the plane passing through points (1, 1, 1), (1,  1, 1), (7, 3, 5) & xz plane.
fcUnqvksa (1, 1, 1), (1,  1, 1), (7, 3, 5) ls xqtjus okys lery vkSj x-z lery ds e/; dks.k Kkr dhft,A
Ans. /2
Sol. Equation of plane passing through (1, 1, 1) is a(x – 1) + b(y – 1) + c(z – 1) = 0 ... (1)
 it passes through (1, – 1, 1) and (– 7, – 3, – 5)
 a.0 – 2.b + 0.c = 0  b=0 and – 8a – 4b – 6c = 0
4a + 2b + 3c = 0  b=0 
4a
 4a + 3c = 0   c= –
3
4
 dr’s of normal to the plane are 1, 0 – and dr’s of the normal to the x-z plane are 0, 1, 0
3
000 
 cos = 0  =
a 2
a 2
1
2

Hindi. (1, 1, 1) ls xqtjus okys lery dk lehdj.k gS a(x – 1) + b(y – 1) + c(z – 1) = 0 ... (1)
 ;g (1, – 1, 1) vkSj (– 7, – 3, – 5) ls xqtjrk gSA
 a.0 – 2.b + 0.c = 0  b = 0 vkSj – 8a – 4b – 6c = 0 4a + 2b + 3c = 0
4a
 b=0  4a + 3c = 0  c = –
3
4
 lery ds vfHkyEc ds fnd~vuqikr 1, 0 – gSaA vkSj x-z lery ds vfHkyEc ds fnd~vuqikr 0, 1, 0 gSaA
3
000 
 cos = 0  =
a 2
a 2
1
2

3y z2
E-9. Find the equation of the plane containing parallel lines (x  4) = = and
4 5
(x  3) =  (y + 2) = z
3y z2
ml lery dk lehdj.k Kkr dhft, ftlesa lekUrj js[kk,¡ (x  4) = = vkSj (x  3) =  (y + 2) =
4 5
z fufgr gaSA
Ans. 11x  y  3z = 35
Sol.
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 32
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x – 4 y 3 z2
  .... (1)
1 4 5
x 3 y 2 z –0
  .... (2)
1 1/  1/ 
Equation of the plane is lery dk lehdj.k gS ––
x–3 y2 z
1 5 2 0  (x – 3) (25 + 8) – (y + 2) (5 – 2) + z(– 4 – 5) = 0
1 –4 5
33x – 99 – 3y – 6 – 9z = 0  33x – 3y – 9z – 105 = 0  11x – y – 3z = 35

E-10. Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line
x3 y2 z
= = [16JM120013]
3 6 2
x3 y2 z
fcUnq (2, 3, 4) dh lery 3x + 2y + 2z + 5 = 0 ls nwjh] js[kk = = ds vuqfn'k Kkr dhft,A
3 6 2
Ans. 7
Sol. The given plane is 3x + 2y + 2z + 5 = 0 ...(i)
x3 y2 z
Line through P(2, 3, 4) and parallel to the line = = is
3 6 2
x2 y3 z4
= = = k(say) ...(ii)
3 6 2
Any point on it is Q(3k + 2, 6k + 3, 2k + 4). Let it lie on (i)
 3(3k + 2) + 2(6k + 3) + 2(2k + 4) + 5 = 0  25k + 25 = 0  k = –1  Q(–1, –3, 2)
  The required distance = PQ = (2  1)2  (3  3)2  (4  2)2 = 49 = 7
Hindi fn;k x;k lery gSA 3x + 2y + 2z + 5 = 0 ...(i)
js[kk tks P(2, 3, 4) ls xqtjs o nh xbZ js[kk ds lekUrj gSA
x3 y2 z x2 y3 z4
= = is = = = k (say) ...(ii)
3 6 2 3 6 2
bl ij dksbZ fcUnq Q(3k + 2, 6k + 3, 2k + 4)
Let it lie on (i)
 3(3k + 2) + 2(6k + 3) + 2(2k + 4) + 5 = 0  25k + 25 = 0  k = –1
 Q(–1, –3, 2)  vHkh"B nwjh = PQ = (2  1)2  (3  3)2  (4  2)2 = 49 = 7 .

E-11. If the acute angle that the vector,  ˆi  ˆj  kˆ makes with the plane of the two vectors
2 ˆi  3 ˆj  kˆ and ˆi  ˆj  2kˆ is cot 1 2 then find the value of  ( + ) –   
    
ˆ ˆ ˆ
lfn'k  i   j  k }kjk nks lfn'kksa 2 ˆi  3 ˆj  kˆ rFkk ˆi  ˆj  2kˆ ds lery ds lkFk cuk;k x;k U;wudks.k
cot 1 2 gS] rks ( + ) – Kkr dhft,A
Ans. 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 33
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol.
ˆi ˆj kˆ
 vector to plane n =
r
2 3 1
1 1 2

n = 5iˆ  5jˆ  5kˆ

( ˆi  ˆj  k)
ˆ (iˆ  ˆj  k)
ˆ
 cos (90 – ) = .
 2  2   2 3
   1
sin  = ( = cot– 1
2  cot  = 2  sin  = )
3    
2 2 2
3
1   
 =  ( + ) –  = 0
3 3  2  2   2
Hindi

ˆi ˆj kˆ
lery ds yEcor~ lfn'k n = 2 3 1
1 1 2

n = 5iˆ  5jˆ  5kˆ

( ˆi  ˆj  k)
ˆ (iˆ  ˆj  k)
ˆ
 cos (90 – ) = .
 2  2   2 3
   1
sin  = ( = cot– 1
2  cot  = 2  sin  = )
3    
2 2 2
3
1   
 =  ( + ) –  = 0
3 3  2  2   2

E-12. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line
x3 y–3 z–2
  . [16JM120014]
2 7 5
ml lery dk lehdj.k Kkr dhft, tks fd fcUnqvksa (3, 4, 1) rFkk (0, 1, 0) ls xqtjrk gS rFkk js[kk
x3 y–3 z–2
  ds lekUrj gSA
2 7 5

Ans. 8x – 13y + 15z + 13 = 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 34
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol.  Equation of plane through (3, 4, 1) is

 (3, 4, 1) ls xqtjus okyk lery gS
a(x – x1) + b(y – y1) + c(z – z1) = 0  a(x – 3) + b(y – 4)+ c(z – 1) = 0 ...(i)
 plane (i) passes through (0, 1, 0)  lery (i) fcUnq (0, 1, 0) ls xqtjrk gS
 –3a – 3b – c = 0  3a + 3b + c = 0 ...(ii)
x3 y–3 z–2
Also the plane (i) is parallel to the line  
2 7 5
x3 y–3 z–2
rFkk lery (i) js[kk   ds lekUrj gSA
2 7 5
 a1 a2 + b1 b2 + c1 c2 = 0  2a + 7b + 5c = 0 ...(iii)
Eliminating a, b, c from (i), (ii) and (iii), we get
(i), (ii) rFkk (iii) ls a, b, c foyqIr djus ij ge izkIr djrs gS
x – 3 y – 4 z –1
3 3 1 = 0   (x – 3) (15 – 7) – (y – 4) (15 – 2) + (z – 1) (21 – 6) = 0
2 7 5
8(x – 3) – 13(y – 4) + 15(z – 1) = 0  8x – 24 – 13y + 52 + 15z – 15 = 0

E-13. Find the vector equation of a line passing through the point with position vector (2iˆ – 3ˆj – 5k)
ˆ and

perpendicular to the plane r.(6iˆ – 3ˆj  5k)

ˆ + 2 = 0.
Also, find the point of intersection of this line and the plane.
,d fcUnq ftldk fLFkfr lfn'k (2iˆ – 3ˆj – 5k)ˆ gS] ls xqtjus okyh rFkk lery r.(6iˆ – 3ˆj  5k)
ˆ + 2 = 0 ds yEcor~
ljy js[kk dk lfn'k lehdj.k Kkr dhft,A
lkFk gh ljy js[kk rFkk lery dk izfrPNsnh fcUnq Hkh Kkr dhft;sA
Ans. ˆ ,  76 , –108 , –170 
r = 2iˆ – 3jˆ – 5kˆ + t (6iˆ – 3ˆj  5k)  35 35 
 35

Sol. The required line is to pass through a = 2iˆ – 3jˆ – 5kˆ and is  to the plane given by
r.(6iˆ – 3ˆj  5k)
ˆ +2=0 ...(1)
Here the vector b = 6iˆ – 3jˆ  5kˆ is normal to the plane (1)
 The reqd. line is along the direction of this vector. Hence its equation is
r = a  tb  r = 2iˆ – 3jˆ – 5kˆ + t (6iˆ – 3ˆj  5k)
ˆ ...(2)
Now line (2) meets the plane (1), when r.[2iˆ – 3jˆ  5kˆ  t(6iˆ – 3jˆ  5k)]
ˆ . (6iˆ – 3ˆj  5k)
ˆ =–2
1
i.e. when 6(2 + 6t) + 3(3 + 3t) + 5(5t – 5) = – 2 i.e., when t =
35
1 ˆ = 1 (76iˆ – 108jˆ – 170k)
Substituting this value of t in (2), we get r = 2iˆ – 3jˆ  5kˆ + (6iˆ – 3jˆ  5k) ˆ
35 35
 76 –108 –170 
 The reqd. point of intersection is  , ,
 35 35 35 

Hindi. vHkh"V js[kk fcUnq a = 2iˆ – 3jˆ – 5kˆ ls xqtjrh gS rFkk fn;s x;s lery ds yEcor~ gS
r.(6iˆ – 3ˆj  5k)
ˆ +2=0 ...(1)
;gk¡ lfn'k b = 6iˆ – 3jˆ  5kˆ lery (1) dk vfHkyEc lfn'k gSA
 vHkh"V js[kk bl lfn'k ds vuqfn'k gS vr% bldk lehdj.k gSA
r = a  tb  r = 2iˆ – 3jˆ – 5kˆ + t (6iˆ – 3ˆj  5k)
ˆ ...(2)
vc js[kk (2) lery (1) dks feyrh gS ;fn
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 35
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

r.[2iˆ – 3jˆ  5kˆ  t(6iˆ – 3jˆ  5k)]

ˆ . (6iˆ – 3ˆj  5k)
ˆ = – 2 tc 6(2 + 6t) + 3(3 + 3t) + 5(5t – 5) = – 2
1
tc t = ; t dk ;g eku (2) esa j[kus ij gesa izkIr gksrk gSA
35
1 ˆ = 1 (76iˆ – 108jˆ – 170k)
r = 2iˆ – 3jˆ  5kˆ + (6iˆ – 3jˆ  5k) ˆ  vHkh"V izfrPNsnh fcUnq gSA
35 35
 76 –108 –170 
 35 , ,
35 
 35

E-14. Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining
the points (1, 4, 2) and (2, 3, 5). Also find the coordinates of the foot of the perpendicular and the
perpendicular distance of the point (4, 0, 3) from the above found plane. [16JM120015]
fcUnq (1, 2, 1) ls xqtjus okys rFkk fcUnqvksa (1, 4, 2) rFkk (2, 3, 5) dks feykus okyh js[kk ds yEcor~ lery dk
lehdj.k Kkr djksA bl lery ij fcUnq (4, 0, 3) ls Mkys x;s yEcikn ds funsZ'kkad rFkk yEc dh yEckbZ Hkh Kkr
dhft,A
Ans. x – y + 3z – 2 = 0 ; (3, 1, 0) ; 11
Sol. Let P(1, 2, 1) Q(1, 4, 2) and R(2, 3, 5). Now d.r.s. of the line QR are 2 – 1, 3 – 4, 5 – 2 = 1, – 1, 3
The reqd. plane is to pass through 'P' and  QR. Its equation is 1(x – 1) – 1 · (y – 2) + 3(z – 1) = 0
 x – y + 3z – 2 = 0 ...(i)
Let B(, , ) be the foot of the  from A(4, 0, 3) on (1).
  –  + 3 – 2 = 0 ...(ii)
and drs. of AB (Normal to the plane (i) are)  – 4,  – 0,  – 3
–4 –0 –3
Hence = = = k (say)   = k + 4,  = – k,  = 3k + 3 ...(iii)
1 –1 3
From (ii) and (iii)
(k + 4) + k + 3(3k + 3) – 2 = 0  k = –1
 The coordinates of the foot of  from A on (1) is B(, , ) = (3, 1, 0)
and  = AB = (4 – 3)2  (0 – 1)2  (3 – 0)2 = 11
Hindi. ekuk P(1, 2, 1) Q(1, 4, 2) rFkk R(2, 3, 5) gS vc js[kk QR ds fn~dvuqikr 2 – 1, 3 – 4, 5 – 2 = 1, – 1, 3
vHkh"V lery fcUnq 'P' ls xqtjsxk rFkk ljy js[kk QR ds yEcor~ gksxkA bl lery dk lehdj.k
1(x – 1) – 1 · (y – 2) + 3(z – 1) = 0  x – y + 3z – 2 = 0 ...(i)
ekuk fcUnq A(4, 0, 3) ls lery (i) ij Mkyk x;k yEc ikn B(, , ) gSA   –  + 3 – 2 = 0 ...(ii)
(lery (i) ds vfHkyEc) ds fn~d vuqikr  – 4,  – 0,  – 3
–4 –0 –3
vr% = = = k (ekuk)   = k + 4,  = – k,  = 3k + 3 ...(iii)
1 –1 3
lehdj.k (ii) rFkk (iii) ls
(k + 4) + k + 3(3k + 3) – 2 = 0  k = –1
 A ls lery (i) ij Mkys x;s yEc ds ikn ds funsZ'kkad B(, , ) = (3, 1, 0)
rFkk yEc dh yEckbZ AB = (4 – 3)2  (0 – 1)2  (3 – 0)2 = 11

E-15. Find the equation of the planes passing through points (1, 0, 0) and (0, 1, 0) and making an angle of
0.25  radians with plane x + y  3 = 0
mu leryksa ds lehdj.k Kkr dhft, tks fcUnqvksa (1, 0, 0) vkSj (0, 1, 0) ls xqtjrs gSa vkSj lery
x + y  3 = 0 ds lkFk 0.25  jsfM;u dk dks.k cukrs gaSA
Ans. x+y ± 2 z = 1
Sol. Equation of plane passing through (1,0,0) is : a(x – 1) + b(y – 0) + c(z – 0) = 0
a(x – 1) + b(y) + c(z) = 0 .............(1)
(1) also passes through (0,1,0)
–a + b + c.0 = 0 .............(2)
angle between plane (1) and x + y – 3 = 0 is

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 36
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 a.1  b.1  c.0 1 ab 1

cos = =  =
4 a 2
1 1 0 2 a b c
2 2 2
2 2

(a + b) = a + b + c
2 2 2 2
2ab = c 2

from equation (2) a = b   2a2 = c2  c= a 2 or c = a 2

So equation of planes are x + y  2 z – 1 = 0
Hindi (1,0,0) ls xqtjus okys lery dk lehdj.k gS a(x – 1) + b(y – 0) + c(z – 0) = 0
a(x – 1) + b(y) + c(z) = 0 .............(1)
lery (1), (0,1,0) ls Hkh xqtjrk gS
–a + b + c.0 = 0 .............(2)
lery (1) rFkk x + y – 3 = 0 ds e/; dks.k gSA
 a.1  b.1  c.0 1 ab 1
cos = =  =
4  a 1 1 0
2
2 a b c 2
2 2 2
2
(a + b)2 = a2 + b2 + c2 2ab = c2
lehdj.k (2) ls a = b  2a2 = c2  c= a 2 or c = a 2
vr% leryksa ds lehdj.k x + y  2 z – 1 = 0 gSaA

E-16. Find the distance between the parallel planes r . (2iˆ  3ˆj  6k)
ˆ = 5 and r . (6iˆ  9ˆj  18k)
ˆ + 20 = 0.

lekUrj leryksa . r . (2iˆ  3ˆj  6k)

ˆ = 5 vkSj r . (6iˆ  9ˆj  18k)
ˆ + 20 = 0 ds chp dh nwjh Kkr dhft,A
5
Ans. unit [16JM120016]
3
20
Sol. Planes are : 2x – 3y + 6z – 5 = 0  2x – 3y + 6z + =0
3
20
–5 –
3 5
Distance between planes is = 
(2)  (– 3)  (6)
2 2 2 3

20
Hindi. lery gSa : 2x – 3y + 6z – 5 = 0 2x – 3y + 6z + =0
3
20
–5 –
3 5
leryksa ds chp dh nwjh = 
(2)  (– 3)  (6)
2 2 2 3

E-17. If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 pass through a straight line, then find
the value of a2 + b2 + c2 + 2abc is :
;fn lery x – cy – bz = 0, cx – y + az = 0 rFkk bx + ay – z = 0 ,d ljy js[kk ls xqtjrs gSa] rks
a2 + b2 + c2 + 2abc dk eku Kkr dhft, &
Ans. 1
1 c b
Sol. c 1 a = 0  (1 – a2) + c (– c – ab) – b(ac + b) = 0 1 – a2 – c2 – abc – abc – b2 = 0
b a 1
a2 + b2 + c2 + 2abc = 1

E-18. (i) If n̂ is the unit vector normal to a plane and p be the length of the perpendicular from the origin
to the
plane, find the vector equation of the plane.
(ii) Find the equation of the plane which contains the origin and the line of intersection of the
planes r . a = p and r . b = q

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 37
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(i) ;fn n̂ ,d lery ds yEcor~ bdkbZ lfn'k gS vkSj ewyfcUnq ls lery ij Mkys x;s yEc dh yEckbZ p gS,
rks lery
dh lfn'k lehdj.k Kkr dhft,A
(ii) leryksa r . a = p vkSj r . b = q dh izfrPNsnu js[kk vkSj ewyfcUnq dks lekfgr djus okys lery dk
lehdj.k Kkr dhft,A
Ans. (i) r . n̂ = ±p (ii) r . (a q  pb ) = 0

Sol.

(i) Projection of OP on n̂
OP dk n̂ ij iz{ksi
r.nˆ  p
(ii) r.a  p  (r.b  q)  0
p
r 0  –p  q  0 
q
p
r .a  p  (r. b  q)  0  r.( aq  pb)  0
q

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 38
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 Marked questions are recommended for Revision.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA

PART - I : ONLY ONE OPTION CORRECT TYPE

Hkkx-I : dsoy ,d lgh fodYi çdkj (ONLY ONE OPTION CORRECT TYPE)

1. Let a, b, c be vectors of length 3, 4, 5 respectively. Let a be perpendicular to b  c , b to c  a and

c to a  b . Then a  b  c is equal to :
ekukfd lfn'kksa a , b , c dh yEckbZ Øe'k% 3, 4, 5 gSA ekuk a lfn'k b  c ds] b lfn'k c  a ds vkSj c lfn'k
a  b ds yEcor gS] rks a  b  c dk eku gS&
(A) 2 5 (B) 2 2 (C) 10 5 (D*) 5 2
Sol. a.(b  c)  0 , b.(c  a)  0 , c.(a  b)  0
 a.b  a.c  0 , b.c  b.a  0 , c.a  c.b  0  a.b  b.c  c.a  0

|abc | = a2  b2  c 2  2(a.b  b.c  c.a) = 9  16  25 = 50

2. Four coplanar forces are applied at a point O. Each of them is equal to k and the angle between two
consecutive forces equals 45º as shown in the figure. Then the resultant has the magnitude equal to :

(A) k 2  2 2 (B) k 3  2 2 (C*) k 4  2 2 (D) k 4  2 2

,d fcUnq O ij pkj leryh; cy yxk;s x, gSA buesa ls izR;sd k ds cjkcj gS vkSj nks Øekxr cyksa ds e/; dks.k
,

45º gS] tSlkfd fp=k esa n'kkZ;k x;k gS] rks ifj.kkeh cy dk ifjek.k gS& [16JM120030]

(A) k 2  2 2 (B) k 3  2 2 (C*) k 4  2 2 (D) k 4  2 2

Sol.

ˆ k ˆ k ˆ ˆ k ˆ k ˆ
These forces can be written in terms of vector as ki, i j, kj and – i j
2 2 2 2

Resultant = kiˆ  (k  2 k)jˆ

magnitude = k 2  (k  2 k)2 = k 4  2 2

Hindi

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 1
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

cyksa dks lfn'k :i esa fuEu izdkj fy[kk tk ldrk gS&

ˆ k ˆ k ˆ ˆ k ˆ k ˆ
ki, i j, kj vkSj – i j
2 2 2 2
ifj.kkeh cy = kiˆ  (k  2 k)jˆ
ifjek.k k 2  (k  2 k)2 = k 4  2 2

1
3. Taken on side AC of a triangle ABC, a point M such that AM = AC . A point N is taken on the side
3
CB such that BN = CB , then for the point of intersection X of AB and MN which of the following
holds good?
1
,d f=kHkqt ABC dh Hkqtk AC ij ,d fcUnq M bl izdkj fy;k tkrk gS fd AM = AC rFkk CB Hkqtk ij ,d
3
fcUnq N bl izdkj fy;k tkrk gS fd BN = CB , rks AB vkSj MN ds izfrPNsn fcUnq X ds fy, fuEu esa ls dkSulk
dFku lR; gS \
1 1 3
(A) XB = AB (B) AX = AB (C*) XN = MN (D) XM = 3 XN
3 3 4
c
Sol. Position vector of M 
3
Position vector of N  (–c  2b) 
  equation of line BC is r  b  (b  c)
 equation of line AB is is r  0  b
c  4c  1 4
 equation of line MN is r  t  – 2    = – 2t, 0 =  t
3  3  3 3
1 b
which gives =  Position vector of X is .
2 2
c
Hindi M dk fLFkfr lfn'k 
3
N dk fLFkfr lfn'k  (–c  2b) 
  js[kk BC dk lehdj.k gS r  b  (b  c)
 js[kk AB dk lehdj.k gS r  0  b
c  4c 
 js[kk MN dk lehdj.k gS r  t  – 2
3  3 
1 4 1 b
  = – 2t, 0 =  t blls izkIr gksrk gS  =  X dk fLFkfr lfn'k gS
3 3 2 2

4. If 3 non zero vectors a, b, c are such that a  b = 2(a  c) , | a | = | c | = 1; | b | = 4 the angle between
1
b and c is cos–1 then b = c  µa where |  | + | µ | is - [16JM120031]
4
;fn rhu v'kwU; lfn'k a, b, c bl izdkj gS fd a  b = 2(a  c) , | a | = | c | = 1; | b | = 4 ,oa b o c ds e/;
1
dks.k cos–1 gS] rks b = c  µa tcfd |  | + | µ | gS&
4
(A*) 6 (B) 5 (C) 4 (D) 0
Sol. a  b = 2 (a  b)   a × (b – 2c) = 0   b – 2c =  a

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 2
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1
squaring (oxZ djus ij) b2 + 4c2 – 4b · c = 2a2   16 + 4 – 4.4.1. = 2 
4
  =4  b = 2c  4a  || + |  | = 6

5. If  is the angle between the vectors p = a î + b ĵ + c k̂ and vector q = b î + c ĵ + a k̂ then range of  is

(A) [0, /3] (B) [/3, 2/3] (C*) [0, 2/3] (D) [0, 5/6]
;fn lfn'k p = a î + b ĵ + c k̂ vkSj q = b î + c ĵ + a k̂ ds e/; dks.k  gS] rc  ds ekuksa dk ifjlj gS&
(A) [0, /3] (B) [/3, 2/3] (C*) [0, 2/3] (D) [0, 5/6]

ab  bc  ca ab  bc  ca 1
Sol.  (a – b) 2
0  1 and (a + b + c)2  0  –
a2  b2  c 2 a b c
2 2 2
2
ab  bc  ca 1 2
Now vc cos =
p.q
= 2  –  cos  10   
| p || q | a  b2  c 2 2 3

6. If the unit vectors e1 and e2 are inclined at an angle 2 and | e1  e2 | < 1, then for  [0, ],  may lie
in the interval :
;fn bdkbZ lfn'kksa e1 vkSj e2 ds chp dks.k 2 gS vkSj | e1  e2 | < 1, rks  [0, ] ds fy,  fuEu esa ls fdl
vUrjky esa gks ldrk gS &
     5    5 
(A*) 0,  (B)  ,  (C)  ,   (D)  , 
 6 6 2  6  2 6 
| e1  e2 | 2 < 1  e1  e2  2e1. e2  1  1 + 1 – 2cos(2) < 1
2 2
Sol.
1    
 2cos2 > 1  cos2 >  2  0,     0,
2  3  6 
7. A line makes angles , , ,  with the four diagonals of a cube, then cos2 + cos2+ cos2+ cos2is
equal to
(A) 1/3 (B) 2/3 (C*) 4/3 (D) 5/3
,d js[kk ?ku ds pkj fod.kksa ds lkFk dks.k , , ,  dks.k cukrh gS rc cos2 + cos2+ cos2+ cos2cjkcj gS&
(A) 1/3 (B) 2/3 (C*) 4/3 (D) 5/3

Sol. Let direction cosines of line is , m, n which makes angle , , ,  with diagonals of cube having
 1 1 1   1 1 1   1 1 1   1 1 1 
direction cosines are  , , , 
  , , ,


 , ,  and 
  , , 

 3 3 3  3 3 3  3 3 3  3 3 3
respectively.
2 2 2
  m n    m n    m n 
So cos2 + cos2+ cos2+ cos2     + 
     + 
    

 3 3 3  3 3 3  3 3 3
2
  m n  4 4
+      = (2 + m2 + n2) =

 3 3 3 3 3

x y z x y z
8. Consider the lines = = and = = , then the equation of the line which
2 3 5 1 2 3
x y z
(A) bisects the angle between the lines is = =
3 3 8
x y z
(B) bisects the angle between the lines is = =
1 2 3
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 3
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(C*) passes through origin and is perpendicular to the given lines is x = y = – z

(D) none of these
x y z x y z
nks js[kk,sa = = vkSj = = gSa, rc ml js[kk dk lehdj.k] tks
2 3 5 1 2 3
x y z
(A) nh xbZ js[kkvksa ds e/; dks.k dks lef}Hkkftr djrh gS] = = gSA
3 3 8
x y z
(B) nh xbZ js[kkvksa ds e/; dks.k dks lef}Hkkftr djrh gS] = = gSA
1 2 3
(C) ewy fcUnq ls xqtjrh gS vkSj nh xbZ js[kkvksa x = y = – z ds yEcor~ gS] gSA
(D) buesa ls dksbZ ugha

Sol.

x y z x y z
  ... (i)   ... (ii)
2 3 5 1 2 3
2iˆ  3ˆj  5kˆ ˆi  2jˆ  3kˆ
â  bˆ    (A) and (B) will be incorrect
38 14
Let the dr’s of line  to (1) and (2) be a, b, c
 2a + 3b + 5c = 0 ... (iii) and a + 2b + 3c = 0 ... (iv)
a b c a b c a b c
        
9 – 10 5  6 4  3 1 1 1 1 1 1
 equation of line passing through (0, 0, 0) and is r to the lines (1) and (2) is
x y z
  Ans.
1 1 1

Hindi.

x y z x y z
  ... (i)   ... (ii)
2 3 5 1 2 3
2iˆ  3ˆj  5kˆ ˆi  2jˆ  3kˆ
â  bˆ    (A) vkSj (B) lgh ugha gaSA
38 14
ekukfd (1) vkSj (2) ds yEcor~ js[kk ds fnd~vuqikr a, b, c gSaA
 2a + 3b + 5c = 0 ... (iii) vkSj a + 2b + 3c = 0 ... (iv)
a b c a b c a b c
        
9 – 10 5  6 4  3 1 1 1 1 1 1
 (0, 0, 0) ls xqtjus okyh js[kk dk lehdj.k tksfd js[kkvksa (1) vkSj (2) ds yEcor~ gSA

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 4
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x y z
  Ans.
1 1 1


9. Given a  xiˆ  yjˆ  2kˆ , b  ˆi  ˆj  kˆ , c  ˆi  2jˆ ; (a ^ b) = , a.c  4 , then
2

fn;k x;k gS fd a  xiˆ  yjˆ  2kˆ , b  ˆi  ˆj  kˆ , c  ˆi  2jˆ ; (a ^ b) = , a.c  4 , rks
2
(A) [a b c]2 | a | (B) [a b c] | a | (C) [a b c]  0 (D*) [a b c] | a |2
Sol. a.b  0  x–y+2=0 ... (1)
a.c  4  x + 2y = 4 ... (2)
0 2 2
 x = 0, y = 2 Hence (vr%)  a  2jˆ  2kˆ  [a b c] = 1 –1 1  8 = | a |2
1 2 0

10. The vectors ˆi  2 ˆj  3kˆ , 2 ˆi  ˆj  kˆ and 3 ˆi  ˆj  4kˆ are so placed that the end point of one vector is the
starting point of the next vector. Then the vectors are : [16JM120032]
(A) Not coplaner (B*) Coplaner but cannot form a triangle
(C) Coplaner but can form a triangle (D) Coplaner & can form a right angled triangle
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
lfn'k i  2 j  3k , 2 i  j  k vkSj 3 i  j  4k bl izdkj gS fd ,d lfn'k dk vfUre fcUnq nwljs lfn'k dk
izkjfEHkd fcUnq gSa] rks lfn'k gSa &
(A) vleryh; (B) leryh; ij f=kHkqt ugha cuk ldrs gS
(C) leryh; ysfdu ,d f=kHkqt cuk ldrs gSaA (D) leryh; vkSj ,d ledks.k f=kHkqt cuk ldrs gSaA
Sol. Clearly triangle is not possible as v1 + v2 + v3  0
Since v 3  v1  v 2
Hence v1, v 2 ,v 3 are coplaner

Hindi Li"Vr% rhuksa ls f=kHkqt cuuk lEHko ugha gS tSlk fd v1 + v2 + v3  0

 v 3  v1  v 2
vr% v1, v 2 ,v3 leryh; gSA

11. Let a, b and c be non-coplanar unit vectors equally inclined to one another at an acute angle . Then
[a b c] in terms of  is equal to: [16JM120033]

(A) (1 + cos ) cos 2  (B) (1 + cos ) 1 2cos2 

(C*) (1  cos ) 1  2 cos  (D) (1  sin ) 1  2 cos 

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 5
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ekuk a , b vkSj c vleryh; bdkbZ lfn'k gSa vkSj izR;sd ,d nwljs ds lkFk leku U;wu dks.k cukrs gS] rks
[a b c] dk eku  ds inksa esa gS&
(A) (1 + cos ) cos 2  (B) (1 + cos ) 1 2cos2 
(C*) (1  cos ) 1  2 cos  (D) (1  sin ) 1  2 cos 

a . a a.b a.c 1 cos  cos 

Sol. [a b 2
c] = b.a b.b b.c = cos  1 cos 
c.a c.b c.c cos  cos  1

12. Consider a tetrahedron with faces f1, f2, f3, f4. Let a1 , a2 ,a3 , a4 be the vectors whose magnitudes are
respectively equal to the areas of f1, f2, f3, f4 and whose directions are perpendicular to these faces in
the outward direction. Then,
(A*) a1  a2  a3  a4 = 0 (B) a1  a3  a2  a4
(C) a1  a2  a3  a4 (D) a1  a2  a3  a4 = 0
,d prq"Qyd ds i`"B f1, f2, f3, f4 gSaA ekukfd pkj lfn'k a1 , a2 ,a3 , a4 gSa ftuds ifjek.k Øe'k% f1, f2, f3, f4 ds
{ks=kQyksa ds cjkcj gSa vkSj fn'kk bu i`"Bksa ds yEcor~ ckgj dh vksj gSA rks]
(A*) a1  a2  a3  a4 = 0 (B) a1  a3  a2  a4
(C) a1  a2  a3  a4 (D) a1  a2  a3  a4 = 0
Sol. a1  AC  AB = (c  a) × (b  a)
a2  DB  DC = b  c
a3  DC  DA = c  a
a4  DA  DB = a  b

13. Let r be a vector perpendicular to a + b + c , where [ a b c ] = 2. If [16JM120034]

r = ( b × c ) + m( c × a ) + n( a × b ), then ( + m + n) is equal to
(A) 2 (B) 1 (C*) 0 (D) –1
ekukfd r ds a + b + c yEcor~ ,d lfn'k gS] tgk¡ [ a b c ] = 2. ;fn
r = ( b × c ) + m( c × a ) + n( a × b ) gks] rks ( + m + n) dk eku gS&
(A) 2 (B) 1 (C) 0 (D) –1
Sol. r.(a  b  c)  0
 [a b c]  m[a b c]  n[a b c]  0 or (;k) ( + m + n) [a b c] = 0
or (;k)  + m + n = 0

14. If a , b , c are three non-coplanar non-zero vectors and r is any vector in space, then
(a  b) × (r  c) + (b  c) × (r  a) + (c  a) × (r  b) is equal to
(A*) 2[ a b c ] r (B) 3[ a b c ] r (C) [ a b c ] r (D) 4[ a b c ] r
;fn a , b , c rhu vleryh;] v'kwU; lfn'k gSa vkSj lef"V esa dksbZ lfn'k r gS] rks
(a  b) × (r  c) + (b  c) × (r  a) + (c  a) × (r  b) dk eku gS&
(A*) 2[ a b c ] r (B) 3[ a b c ] r (C) [ a b c ] r (D) 4[ a b c ] r
Sol. r  xa  yb  zc
Consider (a  b) × (r  c) = [a b c] r – [a b r]c
But [a b r ] = (a  b) . r  z [a b c] above expression = [a b c] (r – zc)
Hence sum = [a b c] (3r – xa  yb  zc) = [a b c] 2r
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 6
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Hindi. r  xa  yb  zc
 (a  b) × (r  c) = [a b c] r – [a b r]c
ysfdu [a b r ] = (a  b) . r  z [a b c] mijksDr O;atd = [a b c] (r – zc)
vr% ;ksx = [a b c] (3r – xa  yb  zc) = [a b c] 2r

15. If b and c are two non-collinear vectors such that a || ( b × c ), then ( a × b ) . ( a × c ) is equal to
(A*) a2 (b.c) (B) b2 (a.c) (C) c 2 (a.b) (D) – a2 (b.c)
;fn b vkSj c nks vlajs[kh; lfn'k bl izdkj gS fd a || ( b × c ), rks ( a × b ) . ( a × c ) dk eku gS&
(A*) a2 (b.c) (B) b2 (a.c) (C) c 2 (a.b) (D) – a2 (b.c)
Sol. a || (b  c)  a = (b  c)
a . a a . c a . a 0
also vkSj (a  b)·(a  c) = = = | a |2 (b.c)
b . a b . c 0 b . c

16. Let 2c  b = 4a  3d . If d c a  and a b d are natural numbers with H.C.F. equal to 1 then how
many statement are true among below six statement.
ekuk 2c  b = 4a  3d ;fn d c a vkSj a b d izkd`r la[;k,sa gS ftudk e-l-i- 1 gS rc uhps fn, N% dFkuksa
esa ls fdrus dFku lgh gSA
(i) a b d  2 (ii) a b c   3
   
(iii) d c b   4 (iv) d c a   1
   
(v)  a  b  ×  c  d = 2 c – 3d (vi) a  b × c  d = 4a – b
(A) 2 (B) 4 (C*) 6 (D) 0
Sol.  2c  b . a  d =  4a  3d . a  d
 2 c a d + b a d = 0  a b d = 2 d c a  .....(i)

   
Now vc 2c  b . c  d = 4a  3d . c  d    d c b  = 4 d c a  .....(ii)

Now vc  2c  b  .  a  c  =  4a  3d  .  a  c   [a b c] = 3 d c a  .....(iii)

From (i) (ii) and (iii) we get d c a  = 1, a b d = 2 , a b c  = 3 , d c b  = 4

vkSj (iii) ls d c a = 1, a b d = 2 , a b c  = 3 , d c b = 4

(i),(ii)

{  H.C.F. of a b d and vkSj d c a  = 1}

Now vc a  b × c  d = a.b.d c – a.b.c  d = 2c – 3d = 4a – b

17. Let u and v are unit vectors and w is a vector such that (u  v)  u = w and w  u  v then the
value of u v w  is equal to
(A*) 1 (B) 2 (C) 0 (D) –1
ekuk u vkSj v bdkbZ lfn'k gS rFkk w ,d lfn'k bl izdkj gS fd (u  v)  u = w rFkk w  u  v ,
u v w  eku cjkcj gS&
(A*) 1 (B) 2 (C) 0 (D) –1

Sol. Given fn;k x;k gS (u  v)  u = w and rFkk w  u  v

 (u  v)  u × u  w  u  u  v  × u  uu = v (as w  u  v )
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 7
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

  u . u v   v . u u  u  u  v
(using u . u = 1 and vkSj u  u = 0, since unit vector ls bdkbZ lfn'k gS)
 v  (v . u) u  v  u . v u  0
  u . v =0 (as u  0) .............(i)

 Nowvc u . (v  w)
   u . (v  ((u  v)  u)) (given fn;k gS w  (u  v) + u)
  u . (v  (u  v)  v  u)  u . ((v . v) u  (v . u) v  v  u)
   u . (| v |2 u  0  v  u) (as u . v  0 from lehdj.k (i) ls)
   | v | (u . u) – u . (v  u)
2

    | v |2 | u |2 – 0 (as u v u = 0)

  1 (as pwafd | u | = | v | = 1)
  u v w  = 1

18 . Find the shortest distance between any two opposite edges of a tetrahedron formed by the planes
y + z = 0, x + z = 0, x + y = 0, x + y + z = 3 a .
(A) a (B) 2a (C) a / 2 (D*) 2 a.
leryksa y + z = 0, x + z = 0, x + y = 0, x + y + z = 3 a ls cus prq"Qyd ds nks foijhr fdukjksa ds e/;
y?kqÙke nqjh gSA
(A) a (B) 2a (C) a / 2 (D*) 2 a.

Sol.
A
, 0) P
O (0, 0

C
(0, 0, Q D
3 a)

Let the equation to one of the pair of opposite edges OA and BC be

y + z = 0, x + z = 0 .....(1) and x + y = 0, x + y + z = 3 a .....(2)
equation (1) and (2) can be expressed in symmetrical form as
x 0 y 0 z0 x 0 y 0 z 3 a
  ..... (3) and,   .......(4)
1 1 1 1 1 0
d. r. of OA and BC are respectively (1, 1, – 1) and (1, – 1, 0).
Let PQ be the shortest distance between OA and BC having direction cosines (, m, n)
 PQ is perpendicular to both OA and BC.
  + m – n = 0 and –m=0
m n
Solving (5) and (6), we get,   = k (say)
1 1 2
also, 2 + m2 + n2 = 1
O
1 A
 k + k + 4k = 1  k = 
2 2 2

6
1 1 2
  =  , m =  , n = 
6 6 6
B C
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 8
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Shortest distance between OA and BC

i.e. PQ = The length of projection of OC on PQ

O P A
90°

90°
C Q B
= | (x2 – x1)  + (y2 – y1) m + (z2 – z1) n |
1 1 2
= 0 . 0 .  3 a . = 2 a.
6 6 6

Sol.
A
, 0) P
O (0, 0

C
(0, 0, Q D
3 a)

ekuk OA rFkk BC foijhr fdukjksa ds ;qXe es ls ,d lehdj.k gSA

y + z = 0, x + z = 0 .....(1) rFkk x + y = 0, x + y + z = 3 a .....(2)
lehdj.k (1) rFkk (2) dks lefer :i esa fy[kus ij
x 0 y 0 z0 x 0 y 0 z 3 a
  ..... (3) rFkk,   .......(4)
1 1 1 1 1 0
OA rFkk BC ds fnd vuqikr (1, 1, – 1) vkSj (1, – 1, 0) gSA
ekuk PQ, OA rFkk BC ds e/; y?kqÙke nqjh gS rFkk fn~ddksT;k,a (, m, n) gSA PQ, OA rFkk BC nksuksa ds
yEcor~ gSA
  + m – n = 0 rFkk –m=0
m n
lehdj.k (5) rFkk (6) ls,   = k (say)
1 1 2
rFkk, 2 + m2 + n2 = 1
O
1 A
 k2 + k2 + 4k2 = 1  k = 
6
1 1 2
  =  , m =  , n = 
6 6 6
B C
OA rFkk BC ds e/; y?kqÙke nqjh

vFkkZr PQ = OC dk PQ ij iz{ksi dh yEckbZ

O P A
90°

90°
C Q B
= | (x2 – x1)  + (y2 – y1) m + (z2 – z1) n |

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 9
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 1 2
= 0 . 0 .  3 a . = 2 a.
6 6 6

19. A plane meets the coordinate axes in A, B, C and () is the centroid of the triangle ABC, then the
equation of the plane is [16JM120036]
x y z x y z 3x 3y 3z
(A*)   3 (B)   1 (C)   1 (D) x +y + z =1
        
,d lery funsZ'kh v{kksa dks A, B, C ij feyrk gS vkSj (), f=kHkqt ABC dk dsUnzd gS] rks lery dk
lehdj.k gS&
x y z x y z 3x 3y 3z
(A*)   3 (B)   1 (C)   1 (D) x +y + z =1
        

Sol.

Let the equation of plane be

ekuk lery dk lehdj.k gS ––
x y z
  1
a b c
as (, , ) is centroid
tSlkfd (, , ) dsUnzd gS ––
a b c
   = ;   and vkSj  =
3 3 3
x y z x y z
 equation of plane   3  lery dk lehdj.k   3
     

x 1 y2 z3
20. Equation of plane which passes through the point of intersection of lines = = and
3 1 2
x3 y 1 z2
= = and at greatest distance from the point (0, 0, 0) is:
1 2 3
(A) 4x + 3y + 5z = 25 (B*) 4x + 3y + 5z = 50
(C) 3x + 4y + 5z = 49 (D) x + 7y – 5z = 2
x 1 y2 z3 x3 y 1 z2
js[kkvksa = = vkSj = = ds izfrPNsn fcUnq ls xqtjus okys vkSj fcUnq (0, 0, 0)
3 1 2 1 2 3
ls vf/kdre nwjh ij fLFkr lery dk lehdj.k gS &
(A) 4x + 3y + 5z = 25 (B) 4x + 3y + 5z = 50
(C) 3x + 4y + 5z = 49 (D) x + 7y – 5z = 2
x 1 y  2 z  3 x  3 y 1 z  2
Sol.   r ... (1)   .... (2)
3 1 2 1 2 3
 coordinates of any point P on line (1)
 P(3r + 1, r + 2, 2r + 3) for point of intersection of (1) and (2)
3r  1  3 r  2  1 2r  3  2 3r  2 r  1 2r  1
      r=1
1 2 3 1 2 3
 point of intersection is (4, 3, 5)  the equation of required plane 
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 10
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

  4(x – 4) + 3(y – 3) + 5(z – 5) = 0  4x + 3y + 52 = 50

x 1 y  2 z  3 x  3 y 1 z  2
Hindi.   r ... (1)   .... (2)
3 1 2 1 2 3
 js[kk (1) ij fLFkr fdlh fcUnq P ds funsZ'kkad gSa ––
 P(3r + 1, r + 2, 2r + 3)
(1) vkSj (2) ds izfrPNsn fcUnq ds fy,
3r  1  3 r  2  1 2r  3  2 3r  2 r  1 2r  1
      r=1
1 2 3 1 2 3
 izfrPNsn fcUnq (4, 3, 5) gSA
 vHkh"V lery dk lehdj.k gS
4(x – 4) + 3(y – 3) + 5(z – 5) = 0  4x + 3y + 52 = 50

21. The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 and
x – 3y + z = 0 = x + 2y + z + 1 are co-planar is :
(A*) – 2 (B) 4 (C) 6 (D) 0
‘a’ dk v'kwU; eku ftlds fy, js[kk,¡ 2x – y + 3z + 4 = 0 = ax + y – z + 2 vkSj x – 3y + z = 0 = x + 2y + z + 1
leryh; gSa] gksxk –&
(A) – 2 (B) 4 (C) 6 (D) 0
Sol. 2x – y + 3z + 4 = 0 = ax + y – z + 2 ... (1)
equation of plane through (1) is (2x – y + 3z + 4) + (ax + y – z + 2) = 0
x(2 + a) + y( – 1) + z(3 – ) + (4 + 2) = 0 .... (2)
x – 3y + z = 0 = x + 2y + 2 + 1 ... (3)
equation of plane passing through (3) is
(x – 3y + z) + (x + 2y + z + 1) = 0  x(1 + ) + y(2 – 3) + z( + 1) +  = 0 .... (4)
2  a  1 3   4  2
if lines (1) and (3) are coplanar, then   
 1 2  3   1 
Solving this we get  = – 1,  = 1  a=–2
Hindi. 2x – y + 3z + 4 = 0 = ax + y – z + 2 ... (1)
(1) ls xqtjus okys lery dk lehdj.k gS ––
(2x – y + 3z + 4) + (ax + y – z + 2) = 0
x(2 + a) + y( – 1) + z(3 – ) + (4 + 2) = 0 .... (2)
x – 3y + z = 0 = x + 2y + 2 + 1 ... (3)
(3) ls xqtjus okys lery dk lehdj.k gS ––
(x – 3y + z) + (x + 2y + z + 1) = 0  x(1 + ) + y(2 – 3) + z( + 1) +  = 0 .... (4)
2  a  1 3   4  2
;fn js[kk (1) vkSj (3) leryh; gSa] rc   
 1 2  3   1 
budks gy djus ij  = – 1,  = 1  a=–2

22. A line having direction ratios 3, 4, 5 cuts 2 planes 2x – 3y + 6z – 12 = 0 and 2x – 3y + 6z + 2 = 0 at

point P & Q, then find length of PQ
35 2 35 2 35 2 35 2
(A*) (B) (C) (D)
12 24 6 8
3, 4, 5 fnd~ vuqikr okyh ,d ljy js[kk nks leryksa 2x – 3y + 6z – 12 = 0 ,oa 2x – 3y + 6z + 2 = 0 dks fcUnqvksa
P ,oa Q, ij izfrPNsn djrh gS] rks PQ dh yEckbZ gS-
35 2 35 2 35 2 35 2
(A*) (B) (C) (D)
12 24 6 8

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 11
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol.

2  12
distance between plane = =2
7
L between line and plane = 
6 – 12  30 24
sin = =
5 2.7 35 2
2 35 2 35 2
sin =  PQ = 2 × =
PQ 24 12

Hindi

2  12
leryksa ds e/; nwjh = =2
7
L js[kk vkSj lery ds e/; dks.k = 
6 – 12  30 24
sin = =
5 2.7 35 2
2 35 2 35 2
sin =  PQ = 2 × =
PQ 24 12

23. A line L1 having dierection ratios 1, 0, 1 lies on xz plane. Now this xz plane is rotated about z-axis by an
angle of 90°. Now the new position of L1 is L2. The angle between L1 & L2 is : [16JM120037]
,d js[kk L1 ds fnd~ vuqikr 1, 0, 1 gS tks xz lery ij fLFkr gSA vc bl xz lery dks z-v{k ds lkis{k 90° dks.k
ls ?kw.kZu djk;k tkrk gSA bl izdkj L1 dh u;h fLFkfr L2 gS] rks L1 ,oa L2 ds e/; dks.k gksxk&
(A) 30° (B*) 60° (C) 90° (D) 45°
1 1
Sol. L1 : 1, 0, 1   = 45°,  = 90°,  = 45°  1 = , m1 = 0, n1 –
2 2
L2 :  = 90°,  = 45°,  = 45°
1 1
So vr% 1 = 0, m1 = , n1 =
2 2
Let angle between L1 & L2 be . ekuk L1 rFkk L2 ds e/; dks.k gS rc
1
so vr% cos = 12 + m1m2 + n1n2  cos = ,  = 60°
2
24._ Focii of ellipse lie on plane 2x + 3y + 6z = 0 are (54, 18, –27) and (–54, –18, 27). If eccentricity of
3
ellipse is then find equation of directrices .
5

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 12
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

lery 2x + 3y + 6z = 0 ij fLFkr nh?kZo`Ùk dh ukfHk;ka (54, 18, –27) ;k (–54, –18, 27) ;fn nh?kZo`Ùk dh mRdsUnzrk
3
gS rc fu;rkvksa ds lehdj.k gS&
5

x – 150 y – 50 z  75 x  150 y  50 z – 75
(A*) = = , = =
3 –6 2 3 –6 2

x  150 y  50 z  75 x  150 y – 50 z  75
(B) = = , = =
3 –6 2 3 –6 2

x – 150 y  50 z  75 x  150 y  50 z – 75
(C) = = , = =
3 –6 2 3 –6 2

x – 150 y – 50 z  75 x  150 y  50 z  75
(D) = = , = =
3 –6 2 3 –6 2

x y z
Sol. Equation of major axis of Ellipse is = = and centre is (0, 0, 0)
6 2 –3

D.Rs of directrices are components of (6 î + 2 ĵ – 3 k̂ ) × (2 î + 3 ĵ + 6 k̂ )  (3, –6, 2)

point of intersection of directrices and major axis are (150, 50, –75) and (–150, –50, 75)

 equation of directories are

x – 150 y – 50 z  75 x  150 y  50 z – 75
= = and = =
3 –6 2 3 –6 2

x y z
Hindi. nh?kZo`Ùk ds nh?kZ v{k dk lehdj.k = = rFkk dsUnz (0, 0, 0) gSA
6 2 –3

(6 î + 2 ĵ – 3 k̂ ) × (2 î + 3 ĵ + 6 k̂ )  (3, –6, 2) ds fu;rkvksa ds fnd~dksT;k,a

fu;rk vkSj nh?kZv{k (150, 50, –75) vkSj (–150, –50, 75) ds izfrPNsn fcUnq  fu;rkvksa ds lehdj.k
x – 150 y – 50 z  75 x  150 y  50 z – 75
= = vkSj = = gSA
3 –6 2 3 –6 2

PART-II: NUMERICAL VALUE QUESTIONS

Hkkx-II : la[;kRed iz'u (NUMERICAL VALUE QUESTIONS)

INSTRUCTION :
 The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto two digit.
 If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal
placed.

funsZ'k :

 bl [k.M esa izR;sd iz'u dk mÙkj la[;kRed eku ds :i esa gS ftlesa nks iw.kk±d vad rFkk nks vad n'keyo ds ckn esa gSA
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 13
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM
vkWQ (truncate/round-off) djsaA

1. Given f2(x) + g2(x) + h2(x)  9 and U(x) = 3f(x) + 4g(x) + 10h(x), where f(x), g(x) and h(x) are continuous
 x  R then maximum value of U(x) is.
fn;k x;k gS fd f2(x) + g2(x) + h2(x)  9 rFkk U(x) = 3f(x) + 4g(x) + 10h(x) tgk¡ f(x), g(x) rFkk h(x) lHkh x  R
ds fy, lrr~ gS] rc U(x) dk vf/kdre eku gS &
Ans. 33.54
Sol. 
ˆ · f(x)iˆ  g(x)jˆ h(x)kˆ }
max {U(x)} = max. { (3iˆ  4ˆj  10k) 
= max.  125 · f(x)2  g(x)2  f(x)2 · cos  =  1125  N = 1125 N – 1000 = 125

2. If in a plane A1, A2, A3,......, A20 are the vertices of a regular polygon having 20 sides and O is its centre
19
and  (OA  OA
i1
i i1 ) =  (OA 2  OA1 ) then || is [16JM120040]

;fn lery esa A1, A2, A3,......, A20 lecgqHkqt ds 'kh"kZ gS ftldh 20 Hkqtk,¡ gS vkSj O bldk dsUnz gS rFkk
19

 (OA  OA
i1
i i1 ) =  (OA 2  OA1 ) rc || dk eku gS
Ans. 19.00
n 1
Sol.  OA  OA
i 1
i i 1 = OA1  OA 2  OA 2  OA 3 .........+ OAn1  OAn = (n – 1) (OA1  OA 2 )

= (1 – n) (OA 2  OA1 ) when tc n = 20   = –19

| PA |2  | PB |2  | PC |2
3. In an equilateral ABC find the value of where P is any arbitrary point lying on
R2
its circumcircle, is
| PA |2  | PB |2  | PC |2
leckgq ABC esa dk eku Kkr dhft, tgka blds ifjxr o`Ùk ij LosPN fcUnq P gS&
R2
Ans. 06.00
A(a)
P (P)

Sol. O (0)
(c)
B C
(b)
Let O (0) be the circumcentre of ABC
ekuk O (0) , ABC dk ifjdsUnz gSA
abc
Given (a) | b || c | R  0 =  abc = 0
3
| PA |2  | PB |2  | PC |2 | P  a |2  | P  b |2  | P  c |2
2
= ? 
R R2
3 | P |  | a |2  | b |2  | c |2 2P.(a  b  c) 6R 2
; =6
R2 R2
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 14
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

4. Let A = 2 î + k̂ , B = î + ĵ + k̂ and C = 4 î  3 ĵ + 7 k̂ . If a vector R   ˆi – ˆj  kˆ satisfies

R x B  C x B and R . A  0 then 2 22is [16JM120039]
ekuk A = 2 î + k̂ , B = î + ĵ + k̂ rFkk C = 4 î  3 ĵ + 7 k̂ gSA ;fn lfn’'k R  ˆi – ˆj  kˆ satisfies
R x B  C x B vkSj R . A  0 dks lUrq"V djrh gS rc 222is
Ans. 69.00
Sol. (R – C)  B  O  R  C  B   A.C + A.B = 0  15 + 3 = 0
  = – 5  R  –iˆ – 8ˆj  2kˆ


x2 y3 zk

5. A line = = cuts the y-z plane and the x-y plane at A and B respectively. If AOB =
1 2 3

, then k, where O is the origin, is
2
x2 y3 zk
,d js[kk = = , y-z lery rFkk x-y lery dks Øe'k% A vkSj B ij izfrPNsn djrh gS ;fn
1 2 3

AOB = rc k dk eku gS tgk¡ O ewy fcUnq gS&
2

Ans. 04.50
x2 y3 zk
Sol. = = =  ( – 2, 2 + 3, 3 + k) for A,  = 2
1 2 3
k  k 2k 
A(0, 7, 6 + k)  for B =–  B  –2  , 3  , 0
3  3 3 
 2k  9
AOB = 90°  AO.OB = 0  7 – 3  =0 or k= 2k = 9
 3  2
x2 y3 zk
Hindi = = =  ( – 2, 2 + 3, 3 + k) A ds fy, =2
1 2 3
k  k 2k 
A(0, 7, 6 + k)  B ds fy,  = –  B  –2  , 3  , 0
3  3 3 
 2k  9
AOB = 90°  AO.OB = 0  7 – 3  =0 ;k k= 2k = 9
 3  2

6. Given four non zero vectors a, b, c and d . The vectors a, b and c are coplanar but not collinear pair
  

by pair and vector d is not coplanar with vectors a, b and c and ( a b )  (b c)  , ( d a )  
3
 
and ( d b )   , if (d c)  cos1(mcos   ncos ) then m – n is :
fn;k gS fd a, b, c ,oa d pkj v'kwU; lfn'k gSaA lfn'k a, b ,oa c leryh; gSa ysfdu ;qXe esa vlajs[kh; gSa vkSj
   

lfn'k d lfn'kksa a, b ,oa c ds lkFk leryh; ugha gSa vkSj ( a b )  (b c )  , ( d a )   , ( d b )  ;fn
3

(d c)  cos1(mcos   ncos ) , rc m – n gS
Ans. 02.00

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 15
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol.

Angle between vector a & b remains same even if we presume them as unit vector. Here for sake of
convinience let a, b, c, d are unit vectors.
 1 1
a.b = cos = .............(1) ; b.c = ............(2)
3 2 2
a . d = cos ...........(3) ; b.d = cos ...........(4)
also b =  (a  c)
Since b is presumed as unit vector
(a  c)  1  2 (a2  c 2  2a.c) =1
or  (1  1  1)  1 
2
=1
 b  (a  c)  c  ba
again d.c  d c cos  = d.(b  a) cos = cos – cos  = cos–1(cos – cos)

Hindi

;fn lfn'k a vkSj b dks bdkbZ lfn'k eku fy;k tk;s rc Hkh buds e/; cuus okyk dks.k vifjofrZr jgrk gSA
vr% lqfo/kk ds fy;s a , b , c , oa d dks ge bdkbZ lfn'k eku ysrs gSA
 1 1
a.b = cos = .............(1) ; b . c = ..........(2)
3 2 2
a.d = cos ...........(3) ; b.d = cos ..........(4)
vkSj b =  (a  c)
tSlk fd lfn'k b dks bdkbZ lfn'k eku fy;k gS
(a  c)  1  2 (a2  c 2  2a.c) = 1
;k 2 (1  1  1)  1  =1
 b  (a  c)  c  ba
iqu% d.c  d c cos  = d.(b  a) cos = cos – cos   = cos–1(cos – cos)

7. If the circumcentre of the tetrahedron OABC is given by

 
a2 b x c  b2  c x a   c 2 a xb  , where

a, b & c are the position vectors of the points A, B, C respectively relative to the origin 'O' such that
[ a b c ] = 36 then  is [16JM120041]

;fn prq"Qyd OABC dk ifjdsUnz

  
a2 b x c  b2  c x a   c 2 a xb  fn;k x;k gS, tgk¡ ewy fcUnq ds lkis{k

a, b rFkk c Øe'k% fcUnq A, B, C ds fLFkfr lfn'k gS tcfd [ a b c ] = 36 rc  gS &
Ans. 72.00
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 16
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. vectors a, b & c are non coplanar so are the vectors a  b, bc
Let position vector of circumcentre r  x(a  b)  y(b  c)  z(c  a)
also OE = AE = EB = EC  | r |  | r a|  | r b|  | r c |
or r  r  a  2r.a = r  b  2r.b  r  c 2  2r.c 2
2 2 2 2 2 2

a2
 2r.a  a2 , 2r.b  b2 , or 2y [a b c]  a2  y=
2[a b c]
Similarly z & x can be obtained
Hindi pw¡fd a, b ,ao c vleryh; gSa] rks lfn'k a  b, b  c Hkh vleryh; gksxsaA
ekuk ifjdsUnz dk fLFkr lfn'k r  x(a  b)  y(b  c)  z(c  a) gSA
,oa OE = AE = EB = EC  | r |  | r a|  | r b|  | r c |
;k r 2  r 2  a2  2r.a = r 2  b2  2r.b  r 2  c 2  2r.c 2

a2
 2r.a  a2 , 2r.b  b2 , ;k 2y [a b c]  a2  y=
2[a b c]
blh izdkj z vkSj x dks Hkh izkIr fd;k tk ldrk gSA

8. Given three point on x – y plane as O(0, 0), A(1, 0) & B(–1, 0). Point P moving on the given plane
satisfying the condition (PA . PB) + 3 (OA . OB) = 0 [16JM120042]
If the maximum & minimum values of | PA | | PB | is M & m respectively then the value of M2 + m2 is
Ans. 34.00

fn, x, rhu fcUnq O(0, 0) A(1, 0) & B(–1, 0) xy lery ij gSA fcUnq P lery ij bl izdkj xfr djrk gS fd
izfrcU/k (PA . PB) + 3 (OA . OB) = 0 gSA ;g | PA | | PB | dk vf/kdre o U;wure eku Øe'k% M rFkk m
gSA rc M2 + m2 dk eku gS&
Sol. P = xi + yj ; AP = (x – 1)i + yj ; BP = (x + 1)i + yj
PA . PB = x2 – 1 + y2 ; OA . OB = –1
Now vc (PA.PB) + 3 (OA.OB) = 0  x2 + y2 – 4 = 0  x2 + y2 = 4
| PA | | PB | = (x  1)2  y 2 (x  1)2  y 2 = 5  2x 5  2x

| PA | | PB | = 25  4x 2
Now from x2 + y2 = 4 ls put x = 2 cos  j[kus ij y = 2 sin 
 | PA | | PB | = 25  16cos  2
; | PA | | PB | max = 25 = M
| PA | | PB | min
= 9 =m ; M2 + m2 = 25 + 9 = 34

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 17
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

9. If the volume of tetrahedron formed by planes whose equations are y + z = 0, z + x = 0, x + y = 0 and
x + y + z = 1 is t cubic unit, then the value of t is
;fn leryksa y + z = 0, z + x = 0, x + y = 0 rFkk x + y + z = 1 }kjk cuk;s x;s prq"Qyd dk vk;ru t ?ku bdkbZ
gS rks t dk eku Kkr dhft,A
Ans. 00.66 or 00.67

Sol.

The planes are

y+z=0 ............(1)
z+x=0 ............(2)
x+y=0 ............(3)
x+y+z=1 ............(4)
Solving above equations we get vertices of the tetrahedron as (0,0,0), (–1,1,1), (1,–1,1) and (1,1,–1)

1 1 1 0 2 1
1 1 4 2 2
Required volume = 1 1 1 = 2 0 1 = = t = 3t = 2
6 6 6 3 3
1 1 1 0 0 1

Hindi

lery gS
y+z=0 ............(1)
z+x=0 ............(2)
x+y=0 ............(3)
x+y+z=1 ............(4)
mijksDr lehdj.kksa dks gy djus ij gesa prq"Qyd ds 'kh"kZ (0,0,0), (–1,1,1), (1,–1,1) rFkk (1,1,–1) feyrs gaS
1 1 1 0 2 1
1 1 4 2 2
vHkhk"V vk;ru = 1 1 1 = 2 0 1 = = t = 3t = 2
6 6 6 3 3
1 1 1 0 0 1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 18
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

10. If r represents the position vector of point R in which the line AB cuts the plane CDE, where position
vectors of points A, B, C, D, E are respectively a  ˆi  2jˆ  kˆ , b  2iˆ  ˆj  2kˆ , c  4ˆj  4kˆ ,
d  2iˆ  2jˆ  2kˆ and e  4iˆ  ˆj  2kˆ , then r is : [16JM120043]
;fn js[kk AB rFkk lery CDE ds izfrPNsn fcUnq R dk fLFkfr lfn'k r gS] tgk¡ fcUnqvksa A, B, C, D, E ds fLFkfr
lfn'k Øe'k% a  ˆi  2jˆ  kˆ , b  2iˆ  ˆj  2kˆ , c  4ˆj  4kˆ , d  2iˆ  2jˆ  2kˆ vkSj e  4iˆ  ˆj  2kˆ , gS] rc r gS:
Ans. 04.24
Sol. Equation of line AB is r  a  (b  a)  r  (iˆ  2jˆ  k)
ˆ  (iˆ  ˆj  k)
ˆ .......(1)
CD  2iˆ  2jˆ – 2kˆ ; CE  4iˆ  5ˆj – 2kˆ ; n  CD  CE = 6iˆ – 4ˆj  2kˆ
So equation of plane CDE is 3x – 2y + z = 12. Solve with line r  (iˆ  2jˆ  k)
ˆ  (iˆ  ˆj  k)
ˆ

3(1  ) – 2(2 – )  1    12  =2 Hence R is 3iˆ  3kˆ

Hindi js[kk AB dk lehdj.k r  a  (b  a)  r  (iˆ  2jˆ  k)
ˆ  (iˆ  ˆj  k)
ˆ .......(1)
CD  2iˆ  2jˆ – 2kˆ ; CE  4iˆ  5ˆj – 2kˆ ; n  CD  CE = 6iˆ – 4ˆj  2kˆ
lery CDE dk lehdj.k 3x – 2y + z = 12 gS &
js[kk r  (iˆ  2jˆ  k)
ˆ  (iˆ  ˆj  k)
ˆ ds lkFk gy djus ij 3(1  ) – 2(2 – )  1    12  =2
vr% fcUnq R, 3iˆ  3kˆ gSA

11. Line L1 is parallel to vector   3iˆ  2jˆ  4kˆ and passes through a point A(7, 6, 2) and line L 2 is parallel
to a vector   2iˆ  ˆj  3kˆ and passes through a point B(5, 3, 4). Now a line L3 parallel to a vector
r  2iˆ  2jˆ  kˆ intersects the lines L1 and L2 at points C and D respectively, then | 4CD | is equal to :
Ans. 36.00 [16JM120044]
js[kk L ] lfn'k    3 i  2 ˆj  4kˆ ds lekUrj gS vkSj fcUnq A(7, 6, 2) ls xqtjrh gS vkSj js[kk L lfn'k
1
ˆ
2

  2 ˆi  ˆj  3kˆ ds lekUrj gS vkSj ,d fcUnq B(5, 3, 4) ls xqtjrh gSA vc ,d js[kk L3 ,d lfn'k

r  2 ˆi  2 ˆj  kˆ ds lekUrj gS vkSj js[kkvksa L1 rFkk L2 dks Øe'k% fcUnqvksa C vkSj D ij izfrPNsn djrh gS] rks
| 4CD | cjkcj gS %
Sol. Equation of line L1 is 7iˆ  6ˆj  2kˆ +  ( 3iˆ  2jˆ  4k)
ˆ

Equation of line L2 is 5iˆ  3ˆj  4kˆ +  (2iˆ  ˆj  3k)

ˆ

CD = 2iˆ  3ˆj  2kˆ +  (3iˆ  2jˆ  4k)

ˆ –  (2iˆ  ˆj  3k)
ˆ . since it is parallel to 2iˆ  2jˆ  kˆ
2  3  2 3  2   2  4  3
 = =  = 2 ,  = 1
2 2 1
 ˆ ˆ ˆ
CD = 6i  6 j  3k  |4 CD | = 36
Hindi. js[kk L1 dk lehdj.k 7iˆ  6ˆj  2kˆ +  (3iˆ  2jˆ  4k)
ˆ gSA

js[kk L2 dk lehdj.k 5iˆ  3ˆj  4kˆ +  (2iˆ  ˆj  3k)

ˆ gSA

CD = 2iˆ  3ˆj  2kˆ +  (3iˆ  2jˆ  4k)

ˆ –  (2iˆ  ˆj  3k)
ˆ

pwafd ;g 2iˆ  2jˆ  kˆ ds lekUrj gSA

2  3  2 3  2   2  4  3
 = =  = 2 ,  = 1
2 2 1
 CD = 6iˆ  6ˆj  3kˆ  |4 CD | = 36

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 19
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

12. L is the equation of the straight line which passes through the point (2, –1, –1); is parallel to the plane
4x + y + z + 2 = 0 and is perpendicular to the line of intersection of the planes 2x + y = 0 = x – y + z. If

the point (3, , ) lies on line L, then absolute value of is

ekuk ljy js[kk dk lehdj.k tks fcUnq (2, –1, –1) ls xqtjrk gS rFkk lery 4x + y + z + 2 = 0 ds lekUrj gS vkSj
leryksa 2x + y = 0 = x – y + z dh izfrPNsnu js[kk ds yEcor~ gS ;fn fcUnq (3, , ) js[kk L ij fLFkr gS rc

dk fujis{k eku gksxk &

Ans. 01.75
x  2 y 1 z 1
Sol.    4a + b + c = 0 ...(i)
a b c
ˆi ˆi kˆ
2x + y = 0 = x – y + z  2 1 0 = ˆi(1  0)  ˆj(2  0)  k(
ˆ 2  1)  ˆi  2jˆ  3kˆ
1 1 1
a – 2b – 3c = 0 ... (ii)
From (i) & (ii) (i) vkSj (ii) ls
a b c a b c
4a + b + c = 0  a – 2b – 3c = 0       
3  2 1  12 8  1 1 13 9
 equation of the line  js[kk dk lehdj.k
x  2 y 1 z 1 32  1  1
   = = = – 14 and  = 8  |  +  | = 6.
1 13 9 1 13 9

x4 y6 z 1
13. The lines = = and 3x – 2y + z + 5 = 0 = 4x + 3y – 4z – 3k are coplanar, then the
3 5 2
value of k is
x4 y6 z 1
js[kk,¡ = = rFkk 3x – 2y + z + 5 = 0 = 4x + 3y – 4z – 3k leryh; gaS] rc k dk eku gS
3 5 2
Ans. 10.66 or 10.67
x  4 y  6 z 1
Sol. L1 :   r ; L2 : 3x – 2y + z + 5 = 0 = 4x + 3y – 4z – 3k
3 5 2
Any point on the first line is (3r – 4, 5r – 6, – 2r + 1)
As lines are coplanar therefore this point must lie on both the planes representing the second line
3(3r – 4) – 2(5r – 6) + (– 2r + 1) + 5 = 0  r=2
hence coordinate of point (2,4,–3)
32
and 4(2) + 3(4) – 4(–3) – 3k = 0  k =
3

x  4 y  6 z 1
Hindi. L1 :    r ; L2 : 3x – 2y + z + 5 = 0 = 4x + 3y – 4z – 3k
3 5 2
izFke js[kk ij dksbZ fcUnq (3r – 4, 5r – 6, – 2r + 1) gSA
tSlkfd js[kk,¡ leryh; gSa] ;g fcUnq mu nksuksa leryksa ij fLFkr gksuk pkfg, tksfd f}rh; js[kk dks iznf'kZr djrsa gSa
3(3r – 4) – 2(5r – 6) + (– 2r + 1) + 5 = 0  r=2
vr% fcUnq ds funsZ'kkad (2,4,–3) gS
32
vkSj 4(2) + 3(4) – 4(–3) – 3k = 0  k =
3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 20
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x 1 y2 z3
14. About the line = = the plane 3x + 4y + 6z + 7 = 0 is rotated till the plane passes
2 3 1
through the origin. Now 4x + y + z = 0 is the equation of plane in new position. The value of 2 + 2
is
x 1 y2 z3
js[kk = = ds lkis{k lery 3x + 4y + 6z + 7 = 0 dks tcrd ?kqek;k tkrk gS rc rd ;g ewy
2 3 1
fcUnq ls ugh xqtjrk gS vc u;h fLFkfr esa lery dh lehdj.k 4x + y + z = 0 gS rc 2 + 2 gS &
Ans. 32.00

Sol. Since 3(2) + 4(–3) + 6(1) = 0 and 3(1) + 4(2) + 6(–3) + 7 = 0

x 1 y2 z3
 the line = = lies in the plane 3x + 4y + 6z + 7 = 0.
2 3 1
In the new position again the line lies in the plane. Let the equation of the new position of the plane be
ax + by + cz = 0, then 2a – 3b + c = 0 and a + 2b – 3c = 0
a b c
 = = i.e. a=b=c 
92 1 6 43
  equation of the required plane is x + y + z = 0
Hindi  3(2) + 4(–3) + 6(1) = 0 vkSj 3(1) + 4(2) + 6(–3) + 7 = 0
x 1 y2 z3
 js[kk = = lery 3x + 4y + 6z + 7 = 0 ij fLFkr gS
2 3 1
iqUk% u;h fLFkfr esa js[kk lery ij fLFkr gS vr% ekukfd u;h fLFkfr es a] lery dk lehdj.k ax + by + cz = 0 gS]
rks
2a – 3b + c = 0 vkSj a + 2b – 3c = 0
a b c
 = = vFkkZr~ a=b=c
92 1 6 43
 vr% vHkh"V lery dk lehdj.k x + y + z = 0

15. The value of tan 3 , where  is the acute angle between the plane faces of a regular tetrahedron, is

tan3 dk eku gksxk tgk¡  leprq"Qyd ds lery Qydksa ds e/; dk U;wudks.k gS& [16JM120046]
Ans. 22.62 or 22.63
Sol. Since tetrahedron is regular AB = BC = AC = DC and angle between two adjcant side = /3
consider planes ABD and DBC vector, normal to plane ABD is = a  b
vector, normal to plane DBC is = b  c angle between these planes is angle between

vectors (a  b) & (b  c)
1 2
 b a c
(a  b).(b  c) 4 1
 cos   = = 
ab bc 3 2
3
a b c
4
 1
Since acute angle is required   cos1    tan 2 2  tan3= 16 2
3
Hindi pawfdfn;k x;k prq"Qyd leprq"Qyd gS vr% AB = BC = AC = DC vkSj nks vklUu Hkqtkvksa ds e/; dks.k /3 gSA
ekuk ABD vkSj DBC nks lery gaSs lery ABD ds yEcor~ lfn'k = a  b
lery DBC ds yEcor~ lfn'k = b  c
Li"Vr% lfn'kksa (a  b) vkSj (b  c) ds e/; cuus okyk dks.k gh
bu leryksa ds e/; cuus okyk dks.k gSaA

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 21
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 2
 b a c
(a  b).(b  c) 4 1
  cos   = = 
ab bc 3 2
3
a b c
4

U;wu dks.k   cos1   

1
  vHkh"V tan 2 2  tan3= 16 2
3
16. R and r are the circumradius and inradius of a regular tetrahedron respectively in terms of the length
k of each edge. If R2 + r2 = k2 then value of  is
R rFkk r Øe'k% leprq"Qyd dh ifjf=kT;k vkSj vUr%f=kT;k izR;sd fdukjksa ds yEckbZ k ds inksa esa gSA ;fn R2 + r2 =
k2 rc  dk eku gksxk &
Ans. 00.41 or 00.42
Sol. circum-radius  distance of circum centre from any of the vertex
abc
 distance of from vertex D (0) [tetrahedron is regular]
4
1 1
Circumradius = abc  a2  b2  c 2  2(a.b  b. c  c.a)
4 4
1 2  k2 k2 k2  1 3
= k  k2  k2  2     = 6k 2 = k
4  2 2 2  4 8
r 1 R k 3 k 5 2
  r= =   R= k &r=  R2 + r 2 = k 
R 3 3 24 8 24 12
 minimum value of p + q = 17
Hindi ifjf=kT;k  ifjdsUnz dh fdlh Hkh 'kh"kZ ls nwjh
abc
 dh 'kh"kZ D (0) ls nwjh [tSlk fd prq"Qyd] ,d leprq"Qyd gS ]
4
1 1
 ifjf=kT;k = abc  a2  b2  c 2  2(a.b  b. c  c.a)
4 4
1 2  k2 k2 k2  1 3
= k  k2  k2  2     = 6k 2 = k
4  2 2 2 4 8
r 1 R k 3 k 5 2
  r= =   R= k &r=  R2 + r 2 = k 
R 3 3 24 8 24 12

 p + q = 17 dk U;wure eku gS

17. A line L on the plane 2x + y – 3z + 5 = 0 is at a distance 3 unit from the point P(1, 2, 3). A spider starts
x 1 y  2 z3
moving from point A and after moving 4 units along the line   it reaches to point P.
2 1 3
and from P it jumps to line L along the shortest distance and then moves 12 units along the line L to
reach at point B. The distance between points A and B is [16JM120047]
lery 2x + y – 3z + 5 = 0 ij js[kk L, fcUnq P(1, 2, 3) ls 3 bdkbZ nwjh ij gSA ,d edM+h fcUnq A ls xfr djrh
x 1 y  2 z3
gqbZ js[kk   ds vuqfn'k 4 bdkbZ nwjh ij fcUnq P ij igq¡prh gS rFkk fcUnq P ls ;g js[kk L ij
2 1 3
y?kqÙke nwjh dh js[kk ds vuqfn'k mNyrh gSA rc js[kk L ds vuqfn'k] 12 bdkbZ xfr djds fcUnq B ij igq¡prh gSA
fcUnqvksa A rFkk B ds e/; U;wure nwjh Kkr dhft,A
Ans. 13.00
Sol. 32  42  122  13

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 22
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

18. The length of edge of a regular tetrahedron D-ABC is 'a'. Point E & F are taken on the edges AD and
BD respectively. Such that E divide DA and F divide BD in the ratio 2 : 1 each. The area of CEF is
3 2
equal to a , then value of k is :
k
leprq"Qyd D-ABC dh Hkqtkvksa dh yEckbZ 'a' gS fcUnq E vkSj F dh Hkqtkvksa AD rFkk BD ij Øe'k% gS rFkk E, DA
3 2
vkSj F, BD dks vuqikr 2 : 1 esa foHkkftr djrk gS CEF dk {ks=kQy a gks rks k dk eku gksxk &
k
Ans. 07.20

D (0)
2
E
1 1
Sol. A C(c )
(a) 2

B(b)
| a | = | b | = | c | = | a b | = | c b | = | a c | = a
On solving we get gy djus ij
2
a
a.b = b.c = a.c =  | a | = | b | = | c | = a
2
 2a  b 1
E  & F =   area of CEF dk {ks=kQy = | CE  CF |
 3  3
  2

1  2a  b  1 2 2 1 1 26–3
=  c c = (a  b)  (c  a)  (b  c) =  a  b
2  3 
  3  2 9 3 3 2  9 

1 5 5 5 5 3 5 3 2
= . | ab |  .| ab |  . | a | | b | sin 60  . a2 . = a
2 9 18 18 18 2 36

y z x z
19. If 'd' be the shortest distance between the lines + = 1; x = 0 and – = 1; y = 0 and if
b c a c
 1 1 1
2
=2
+ 2 + 2 then  is
d a b c
y z x z  1 1 1
;fn js[kkvksa + = 1; x = 0 rFkk – = 1, y = 0 ds e/; y?kqÙke nwjh 'd' gS rFkk ;fn 2 = 2 + 2 + 2
b c a c d a b c
rc  dk eku gS &
Ans. 04.00
x y zc x y zc
Sol. L1 : = = =r ; L2 : = = =
0 b c a 0 c
Dr's of AB are –a, br, –cr – c + 2c   AB is perpendicular to both the lines
0(–a) + b. br + (–c) (–cr– c + 2c) = 0  (b2 + c2) r + c2 = 2c2 .....(1)
and a(–a) + 0(br) + c (–cr– c + 2c) = 0  –(a2 + c2) – c2r + 2c2 = 0
(a2 + c2) + c2r = 2c2 .....(2)
from (1) & (2)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 23
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2b2c 2 2a2c 2
 , r 
a 2b 2  b 2 c 2  c 2 a 2 a2 b2  b2 c 2  c 2 a 2
 2a2bc 2  a2b2  b2c 2 – c 2a2 
A 0 , , c   
 a 2b 2  b 2 c 2  c 2 a 2  a b b c c a
2 2 2 2 2 2
 
 2ab2c 2  b2 c 2  a2b2 – c 2 a 2 
B 2 2 , 0 , c  2 2  
 a b  b2 c 2  c 2 a2
a b b c c a
2 2 2 2
 
4a2b4c 4 4a4b2c 4 4c 2 (a 4b 4 )
d2 = + +
(a2b2  b2c 2  c 2a2 )2 (a2b2  b2c 2  c 2a2 )2 (a2b2  b2c 2  c 2a2 )2
4 (a2b2  b2c 2  c 2a2 )2 a2b2  b2 c 2  c2 a2 4 1 1 1
= =  2 = 2  2  2
d 2
a b c a b c a b c
2 4 4 4 2 4 4 4 2 2 2 2
abc d a b c
x y zc x y zc
Hindi L1 : = = =r ; L2 : = = =
0 b c a 0 c
AB ds fnd~ vuqikr gS –a, br, –cr – c + 2c
AB nksuksa js[kkvksa ds yEcor~ gS
0(–a) + b. br + (–c) (–cr– c + 2c) = 0
(b2 + c2) r + c2 = 2c2 ................(1)
rFkk a(–a) + 0(br) + c (–cr– c + 2c) = 0
–(a2 + c2) – c2r + 2c2 = 0
(a2 + c2) + c2r = 2c2 ...............(2)
(1) vkSj (2) ls
2b2c 2 2a2c 2
 , r 
a 2b 2  b 2 c 2  c 2 a 2 a2 b2  b2 c 2  c 2 a 2
 2a2bc 2  a2b2  b2c 2 – c 2a2 
A 0 , , c   
 a 2b 2  b 2 c 2  c 2 a 2  a b b c c a
2 2 2 2 2 2
 
 2ab2c 2  b2 c 2  a2b2 – c 2 a 2 
B 2 2 , 0 , c  2 2  
 a b  b2 c 2  c 2 a2
a b b c c a
2 2 2 2
 
4a2b4c 4 4a4b2c 4 4c 2 (a 4b 4 )
d2 = + +
(a2b2  b2c 2  c 2a2 )2 (a2b2  b2c 2  c 2a2 )2 (a2b2  b2c 2  c 2a2 )2
4 (a2b2  b2c 2  c 2a2 )2 a2b2  b2 c 2  c2 a2 4 1 1 1
= =  2 = 2  2  2
d2 a b c a b c a b c
2 4 4 4 2 4 4 4 2 2 2 2
abc d a b c

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 24
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

Hkkx - III : ,d ;k ,d ls vf/kd lgh fodYi çdkj
1. A vector a has components 2p and 1 with respect to a rectangular cartesian system. The system is
rotated through a certain angle about the origin in the counterclockwise sense. If with respect to the
new system, a has components p + 1 and 1, then
1 1
(A*) p =  (B*) p = 1 (C) p =  1 (D) p =
3 3
dkrhZ; funsZ'kkad fudk; ds lkis{k ,d lfn'k a ds ?kVd Øe'k% 2p ,oa 1 gSaA fudk; dks ewyfcUnq ds lkis{k okekorZ
fn'kk esa ,d fuf'pr dks.k ls ?kqek;k tkrk gSA ;fn u;s fudk; ds lkis{k lfn'k a ds ?kVd p + 1 ,oa 1 gSa] rks
1 1
(A*) p =  (B*) p = 1 (C) p =  1 (D) p =
3 3
Sol. Before rotation a  2piˆ  ˆj after rotation a  (p  1)iˆ  ˆj . Since length of vector remains unaltered

4p2  1 = (p  1)2  1  4p2 = (p + 1)2

1
p + 1 = ± 2 p p = 1 or 
3
Hindi fudk; ds ?kw.kZu ls iwoZ a  2piˆ  ˆj . ?kw.kZu ds ckn a  (p  1)iˆ  ˆj . pawfd ?kw.kZu ds ckn Hkh lfn'k dh yEckbZ rks
vifjofrZr gh jgrh gS
4p2  1 = (p  1)2  1  4p2 = (p + 1)2
1
p + 1 = ± 2 p p = 1 or 
3

2. If z1  a ˆi  b ˆj and z2  c ˆi  d ˆj are two vectors in ˆi and ˆj system, where z1  z2 = r and

z1 . z2 = 0, then w1  a ˆi  c ˆj and w 2  b ˆi  d ˆj satisfy : [16JM120048]

(A*) w1 = r (B*) w 2 = r (C*) w1 .w 2 = 0 (D) | w1 |  |w 2 |

;fn î rFkk ĵ fudk; esa nks lfn'k z1  a ˆi  b ˆj vkSj w 2  b ˆi  d ˆj gSa] tgk¡ z1  z2 = r vkSj z1 . z2 = 0. rks
w1  a ˆi  c ˆj vkSj w 2  b ˆi  d ˆj larq"V djrs gSa &
(A*) w1 = r (B*) w 2 = r (C*) w1 .w 2 = 0 (D) | w1 |  |w 2 |
a d
Sol. As given a2 + b2 = c2 + d2, ac + bd = 0 from second relation ac = – bd  – 
b c
using in first relation 2b2 + b2 = c2 + 2c2
 b2 = c2 & a2 = d2  (b = c, a = – d) or (b = – c, a = d).
Now | w1 |  a2  c 2  a2  b2 | w 2 |  b2  d2  c 2  d2
w1.w 2 = ab + cd = ab + b(– a) = 0
a d
Hindi tSlk fd fn;k gS a2 + b2 = c2 + d2, ac + bd = 0 f}rh; lEcU/k ls ac = – bd  – 
b c
izFke lEcU/k esa mi;ksx djus ij 2b2 + b2 = c2 + 2c2
 b2 = c2 & a2 = d2  (b = c, a = – d) or (b = – c, a = d)
vc | w1 |  a c  a b
2 2 2 2
, | w2 |  b2  d2  c 2  d2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 41
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

w1.w 2 = ab + cd = ab + b(– a) = 0
3. If a, b, c, x, y, z  R such that ax + by + cz = 2, then which of the following is always true
;fn a, b, c, x, y, z  R bl izdkj gS fd ax + by + cz = 2 rks fuEu esa ls dkSuls lnSo lR; gS
(A*) (a2 + b2 + c2)(x2 + y2 + z2)  4 (B*) (x2 + b2 + z2)(a2 + y2 + c2)  4
(C*) (a + y + z )(x + b + c )  4
2 2 2 2 2 2
(D*) (a2 + b2 + z2)(x2 + y2 + c2)  4
Sol. Let A  aiˆ  bjˆ  ckˆ & B  xiˆ  yjˆ  zkˆ
given that A.B = 2  | A | | B | cos = 2
 x2  y2  z2 . a2  b2  c 2  2
 (a2 + b2 + c2)(x2 + y2 + z2)  4
Similarly the others.
ekuk A  aiˆ  bjˆ  ckˆ ,oa B  xiˆ  yjˆ  zkˆ
fn;k x;k gS A.B = 2
 | A | | B | cos = 2  x2  y2  z2 . a2  b2  c 2  2
 (a2 + b2 + c2)(x2 + y2 + z2)  4
blh izdkj vU;

4. The direction cosines of the lines bisecting the angle between the lines whose direction cosines are 
1, m1, n1 and 2, m2, n2 and the angle between these lines is , are [16JM120049]
mu js[kkvksa dh fnd~ dksT;k,sa tks 1, m1, n1 vkSj 2, m2, n2 fnd~dksT;kvksa okyh js[kkvksa ds e/; dks.k dks lef}Hkkftr
djrh gSa tcfd bu js[kkvksa ds e/; dks.k gS] gkasxh &
2 m1 – m2
– n – n2  2 m  m2 n  n2
(A) 1
, , 1 (B*) 1 , 1 , 1
     
cos cos cos 2cos 2cos 2cos
2 2 2 2 2 2
 m  m n  n  m  m n  n
(C*) 1 2
, 1 2
, 1 2
(D*) 1 2 , 1 2
, 1 2
     
2 sin 2sin 2 sin 2 sin 2sin 2 sin
2 2 2 2 2 2
Sol. cos = 12 + m1m2 + n1n2 ; Dc's of 1 (bisector)
1  2 1  2 1  2 1 2
= = =
(  2
 (m1  m2 )2  (n1  n2 )2 2  2(  m1m2  n1n2 ) 2  2cos  2cos  / 2
1 2) 1 2

m1  m2 n1  n2
Similarly ,
2cos  / 2 2cos  / 2
Similarly Dc's for bisector 2
1 2 m  m2 n1  n2
, 1 ,
  
2 sin 2sin 2 sin
2 2 2

Hindi cos = 12 + m1m2 + n1n2. 1 (v/kZd) dh fnd~dksT;k,¡ gaS

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 42
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1  2 1  2 1  2 1 2
= = =
(  2
 (m1  m2 )2  (n1  n2 )2 2  2(  m1m2  n1n2 ) 2  2cos  2cos  / 2
1 2) 1 2

m1  m2 n1  n2
blh izdkj ,
2cos  / 2 2cos  / 2
blh izdkj v)Zd 2 dh fnd~dksT;k,¡ gSa
1 2 m  m2 n1  n2
, 1 ,
  
2 sin 2sin 2 sin
2 2 2

5. The value(s) of  [0, 2] for which vector a  ˆi  3jˆ  (sin2) kˆ makes an obtuse angle with the
 ˆ  ˆ
z-axis and the vectors b  (tan ) ˆi  ˆj  2 sin k and c = (tan ) î + (tan ) ĵ – 3 cosec k are
2 2
orthogonal, is/are
 [0, 2] ds ¼dk½ og eku ftuds fy, lfn'k a  ˆi  3jˆ  (sin2) kˆ , z-v{k ds lkFk vf/kd dks.k cukrk gS vkSj
 ˆ 
lfn'k b  (tan ) ˆi  ˆj  2 sin k ,oa c   tan   ˆi   tan   ˆj  3 cosec kˆ ijLij yEcor~ gS] gksxk&
2 2
(A) tan 3 1
(B*)  tan 1 2 (C) + tan 1
3 (D*) 2  tan 1 2
Sol. Since a makes obtuse angle with z-axis
sin2
 <0 i.e. sin 2 < 0
1  9  sin2 2
 3
 either <  <  or < < 2 ......(i)
2 2
since b and c are orthogonal
 tan2 – tan – 6 = 0 i.e. tan = 3, –2 ......(ii)
from (i) and (ii), we get
tan  = – 2    =  – tan–1 2 or  = 2 – tan–12

Hindi pwafd a , z-v{k ds lkFk vf/kd dks.k cukrk gSA

sin2
 <0 vFkkZr~ sin 2 < 0
1  9  sin2 2
 3
 ;k rks <  <  ;k < < 2 ......(i)
2 2
pwafd b vkSj c ykfEcd gaSA
 tan2 – tan – 6 = 0 vFkkZr~ tan = 3, –2 ......(ii)
lehdj.k (i) vkSj (ii) ls tan  = – 2  =  – tan 2 ;k  = 2 – tan–12
–1

11
6. The vector ˆi  xjˆ  3kˆ is rotated through an angle of cos–1 and doubled in magnitude, then it
14
becomes 4iˆ  (4x  2)jˆ  2kˆ . The value of ' x ' CANNOT be : [16JM120050]

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 43
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

11
,d lfn'k ˆi  xjˆ  3kˆ dks dks.k cos–1 ls ?kqek;k tkrk gS] rFkk fQj bldk ifjek.k nqxuk dj fn;k tkrk gS] rks
14
ˆ ˆ ˆ
;g lfn'k 4i  (4x  2)j  2k cu tkrk gSA ' x ' dk eku ugh gks ldrk gS&
2 2 20
(A*)  (B*) (C*)  (D) 2
3 3 17
Sol. | AC |2 = | 2AB |2  | 4iˆ  (4x  2) ˆj  2kˆ |2 = 4 | (iˆ  xjˆ  3k)
ˆ |2
 16 + (4x – 2)2 + 4 = 4 (1 + x2 + 9)  20 + 16x2 + 4 – 16x = 4 + 4x2 + 36
 12x – 16x – 16 = 0
2
 3x – 4x – 4 = 0
2

2
 x=2,– ...(i)
3
AB . AC AB . AC
angle between AB and AC is cos  = =
| AB | | AC | 2 | AB |2
11 (iˆ  xjˆ  3k)
ˆ . (4iˆ  (4x  2) ˆj  2k)ˆ 4  x(4x  2)  6
 = =
14 2(1  x  9)
2
2x 2  20
 11 x + 110 = 70 + 28 x – 14x
2 2

20
  17x2 – 14x – 40 = 0   x = 2, – .......(ii)
17
from (i) and (ii)
x=2

Hindi | AC |2 = | 2AB |2  | 4iˆ  (4x  2) ˆj  2kˆ |2 = 4 | (iˆ  xjˆ  3k)

ˆ |2 
  16 + (4x – 2) + 4 = 4 (1 + x + 9)
2 2
 20 + 16x + 4 – 16x = 4 + 4x2 + 36
2

  12x – 16x – 16 = 0
2
 3x – 4x – 4 = 0
2

2
 x=2,– ........(i)
3
AB . AC AB . AC
AB vkSj AC ds chp dks.k cos  = =
| AB | | AC | 2 | AB |2
11 (iˆ  xjˆ  3k)
ˆ . (4iˆ  (4x  2) ˆj  2k)
ˆ 4  x(4x  2)  6
 = =
14 2(1  x  9) 2
2x 2  20
 11 x2 + 110 = 70 + 28 x2 – 14x
20
 17x2 – 14x – 40 = 0  x = 2, – ......(ii)
17
lehdj.k (i) vkSj (ii) ls x=2

7. The vertices of a triangle are A (1, 1, 2), B(4, 3, 1) and C(2, 3, 5). A vector representing the bisector of
the angle A is :
,d f=kHkqt ds 'kh"kZ A (1, 1, 2), B(4, 3, 1) vkSj C(2, 3, 5) gSaA dks.k A ds lef}Hkktd dks fu:fir djus okyk
lfn'k gS &
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 44
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(A*) 2iˆ – 4kˆ (B*) –2iˆ  4kˆ (C*) –2iˆ – 2jˆ  kˆ (D*) 2iˆ  2jˆ  kˆ
Sol. AB = 9  4  1  14 ; AC = 1  4  9  14

M (3, 3, 3) ; AM = 2iˆ  2jˆ  kˆ

8. The vector c , parallel to the internal bisector of the angle between the vectors a  7 ˆi  4 ˆj  4kˆ and
b   2 ˆi  ˆj  2kˆ with  c  = 5 6 , is :
,d lfn'k c dh fn'kk lfn'kksa a  7 ˆi  4 ˆj  4kˆ vkSj b   2 ˆi  ˆj  2kˆ ds vUr% dks.k lef)Hkktd ds vuqfn'k
gS ;fn  c  = 5 6 gS rks gS&
(A*)
5 ˆ
3

i  7 ˆj  2kˆ  (B)
3

5 ˆ
i  7 ˆj  2kˆ  (C*)
3

5 ˆ
 i  7 ˆj  2kˆ  (D)
3

5 ˆ
 i  7 ˆj  2kˆ 
 ˆ ˆ ˆ ˆ ˆ ˆ
Sol. ˆ =   7i – 4 j – 4k  –2i – j  2k  =  7iˆ – 4ˆj – 4kˆ  3(–2iˆ – ˆj  2k)
any such vector c =  (aˆ  b) ˆ 
 9 3  9  
 

 ˆ 
= i – 7ˆj  2kˆ    | c |= 5 6  1  49  4 = 5 6 
9   9

 95 6 15 ˆ ˆ =  5 (iˆ – 7ˆj  2k)

  54 = 5 6   =  = 15  c =  (i – 7ˆj  2k) ˆ
9 54 9 3

Hindi ,sls lfn'k dks fuEu :i esa fy[kk tk ldrk gS&

 ˆ ˆ ˆ ˆ ˆ ˆ
ˆ =   7i – 4 j – 4k  –2i – j  2k  =  7iˆ – 4ˆj – 4kˆ  3(–2iˆ – ˆj  2k)
c =  (aˆ  b) ˆ 
 9 3  9  
 

 ˆ 
= i – 7ˆj  2kˆ    | c |= 5 6  1  49  4 = 5 6 
9   9

 95 6 15 ˆ ˆ =  5 (iˆ – 7ˆj  2k)

  54 = 5 6   =  = 15  c =  (i – 7ˆj  2k) ˆ
9 54 9 3
9. A line passes through a point A with position vector 3 ˆi  ˆj  kˆ and is parallel to the vector
2 ˆi  ˆj  2kˆ . If P is a point on this line such that AP = 15 units, then the position vector of the point P
is/are [16JM120051]
,d js[kk fcUnq A ls xqtjrh gS ftldk fLFkfr lfn'k 3 ˆi  ˆj  kˆ gS vkSj lfn'k 2 ˆi  ˆj  2kˆ ds lekUrj gSA ;fn
bl js[kk ij ,d fcUnq P bl izdkj gS fd AP = 15 bdkbZ] rks fcUnq P dk fLFkfr lfn'k gS@gSa&
(A) 13 ˆi  4 ˆj  9kˆ (B*) 13 ˆi  4 ˆj  9kˆ (C) 7 ˆi  6 ˆj  11kˆ (D*)  7 ˆi  6 ˆj  11kˆ
Sol. line is r  (3iˆ  ˆj  k)
ˆ  (2iˆ  ˆj  2k)
ˆ . Let position vector of point p is  p = (3iˆ  ˆj  k)
ˆ  (2iˆ  ˆj  2k)
ˆ

given that | p  a |  15  | p  (3iˆ  ˆj  k)

ˆ |  15 or | (2iˆ  ˆj  2k)
ˆ |  15 
  |3| = 15  =±5  p  (3iˆ  ˆj  k)
ˆ  5(2iˆ  ˆj  2k)
ˆ  
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 45
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Hindi js[kk r  (3iˆ  ˆj  k)

ˆ  (2iˆ  ˆj  2k)
ˆ gSA ekuk fcUnq dk fLFkfr lfn'k p gS 
  p = (3iˆ  ˆj  k)
ˆ  (2iˆ  ˆj  2k)
ˆ fn;k gS fd | p  a |  15 
| p  (3iˆ  ˆj  k)
ˆ |  15 or | (2iˆ  ˆj  2k)
ˆ |  15  |3| = 15  = ± 5
 p  (3iˆ  ˆj  k)
ˆ  5(2iˆ  ˆj  2k)
ˆ

x 1
y 1 z x 1 y3 z 1
10. Acute angle between the lines = =and = = where  > m > n, and 
m n m n
, m, n are the roots of the cubic equation x3 + x2 – 4x = 4 is equal to :
x 1 y 1 z x 1 y3 z 1
js[kkvksa = = rFkk= = ds e/; U;wudks.k Kkr dhft;s tgka  > m > n,
m n m n
rFkk, m, n f=k?kkrh; lehdj.k x3 + x2 – 4x = 4 ds ewy gSaA
3 65 13 2
(A) cos–1 (B*) sin–1 (C*) 2cos–1 (D) tan–1
13 9 18 3

m  mn  n
Sol. Cos = .......(i)
 m2  n2
2

x3 + x2 – 4x – 4 = 0  +m+n=–1   m + mn + n= – 4
( + m + n)2 = 2 + m2 + n2 + 2 (– 4)  2 + m2 + n2 = 1 + 8 = 9 
4 4
  cos  = –  acute angle between the lines is cos–1
9 9
m  mn  n
Hindi. Cos = .......(i)
 m2  n2
2

x3 + x2 – 4x – 4 = 0  +m+n=–1  m + mn + n= – 4
( + m + n)2 = 2 + m2 + n2 + 2 (– 4)  2 + m2 + n2 = 1 + 8 = 9 
4 4
  cos  = –  js[kkvksa ds e/; U;wudks.k cos–1 gSA 
9 9

x  2 y 1 z 1
11. The line = = intersects the curve xy = c2, z = 0 if c is equal to : [16JM120053]
3 2 1
x  2 y 1 z 1
js[kk = = , oØ xy = c2, z = 0 dks izfrPNsn djrh gS rks c dk eku gS&
3 2 1
(A) –1 (B*) – 5 (C*) 5 (D) 1

Sol. Any pt. on line is (3 + 2, 2 – 1, 1 – )

but it lies on the curve xy = c2 & z = 0  (3 + 2) (2 – 1) = c2 & 1 –  = 0
 (3 + 2) (2 – 1) = c &  = 1
2
 c2 = 5 c = ± 5
Hindi js[kk ij dksbZ fcUnq (3 + 2, 2 – 1, 1 – ) gSA
ijUrq ;g oØ xy = c2 o z = 0 ij fLFkr gS  (3 + 2) (2 – 1) = c2 o 1–=0
 (3 + 2) (2 – 1) = c 2
o =1  c =5
2
c = ± 5

x  1 y  2 z  3 x  1 2y  4 3z  9 x  2 y  2 z  3
12. Three distinct lines   ,   ,   are
3 2 1 5 3 1 3 2 
concurrent the value of  may be :

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 46
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x  1 y  2 z  3 x  1 2y  4 3z  9 x  2 y  2 z  3
rhu fHkUu js[kk,¡   ,   ,   laxkeh gS] rks  dk
3 2 1 5 3 1 3 2 
eku gks ldrk gS
(A) 1 (B*) –1 (C) 2 (D) –2
Sol. First two lines pass through the point (1, 2, 3) the third line must pass through the same point therefor 
2 = 1   = ±1. But  cannot be 1 as the third line consids with the first.
izFke nks js[kk,¡ fcUnq (1, 2, 3) ls xqtjrh gSA rhljh js[kk Hkh blh fcUnq ls xqtjrh gS blfy, 2 = 1   = ±1 ijUrq
 dk eku 1 ugha gksxk D;ksafd izFke] rhljh js[kk ij laikrh gSA

x6 y  10 z  14
13. The line = = is the hypotenuse of an isosceles right angle triangle whose opposite
5 3 8
vertex is (7, 2, 4) Then the equation of remaining sides is/are -
x6 y  10 z  14
js[kk = = lef}ckgq ledks.k f=kHkqt dk d.kZ gS ftldk foijhr 'kh"kZ (7, 2, 4) gSA rc 'ks"k
5 3 8
Hkqtkvksa ds lehdj.k gksxsa &

x7 y2 z4 x7 y2 z4

(A*) = = (B*) = =
3 6 2 2 3 6
x7 y2 z4 x7 y2 z4
(C) = = (D) = =
3 6 2 2 3 6
Sol.
A

C
B(7,2,4)

P = (5k – 6, 3k – 10, 8k – 14)

Now vc BP . AC = 0
5(5k – 13) + 3 (3k – 12) + 8 (8k – 18) = 0
245 5
k= =
98 2
 13 5  98
P =  ,  ,6  , Now vc length BP dh yEckbZ =
 2 2  2
 13 5 98 5 3 98 8 98 
Now vc A,C =   . ,  ,6  , 
 2 98 2 2 98. 2 98 2 

A = (9, –1, 10)
C = (4, –4, 2)
x7 y2 z4
Equation of BC dk lehdj.k = =
3 6 2
x7 y2 z4
Equation of AC dk lehdj.k = =
2 3 6
14. Two lines are nks js[kk,sa gS
x 1 y 1 z 1 x 1 y 1 z 1
L1 :   ; L2 :  
1 2 2 2 1 2
Equation of line passing through (2, 1, 3) and equally inclined to L1 & L2 is/are

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 47
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

fcUnq (2, 1, 3) ls xqtjus okyh js[kk dk lehdj.k tks L1 rFkk L2 ls cjkcj >qdh gqbZ gS
x  2 y 1 z  3 x 3 y 2 z5
(A*)   (B*)  
2 2 3 1 1 2
x  2 y 1 z  3 x y 1 z  6
(C*)   (D*)  
1 1 3 2 2 3
Sol. Any line which is perpendicular to direction1 + 2, 2 + 1, 2 + 2 or 1 – 2, 2 – 1, 2 – 2 i.e., 3, 3, 4 or –1, 1, 0
is always equally inclined to L1 & L2.
dksbZ js[kk tks 1 + 2, 2 + 1, 2 + 2 ;k 1 – 2, 2 – 1, 2 – 2 vFkkZr~ 3, 3, 4 ;k –1, 1, 0 ds yEcor~ rFkk L1 rFkk L2 ds
lkFk cjkcj >qdh gqbZ gSA


15. Which of the followings is/are correct : [16JM120052]
fuEu esa ls dkSulk lgh gSA
(A*) The angle between the two straight lines r = 3 î – 2 ĵ + 4 k̂ +  (– 2 î + ĵ + 2 k̂ ) and
 4 
r = î + 3 ĵ – 2 k̂ +  (3 î – 2 ĵ + 6 k̂ ) is cos–1  
 21 
nks ljy js[kkvksa r = 3 î – 2 ĵ + 4 k̂ +  (– 2 î + ĵ + 2 k̂ ) vkSj
chp dks.k cos–1   gSaA
4
r = î + 3 ĵ – 2 k̂ +  (3 î – 2 ĵ + 6 k̂ ) ds

21 
ˆ (iˆ  r )  (r.j)
(B*) (r.i) ˆ ( ˆj  r )  (r.k)
ˆ (kˆ  r ) = 0 .
(C*) The force determined by the vector r = (1, 8, 7) is resolved along three mutually perpendicular
directions, one of which is in the direction of the vector a  2 ˆi  2 ˆj  kˆ . Then the vector component of
7 ˆ
the force r in the direction of the vector a is  (2i  2jˆ  k)
ˆ
3
lfn'k r = (1, 8, 7) }kjk fu:fir cy dks rhu ijLij yEcor~ fn'kkvksa esa fo;ksftr fd;k tkrk gS ftuesa ls ,d]
7
lfn'k a  2 ˆi  2 ˆj  kˆ dh fn'kk esa gSA rc lfn'k a dh fn'kk esa cy r dk lfn'k ?kVd  (2iˆ  2jˆ  k)
ˆ gSA
3
1
(D*) The cosine of the acute angle between any two diagonals of a cube is .
3
1
,d ?ku ds fdUgha nks fod.kksZ ds U;wu dks.k dk dksT;k (cosine) gS.
3
Sol. S1 : Angle between the given lines is = angle between the vectors ( 2iˆ  ˆj  2k)
ˆ and (3iˆ  2jˆ  6k)
ˆ

nh xbZ js[kkvksa ds chp dks.k = lfn'kksas (2iˆ  ˆj  2k)

ˆ vkSj (3iˆ  2jˆ  6k)
ˆ ds chp dk dks.k

( 2iˆ  ˆj  2k).(3i
ˆ ˆ  2jˆ  6k)
ˆ  4 
  cos = = cos–1  
ˆ ˆ ˆ ˆ ˆ
|  2i  j  2k | | 3i  2j  6k | ˆ  21 
S2 : Let (ekukfd) ( r  xiˆ  yjˆ  2kˆ ) r  xiˆ  yjˆ  2kˆ
given expression is x[ykˆ  zj]
ˆ  y[ziˆ  xk]
ˆ  z[xjˆ  yi]
ˆ 0

fn;k x;k O;atd x[ykˆ  zj]

ˆ  y[ziˆ  xk]
ˆ  z[xjˆ  yi]
ˆ  0 gSA

r.a
S3 : Component of r in direction of a = â
|a|

 (iˆ – 8ˆj  7k)

 ˆ .(2iˆ  2jˆ  k)
ˆ  (2iˆ  2ˆj  k)
 ˆ 21 ˆ
=  =  ˆ =  7 (2iˆ  2jˆ  k)
(2i  2jˆ  k) ˆ

 3 
 3 9 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 48
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

r.a
lfn'k a dh fn'kk esa lfn'k r dk ?kVd = â
|a|

 (iˆ – 8ˆj  7k)

 ˆ .(2iˆ  2jˆ  k)
ˆ  (2iˆ  2ˆj  k)
 ˆ 21 ˆ ˆ =  7 (2iˆ  2jˆ  k)
=  =  (2i  2jˆ  k) ˆ

 3 
 3 9 3

S4 : Dr's of diagonal BD = a, – a, a or 1,–1,1

Dr's of diagonal AF = – a, a, a or –1, 1, 1
1  1  1 1
Angle between above diagonals cos   =
3 3 3

fod.kZ BD ds fnd~ vuqikr = a, – a, a ;k 1,–1,1

fod.kZ AF ds fnd~ vuqikr = –a, a, a ;k –1, 1, 1

1  1  1 1
mijksDr fod.kksZ ds e/; dks.k cos   =
3 3 3

16. If the distance between points (, 5, 10) from the point of intersection of the line.
r = (2 î  ĵ + 2 k̂ ) +  (2 î  4 ĵ + 12 k̂ ) and plane r . ( î  ĵ + k̂ ) = 5 is 13 units, then value of  may be
;fn js[kk r = (2 î  ĵ + 2 k̂ ) +  (2 î  4 ĵ + 12 k̂ ) vkSj lery r . ( î  ĵ + k̂ ) = 5 ds izfrPNsn fcUnq dh fcUnq
(, 5, 10) ds e/; nwjh 13 bdkbZ gS rc  dk eku gks ldrk gS
80
(A) 1 (B*) – 1 (C) 4 (D*)
63
x  2 y 1 z2 x  2 y 1 z  2
Sol. Line :  =    = r (Let) 
2 4 12 1 2 6
  x=r+2  y = 2r – 1  z = 6r + 2
plane : x – y + z = 5  r + 2 – (2r – 1) + 6r + 2 = 5  5r = 0  r=0
 point of intersection is (2, – 1, 2)   ( – 2)2 + (5 + 1)2 + (10 – 2)2 = 169
80
1262 – 34 – 160 = 0  632 – 17 – 80 = 0   = – 1, .
63
x  2 y 1 z2 x  2 y 1 z  2
Hindi. js[kk :  =     =r (ekukfd) 
2 4 12 1 2 6
  x=r+2  y = 2r – 1  z = 6r + 2
lery : x – y + z = 5  r + 2 – (2r – 1) + 6r + 2 = 5  5r = 0  r=0
 izfrPNsn fcUnq (2, – 1, 2) gSA ( – 2) + (5 + 1) + (10 – 2) = 169
2 2 2

80
  1262 – 34 – 160 = 0  632 – 17 – 80 = 0   = – 1, .
63

17.  
A vector v =  a ˆj  bkˆ is coplanar with the vectors $i  $j  2 k$ and ˆi  2 ˆj  kˆ and is orthogonal to

the vector  2 ˆi  ˆj  kˆ . It is given that the length of projection of v along the vector ˆi  ˆj  kˆ is equal to
6 3 . Then the value of 2ab may be [16JM120054]
 
,d lfn'k v =  a ˆj  bkˆ lfn'kksa $i  $j  2 k$ vkSj ˆi  2 ˆj  kˆ ds lkFk leryh; gS tks lfn'k  2 ˆi  ˆj  kˆ ds
yEcor~ gSA ;fn fn;k x;k gS fd v ds iz{ksi dh yEckbZ lfn'k ˆi  ˆj  kˆ ds vuqfn'k 6 3 ds cjkcj gS rc 2ab
dk eku cjkcj gS
(A) 81 (B) 9 (C) –9 (D*) –81

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 49
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

( ˆi – ˆj ˆk )· V
Sol. V =  (–2iˆ  ˆj  k)
ˆ  (iˆ  ˆj – 2k)
ˆ  (iˆ – 2jˆ  k)
ˆ  9(– ˆj  k)
ˆ    6 3  =±1
| ˆi – ˆj  kˆ |

18. â and b̂ are two given unit vectors at right angle. The unit vector equally inclined with â , b̂ and
â × b̂ will be:
â ,oa bˆ nks ijLij yEcor~ bdkbZ lfn'k gSa rks â , bˆ vkSj â  bˆ ds lkFk leku dks.k cukus okyk bdkbZ lfn'k
gksxk&
(A*) –
1
3
aˆ  bˆ  aˆ  bˆ  (B*)
1
3
aˆ  bˆ  aˆ bˆ 
(C)
1
3
aˆ  bˆ  aˆ bˆ  (D) –
1
3
aˆ  bˆ  aˆ  bˆ 
. aˆ . bˆ ˆ
 (aˆ  b)
Sol. Let vector is   1aˆ  2bˆ  3 ( aˆ  bˆ ) also cos =  
 aˆ  bˆ  aˆ  bˆ

 . aˆ = . bˆ = .(aˆ  b)
ˆ [ â  bˆ = â b̂ sin90º = 1]

 1 = 2 = 3 = (let)   =  (aˆ  bˆ  aˆ  b)
ˆ

  aˆ  bˆ   2 aˆ  bˆ 
2
 =  aˆ 2  bˆ 2  aˆ  bˆ  2aˆ . bˆ  2bˆ . . aˆ =1

1 1
  1 1 1  1     =  (aˆ  bˆ  aˆ  b)
ˆ  
3 3

. aˆ . bˆ ˆ
 (aˆ  b)
Hindi ekuk vHkh"B lfn'k   1aˆ  2bˆ  3 ( aˆ  bˆ ) vkSj cos =  
 aˆ  bˆ  aˆ  bˆ

 . aˆ = . bˆ = .(aˆ  b)
ˆ [ â  bˆ = â b̂ sin90º = 1]

 1 = 2 = 3 = (let)   =  (aˆ  bˆ  aˆ  b)
ˆ

  aˆ  bˆ   2 aˆ  bˆ 
2
 =  aˆ 2  bˆ 2  aˆ  bˆ  2aˆ . bˆ  2bˆ . . aˆ =1

1 1
  1 1 1  1     =  (aˆ  bˆ  aˆ  b)
ˆ  
3 3


19. Let a = 2iˆ – ˆj  kˆ , b = ˆi  2jˆ – kˆ and c = ˆi  ˆj – 2kˆ be three vectors. A vector in the plane of b and c
2
whose length of projection on a is of , is [16JM120055]
3
ekuk a = 2iˆ – ˆj  kˆ , b = ˆi  2jˆ – kˆ rFkk c = ˆi  ˆj – 2kˆ rhu lfn'k gSA b ,oa c ds lery esa ,d lfn'k
2
ftldk a ij iz{ksi dh yEckbZ] ifjek.k dk gS] gksxk&
3
(A*) 2i + 3j – 3k (B) 2i + 3j + 3k (C*) – 2i – j + 5k (D) i – 5j + 3k

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 50
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x y z
Sol. Let r = xi + yj + zk, then [r b c] = 0  1 2 –1 = 0, –3x + y – z = 0 ....(1)
1 1 –2
r.a 2 2x – y  3 2
=   =±   2x – y + z = ± 2 ...(2)
|a| 3 6 3
from (1) and (2) x = 2 ;y – z = 6 there fore r = 2i + yj +(y ± 6)k (A) & (C) are answer
x y z
Hindi ekuk r = xi + yj + zk rc [r b c] = 0 1 2 –1 = 0, –3x + y – z = 0 ....(1)
1 1 –2
r.a 2 2x – y  3 2
=  =±  2x – y + z = ± 2 ...(2)
|a| 3 6 3
(1) o (2) ls x = 2 ; y–z= 6
blfy;s r= 2i + yj +(y ± 6)k  (A) o (C) lgh mÙkj gSA

20. ˆ , B(3iˆ  ˆj  5k)

The position vectors of the angular points of a tetrahedron are A(3iˆ  2jˆ  k) ˆ , C(4iˆ  3k)
ˆ
ˆ . Then the acute angle between the lateral face ADC and the base face ABC is :
and D(i)
5 2 2 5
(A*) tan–1 (B) tan–1 (C) cot–1 (D*) cot–1
2 5 5 2
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ gS] rks <kyq
fdlh prq"Qyd ds 'kh"kZ fcUnqvksa ds fLFkfr lfn'k A(3i  2j  k) , B(3i  j  5k) , C(4i  3k) rFkk D(i)
Qyd (lateral face) ADC vkSj vk/kkj Qyd ABC ds chp U;wu dks.k Kkr dhft, \
5 2 5 2
(A*) tan–1 (B) tan–1 (C) cot–1 (D*) cot–1
2 5 2 5

Sol. AD  2iˆ  2jˆ  kˆ , AC  ˆi  2jˆ  2kˆ

ˆi ˆj kˆ
 vector perpendicular to the face ADC is = 2 2 1 = 6iˆ  3ˆj  6kˆ  AB  3ˆj  4kˆ
1 2 2

ˆi ˆj kˆ
 A vector perpendicular to the face ABC is = 0 3 4 = 2iˆ  4jˆ  3kˆ
1 2 2

12  12  18 2
 acute angle between the two faces is given by cos  = =
36  9  36 4  16  9 29
5 5
 tan  =   = tan–1
2 2
Hindi AD  2iˆ  2jˆ  kˆ , AC  ˆi  2jˆ  2kˆ
ˆi ˆj kˆ
  Qyd ADC ds yEcor~ lfn'k = 2 2 1 = 6iˆ  3ˆj  6kˆ  AB  3ˆj  4kˆ
1 2 2

ˆi ˆj kˆ
 Qyd ABC ds yEcor~ lfn'k = 0 3 4 = 2iˆ  4jˆ  3kˆ
1 2 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 51
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

12  12  18 2
 nksuksa Qydksa ds chp U;wudks.k cos  = =
36  9  36 4  16  9 29
5 5
 tan  =   = tan–1
2 2

3
21. If a, b, c and d are unit vectors such that (a  b) . (c  d) =  and a . c = , then
2
[16JM120056]
(A*) a , b, c are coplanar if  = 1 (B) Angle between b and d is 30° if  = – 1
(C*) angle between b and d is 150° if  = – 1 (D) If  = 1 then angle between b and c is 60°
3
;fn bdkbZ lfn'k a, b, c ,oa d bl izdkj gS fd (a  b) . (c  d) =  rFkk a . c = , rc
2
(A*) a , b, c leryh; gksxsa ;fn  = 1 (B) b o d ds e/; dks.k 30° gksxk ;fn  = – 1
(C*) b ,oa d ds e/; dks.k 150° gksxk ;fn  = – 1 (D) ;fn  = 1 rc b o c ds e/; dks.k 60° gksxkA
Sol. If  = – 1 then a  b , c  d and angle between a  b , c  d is 
 between b and d = 360° – (90° + 90° + 30°) = 150°

If (a  b) . (c  d) = 1, then following figure is possible then  between b and d = 30°

Hindi ;fn  = – 1 rc a  b , c  d rFkk a  b , c  d ds e/; dks.k  gSA

 b o d ds e/; dks.k = 360° – (90° + 90° + 30°) = 150°

;fn (a  b) . (c  d) = 1,rc fuEu fp=k lEHko gS rc  b o d ds e/; dks.k = 30°

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 52
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

22. The volume of a right triangular prism ABCA1B1C1 is equal to 3. If the position vectors of the vertices of
the base ABC are A(1, 0, 1), B(2,0, 0) and C(0, 1, 0), then position vectors of the vertex A 1 can be:
,d ledks.k f=kHkqtkdkj fizTe ABCA1B1C1 dk vk;ru 3 gSA ;fn vk/kkj ABC ds 'kh"kks± ds fLFkfr lfn'k
A(1, 0, 1); B(2,0, 0) vkSj C(0, 1, 0) gSa] rks 'kh"kZ A1 dk fLFkfr lfn'k gks ldrk gS& [DRN2323]
(A*) (2, 2, 2) (B) (0, 2, 0) (C) (0,  2, 2) (D*) (0,  2, 0)
6
Sol. Volume of prism = Area of base ABC × height or 3 = h
2
 h= 6
Required point A1 should be just above point A
i.e. line AA1 is normal to plane ABC and AA1 = 6

Hindi fizTe dk vk;ru = vk/kkj ABC dk {kS=kQy × Å¡pkbZ

6
or 3= h  h= 6
2
vHkh"V fcUnq A1 , fcUnq A ds Bhd Åij gksuk pkfg,
vFkkZr~ js[kk AA1 lery ABC ds yEcor~ gS vkSj AA1 = 6

23. The coplanar points A , B , C , D are (2  x , 2 , 2) , (2 , 2  y , 2) , (2 , 2 , 2  z) and (1 , 1 , 1)

respectively, then [16JM120057]
1 1 1 x 1 y 1 z 1
(A*)   =1 (B*)   =2
x y z x y z
1 1 1 x y z
(C) + + =1 (D) + + +2=0
1 x 1 y 1 z 1 x 1 y 1  z
;fn pkj fcUnq A , B , C , D Øe'k% (2  x , 2 , 2) , (2 , 2  y , 2) , (2 , 2 , 2  z) vkSj (1 , 1 , 1) leryh; gSa] rks
&

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 53
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 1 1 x 1 y 1 z 1
(A*)   =1 (B*)   =2
x y z x y z
1 1 1 x y z
(C) + + =1 (D) + + +2=0
1 x 1 y 1 z 1 x 1 y 1 z
Sol. A (2  x , 2 , 2) , B (2 , 2  y , 2), C (2 , 2 , 2  z), D(1 , 1 , 1)
DA  (1  x) ˆi  ˆj  kˆ ; DB  ˆi  (1  y) ˆj  kˆ ; DC  ˆi  ˆj  (1  z) kˆ
If four points are coplanar then [DA, DB, DC] = 0
;fn pkj fcUnq leryh; gaS] rks [DA, DB, DC] = 0
1 x 1 1
 1 1 y 1 =0
1 1 1 z
c1  c1 – c2 and (rFkk) c2 c2 – c3
1 x 1 1
 1 1  y 1 =0
1 1 1 z
–x (–y + yz –z) + 1 (+yz) = 0 xy – xyz + xz + yz = 0
1 1 1
xy + yz + zx = xyz    = 1.
x y z

24. Which of the following statement(s) is/are correct :

(A*) If a,b,c are non–coplanar and d is any vector, then
[d b c] a  [d c a] b  [d a b] c  [a b c] d  0
(B*) If  is incentre of  ABC then | BC | A  | CA |  B | AB |  C = 0
(C*) Any vector in three dimension can be written as linear combination of three non–coplanar
vectors.
abc
(D) In a triangle, if position vector of vertices are a,b,c , then position vector of incentre is
3
.
fuEu esa ls dkSulk dFku lgh gS
(A*) ;fn a,b,c vleryh; gS rFkk d dksbZ lfn'k gS rc
[d b c] a  [d c a] b  [d a b] c  [a b c] d  0
(B*) ;fn   ABC dk vUr%dsUnz gS rc | BC | A | CA |  B | AB |  C = 0
(C*) f=kfoeh; esa dksbZ lfn'k] rhu vleryh; lfn'kksa ds ,d ?kkr lap; ds :i esa fy;k tk ldrk gSA
abc
(D) f=kHkqt esa ;fn 'kh"kksZa ds fLFkfr lfn'k a,b,c gS] rc vUrdsUnz dk fLFkfr lfn'k gSA
3
Sol. (A) [d b c] a + [d c a] b + [d a b] c – [a b c] d
= ([ b c d ] a [ b c a] d ) + ([d a b] c  [d a c] b)
= (b  c)  (a d ) + (d  a)  (c  b)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 54
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

= (d  a)  (b  c) – (d  a)  (b  c) = 0

aa  bb  cc
(B)  
abc

25. Let V = 2 ˆi  ˆj  kˆ & W = ˆi  3kˆ . If U is a unit vector, then the value of the scalar triple product
 U V W  may be : [16JM120058]
 
ekuk V = 2 ˆi  ˆj  kˆ rFkk W = ˆi  3kˆ gS ;fn U ,d bdkbZ lfn'k gS rc f=kd xq.ku  U V W  gks ldrk
gS
(A*) – 59 (B*) 10  6 (C*) 59 (D) 60
Hint : Use formula of scalar triple product vfn'k f=kd xq.ku dk mi;ksx
Sol. V  2iˆ  ˆj  kˆ  W = ˆi  3kˆ  [ u v w] = u . [ (2iˆ  ˆj  k)
ˆ × (iˆ  3k)
ˆ ]

= u . (3iˆ  7ˆj  k)
ˆ | u | | 3iˆ  7ˆj  kˆ | cos =  59  [uv w]  59

26. If A  B = a , A . a = 1 and A  B = b , then

;fn A  B = a , A . a = 1 rFkk A  B = b , rc


a  b  a | a |2 1 
(A*) A = a  b  a (B) B = 2
2
|a| |a|

b  a  a | a |2 1 
(C) A = b  a  a (D*) B = 2
2
|a| |a|
Sol. Given fn;k x;k gS A B  a .....(i)
 a.(A  B) = a.a 
   a. A  a . = a.a  1 + a.B = | a |2   a . B = |a| – 1
2
.....(ii)
Given fn;k x;k gS A B  b   
a A B = a ×b  

   (a.B) A – (a. A) B  a  b  (| a |2 1) A  B = a  b .....(iii)

[Using equation (ii)] lehdj.k (ii) ls

b  a  a | a |2 1 
solving equation (i) and (iii) simultaneously, we get A = a  b  a and B =
2
| a |2 |a|

x4 y2 z  k2
27. The line = = lies in the plane 2 x  4 y + z = 1 . Then the value of k
k 1 2
cannot be :
(A) 1 (B*)  1 (C*) 2 (D*) –2
x4 y2 zk 2
js[kk = = lery esa fLFkr 2 x  4 y + z = 1 gS] rc k dk eku ugha gks ldrk gS
k 1 2
(A) 1 (B*)  1 (C*) 2 (D*) –2
Sol. (4, 2, k 2 ) lies on 2x – 4y + z = 1  k = ±1
Also (kiˆ  ˆj  2k)
ˆ . (2iˆ  4ˆj  k)
ˆ 0  k=1
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 55
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(4, 2, k 2 ) lery ij fLFkr gSA 2x – 4y + z = 1  k = ±1

vr% (kiˆ  ˆj  2k)
ˆ . (2iˆ  4ˆj  k)
ˆ 0  k=1

x  x2 y  y2 z  z2
28. Equation of the plane passing through A(x 1, y1, z1) and containing the line = = is
d1 d2 d3

x  x2 y  y2 z  z2
js[kk = = dks lekfgr djus okys vkSj fcUnq A(x1, y1, z1) ls xqtjus okys lery dk
d1 d2 d3
lehdj.k gS &

x  x1 y  y1 z  z1 x  x2 y  y2 z  z2
(A*) x 2  x1 y 2  y1 z2  z1 = 0 (B*) x1  x 2 y1  y 2 z1  z2 = 0
d1 d2 d3 d1 d2 d3
x  d1 y  d2 z  d3 x y z
(C) x1 y1 z1 = 0 (D) x1  x 2 y1  y 2 z1  z2 = 0
x2 y2 z2 d1 d2 d3

Sol. Vectors AR, AB & C are coplanar

x – x1 y – y1 z – z1 x – x2 y – y2 z – z2
Equation of the required plane x 2 – x1 y 2 – y1 z2 – z1  0 or x1 – x 2 y1 – y 2 z1 – z2  0
d1 d2 d3 d1 d2 d3

x –1 y–2 z–3
29. A line = = intersects the plane x – y + 2z + 2 = 0 at point A. The equation of the
2 3 4
straight line passing through A lying in the given plane and at minimum inclination with the given line
is/are
x –1 y–2 z–3
,d js[kk = = lery x – y + 2z + 2 = 0 dks fcUnq A ij izfrPNsn djrh gSA A ls xqtjus
2 3 4
okyh ljy js[kk dk lehdj.k gS tks fn, x, lery esa fLFkr gS vkSj nh xbZ js[kk ds lkFk U;wure >qdko ij gSA
x 1 y 1 z 1
(A*) = = (B*) 5x – y + 4 = 0 = 2y – 5z – 3
1 5 2
x2 y6 z3
(C*) 5x + y – 5z + 1 = 0 = 2y – 5z – 3 (D*) = =
1 5 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 56
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol.

2r + 1 – (3r + 2) + 2 (4r + 3) + 2 = 0  7r + 7 = 0  r=–1  

  A(–1, –1, – 1) required line will be projection of given line in the plane foot of  of P will be on
D
(vHkh"V js[kk] nh xbZ js[kk dk lery esa iz{ksi gksxkA P ds yEc dk ikn D ij gksxkA)
x –1 y–2 z–3  1– 2  2.3  2  x –1 y – 2 z–3 –7
= = = – 2  ; = = =
–1  –1
 1  (–1)  2 
1 2 2 2 1 2 6
–1 19 4 x 1 y 1 z 1 x 1 y 1 z 1
x= ; y= ;z=  = =   1= +1 = +1
6 6 6 2 5 2 2 5 2
x 1 y 1 z 1 x 1 y 1 y 1 z 1
Also blfy,, = =  = and rFkk =
2 5 2 2 5 5 2
 5x – y + 4 = 0 = 2y – 5z – 3
alsovr% (5x – y + 4) + (2y – 5z – 3) = 0 = 2y – 5z – 3  5x + y – 5z + 1 = 0 = 2y – 5z – 3

30. If the -plane 7x + ( + 4)y + 4z – r = 0 passing through the points of intersection of the planes
2x  3y z +1= 0 and x  y  2z + 3 = 0 and is perpendicular to the plane 3x  y  2z = 4 and
 12 78 57 
 , ,  is image of point (1, 1, 1) in plane, then [16JM120059]
    
;fn -lery 7x + ( + 4)y + 4z – = 0 leryksa 2x  3y z +1= 0 vkSj x  y  2z + 3 = 0 ds izfrPNsn fcUnqvksa
ls xqtjrk gS vkSj lery 3x  y  2z = 4 ds yEcor gS rFkk fcUnq (1, 1, 1) dk lery esa izfrfcEc
 12 78 57 
 , ,  gS rc &
    
(A*)  = 9 (B)  = – 117 (C)  = – 9 (D*)  = 117
Sol. Let the plane is ekuk lery gS&
(2x  3y z) + 1 +  (x  y  2z + 3) = 0 ....(1)
(2 + ) x + (3 + ) y – (1 + 2) z + 1 + 3 = 0  3(2 + ) – (3 + ) + 2(1 + 2) = 0
+ 6 + 5 = 0   = – 5/6
Putting value of  in (1) (1) esa  dk eku j[kus ij
7x + 13y + 4z – 9 = 0  =9
Now image of (1, 1, 1) in plane is vc lery  esa (1,1,1) dk izfrfcEc gS
x 1 y 1 z 1  7  13  4  9  x 1 y 1 z 1 15
= = = –2   ; = = =–
7 13 4  49  169  16  7 13 4 117
12 78 57
x , y= , z=   = 117.
117 117 117

31. The planes 2x – 3y – 7z = 0, 3x – 14y – 13z = 0 and 8x – 31y – 33z = 0 [16JM120060]

(A*) pass through origin (B*) intersect in a common line
(C) form a triangular prism (D*) pass through infinite the many points
lery 2x – 3y – 7z = 0, 3x – 14y – 13z = 0 vkSj 8x – 31y – 33z = 0
(A) ewy fcUnq ls xqtjrs gSaA (B) ,d mHk;fu"B js[kk ij dkVrs gSaA
(C) ,d f=kdks.kh; fizTe cukrs gSaA (D) vuUr fcUnqvksa ls xqtjrs gaAS

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 57
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2x  3y  7z  0 

Sol. 3x  14y  13z  0  obviously all the three planes pass through origin
8x  31y  33z  0 
2 3 7
D = 3 14 13 = 2(462 – 403) + 3(– 99 + 104) – 7(– 93 + 112) = 118 + 15 – 133 = 0
8 31 33
From the theory of system of equations D = D1 = D2 = D3 = 0
 System of equations has infinite solutions
 hence three planes intersect in a common line
2x  3y  7z  0 

Hindi. 3x  14y  13z  0  Li"Vr;k rhuksa lery ewy fcUnq ls xqtjrs gSaA
8x  31y  33z  0 
2 3 7
D = 3 14 13 = 2(462 – 403) + 3(– 99 + 104) – 7(– 93 + 112) = 118 + 15 – 133 = 0
8 31 33
lehdj.k fudk;ksa ds fl)kUr ls D = D1 = D2 = D3 = 0  lehdj.kksa ds fudk; ds vuUr gy gSaA
 vr% rhuksa lery ,d mHk;fu"B js[kk ij dkVrs gSaA

32. If a,b,c and d are the position vectors of the points A, B, C and D respectively in three dimensional
space no three of A, B, C, D are collinear and satisfy the relation 3a  2b  c  2d = 0 , then :
(A*) A, B, C and D are coplanar [16JM120061]
(B) The line joining the points B and D divides the line joining the point A and C in the ratio 2 : 1.
(C*) The line joining the points A and C divides the line joining the points B and D in the ratio 1 : 1.
(D*) the four vectors a,b,c and d are linearly dependents.
;fn f=kfoe esa a,b,c rFkk d Øe'k% fcUnqvksa A, B, C ,oa D ds fLFkfr lfn'k gSaA fcUnqvksa A, B, C ,oa D esa ls dksbZ Hkh
rhu lajs[kh; ugha gS vkSj lEcU/k 3a  2b  c  2d = 0 dks lUrq"V djrs gSa rks
(A) A, B, C vkSj D leryh; gSA
(B) B vkSj D dks tksM+us okyh js[kk fcUnq A rFkk C dks tksM+us okyh js[kk dks 2 : 1 esa foHkkftr djrh gSA
(C) fcUnq A vkSj C dks tksM+us okyh js[kk B vkSj D dks tksM+us okyh js[kk dks 1 : 1 esa foHkkftr djrh gSA
(D) pkjksa lfn'k a,b,c vkSj d js[kh; ijrU=k gSaA
Sol. 3a  2b  c  2d  0 sum of coefficient = 0 
  a,b,c,d are coplanar
b  d 3a  c
Also 2b  2d = 3a  c  =
2 4

Hindi 3a  2b  c  2d  0
xq.kkadksa dk ;ksxQy = 0  a,b,c,d leryh; gSA
b  d 3a  c
iqu% 2b  2d = 3a  c  =
2 4

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 58
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

PART - IV : COMPREHENSION

Hkkx - IV : vuqPNsn (COMPREHENSION)

Comprehension # 1
vuqPNsn # 1
In a parallelogram OABC, vectors a, b, c are respectively the position vectors of vertices A, B, C with
reference to O as origin. A point E is taken on the side BC which divides it in the ratio of 2 : 1 internally.
Also, the line segment AE intersect the line bisecting the angle O internally in point P. If CP, when
extended meets AB in point F. Then
,d lekUrj prqHkqZt OABC esa 'kh"kksZ A, B, C ds fLFkfr lfn'k Øe'k% a, b, c gS rFkk O ewy fcUnq gSA BC Hkqtk ij
,d fcUnq E bl izdkj fy;k tkrk gS fd og bls 2 : 1 ds vuqikr esa vUr%foHkkftr djrk gS] lkFk gh js[kk [k.M AE
dks.k O dks vUr%lef}Hkkftr djus okyh js[kk dks P fcUnq ij feyrk gSA ;fn CP dks c<+k;k tk;sa rks og AB dks F
fcUnq ij feyrh gS] rc
1. The position vector of point P, is
fcUnq P dk fLFkfr lfn'k gS&

3a c a c 
  a c 

a c 
(A*)    (B)   
3c 2a 
a c 
 3c 2a 
a c 

2a c 
a c 
 3a c 
a c 

(C)    (D)   
3c 2a 
a c 
 3c 2a 
a c 


2. The position vector of point F, is

fcUnq F dk fLFkfr lfn'k gS&
1 a a 2a a
(A*) a  c (B) a  c (C) a  c (D) a  c
3 c c c c

3. The vector AF , is given by

lfn'k AF fuEu }kjk fn;k tkrk gS&
a a 2a 1 a
(A) – c (B) c (C) c (D*) c
c c c 3 c

Sol. for 1 to 3
2c  b
E =
3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 59
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 a c 
equation of OP r =     ...(1)
 a c 
 
 2c  b 
 µa  
Let P divide EA in µ : 1  P 3 
 µ 1 
 
P lies on (1)
1 E 2
C(c) B(b)

F
A(a)
O(0)

2c  b 3c  a
µa   a  µa   
3 c 3 =  a c
=     ac b   
µ 1  a c  µ 1  a c 
   
Comparing coefficient of a and c
1
µ
3 =  ...(2) ; and
1
=

...(3)
µ 1 a µ 1 c
divided (2) by (3)
1 c c 1
µ+ =  µ= 
3 a a 3

1  3 a c
Put in (3)  =  =
c 2 c c c 2 a

a 3

3 a c  a c 
So position vector of P r =    . Now for solution of 4 equation of AB
3 c 2 a  a c 

r = a  (b  a)  a  (c) ...(4)
 3|c|a 3|a|c 
equation of CP  r = c + µ  c
 3 c  2 | a | 3 | c | 2 a 
 
3 | c | a  3 | a | c 3 | c | c 2 | a | c   3 | c | a | a | c  3 | c | c 
r = c µ   r = c  µ   ..(5)
 3 | c | 2 | a |   3 | c | 2 | a | 
Comparing (4) and (5)
µ | a | 3µ | c | 3 c  2 a  µ | a | 3µ | c |
=1+   = ...(6)
3 | c | 2 | a | 3 c 2a
3c 2 a
µ=
3c
Put value of µ in equation 6
µ | a | 3 | c | | a | 3 | c | 1 | a |
=1+  =1+ =
3 | c | 2 | a | 3|c| 3|c|

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 60
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1|a|
So position vector of F is = a  c  Solution – 5
3|c|
1 |a| 1|a|
A F = p.v. of F – p.v. of A = a + c a= c
3 |c| 3 |c|
Sol. for 1 to 3
2c  b  a c 
E =  OP dk lehdj.k r =     ...(1)
3  a c 
 
ekuk P, EA dks µ : 1 esa foHkkftr djrk gSA
 2c  b  2c  b
 µa   µa   a 
3   P, (1) ij fLFkr gS  3 c
P =   
 µ 1  µ 1  a c 
 
 

3c  a
µa   a 
3 c
a  c  b  =     a rFkk c ds xq.kkad dh rqyuk djus ij
µ 1  a c 
 
1
µ
3 =  1 
...(2) rFkk = ...(3)
µ 1 a µ 1 c
(2) dks (3) ls foHkkftr djus ij
1 c c 1
µ+ =  µ= 
3 a a 3

1  3 a c
(3) esa j[kus ij  =  =
c 2 c c c 2 a

a 3

3 a c  a c 
blfy, P dk fLFkfr lfn'k  r=   
3 c 2 a  a c 

vc 4 ds gy ds fy, AB dk lehdj.k r = a  (b  a)  a  (c) ...(4)
CP dk lehdj.k
 3|c|a 3|a|c  3 | c | a  3 | a | c  3 | c | c  2 | a | c 
r =c + µ    c   r = c  µ  
 3 c  2 | a | 3 | c | 2 a   3 | c | 2 | a | 
 
 3 | c | a | a | c  3 | c | c 
r = c µ  ..(5)
 3 | c | 2 | a | 
(4) vkSj (5) dh rqyuk djus ij
µ | a | 3µ | c | 3 c  2 a  µ | a | 3µ | c |
=1+  = ...(6)
3 | c | 2 | a | 3 c 2a
3 c 2 a
µ=
3 c
lehdj.k 6 esa µ dk eku j[kus ij
µ | a | 3 | c | | a | 3 | c | 1 | a |
=1+   = 1 + =
3 | c | 2 | a | 3|c| 3|c|
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 61
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1|a|
blfy, F dk fLFkfr lfn'k = a  c  gy – 5
3|c|
1 |a| 1|a|
AF = p.v. of F – p.v. of A = a + c a = c
3 |c| 3 |c|

Comprehension # 2
Let a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 be two planes, where d1, d2 > 0. Then origin lies
in acute angle if a1a2 + b1b2 + c1c2 < 0 and origin lies in obtuse angle if a1a2 + b1b2 + c1c2 > 0.
Further point (x1, y1, z1) and origin both lie either in acute angle or in obtuse angle ,
if (a1x1 + b1y1 + c1z1 + d1) (a2x1 + b2y1 + c2z1 + d2) > 0, one of (x1, y1, z1) and origin lie in acute angle and
the other in obtuse angle, if (a1x1 + b1y1 + c1z1 + d1) (a2x1 + b2y1 + c2z1 + d2) < 0

4. Given the planes 2x + 3y – 4z + 7 = 0 and x – 2y + 3z – 5 = 0, if a point P is (1, – 2, 3) and O is origin,

then
(A) O and P both lie in acute angle between the planes
(B*) O and P both lie in obtuse angle between the planes
(C) O lies in acute angle, P lies in obtuse angle.
(D) O lies in obtue angle, P lies in acute angle.
Sol. (B)
Equation of the second plane is –x + 2y –3z + 5 = 0  2 (–1) + 3 . 2 + (– 4) (–3) > 0
 O lies in obtuse angle (2 × 1 + 3(–2) – 4 × 3 + 7) (–1 + 2 (–2) – 3 × 3 + 5)
= (2 – 6 – 12 + 7) (–1 – 4 – 9 + 5) > 0  P lies in obtuse angle.
5. Given the planes x + 2y – 3z + 5 = 0 and 2x + y + 3z + 1 = 0. If a point P is (2, –1, 2) and O is origin,
then
(A) O and P both lie in acute angle between the planes
(B) O and P both lie in obtuse angle between the planes
(C*) O lies in acute angle, P lies in obtuse angle.
(D) O lies in obtue angle, P lies in acute angle.
Sol. (C)
1 × 2 + 2 × 1 – 3 × 3 < 0 O lies in acute angle.
Also (2 + 2 (–1) – 3(2) + 5) (2 × 2 – 1 + 3 × 2 + 1) = (–1) (10) < 0  P lies in obtuse angle.

6. Given the planes x + 2y – 3z + 2 = 0 and x – 2y + 3z + 7 = 0, if the point P is (1, 2, 2) and O is origin,
then
(A*) O and P both lie in acute angle between the planes
(B) O and P both lie in obtuse angle between the planes
(C) O lies in acute angle, P lies in obtuse angle.
(D) O lies in obtue angle, P lies in acute angle.
Sol. (A)
1–4–9<0  O lies in acute angle.
Further (1 + 4 – 6 + 2) (1 – 4 + 6 + 7) > 0  The point P lies in acute angle.

vuqPNsn # 2
ekuk fd a1x + b1y + c1z + d1 = 0 rFkk a2x + b2y + c2z + d2 = 0 nks lery gSa tgk¡ d1, d2 > 0 gSA rc ewy fcUnq
U;wudks.k esa fLFkr gksxk ;fn a1a2 + b1b2 + c1c2 < 0 rFkk ewy fcUnq vf/kd dks.k esa fLFkr gksxk ;fn a1a2 + b1b2 +
c1c2 > 0 gSA
blds vykok fcUnq (x1, y1, z1) rFkk ewy fcUnq nksuksa] ;k rks U;wudks.k esa ;k vf/kd dks.k esa gksxsa ;fn
(a1x1 + b1y1 + c1z1 + d1) (a2x1 + b2y1 + c2z1 + d2) > 0,
iqu% fcUnq (x1, y1, z1) rFkk ewy fcUnq esa ls dksbZ ,d] U;wUkdks.k esa rFkk nwljk vf/kd dks.k esa gksxk] ;fn (a1x1 + b1y1
+ c1z1 + d1) (a2x1 + b2y1 + c2z1 + d2) < 0.

4. lery 2x + 3y – 4z + 7 = 0 rFkk x – 2y + 3z – 5 = 0 fn;s x;s gaS ;fn fcUnq P(1, – 2, 3) gS vkSj O ewy fcUnq gS,
rc&
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 62
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(A) O rFkk P nksuksa leryksa ds e/; U;wudks.k esa fLFkr gSa

(B*) O rFkk P nksuksa leryksa ds e/; vf/kd dks.k esa fLFkr gSa
(C) O U;wudks.k esa rFkk P vf/kd dks.k esa fLFkr gSa
(D) O vf/kd dks.k esa rFkk P U;wudks.k esa fLFkr gSa
Sol. (B)
nwljs lery dk lehdj.k –x + 2y –3z + 5 = 0 gS 2 (–1) + 3 . 2 + (– 4) (–3) > 0
 O vf/kd dks.k esa fLFkr gSA (2 × 1 + 3(–2) – 4 × 3 + 7) (–1 + 2 (–2) – 3 × 3 + 5)
= (2 – 6 – 12 + 7) (–1 – 4 – 9 + 5) > 0  P vf/kd dks.k esa fLFkr gSA
5. lery x + 2y – 3z + 5 = 0 rFkk 2x + y + 3z + 1 = 0 fn;s x;s gaSA ;fn ,d fcUnq P(2, –1, 2) gS vkSj O ewy fcUnq
gS, rc&
(A) O rFkk P nksuksa leryksa ds e/; U;wudks.k esa fLFkr gSaA
(B) O rFkk P nksuksa leryksa ds e/; vf/kd dks.k esa fLFkr gSaA
(C*) O U;wudks.k esa rFkk P vf/kd dks.k esa fLFkr gaSA
(D) O vf/kd dks.k esa rFkk P U;wudks.k esa fLFkr gaSA
Sol. (C)
1 × 2 + 2 × 1 – 3 × 3 < 0 O U;wudks.k esa fLFkr gSA
,oa (2 + 2 (–1) – 3(2) + 5) (2 × 2 – 1 + 3 × 2 + 1) = (–1) (10) < 0  P vf/kd dks.k esa fLFkr gSA
6. lery x + 2y – 3z + 2 = 0 rFkk x – 2y + 3z + 7 = 0 fn;s x,s gSa ;fn fcUnq P(1, 2, 2) gS vkSj O ewy fcUnq gS, rc&
(A*) O rFkk P nksuksa leryksa ds e/; U;wudks.k esa fLFkr gSa
(B) O rFkk P nksuksa leryksa ds e/; vf/kd dks.k esa fLFkr gSa
(C) O U;wudks.k esa rFkk P vf/kd dks.k esa fLFkr gaS
(D) O vf/kd dks.k esa rFkk P U;wudks.k esa fLFkr gSa
Sol. (A)
1–4–9<0  O U;wudks.k esa fLFkr gSA
iqu% (1 + 4 – 6 + 2) (1 – 4 + 6 + 7) > 0  fcUnq P U;wudks.k esa fLFkr gSA

Comprehension # 3
If a, b, c & a',b',c' are two sets of non-coplanar vectors such that a.a'=b.b'=c.c' = 1 , then the two
b x c c x a a x b
systems are called Reciprocal System of vectors and a= , b  and c  .
[a b c ] [a b c ] [a b c ]
;fn a, b, c o a',b',c' nks vleryh; lfn'k ds nks leqPp; bl izdkj gS fd a.a'=b.b'=c.c' = 1 , rc nksuksa fudk;ks
b x c c x a a x b
dks lfn'kkas dk O;qRØe fudk; dgrs gSA rFkk a= , b  vkSj c  .
[a b c ] [a b c ] [a b c ]
7. Find the value of a  a  b  b c  c  .
a  a  b  b c  c  dk eku gksxk
(A*) 0 (B) a  b  c (C) a  b  c (D) a  b  c

1
Sol. We have ge tkurs gS : a =  (b  c) , b =  (c  a) and rFkk c  =  (a  b) , where tgk¡  =
[a b c]
a  a  a  (b  c)  {a  (b  c)}  {(a . c) b  (a . b) c}
b  b  b  (c  a)  {b  (c  a)}  {(b . a) c  (b . c) a}
rFkk and c  c  c  (a  b)  {c  (a  b)}  {(c . b) a  (c . a) b}
 a  a  b  b  c  c =  {(a . c) b  (a . b) c}  {(b . a) c  (b . c) a}  {(c . b) a  (c . a) b}
=  [(a . c) b  (a . b) c  (b . a) c  (b . c) a  (c . b) a  (c . a) b] 
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 63
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

=  [(a . c) b  (a . b) c  (a . b) c  (b . c) a  (b . c) a  (a . c) b] = 0  0

abc
8. Find value of  such that a  b + b  c  + c   a =  .
[abc]
abc
 dk eku Kkr dhft, tcfd a  b + b  c + c  a =  .
[abc]

(A) – 1 (B*) 1 (C) 2 (D) – 2

(b  c)  (c  a) c abc
Sol. a  b = 2
=  a  b + b  c  + c   a = so blfy,  = 1
[abc] [abc] [abc]

9. If [(a × b) × (b × c) (b × c) × (c × a) (c × a) × (a × b)] = [abc] n , then find n.
;fn [(a × b) × (b × c) (b × c) × (c × a) (c × a) × (a × b)] = [abc] n, gks rks n dk eku Kkr dhft,A
(A*) n = – 4 (B) n = 4 (C) n = – 3 (D) n = 3
ca  c  a a  b b  c  [abc] 2
Sol. (a × b) × (b × c) = 2
  2 2 2
= 6
= [abc]4  n=–4
[abc]  [abc] [abc] [abc]  [abc]

Comprehension # 4
The vertices of square pyramid are A(0, 0, 0), B(4, 0, 0), C(4, 0, 4), D(0, 0, 4) and E(2, 6, 6)

10. Volume of the pyramid is : [16JM120062]

(A*) 32 (B) 16 (C) 8 (D) 4

11. Centroids of triangular faces of square pyramid are

(A) Non-coplanar (B) Coplanar but the plane is not parallel to base plane
(C*) Coplanar & plane is parallel to base plane (D) Co-linear

12. The distance of the plane EBC from ortho-centre of ABD is : [16JM120064]
12
(A) 2 (B) 5 (C*) (D) 10
10
1 1
Sol. (10) Volume of pyramid = × base area × height = × 16 × 6 = 32
3 3
 10 10   14 
(11) Co-ordinates of centroid of faces EAB, EBC, ECD & EDA are (2, 2, 2)  ,2,  ,  2,2,  ,
 3 3   3 
 2 10 
 3 ,2, 3  . y-co-ordinate of each point is 2. hence these points are co-planar & plane is parallel
 
to base plane.
x4 y z
(12) ortho-centre of ABD is (0, 0, 0) equation of plane EBC is 2 6 6 = 0
0 0 4
4.((x – 4).6 + 2y) = 0  6x – 24 + 2y = 0  3x + y – 12 = 0
0  0  21 12
d= =
3 1
2 10

vuqPNsn # 4

oxZ fijkfeM ds 'kh"kZ A(0, 0, 0), B(4, 0, 0), C(4, 0, 4), D(0, 0, 4) vkSj E(2, 6, 6) gSA
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 64
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

10. fijkfeM dk vk;ru

(A*) 32 (B) 16 (C) 8 (D) 4

11. oxZ fijkfeM ds f=kHkqtkdkj Qydksa dk dsUnzd gS &

(A) vleryh;
(B) leryh; ijUrq lery] vk/kkj lery ds lekUrj ugha gSA
(C) leryh; vkSj lery] vk/kkj lery ds lekUrj gSA (D) lajs[kh;

12. ABD ds yEcdsUnz ls lery EBC dh nwjh gS &

12
(A) 2 (B) 5 (C*) (D) 10
10
1 1
Sol. (10) fijkfeM dk vk;ru = × vk/kkj dk {ks=kQy × Å¡pkbZ = × 16 × 6 = 32
3 3
 10 10   14   2 10 
(11) Qydksa EAB, EBC, ECD rFkk EDA ds dsUnzd Øe'k% (2, 2, 2)  ,2,  ,  2,2,  ,  ,2, 
 3 3   3  3 3 
gksxsaA izR;sd fcUnq dk y-funsZ'kkad 2 gS vr% ;s fcUnq leryh; gS rFkk lery vk/kkj lery ds lekUrj gSA
x4 y z
(12) ABD dk yEcdsUnz (0, 0, 0) gSA lery EBC dk lehdj.k 2 6 6 = 0 gSA
0 0 4
4.((x – 4).6 + 2y) = 0  6x – 24 + 2y = 0  3x + y – 12 = 0
0  0  21 12
d= =
3 1
2 10

Comprehension # 5
General equation of a sphere is given by x 2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, where (u, v, w) is
the centre and u2  v 2  w 2  d is the radius of the sphere.
Let P be a any plane and F is the foot of perpendicular from centre(C) of the sphere to this plane.
If CF > u2  v 2  w 2  d then plane P neither touches nor cuts the sphere.
If CF = u2  v 2  w 2  d then plane P touches the sphere.
If CF < u2  v 2  w 2  d then intersection of plane P and sphere is a circle with
u2  v 2  w 2 – d –  CF 
2
radius =
vuqPNsn # 5
xksys dk lehdj.k tks x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 ls fn;k tkrk gS] tgk¡ dsUnz (u, v, w) vkSj
xksys fd f=kT;k u2  v 2  w 2  d gksrh gSA
ekuk P dksbZ lery gS rFkk F, xksys ds dsUnz ls lery ij yEcikn ds funs’'Z kkad gSA
;fn CF > u2  v 2  w 2  d rc lery P, xksys dks u rks Li'kZ djrk gS vkSj u gh izfrPNsn djrk gSA
;fn CF = u2  v 2  w 2  d rc lery P, xksys dks Li'kZ djrk gSA
;fn CF < u2  v 2  w 2  d rc lery P, rFkk xksys ds izfrPNsnu ls ,d o`Ùk gS ftldh
9f=kT;k = u2  v 2  w 2 – d – CF 
2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 65
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

13. Find the equation of the sphere having centre at (1, 2, 3) and touching the plane
x + 2y + 3z = 0.
lery x + 2y + 3z = 0 dks Li'kZ djus okys ml xksys dk lehdj.k Kkr dhft, ftldk dsUnz (1, 2, 3) gSA
(A*) x2 + y2 + z2 – 2x – 4y – 6z = 0
(B) x2 + y2 + z2 – 2x + 4y – 6z = 0
(C) x2 + y2 + z2 – 2x – 4y + 6z = 0
(D) x2 + y2 + z2 + 2x – 4y – 6z = 0
Solution : Given plane is x + 2y + 3z = 0 ..... (1)
Let H be the centre of the required sphere.

P
Given H  (1, 2, 3)
Radius of the sphere,
HP = length of perpendicular from H to plane (1)
| 1 2  2  3  3 |
= = 14
14
Equation of the required sphere is (x – 1)2 + (y – 2)2 + (z – 3)2 = 14
or x2 + y2 + z2 – 2x – 4y – 6z = 0
gy fn;k x;k lery x + 2y + 3z = 0 gS ..... (1)
ekuk vHkh"V xksys dk dsUnz H gSA

P
fn;k gS] H  (1, 2, 3)
xksys dh f=kT;k HP,
| 1 2  2  3  3 |
= H dh lery (1) ls yEcor nwjh = = 14
14
vr% vHkh"V xksys dk lehdj.k gksxkA
(x – 1)2 + (y – 2)2 + (z – 3)2 = 14 ;k x2 + y2 + z2 – 2x – 4y – 6z = 0

14. A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn
from origin to this plane is:
(A*) x2 + y2 + z2  x  2y  3z = 0 (B) x2 + 2y2 + 3z2  x  2y  3z = 0
(C) x + 4y + 9z + x + 2y + 3 = 0
2 2 2
(D) x2 + y2 + z2 + x + 2y + 3z = 0
,d pj lery ,d fLFkj fcUnq (1, 2, 3) ls xqtjrk gSA ewy fcUnq ls bl lery ij Mkys x, yEc ds ikn dk
fcUnqiFk gS &
(A) x2 + y2 + z2  x  2y  3z = 0 (B) x2 + 2y2 + 3z2  x  2y  3z = 0
(C) x2 + 4y2 + 9z2 + x + 2y + 3 = 0 (D) x2 + y2 + z2 + x + 2y + 3z = 0
Sol. OP  AP
( – 1) + ( – 2) + ( – 3) = 0
 Locus of P(, , ) is
 P (, , ) dk fcUnqiFk gS
x2 + y2 + z2 – x – 2y – 3z = 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 66
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x 1 y  2 z  3
15. Find the length of the chord intercepted on the line   by the sphere
1 2 3
22
x2 + y2 + z2 – 2x – 2y – z = 0.
3
x 1 y  2 z  3 22
js[kk   }kjk xksys x2 + y2 + z2 – 2x – 2y – z = 0. ij dkVh x;h thok dh yEckbZ Kkr
1 2 3 3
dhft,A
(A*) 56 (B) 54 (C) 9 (D) 6
x y z
Sol. The given line can be written as   = x = , y = 2, z = 3
1 2 3
subtitute these values in sphere
22
2 + 42 + 92 – 2 – 4 –  3 = 0   = 0, 2
3
so the points are (0, 0, 0) and (2, 4, 6) so length of chord = 56

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 67
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

* Marked Questions may have more than one correct option.

* fpfUgr iz'u ,d ls vf/kd lgh fodYi okys iz'u gS -

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

Hkkx - I : JEE (ADVANCED) / IIT-JEE ¼fiNys o"kksZ½ ds iz'u

1. Let P, Q, R and S be the points on the plane with position vectors – 2 î – ĵ , 4 î , 3 î + 3 ĵ and
– 3 î + 2 ĵ respectively. The quadrilateral PQRS must be a [IIT-JEE-2010, Paper-1, (3, –1), 84]
(A*) parallelogram, which is neither a rhombus nor a rectangle
(B) square
(C) rectangle, but not a square
(D) rhombus, but not a square
ekuk fd ,d lery esa fcUnq P, Q, R rFkk S gSaA ftuds fLFkfr lfn'k Øe'k% – 2 î – ĵ , 4 î , 3 î + 3 ĵ rFkk
– 3 î + 2 ĵgSaA rks prqHkqZt PQRS gksxk
(A) ,d lekukUrj prqHkqZt tks u leprqHkZqt gS vkSj u gh ,d vk;r gS
(B) ,d oxZ
(C) ,d vk;r ijUrq oxZ ugha
(D) ,d leprqHkZqt ijUrq oxZ ugha
Sol. PQ = 36  1 = 37 = RS,
PS = 1  9 = 10 = QR,
PQ  PS
1
slope of PQ = , slope of PS = – 3
6

PQ is not  to PS
So it is parallelogram, which is neither a rhombus nor a rectangle

Hindi PQ = 36  1 = 37 = RS,
PS = 1  9 = 10 = QR,
PQ  PS
1
PQ dh izo.krk = , PS = – 3 dh izo.krk
6

PQ, PS dsyEcor~ ugha gS

blfy, ;g lekUrj prqHkqZt tks u rks leprqHkqZt vkSj u gh vk;r

ˆi – 2ˆj 2iˆ  ˆj  3kˆ

2. If a and b are vectors in space given by a = and b = , then the value of
5 14
 2a  b . a  b  a – 2b is [IIT-JEE-2010, Paper-1, (3, 0), 84]

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 68
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ˆi – 2ˆj 2iˆ  ˆj  3kˆ

;fn rFkk f=kfoe (space) esa nks lfn'k bl izdkj gS fd] a = rFkk b = rc
5 14
 2a  b . a  b  a – 2b dk eku Kkr dhft,A
Ans. 5
ˆi – 2ˆj 2iˆ  ˆj  3kˆ
Sol. a = , b =
5 14
|a | = 1 , |b | = 1
a . b =0

(2a  b) .  (a  b) (a – 2b) 

= (2a  b) .  (a .(a – 2b)) b – (b.(a – 2b) ) a 

= (2a  b) .  (a. a – 2a . b) b – (b. a – 2b. b) a 

= (2a  b) .  (1– 0) b – (0  2) a 
= (2a  b) .  b  2a 
= 2 (a . b)  4(a . a)  b. b  2(b. a) = 0 + 4 + 1 + 0 = 5

3. Two adjacent sides of a parallelogram ABCD are given by [IIT-JEE-2010, Paper-2, (5, –2), 79]
ˆ ˆ ˆ ˆ ˆ ˆ
AB = 2i  10j  11k and AD =  i  2j  2k . The side AD is rotated by an acute angle  in the plane
of the parallelogram so that AD becomes AD . If AD makes a right angle with the side AB, then the
cosine of the angle  is given by
,d lekukUrj prqHkqZt ABCD dh nks layXu Hkqtk,¡ fuEu izdkj ls nh x;h gSaA
AB = 2iˆ  10jˆ  11kˆ rFkk AD = ˆi  2jˆ  2kˆ lekukUrj prqHkqZt ds lery esa Hkqtk AD dks ,d U;wu dks.k  ls
?kqek;k tkrk gS ftlls fd AD, AD cu tkrh gSA ;fn AD , AB ds lkFk ledks.k cukrh gS] rks cos () dk eku
fuEu gS

8 17 1 4 5
(A) (B*) (C) (D)
9 9 9 9
2  20  22 8
Sol. cos  = = [Using dot product] [vfn'k xq.ku ls ]
15  3 9
 +  = 90º
 = 90º – 

17
cos  = sin  =
9

x y z
4. =
Equation of the plane containing the straight line = and perpendicular to the plane
2 3 4
x y z x y z
containing the straight lines = = and = = is
3 4 2 4 2 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 69
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x y z x y z x y z
ljy js[kk = = dks vUrZfo"V djus okys rFkk] js[kkvksa = = ,oa = = dks vUrZfo"V
2 3 4 3 4 2 4 2 3
djus okys (containing) lery ds yEcor~ lery dk lehdj.k fuEu gS [IIT-JEE-2010,
Paper-1, (3, –1), 84]
(A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0 (C*) x – 2y + z = 0 (D) 5x + 2y – 4z = 0
Sol. Direction ratio's of normal to plane containing the straight line
js[kk dks lekfgr djus okys lery ds vfHkyEc ds fn~dvuqikr
ˆi ˆj kˆ
3 4 2 = 8 î – ĵ – 10 k̂
4 2 3
x 0 y 0 z0
Required planevHkh"V lery 2 3 4 = 0 – 26x + 52y – 26z = 0  x – 2y + z = 0
8 1 10

x  1
5. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A  y  = 0 
 
 z  0 
has exactly two distinct solutions, is

 x   1
,sls 3 × 3 ds vkO;wgksa A dh la[;k] ftudh izfof"V;k¡ 0 ;k 1 gSa rFkk ftuds fy, fudk; A  y  = 0  ds Bhd
 z  0 
(exactly) nks fHkUu gy gSa] fuEu gS [IIT-JEE-2010, Paper-1, (3, –1), 84]

(A*) 0 (B) 29 – 1 (C) 168 (D) 2

Sol. a1x + b1y + c1z = 1
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0
No three planes can meet at two distinct points. So number of matrices is 0
dksbZ Hkh rhu lery fdUgh nks fHkUu fcUnqvksa ij ugha fey ldrs blfy, vkO;wgksa dh la[;k 0 gSA

6. If the distance between the plane Ax – 2y + z = d and the plane containing the lines
x –1 y–2 z–3 x–2 y–3 z–4
= = and = = is 6 , then |d| is
2 3 4 3 4 5
x –1 y–2 z–3 x–2 y–3 z–4
;fn lery Ax – 2y + z = d rFkk js[kkvksa = = rFkk = = dks
2 3 4 3 4 5
vUrfoZ"V djus okys lery ds chp dh nwjh 6 gks] rks |d| dk eku gSA [IIT-JEE-2010, Paper-1, (3, 0), 84]
Ans. 6
x –1 y – 2 z – 3
Sol. Equation of plane is 2 3 4 =0
3 4 5
x – 2y + z = 0 ..........(1)
Ax – 2y + z = d ..........(2)
A –2 1
Compare = =  A = 1
1 –2 1
d
Distance between planes is  6
1 1 4
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 70
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 |d| = 6

x –1 y – 2 z – 3
Hindi lery dk lehdj.k 2 3 4 =0
3 4 5
x – 2y + z = 0 ..........(1)
Ax – 2y + z = d ..........(2)
A –2 1
rqyuk djus ij = =  A = 1
1 –2 1
d
leryksa ds e/; nwjh  6  |d| = 6
1 1 4

7. If the distance of the point P(1, –2, 1) from the plane x + 2y – 2z = , where  > 0, is 5, then the foot of
the perpendicular from P to the plane is
;fn fcUnq P(1, –2, 1) ls lery x + 2y – 2z = , tgk¡  > 0 dh nwjh 5 gS] rks fcUnq P ls lery ij Mkys x, yEc
dk ikn (foot) fuEu gS [IIT-JEE-2010, Paper-2, (5, –2), 79]
8 4 7 4 4 1  1 2 10  2 1 5
(A*)  , ,   (B)  ,  ,  (C)  , ,  (D)  ,  , 
 3 3 3   3 3 3   3 3 3   3 3 2
1 4  2  
Sol. D= =5
3
 + 5 = 15 (  > 0)
  = 10
 plane is x + 2y – 2z – 10 = 0
Let foot of perpendicular is (, , )
 1   2  1  1  4  2  10  5 8 4 7
= = =–   =  = ,  = ,  = –
1 2 2  9  3 3 3 3

1 4  2  
Hindi D= =5
3
 + 5 = 15 (  > 0)
  = 10
 lery x + 2y – 2z – 10 = 0
ekuk yEcikn (, , ) gSA
 1   2  1  1  4  2  10  5 8 4 7
= = =–   =  = ,= ,=–
1 2 2  9  3 3 3 3

8. Match the statements in Column-I with those in Column-II. [IIT-JEE-2010, Paper-2, (8, 0), 79]

Column-I Column-II

(A) A line from the origin meets the lines (p) –4

8
x–
x – 2 y –1 z 1 3 = y 3 = z –1
= = and
1 –2 1 2 –1 1
at P and Q respectively. If length PQ = d, then d2 is

(B) The values of x satisfying (q) 0

3
tan–1(x + 3) – tan–1(x – 3) = sin–1   are
5
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 71
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(C) Non-zero vectors a , b and c satisfy a . b = 0, (r) 4

( b – a).(b  c)  0 and 2 | b  c | | b – a | . If a  µb  4c
then possible value of µ are

(D) Let f be the function on [–, ] given by (s) 5

 9x 
sin  
f(0) = 9 and f(x) =  2  for x  0. The value (t) 6
x
sin  
2

2
 –
of f(x) dx is

dkWye -I esa fn, x, oDrO;ksa dk dkWye -II esa fn, oDrO;ksa ls lqesy djsaA
dkWye -I dkWye -II

(A) ewy fcUnq ls [khaph x;h ,d ljy js[kk] ljy js[kkvksa (p) –4
8
x–
x – 2 y –1 z 1 3 = y 3 = z –1
= = rFkk
1 –2 1 2 –1 1
dks Øe'k% P ,oa Q ij dkVrh gSA ;fn yEckbZ PQ = d
gks rks d2 gS&

3
(B) tan–1(x + 3) – tan–1(x – 3) = sin–1   dks larq"V djus okys (q) 0
5
x ds eku gSa&

(C) rhu v'kqU; lfn'k a , b rFkk c lehdj.k a . b = 0, (r) 4

( b – a).(b  c)  0 rFkk 2 | b  c | | b – a | dks larq"V djrs gSaA
;fn a  µb  4c gks] rks  ds lEHko eku gS&
 9x 
sin  
(D) ekukfd [–, ] ij Qyu f ; f(0) = 9 rFkk f(x) =  2  (s) 5
x
sin  
2

2
 –
for x  0. The value of f(x) dx dk eku gS& (t) 6

Ans. (A)  (t), (B)  (p, r), (C)  (q) (JEE given q, s) (D)  (r)
x y z
Sol. (A) Let the line through origin is = =
 µ 1
 x = z , y = µz ...........(1)
x–2 y –1 z 1
To find point of intersection of line (1) and line = = ..........(2)
1 –2 1
z – 2 µz – 1
we have = =z+1
1 –2
3
 z= =
 –1
 + 3µ + 5 = 0 ..........(3)
To find point of intersection of line (1) and line = = ........(4)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 72
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

8
z –
we have 3 = z  3 = z – 1
2 –1 1
2 –2
 z= =
3( – 2) µ 1
 3+ µ = 5 ............(5)
5 5
Solving (3) and (5),  = and µ = –
2 2
 z = 2, x = 5, y = – 5 for point P
4 10 10
and z = , x = , y= – for point Q
3 3 3
4 25 25
 PQ2 = + + =6
9 9 9

(B) tan–1 (x + 3) – tan–1 (x – 3) = sin–1 (3/5)

 x  3 – x  3
 tan–1  –1
 = tan (3/4)
 1 x – 9 
2

6 3
 2 =  x2 = 16
x –8 4
 x=±4

(C) Since . a b = 0
 Let b = 1 î , a = 2 ĵ
 2 ĵ – 1 µb
Now 2| b + c | = | b – a | & a = µ b + 4 c  2 1ˆi  = | 1ˆi – 2 ˆj |
4
| 1(4 – µ) ˆi  2 ˆj | = 2 | 1 ˆi  2 ˆj |
 squaring

1 (4 – µ)2  2 = 41  42  3 2 = (12 + µ2 – 8µ) 1 .........(1)

2 2 2 2 2 2

Also ( b – a ).( b + c ) = 0
  ˆj – 1 µiˆ  1 (4 – µ) –  2
2 2

 (1ˆi – 2 ˆj) .  1 î  2  = 0  =0

 4  4
 2  1 (4 – µ) ..............(2)
2 2

from (1) & (2)

12 + µ2 – 8µ = 12 – 3µ
 µ2 – 5µ = 0  µ = 0, 5 (µ = 5 reject)

9x 9x x 9x x
 sin  sin cos  sin cos
2 4 8
  0 
(D) = 2 dx = 2 2 dx = 2 2 dx
 
x x x  sin x
sin sin cos 0
2 2 2

4 sin5x  sin 4x
=
 
0
sin x
dx ......(i)

b b
(using  f(x)dx
0
=  f(a  b – x)dx )
0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 73
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry


4 sin5x  sin4x
 0
= dx ......(ii)
sin x
Add (i) and (ii)

4 sin5x
= 
 0 sin x
dx

Consider
 
4 sinkx  sin(k  2) x 8 cos(k  1)x sin x
k – k–2 =
 
0
sin x
dx =
 
0
sin x
dx

k = k–2

4
so 5 = 3  5 = 1 =
 
0
dx = 4

Aliter

2 sin (9x / 2)
Let  =
 
–
sin(x / 2)
dx

4 sin(9x / 2)
=
 
0
sin(x / 2)
dx .......(1) ( f(x) is even function)

4 cos(9x / 2)
=
 
0
cos(x / 2)
dx .......(2)
b b
(using  f(x)dx =  f(a  b – x)dx
0 0
)

Add (1) & (2)

 
4 sin5x 4 sin5x
=
 
0
2sin(x / 2)cos(x / 2)
dx =
 
0
sin x
dx
/2
8 sin5x
  =
 
0
sin x
dx

/2
8  16 sin5 x – 20 sin3 x  5 sin x 
 =
 
0

 sin x
 dx

/2
8
 =
 
0
(16sin4 x – 20sin2 x  5 ) dx

8  3  1  1  5 
  = 16x – 20  
  422 2  2 2 
8  5 
  = 3  – 5  
  2 
  =4

x y z
Hindi (A) ekuk ewy fcUnq ls tkus okyh js[kk = =
 µ 1
 x = z , y = µz ...........(1)
x–2 y –1 z 1
js[kk (1) vkSj js[kk = = ds izfrPNsnu fcUnq ds fy, ..........(2)
1 –2 1
z – 2 µz – 1
= =z+1
1 –2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 74
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

3
 z= =
 –1
 + 3µ + 5 = 0 ..........(3)

8
x–
js[kk (1) vkSj js[kk 3 = y3 = z –1
ds izfrPNsnu ds fy, ........(4)
2 –1 1
8
z –
3 = z  3 =
z –1
2 –1 1
2 –2
 z= =
3( – 2) µ 1
 3+ µ = 5 ............(5)

5 5
(3) vkSj (5) ls ,  = vkSj µ = –
2 2
 fcUnq P ds fy, z = 2, x = 5, y = – 5
4 10 10
vkSj z = , x = , y= –
3 3 3
4 25 25
 vkSj fcUnq Q ds fy, PQ2 = + + =6
9 9 9

(B) tan–1 (x + 3) – tan–1 (x – 3) = sin–1 (3/5)

 x  3 – x  3
 tan–1  –1
 = tan (3/4)
 1 x – 9 
2

6 3
 2 =  x2 = 16
x –8 4
 x=±4

(C) pawfd a b = 0
 ekuk b = 1 î , a = 2 ĵ
 2 ĵ – 1 µb
vc 2| b + c | = | b – a | vkSj a = µ b + 4 c  2 1ˆi  = | 1ˆi – 2 ˆj |
4
| 1(4 – µ) ˆi  2 ˆj | = 2 | 1 ˆi  2 ˆj |
 oxZ djus ij
1 (4 – µ)2  2 = 41  42  3 2 = (12 + µ2 – 8µ) 1 .........(1)
2 2 2 2 2 2

rFkk ( b – a ).( b + c ) = 0
  ˆj – 1 µiˆ  1 (4 – µ) –  2
2 2

 (1ˆi – 2 ˆj) .  1 î  2  = 0  =0

 4  4
 2  1 (4 – µ)
2 2
..............(2)
(1) o (2) ls
12 + µ2 – 8µ = 12 – 3µ
 µ2 – 5µ = 0  µ = 0, 5 (µ = 5 reject)

9x 9x x 9x x
 sin  sin cos  sin cos
2 2 dx = 4 8
  
(D) = 2 2 dx = 2 2 dx
 
x  x x  sin x
sin 0 sin cos 0
2 2 2
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 75
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry


4 sin5x  sin 4x
=
 
0
sin x
dx ......(i)

b b
(  f(x)dx =  f(a  b – x)dx ds mi;ksx ls)
0 0

4 sin5x  sin4x
=
 
0
sin x
dx ......(ii)

(i) vkSj (ii) dks tksM+us ij


4 sin5x
=
 
0
sin x
dx

ekukfd
 
4 sinkx  sin(k  2) x 8 cos(k  1)x sin x
k – k–2 =
  0
sin x
dx =
 
0
sin x
dx

k = k–2

4
vr% 5 = 3  5 = 1 =
 
0
dx = 4

oSdfYid

2 sin (9x / 2)
ekuk  =
 
–
sin(x / 2)
dx

4 sin(9x / 2)
=
 
0
sin(x / 2)
dx .......(1) ( f(x) le Qyu gS)

4 cos(9x / 2)
=
 
0
cos(x / 2)
dx .......(2)
b b
(  f(x)dx =  f(a  b – x)dx ds mi;ksx ls)
0 0

(1) o (2) dks tksM+us ij

 
4 sin5x 4 sin5x
=
 
0
2sin(x / 2)cos(x / 2)
dx =
 
0
sin x
dx
/2
8 sin5x
  =
 
0
sin x
dx

/2
8  16 sin5 x – 20 sin3 x  5 sin x 
 =
 
0

 sin x
 dx

/2
8
 =
 
0
(16sin4 x – 20sin2 x  5 ) dx

8  3  1  1  5 
  = 16x – 20  
  422 2  2 2 
8  5 
  = 3  – 5  
  2 
  =4

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 76
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

9. Let a  ˆi  ˆj  kˆ , b  ˆi  ˆj  kˆ and c  ˆi  ˆj  kˆ be three vectors. A vector  in the plane of a and b ,

1
whose projection on c is , is given by
3
ekuk fd a  ˆi  ˆj  kˆ , b  ˆi  ˆj  kˆ vkSj c  ˆi  ˆj  kˆ rhu lfn'k gSA ,d lfn'k  tks a vkSj b lery esa fLFkr
1
gS] vkSj ftldk c ij iz{ksi gS, fuEu gS& [IIT-JEE 2011, Paper-1, (3, –1), 80]
3
(A) ˆi  3ˆj  3kˆ (B) 3iˆ  3ˆj  kˆ (C*) 3iˆ  ˆj  3kˆ (D) ˆi  3ˆj  3kˆ
Ans. (C)
Sol. Let  = a  b
  = (  ) î + (  ) + (  )kˆ
1
Now . cˆ =
3
(   )  (   )  (   ) 1
 =
3 3
–  
 =  
   = (2+ 1) i – ĵ + (2) k̂
For 1,  = 3lˆ  ˆj  3kˆ

Hindi ekuk  = a  b
  = (  ) î + (  ) + (  )kˆ
1
vc . cˆ =
3
(   )  (   )  (   ) 1
 =
3 3
–  
 =  
   = (2+ 1) i – ĵ + (2) k̂
1,  = 3lˆ  ˆj  3kˆ ds fy,

10*. The vector(s) which is/are coplanar with vectors ˆi  ˆj  2kˆ and ˆi  2jˆ  kˆ , and perpendicular to the
vector ˆi  ˆj  kˆ is/are
og (os) lfn'k tks lfn'kksa ˆi  ˆj  2kˆ rFkk ˆi  2jˆ  kˆ }kjk cus ry esa fLFkr gS(gS)a , vkSj lfn'k ˆi  ˆj  kˆ ds yEcor
gS(gSa), fuEu gS(gSa) [IIT-JEE 2011, Paper-1, (4, 0), 80]
(A*) ˆj  kˆ (B) ˆi  ˆj (C) ˆi  ˆj (D*) ˆj  kˆ
Ans. (A, D)
Sol. a = î + ĵ + 2 k̂
b = î + 2 ĵ + k̂
c = î + ĵ + k̂
Required vector is  c × ( a × b )
[( c . b ) a – ( c . a ) b ]
[ (1+2+1) – ( î + ĵ +2 k̂ ) – (1+1+2) ( î +2 ĵ + k̂ ) ]
 [–4 ĵ + 4 k̂ ]

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 77
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

so our vector in parallel – ĵ + k̂

Hindi a = î + ĵ + 2 k̂
b = î + 2 ĵ + k̂
c = î + ĵ + k̂
vHkh"V lfn'k  c × ( a × b ) gS
[( c . b ) a – ( c . a ) b ]
[ (1+2+1) – ( î + ĵ +2 k̂ ) – (1+1+2) ( î +2 ĵ + k̂ ) ]
 [–4 ĵ + 4 k̂ ]

lfn'k – ĵ + k̂ ds lekUrj gSA

11. Let a  ˆi  kˆ , b  ˆi  ˆj and c  ˆi  2jˆ  3kˆ be three given vectors. If r is a vector such that
r  b  c  b and r . a  0 , then the value of r . b is [IIT-JEE 2011, Paper-2, (4, 0), 80]

ekuk fd rhu lfn'k a  ˆi  kˆ , b  ˆi  ˆj vkSj c  ˆi  2jˆ  3kˆ fn;s x;s gSA ;fn ,d lfn'k r bl izdkj dk gS
fd r  b  c  b vkSj r . a  0 ekU; gSa] rc r . b dk eku gSA
Ans. 9
Sol. (r  c)  b = 0
r – c = b  r = c + b R
r . a =0
 (c + b ) . a = 0
 (( î + 2 ĵ + 3 k̂ ) + (– î + ĵ )) .(– î – k̂ ) = 0
 ((1 – ) î +(2+) ĵ + 3 k̂ ) . (– î – k̂ ) = 0
 –1–3=0
 =4
so vr% r . b = ( – 3 î + 6 ĵ + 3 k̂ ) . (– î + k̂ )
=3+6=9

12. Match the statements given in Column-I with the values given in Column-II
Column-I Column-II

(A) If a  ˆj  3 kˆ , b  ˆj  3 kˆ and c  2 3 kˆ form a triangle, (p)
6
then the internal angle of the triangle between a and b is
 2
b
(B) If  (f(x)  3x) dx = a2 – b2, then the value of f   is (q)
a 6 3
2 
5/6

ln3 7/ 6
(C) The value of sec ( x) dx is (r)
3
 1 
(D) The maximum value of Arg   for |z| = 1, z  1 is given by (s)  
 1 z 

(t)
2
dkWye I esa fn;s x;s oDrO;ksa dk dkWye II esa fn;s x;s ekuksa ls lqesy djsa [IIT-JEE 2011, Paper-2, (8, 0), 80]
dkWye-I dkWye-II
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 78
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry


(A) ;fn a  ˆj  3 kˆ , b  ˆj  3 kˆ vkSj c  2 3 kˆ ,d (p)
6
f=kHkqt fufeZr djrs gS] rks f=kHkqt dk vkUrfjd dks.k tks a vkSj b
ds chp fLFkr gS
 2
b
(B) ;fn  (f(x)  3x)
a
dx = a2 – b2, rks f   dk eku gS
6
(q)
3
2 5/6

ln3 7/ 6
(C) sec ( x) dx dk eku gS (r)
3
 1 
(D) ;fn |z| = 1 vkSj z  1 rks Arg   dk mPpre eku gS (s) 
 1 z 

(t)
2
Ans. (A)  (q), (B)  (p), (C)  (s), (D)  (t)
a . b 1  3 2
Sol. (A) cos( – ) = = =
|a | |b| 1 3 1 3 4

1
– cos =
2
2
 =
3

(B) Using Leibeintz Theorem

f(b) – 3b = – 2b
f(b) = b

5/6
2 2  n | sec x  tan x | 
5/6
(C)
n3 7/6
 (sec x) dx = 
n3  

7 / 6
  2 1  2 1   3 3 
 n     n     n  
  2
 3 3 3 3   
2
3 1 
=   =   =
n3    n3   
   
   

(D) z (z  1) lies on circle with center 0, radius 1

 1 
Arg   = Arg 1 – Arg (1 – z) = angle between OA and BA
 1 z 


Absolute value of angle between OA and BA tends to as B tends to A.
2
Alter # 1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 79
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 1 
arg   = |arg 1 – arg (1 – z)| = |arg (1 – z)|
 1 z 
as |z| = 1 i.e. z lies on circle

 – z lies on circle

 1 – z lies on circle


 max |arg (1 – z)| =
2
Alter # 2
z = ei
1 1 1    1 1 
= =  sin  icos  = +i cot
1 z  
2 sin 
2 2 2 2 2
2sin2  isin 
2 2

1 1
Locus is is x =
1 z 2


Maximum value of  tends to
2

a . b 1  3 2
Hindi (A) cos( – ) = = =
|a | |b| 1 3 1 3 4

1
– cos =
2
2
 =
3

(B) Leibeintz Theorem ls

f(b) – 3b = – 2b
f(b) = b

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 80
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

5/6
2 2  n | sec x  tan x | 
5/6
(C)
n3  (sec x)
7/6
dx = 
n3  

7 / 6
  2 1  2 1   3 3 
 n     n     n  
 2
 3 3 3 3   
2
3 1 
=   =   =
n3    n3   
   
   

(D) z (z  1) o`Ùk ij fLFkr gksaxs ftldk dsUnz 0 vkSj f=kT;k 1 gSA

 1 
Arg   = Arg 1 – Arg (1 – z) = OA vkSj BA dk e/; dk dks.k
 1 z 


OA vkSj BA ds e/; ds dks.k dk fujis{k eku dh vksj vxzlj gS tcfd B, A dh vksj vxzlj gSA
2
Alter # 1
 1 
arg   = |arg 1 – arg (1 – z)| = |arg (1 – z)|
 1 z 
pwafd |z| = 1 i.e. z o`Ùk ij fLFkr gSA

 – z o`Ùk ij fLFkr gSA

 1 – z o`Ùk ij fLFkr gSA


 max |arg (1 – z)| =
2
Alter # 2
z = ei

1 1 1    1 1 
= =  sin  icos  = +i cot
1 z  
2 sin 
2 2 2 2 2
2sin2  isin 
2 2

1 1
fcUnqiFk ,x=
1 z 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 81
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry


 dk vf/kdre eku dh vksj gSA
2

13. The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1, –1, 4) with the
plane
5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2, 1,4) to QR, then the length
of the line segment PS is [IIT-JEE 2012, Paper-1, (3, –1), 70]
fcUnq P fcanqvksa Q(2, 3, 5) vkSj R(1, –1, 4) ls xqtjus okyh ljy js[kk ,oa lery 5x – 4y – z = 1 dk izfrPNsnh
fcanq gSA ;fn fcanq T(2, 1, 4) ls QR ij Mkys x;s yEc dk yEc&ikn S gS rks js[kk&[k.M PS dh yEckbZ fuEu gS&
1
(A*) (B) 2 (C) 2 (D) 2 2
2
Sol. Ans. (A)
Equation of QR is
x–2 y–3 z–5
= =
1 4 1
Let P  (2 + , 3 + 4, 5 + )
10 + 5 – 12 – 16 – 5 –  = 1
– 7 – 12 = 1
–2
 =
3
 4 1 13 
then P   , ,
3 3 3 
Let S = (2 + µ, 3 + 4µ, 5 + µ)

TS = (µ)iˆ  (4µ  2)jˆ  (µ  1)kˆ

TS . (iˆ  4ˆj  k)
ˆ =0
µ + 16µ + 8 + µ + 1 = 0
1
µ= –
2
3 9
S =  , 1,
 2 2 
2 2
4 3 4  13 9  1 4 1 1 4 9 1
PS = 3 – 2  9  3 – 2 =   =  = =
    36 9 36 18 9 18 2

Hindi Ans. (A)

QR dk lehdj.k gksxk
x–2 y–3 z–5
= =
1 4 1
ekuk P  (2 + , 3 + 4, 5 + )
10 + 5 – 12 – 16 – 5 –  = 1
– 7 – 12 = 1
–2
 =
3

rks P   , ,
4 1 13 
 3 3 3 
ekuk S = (2 + µ, 3 + 4µ, 5 + µ)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 82
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry


TS = (µ)iˆ  (4µ  2)jˆ  (µ  1)kˆ

TS . (iˆ  4ˆj  k)
ˆ =0
µ + 16µ + 8 + µ + 1 = 0
1
µ= –
2
3 9
S =  , 1,
2 2 
2 2
4 3 4  13 9 
PS = 3 – 2  9  3 – 2
   
1 4 1
=  
36 9 36
1 4 9 1
=  = =
18 9 18 2

14. If a, b and c are unit vectors satisfying | a – b |2 + | b – c |2 + | c – a |2 = 9, then |2 a + 5 b + 5 c | is

[IIT-JEE 2012, Paper-1, (4, 0), 70]
;fn a, b vkSj c bdkbZ lfn'k (unit vectors) gSa tks | a – b |2 + | b – c |2 + | c – a |2 = 9 dks larq"V djrs gSa
rc |2 a + 5 b + 5 c | dk eku gSA
Sol. Ans 3
6 – 2a·b – 2b·c – 2c·a = 9
–3
a·b  b·c  c·a  = 2
2
abc 0

a·b  b·c  c·a  3 + 2  0

–3
a·b  b·c  c·a 
2
–3
Since pwafd = a·b  b·c  c·a =
2
 abc = 0  abc= 0
 2a  5(–a) = | 3a |  3

15. The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and
2
x – y + z = 3 and at a distance from the point (3, 1, –1) is [IIT-JEE 2012, Paper-2, (3, –1), 66]
3
,d lery] tks leryksa x + 2y + 3z = 2 vkSj x – y + z = 3 dh izfrPNsnh js[kk ls xqtjrk gS vkSj fcUnq (3,1, –1)
2
ls dh nwjh ij gS] dk lehdj.k fuEu gSA
3
(A*) 5x – 11y + z = 17 (B) 2x  y  3 2 – 1
(C) x + y + z = 3 (D) x – 2y = 1 – 2
Sol. Ans. (A)
Equation of required plane
(x + 2y + 3z – 2) + (x – y + z – 3) = 0
 (1 + )x + (2 – )y + (3 + )z – (2 + 3) = 0
distance from point (3, 1, – 1)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 83
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

3  3  2 –  – 3 –  – 2 – 3 2
= =
(1   )  (2 –  )  (3   )
2 2 2
3
–2 2
 =
3  4  14
2
3
 3 = 32 + 4 + 14
2

7
 = –
2
equation of required plane
5x – 11y + z – 17 = 0

16. If a and b are vectors such that a  b  29 and a  (2iˆ  3jˆ  4k)
ˆ = (2iˆ  3jˆ  4k)
ˆ × b , then a

possible value of (a  b) . (–7iˆ  2jˆ  3k)

ˆ is [IIT-JEE 2012, Paper-2, (3, –1), 66]

;fn lfn'k a vkSj b ds fy, a  b  29 vkSj a  (2iˆ  3jˆ  4k)

ˆ = (2iˆ  3jˆ  4k)
ˆ × b gS] rc (a  b) .

(–7iˆ  2jˆ  3k)

ˆ dk ,d lEHkkfor eku fuEu gksxk&
(A) 0 (B) 3 (C*) 4 (D) 8
Hindi. Ans. (C)
Let ekuk c = 2iˆ  3jˆ  4kˆ
ac = c b
 (a  b)  c = 0
 (a  b) || c
Let ekuk (a  b) = c
 |ab| = |  | |c |
 29 = |  | . 29
 =1
 a  b = ± (2iˆ  3jˆ  4k)ˆ
Now vc (a  b).(–7iˆ  2jˆ  3k)
ˆ = ± (– 14 + 6 + 12)
=±4

x 1 y 1 z x 1 y 1 z
17*. If the straight lines = = and = = are coplanar, then the plane(s)
2 k 2 5 2 k
containing these two lines is(are) [IIT-JEE 2012, Paper-2, (4, 0), 66]

(A) y + 2z = –1 (B*) y + z = –1 (C*) y – z = –1 (D) y – 2z = –1

x 1 y 1 z x 1 y 1 z
;fn ljy js[kk;sa = = vkSj = = leryh; (coplanar) gSa rks og (os) lery ftlesa
2 k 2 5 2 k
;s nksuksa js[kk;sa fLFkr gSa] fuEu gS (gSa)
(A) y + 2z = –1 (B*) y + z = –1 (C*) y – z = –1 (D) y – 2z = –1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 84
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. Ans. (B,C)

For co-planer lines [a – c b d]  0
a  (1, – 1, 0) , = (–1, – 1, 0)
b = 2iˆ  kjˆ  2kˆ d = 5iˆ  2jˆ  kkˆ
2 0 0
Now 2 k 2 = 0  k=±2
5 2 k
n1 = b1  d1 = 6ˆj – 6kˆ for k = 2
n2 = b2  d2 = 14ˆj  14kˆ for k = – 2
so the equation of planes are  r – a.n1  0  y – z = – 1 ...... (1)
 r – a.n2  0  y + z = – 1 ...... (2)
so answer is (B,C)

Hindi Ans. (B,C)

nks leryh; js[kkvksa ds fy, [a – c b d]  0
a  (1, – 1, 0) , = (–1, – 1, 0)
b = 2iˆ  kjˆ  2kˆ d = 5iˆ  2jˆ  kkˆ
2 0 0
vc 2 k 2 = 0  k=±2
5 2 k
n1 = b1  d1 = 6ˆj – 6kˆ k = 2 ds fy,
n = b  d = 14ˆj  14kˆ
2 2 2 k = – 2 ds fy,
blfy, lery dh lehdj.k gS  r – a .n1  0  y – z = – 1 ...... (1)
 r – a.n2  0 y+z=–1 ...... (2)
blfy, mÙkj (B,C) gSA

18. Let PR  3iˆ  ˆj – 2kˆ and SQ  ˆi – 3jˆ – 4kˆ determine diagonals of a parallelogram PQRS and
PT  ˆi  2jˆ  3kˆ be another vector. Then the volume of the parallelepiped determined by the vectors
PT, PQ and PS is [JEE (Advanced) 2013, Paper-1, (2, 0)/60]
ekukfd PR  3iˆ  ˆj – 2kˆ rFkk SQ  ˆi – 3jˆ – 4kˆ ,d lekUrj prqHkqZt PQRS ds fod.kZ fu/kkZfjr djrs gS vkSj
PT  ˆi  2jˆ  3kˆ ,d vU; lfn'k gS] rc lfn'kksa PT, PQ rFkk PS }kjk fu/kkZfjr lekUrj "kV~Qyd dk vk;ru
gS& [JEE (Advanced) 2013, Paper-1, (2, 0)/60]
(A) 5 (B) 20 (C*) 10 (D) 30

Sol. (C)
PR  PQ  PS
SQ  PQ – PS

PR – SQ
PS 
2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 85
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

V= PQ PS PT 
 

1 PR SQ , PR – SQ , PT 
V=
4  
1 PR , SQ , PT 
V=
2  
3 1 –2
1
1 –3 –4
2
1 2 3
1
(– 3 – 7 – 10) = 10
2

x  2 y 1 z
19. Perpendicular are drawn from points on the line   to the plane x + y + z = 3. The feet of
2 –1 3
perpendiculars lie on the line [JEE (Advanced) 2013, Paper-1, (2, 0)/60]
x  2 y 1 z
ry x + y + z = 3 ij js[kk   ij fLFkr fcUnqvksa ls yEc Mkys tkrs gSA yEc -ikn fuEu js[kk ij
2 –1 3
fLFkr gS&
x y –1 z – 2 x y –1 z – 2 x y –1 z – 2 x y –1 z – 2
(A)   (B)   (C)   (D*)  
5 8 –13 2 3 –5 4 3 –7 2 –7 5
Sol. (D)
x2 y 1 z
Any point on line = = =
2 1 3
Let any two points on this line are
A(– 2, – 1, 0), B (0, – 2, 3) Put ( = 0, 1)
Let foot of perpendicular from A( – 2, – 1, 0) on plane is (, , )
2  1 0
 = = = µ (say)
1 1 1
Also,  +  +  = 3
 µ–2+µ–1+µ=3µ=2
 M(0, 1, 2)
 2 4 11 
Similarly foot of perpendicular from B(0, – 2, 3) on plane is N  , , 
3 3 3 
x0 y 1 z2
So, equation of MN is = = .
2 7 5
3 3 3

Hindi. (D)
x2 y 1 z
js[kk = = =  ij dksbZ fcUnq
2 1 3
ekuk bl js[kk ij dksbZ nks fcUnq gS
A(– 2, – 1, 0), B (0, – 2, 3) ( = 0, 1) j[kus ij
ekuk A( – 2, – 1, 0) ls lery ij yEcikn ds funsZ'kkad (, , ) gSA
2  1 0
 = = = µ (ekuk)
1 1 1
rFkk, ++=3
 µ–2+µ–1+µ=3µ=2
 M(0, 1, 2)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 86
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2 4 11
blh izdkj B(0, – 2, 3) ls lery ij yEcikn ds funsZ'kkad N  , ,  gSA
3 3 3 
x0 y 1 z2
blfy, MN dk lehdj.k = = .
2 7 5
3 3 3

20.* A line l passing through the origin is perpendicular to the lines

l1 : (3 + t) î + (– 1 + 2t) ĵ + (4 + 2t) k̂ , –  < t < 
 l2 : (3 + 2s) î + (3 + 2s) ĵ + (2 + s) k̂ , –  < s < 
Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1
is(are) [JEE (Advanced) 2013, Paper-1, (4, – 1)/60]
7 7 5 7 7 8
(A)  , ,  (B*) (–1, ,–1, 0) (C) (1, 1, 1) (D*)  , , 
 3 3 3  9 9 9
,d js[kk l ] tks ewyfcUnq ls xqtjrh gS] js[kkvksa
l1 : (3 + t) î + (– 1 + 2t) ĵ + (4 + 2t) k̂ , –  < t < 
 l2 : (3 + 2s) î + (3 + 2s) ĵ + (2 + s) k̂ , –  < s < 
ij yEcor gSA rc, l2 ij fLFkr fcUnq ¼fcUnqvksa½ ds funsZ'kkad] tks js[kkvksa l rFkk l1 ds çfrPNsn fcUnq ls 17 dh nwjh
ij gSa ¼gSa½] fuEu gS ¼gSa½ % [JEE (Advanced) 2013, Paper-1, (4, – 1)/60]
7 7 5 7 7 8
(A)  , ,  (B*) (–1, ,–1, 0) (C) (1, 1, 1) (D*)  , , 
3 3 3 9 9 9
Sol. (B,D)
Let equation of line  is
x0 y0 z0
: = = =k
a b c
This line  is perpendicular to given line 1 and 2.
Hence a + 2b + 2c = 0
2a + 2b + c = 0
a b c
= =
2 3 2

x y z
Hence equation of  is = = = k 1, k2
2 3 2

      for 1 for 2
Now A(–2k1, 3k1, –2k1) B(–2k2, 3k2, –2k2)
Point A satisfied 1

–2k1 î + 3k1 ĵ – 2k1 k̂ = (3 + t) î + (–1 + 2t) ĵ + (4 + 2t) k̂

3 + t = –2k1 .......(1)
–1 + 2t = 3k1 .......(2)
4 + 2t = –2k1 .......(3)
(2) & (3) –5 = 5k1  k1 = –1  A (2, –3, 2)
Let any point on 2 (3 + 2S, 3 + 2S, 2 + S)

Given (1  2S)2  (6  2S)2  (S)2 = 17

9S + 28S + 37 = 17
2

9S2 + 28S + 20 = 0
9S2 + 18S + 10S + 20 = 0
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 87
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

9S(S + 2) + 10 (S + 2) = 0
S = –2, –10/9
Hence (–1, –1, 0) , (7/9, 7/9, 8/9)
Ans. (B) & (D)
Hindi (B,D)
ekuk js[kk  dk lehdj.k
x0 y0 z0
: = = =k
a b c
;g js[kk  nh xbZ js[kk 1 rFkk 2 ds yEcor~ gS
vr% a + 2b + 2c = 0
2a + 2b + c = 0
a b c
= =
2 3 2

x y z
vr%  dk lehdj.k = = = k 1, k2
2 3 2

      for 1 for 2
vc A(–2k1, 3k1, –2k1) B(–2k2, 3k2, –2k2)
fcUnq A, 1 ij larq"B gksrk gS

–2k1 î + 3k1 ĵ – 2k1 k̂ = (3 + t) î + (–1 + 2t) ĵ + (4 + 2t) k̂

3 + t = –2k1 .......(1)
–1 + 2t = 3k1 .......(2)
4 + 2t = –2k1 .......(3)
(2) & (3) –5 = 5k1  k1 = –1  A (2, –3, 2)
ekuk 2 (3 + 2S, 3 + 2S, 2 + S) ij dksbZ fcUnq
fn;k x;k gS (1  2S)2  (6  2S)2  (S)2 = 17
9S + 28S + 37 = 17
2

9S2 + 28S + 20 = 0
9S2 + 18S + 10S + 20 = 0
9S(S + 2) + 10 (S + 2) = 0
S = –2, –10/9
vr% (–1, –1, 0) , (7/9, 7/9, 8/9)
Ans. (B) & (D)

21.  
Consider the set of eight vectors V = aiˆ  bjˆ  ckˆ : a,b,c  –1,1 . Three non-coplanar vectors can be
chosen from V in 2 ways. Then p is
p
[JEE (Advanced) 2013, Paper-1, (4, – 1)/60]
vkB lfn'kksa dk leqPp; V = aiˆ  bjˆ  ckˆ : a,b,c  –1,1 . yhft, A V ls rhu vleryh; lfn'k 2p çdkj ls
pqus tk ldrs gSA rc p dk eku gS% [JEE (Advanced) 2013, Paper-1, (4, – 1)/60]
Ans. 8
C3 – 24 = 32
Sol. Among set of eight vectors four vectors form body diagonals of a cube, remaining four will be
parallel (unlike) vectors.
Numbers of ways of selecting three vectors will
be 4C3 × 2 × 2 × 2 = 25
Hence p = 5
Alternative
Eight vectors
x  ˆi ˆj kˆ
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 88
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

y  ˆi ˆj kˆ 
 z  ˆi ˆj kˆ

  ˆi ˆj kˆ
x '  ˆi ˆj kˆ
y '  ˆi ˆj kˆ
z'  ˆi ˆj –kˆ
 '  ˆi ˆj kˆ
If we take x , x ' and any one of remaining sin x, vectors will always be coplaner
 No. of coplaner vectors = 6
similarly on taking y , y ' = 6
z, z' = 6
,  ' = 6  No. of set of coplaner vectors = 24
Alternative

A(0, 0, 0)
B(1, 0, 0)
C(1, 0, 1)
D(0, 0, 1)
E(0, 1, 1)
F(0, 1, 0)
G(1, 1, 0)
H(1, 1, 1)
AH  ˆi  ˆj  kˆ
BE  –iˆ  ˆj  kˆ
CF  –iˆ  ˆj – kˆ
Non-coplaner

Hindi. vkB lfn'kksa ds leqPp;ksa ls pkj lfn'k] ?ku ds fod.kZ ds 'kh"kZ gSA 'ks"k] lHkh lekUrj lfn'k gSA
rhu lfn'kksa ds p;u djus ds Øep; gksxsa&
C3 × 2 × 2 × 2 = 2 5
4

vr% p = 5
Alternative
vkB lfn'k
x  ˆi ˆj kˆ
y  ˆi ˆj kˆ 
 z  ˆi ˆj kˆ

  ˆi ˆj kˆ
x '  ˆi ˆj kˆ
y '  ˆi ˆj kˆ
z'  ˆi ˆj –kˆ

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 89
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 '  ˆi ˆj kˆ

;fn ge x , x ' rFkk sin x es ls ,d 'ks"k lfn'k] lnSo leryh; gksxsaA

 lery lfn'kksa dh la[;k = 6
blh izdkj y , y ' = 6 ysus ij
z, z' = 6
,  ' = 6  leryh; lfn'kksa ds leqPp;ksa dh la[;k = 24
Ans. 8C3 – 24 = 32

Alternative
A(0, 0, 0)
B(1, 0, 0)

C(1, 0, 1)
D(0, 0, 1)
E(0, 1, 1)
F(0, 1, 0)
G(1, 1, 0)
H(1, 1, 1)
AH  ˆi  ˆj  kˆ
BE  –iˆ  ˆj  kˆ
CF  –iˆ  ˆj – kˆ vleryh;
y z y z
22.* Two lines L1 : x = 5, = and L2 : x = , = are coplanar. Then  can take value(s)
3– –2 –1 2–
y z y z
nks js[kk,a L1 : x = 5, = rFkk L2 : x = , = leryh; gSA rc  dk eku gks ldrk gS&
3– –2 –1 2–
[JEE (Advanced) 2013, Paper-2, (3, –1)/60]
(A*) 1 (B) 2 (C) 3 (D*) 4
Sol. (A, D)
x–5 –y z
 
0  – 3 –2
x– y z
 
0 –1 2 – 

5– 0 0
0 3– –2 =0
0 –1 2 – 
(5 – ) ((3 – ) (2 – ) – 2) = 0
(2 – 5 + 6 – 2) = 0
( – 5)(2 – 5 + 4) = 0
 = 1, 4, 5

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 90
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

23. Match List I with List II and select the correct answer using the code given below the
lists :
[JEE (Advanced) 2013, Paper-2, (3, –1)/60]
List - I List - II
P. Volume of parallelopiped determined by vectors 1. 100
a,b and c is 2. Then the volume of the parallelepiped
determined by vectors 2(a  b),3(b  c) and (c  a) is

Q. Volume of parallelepiped determined by vectors a,b 2. 30

and c is 5. Then the volume of the parallelepiped
determined by vectors 3(a  b),(b  c) and 2 (c  a) is

R. Area of a triangle with adjacent sides determined by 3. 24

vectors a and b is 20. Then the area of the triangle
with adjacent sides determined by vectors  2a  3b
and (a – b) is

S. Area of a paralelogram with adjacent sides determined by 4. 60

vectors a and b is 30. Then the area of the parallelogram
with adjacent sides determined by vectors (a  b) and a is
lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s %
[JEE (Advanced) 2013, Paper-2, (3, –1)/60]
lwph- I lwph - II
P. lfn'kksa a,b rFkk c }kjk fu/kkZfjr lekarj "kV~Qyd dk 1. 100
vk;ru 2 gSA rc lfn'kksa 2(a  b),3(b  c) rFkk (c  a)
}kjk fu/kkZfjr lekarj "kV~Qyd dk vk;ru gS

Q. lfn'kksa a,b rFkk c }kjk fu/kkZfjr lekarj "kV~Qyd dk 2. 30

vk;ru 5 gSA rc lfn'kksa 3(a  b),(b  c) rFkk 2 (c  a)
}kjk fu/kkZfjr lekarj "kV~Qyd dk vk;ru gS

R. ,d f=kHkqt dk {ks=kQy] ftldh layXu Hkqtk,¡ lfn'kksa a rFkk b 3. 24

}kjk fu/kkZfjr gS] 20 gSA rc lfn'kksa  2a  3b  rFkk (a – b) }kjk
fu/kkZfjr layXu Hkqtkvksa okys f=kHkqt dk {ks=kQy gS

S. ,d lekarj prqHkqZt dk {ks=kQy] ftldh layXu Hkqtk,¡ lfn'kksa 4. 60

a rFkk b }kjk fu/kkZfjr gSa 30 gSA rc lfn'kksa (a  b) rFkk a }kjk
fu/kkZfjr layXu Hkqtkvksa okys lekarj prqHkZqt dk {ks=kQy gS
Codes :
P Q R S
(A) 4 2 3 1
(B) 2 3 1 4
(C*) 3 4 1 2
(D) 1 4 3 2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 91
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. (C)
(P) [a b c] = 2

2 a  b ,3   
b  c , c  a

6 [ a  b b  c c  a ] = 6 [a b c]2
= 6 × 4 = 24
P3

(Q) [a b c] = 5
[3 (a  b) (b  c) 2 (c  a) ]
= 6 × 2 [a b c]
= 12 × 5 = 60
Q4

1
(R) | a  b | = 20
2
1
1 = | (2a  3b) × (a  b) |
2
= | 2a  b – 3(a  b) |
5
= | ab |
2
= 5 × 20 = 100
R1

(S) | a  b | = 30
| (a  b) × a | = | b  a| = 30
S 2

x –1 y z3 x – 4 y3 z3

24. Consider the lines L1 :   , L2   : and the planes P1 : 7x + y + 2z = 3,
2 –1 1 1 1 2
P2 : 3x + 5y – 6z = 4. Let ax + by + cz = d the equation of the plane passing through the point of
intersection of lines L1 and L2, and perpendicular to planes P1 and P2.
Match List - I with List- II and select the correct answer using the code given below the lists :
[JEE (Advanced) 2013, Paper-2, (3, –1)/60]
x –1 y z3 x – 4 y3 z3
js[kk,a L1 :   , L2   : rFkk lery P1 : 7x + y + 2z = 3, P2 : 3x + 5y – 6z
2 –1 1 1 1 2
= 4 yhft,A ekuk fd ax + by + cz = d, js[kkvksa L1 o L2 ds izfrPNsn fcUnq ls xqtjus okyk rFkk lery P1 o P2 ds
yEcor~] lery dk lehdj.k gSA
lwph - I dks lwph - II ls lqefs yr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk iz;ksx djds lgh mÙkj pqfu;s :
[JEE (Advanced) 2013, Paper-2, (3, –1)/60]
List- I / lwph I List- II / lwph II
P. a= 1. 13
Q. b= 2. –3
R. c= 3. 1
S. d= 4. –2
Codes :
P Q R S
(A*) 3 2 4 1
(B) 1 3 4 2
(C) 3 2 1 4

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 92
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(D) 2 4 1 3
Sol. (A)
x 1 y z3
L1 : = =
2 1 1
ˆi ˆj kˆ
Normal of plane P : n  7 1 2
3 5 6
= î (–16) – ĵ (–42 – 6) + k̂ (32)
= –16 î + 48 ĵ + 32 k̂
 n = ˆi  3jˆ  2kˆ
Point of intersection of L1 and L2
2k1 + 1 = k2 + 4
–k1 = k2 – 3
1 = 3k2 – 2
k2 = 1
Point of intersection (5, –2, –1)
Plane (x – 5) – 3 (y + 7) – 2(z + 1) = 0
x – 3y – 2z – 5 – 6 – 2 = 0
x – 3y – 2z = 13
 a = 1, b = 3, c = –2, d = 13

Hindi (A)
x 1 y z3
L1 : = =
2 1 1
ˆi ˆj kˆ
lery P dk vfHkyEc : n  7 1 2
3 5 6
= î (–16) – ĵ (–42 – 6) + k̂ (32)
= –16 î + 48 ĵ + 32 k̂
 n = ˆi  3jˆ  2kˆ
L1 rFkk L2 dk izfrPNsn fcUnq
2k1 + 1 = k2 + 4
–k1 = k2 – 3
1 = 3k2 – 2
k2 = 1
izfrPNsn fcUnq (5, –2, –1)
lery (x – 5) – 3 (y + 7) – 2(z + 1) = 0
x – 3y – 2z – 5 – 6 – 2 = 0
x – 3y – 2z = 13
 a = 1, b = 3, c = –2, d = 13

25*. Let x,y and z be three vectors each of magnitude 2 and the angle between each pair

of them is . If a is a nonzero vector perpendicular to x and y  z and b is a nonzero
3
vector perpendicular to y and z  x , then [JEE (Advanced) 2014, Paper-1, (3, 0)/60]

(A*) b  (b.z)(z – x) (B*) a  (a.y)(y – z) (C*) a.b  –(a.y)(b.z) (D) a  (a.y)(z – y)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 93
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ekuk fd lfn'kks a (vectors) x,y rFkk z es a iz R ;s d dk ifjek.k 2 gS a rFkk iz R ;s d ;q X e (pair) ds


e/; dk dks . k gS A ;fn 'kw U ;s r j (non-zero) lfn'k a lfn'kks a x rFkk y  z ds yEcor~
3
(perpendicular) gS ,oa 'kw U ;s r j lfn'k b lfn'kks a y rFkk z  x ds yEcor~ gS ] rc
(A*) b  (b.z)(z – x) (B*) a  (a.y)(y – z) (C*) a.b  –(a.y)(b.z) (D) a  (a.y)(z – y)
Ans. (ABC)
Sol. x  y  z  2

=
3
a  x   y  z 
b  y   z  x 
a  x.z  y –  x.y  z 
 1 1 
a  2 y – 2 z
 2 2 
a    y – z
b  z – x
Similarly blh iz d kj
 1
a.y    2 – 2   = 
 2
 a   a.y   y – z   (B)
 1
b.z    2 – 2  
 2
 = b.z

 
 b  b.z  z – x   (A)

(A)  
a.b   a.y   y – z . b.y  z – x 

 
=  a.y  b.z  yz – yx – 2  xz 

=  a.y  b.z 

= –  a.y  b.z   (C)

26. From a point P(,,), perpendiculars PQ and PR are drawn respectively on the lines y =
x, z = 1 and y = –x, z = –1. If P is such that QPR is a right angle, then the possible
value(s) of  is(are)
[JEE (Advanced) 2014, Paper-1, (3, 0)/60]
(A) 2 (B) 1 (C*) –1 (D) – 2
fcUnq P(,,) ls js [ kkvks a y = x, z = 1 rFkk y = –x, z = –1 ij Mkys x;s yEc (perpendicular)
Øe'k% PQ rFkk PR gS a A ;fn QPR ledks . k (right angle) gS ] rks  dk ( ds ) la H kkfor eku gS ( gS a )
(A) 2 (B) 1 (C*) –1 (D) – 2
Ans. (C)
Sol. Line is
x0 y0 z0
= = = ......(1)
1 1 1
Q(, , 1)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 94
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Direction ratio of PQ are

 – ,  – ,  – 1
Since PQ is perpendicular to (1)
 –+–+0=0
=
 Direction ratio of PQ are
0, 0,  – 1
Another line is
x0 y0 z 1
= = = ......(2)
1 1 0
 R ( – , , – 1)
 Direction ratio of PR are
 + ,  – ,  + 1
Since PQ is perpendicular to (ii)
 ––+–=0
=0
 R(0, 0, – 1)
and Direction ratio of PQ are , ,  + 1
Since PQ  PR
 0 + 0 + 2 – 1 = 0  = ± 1  B, C
For  = 1 the point is on the line so it will be rejected.
  = – 1.

Hindi. js[kk
x0 y0 z0
= = = ......(1)
1 1 1
Q(, , 1)
PQ ds fn~d vuqikr
 – ,  – ,  – 1
pawfd PQ, (1) ds yEcor~ gS
 –+–+0=0
=
 PQ ds fn~d vuqikr
0, 0,  – 1
vU; js[kk gS
x0 y0 z 1
= = = ......(2)
1 1 0
 R ( – , , – 1)
 PR ds fn~d vuqikr
 + ,  – ,  + 1
pwafd PQ, (ii) ds yEcor~ gS
 ––+–=0
=0
 R(0, 0, – 1) vkSj PQ ds fnd~ vuqikr , ,  + 1 gS
pwafd PQ  PR  0 + 0 + 2 – 1 = 0  = ± 1  B, C
 = 1 ds fy, fcUnq js[kk ij gS blfy, ;g vLohdk;Z gksxkA
  = – 1.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 95
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

27. Let a, b and c be three non-coplanar unit vectors such that the angle between every

pair of them is . If a  b  b  c  pa  qb  rc , where p,q and r are scalars, then the value
3
p2  2q2  r 2
of is
q2
ekuk fd a, b rFkk c rhu vleryh; (non-coplanar) bdkbZ lfn'k gS ] ftuds iz R ;s d ;q X e ds

e/; dk dks . k gS A ;fn a  b  b  c  pa  qb  rc tgk¡ p,q ,oe~ r vfn'k (scalars) gS , rc
3
p2  2q2  r 2
dk eku gS :
q2
[JEE (Advanced) 2014, Paper-1, (3, 0)/60]
Ans. (4)
Sol. pa + qb+ rc = a × b + b × c
Taking dot product with a , b , c we get ( a , b , c ds lkFk vfn'k xq.ku ysus ij½
q r
p + + = [a b c] ......(1)
2 2
p r
+ q + = 0 ......(2)
2 2
p q
+ + r = [a b c] ......(3)
2 2
(1) & (3)  p = r & q = –p (1) vkSj (3)  p = r & q = –p
p2  2q2  r 2 p2  2p2  p2
2
= = 4 Ans.
q p2

28. List I List II

[JEE (Advanced) 2014, Paper-2, (3, –1)/60]

3
P. Let y(x) = cos(3 cos – 1 x), x  [–1, 1], x  ± . Then 1. 1
2
1  2 d2 y(x) dy(x) 
y(x) 

 x –1
dx 2
x
dx 
 equals

Q. Let A 1 , A 2 ,......, A n (n > 2) be the vertices of a regular polygon of n 2. 2

sides with its centre at the origin. Let ak be the position vector of
n–1 n–1
the point A k , k = 1, 2,...., n. If  (a
k 1
k  ak 1 )   (a .a
k 1
k k 1 ) , then

the minimum value of n is

x2 y2
R. If the normal from the point P(h, 1) on the ellipse   1 is 3. 8
6 3
perpendicular to the line x + y = 8, then the value of h is
S. Number of positive solutions satisf ying the equation 4. 9
 1   1   2
is tan–1    tan–1    tan–1  2 
 2x  1  4x  1 x 

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 96
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

lwph- I lwph - II

[JEE (Advanced) 2014, Paper-2, (3, –1)/60]

3
P. ekukfd y(x) = cos(3 cos – 1 x), x  [–1, 1], x  ± , rks 1. 1
2
1  2 d2 y(x) dy(x) 
 x –1
y(x) 
dx 2
x 
dx 
 dk eku gS&

Q. ekukfd A 1 , A 2 ,......, A n (n > 2) ,d n Hkqth; lecgqHkqt (regular polygon) 2.

2
ds 'kh"kZ (vertices) gS ftldk dsUnz ewyfcUnq esa gSA ekuk fd ak fcUnq
A k , k = 1, 2,...., n dk fLFkfr lfn'k (position vector) gSA ;fn
n–1 n–1

 (a
k 1
k  ak 1 )   (a .a
k 1
k k 1 ) gS] rc n dk U;wure eku gS&

x2 y2
R. ;fn nh?kZo`Ùk (ellipse)   1 ij fcUnq P(h, 1) ls [khapk x;k 3. 8
6 3
vfHkyEc js[kk x + y = 8 ij yEcor~ gS] rks h dk eku gS&

lehdj.k tan–1 
1   1   2
S.   tan–1    tan–1  2  dks larq"V djus okys 4. 9
 2x  1  4x  1 x 
/kukRed gyksa dh la[;k gS&

P Q R S
(A) 4 3 2 1
(B) 2 4 3 1
(C) 4 3 1 2
(D) 2 4 1 3
Ans. (A)
Sol. (P) y = 4x3 – 3x where cos = x
dy
= 12x2 – 3
dx
d2 y dy
x = (x2 – 1) . 24x + x(12x2 – 3)
dx 2 dx
= 36x3 – 27x = 9(4x3 – 3x) = 9y
1 dy 
Hence
y
 2
 x –1


d2 y
2

x  =9
dx 

dx 

(Q) a1  a2  a2  a3  . . .  an–1  an
= a1 . a2  a2 . a3  . . .  an–1 .an
Let = a1  a2 . . . . . = an =  (as centre is origin)
2
More over angle between 2 consecutive ai ' s is
n

Hence given equation reduces to

 2   2 
(n – 1)2 sin   = (n – 1)2 cos  
 n   n 
 2  2 
 tan  =1    n=8
 n  n 4
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 97
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

6x 3y  a2 x b 2 y 
(R) Equation of normal – =3  Equation of normal is –  a2 – b 2 
h 1 x1 y1 
 
6
slope = = 1 (as it is perpendicular to z + y = 1)  h=2
3h

 1   1   2 
(S) tan–1   + tan–1   + tan–1  2 
 2x  1   4x  1  x 
1 1

6x  2
 2x  1 4x  1  2
2 2
  2
1–
1 x 8x  6x x
2

 2x  1 4x  1
 3x3 + x2 = 8x2 + 6x  3x3 – 7x2 – 6x = 0
 3x2 – 7x + 6 = 0 (as x  0)
2  2 
 (x – 3) (3x + 2) = 0  x=– ,3   3 is rejected 
3  

Hindi (P) y = 4x3 – 3x tgk¡ cos = x

dy
= 12x2 – 3
dx
d2 y dy
x = (x2 – 1) . 24x + x(12x2 – 3)
dx 2 dx
= 36x3 – 27x = 9(4x3 – 3x) = 9y
1 dy 

y
 d2 y
dx

vr%  x2 – 1 2  x  = 9

dx 

(Q) a1  a2  a2  a3  . . .  an–1  an
= a1 . a2  a2 . a3  . . .  an–1 .an
ekuk = a1  a2 . . . . . = an =  (pawfd dsUæ] ewy fcUnq gS)
2
nks Øekxr ai ' s ds e/; dks.k gS
n

vr% fn, x,s lehdj.k gksrh gS

 2   2 
(n – 1)2 sin   = (n – 1)2 cos  
 n   n 
 2  2 
 tan  =1    n=8
 n  n 4

6x 3y  a2 x b 2 y 
(R) vfHkyEc dk lehdj.k – =3  Equation of normal is –  a2 – b 2 
h 1 x1 y1 
 
6
izo.krk = = 1 (pawfd ;g z + y = 1 ds yEcor~ gS)  h=2
3h

 1   1   2 
(S) tan–1   + tan–1   + tan–1  2 
 2x  1   4x  1  x 

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 98
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 1

6x  2
 2x  1 4x  1  2
2 2
 
1–
1 x 8x  6x
2
x2
 2x  1 4x  1
 3x3 + x2 = 8x2 + 6x  3x3 – 7x2 – 6x = 0
 3x2 – 7x + 6 = 0 (pawfd x  0)
2  2 
 (x – 3) (3x + 2) = 0  x=– ,3   3 is rejected 
3  

29*. In R 3 , consider the planes P 1 : y = 0 and P 2 : x + z = 1. Let P 3 be a plane, different from

P 1 and P 2 , which passes through the intersection of P 1 and P 2 . If the distance of the
point (0, 1, 0) from P 3 is 1 and the distance of a point ( ,  , ) from P 3 is 2, then which of
the following relation is (are) true ?
eku yhft, fd R 3 esa P 1 : y = 0 vkSj P 2 : x + z = 1 nks lery gSA ekuk fd P 3 ,d lery gS tks lery P 1
,oa P 2 ls fHkUu gS rFkk P 1 ,oa P 2 ds izfrPNsnu (intersection) ls tkrk gSA ;fn fcUnq (0, 1, 0) ls P 3 dh nwjh
,d (1) gS rFkk fcUnq (,  , ) ls P 3 dh nwjh nks (2) gS, rc fuEufyf[kr lEca/k ( lEca/kksa) esas dkSu lk ( ls) larqf"Vr
gksrs gSA ( gSa) ?
[JEE (Advanced) 2015, P-1 (4, –2)/ 88]
(A) 2 +  + 2 + 2 = 0 (B*) 2 –  + 2 + 4 = 0
(C) 2 +  – 2 – 10 = 0 (D*) 2 –  + 2 – 8 = 0
Ans. (B,D)
Sol. Let P3 be P2 + P1 = 0  x + y + z – 1 = 0
Distance from (0, 1, 0) is 1
0    0 1
  1
1  2  1
1
= 
2
 Equation of P3 is 2x – y + 2z – 2 = 0
Dist. from () is 3
2    2  2
 2  2 –  + 2 = 2 ± 6
3
 option (B, D) are correct.

30*. In R 3 , let L be a straight line passing through the origin. Suppose that all the points on L
are at a constant distance from the two planes P 1 : x + 2y – z + 1 = 0 and P 2 : 2x – y + z
– 1 = 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L
to the plane P 1 . W hich of the following points lie(s) on M?
ekuk fd R 3 esa L ,d ljy js[kk gS tks fd ewy fcUnq ls tkrh gSA ekuk fd L ds lHkh fcUnq leryksas P 1 : x + 2y –
z + 1 = 0 rFkk P 2 : 2x – y + z – 1 = 0 ls fLFkj nwjh ij gSA ekuk fd L ds fcUnqvksa ls lery P 1 ij Mkys
x, yEcksa ds iknksa (feet of the perpendiculars) dk iFk (locus) M gSA fuEufyf[kr fcUnqvksa es ls dkSu lk
( ls) fcUnq iFk M ij fLFkr gS gS ( gS) ? [JEE (Advanced) 2015, P-1 (4, –2)/ 88]
 5 2  1 1 1  5 1  1 2
(A*)  0, – , –  (B*)  – , – ,  (C)  – , 0,  (D)  – , 0, 
 6 3  6 3 6  6 6  3 3
Ans. (A,B)
Sol. Let v be the vector along L
ˆi ˆj kˆ
then v = 1 1 = ˆi  3jˆ  5kˆ
2
2 1 1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 99
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

So any point on line L is A(, – 3, –5)

Foot of perpendicular from A to P, is
h   k  3  5    6  5  1 1
= = =  =–
1 2 1 1 4  1 6
1 1 1
h =  – , k = –3 – ,  = –5 +
6 3 6
 1 1 1
so foot is    , 3  , 5  
 6 3 6
So (A, B)

Hindi. ekuk L ds vuqfn'k lfn'k v L

ˆi ˆj kˆ
rc v = 1 2 1 = ˆi  3jˆ  5kˆ
2 1 1
vr% L ij dksbZ fcUnq A(, – 3, –5) gS
A ls P ij yEcikn
h   k  3  5    6  5  1 1
= = =  =–
1 2 1 1 4  1 6
1 1 1
h =  – , k = –3 – ,  = –5 +
6 3 6

vr% yEcikn    , 3  , 5   gSA vr% (A, B) fodYi lgh gSA
1 1 1
 6 3 6

31*. Let PQR be a triangle. Let a  QR, b  RP and c  PQ . If | a | = 12, | b | 4 3 and

b.c  24 , then which of the following is(are) true?
| c |2 | c |2
(A*)  | a | 12 (B)  | a |  30
2 2
(C*) | a  b  c  a | 48 3 (D*) a . b   72
ekuk fd PQR ,d f=kHkqt gSA ekuk fd a  QR, b  RP vkSj c  PQ gSA ;fn | a | = 12 | b | 4 3 vkSj
b.c  24 , rc fuEufyf[kr esa ls dkSu lk ¼ls½ lgh gS ¼gSa½? [JEE (Advanced) 2015, P-1 (4, –2)/ 88]
| c |2 | c |2
(A)  | a | 12 (B)  | a |  30
2 2
(C) | a  b  c  a | 48 3 (D) a. b  72
Ans. (A,C,D)

Sol. abc = 0
 b  c  –a
 48 + + 48 = 144
 c 2 = 48
 c2  4 3
2
c
 – a = 24 – 12 = 12 Ans. (A)
2

Further ;|fi

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 100
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

a  b  –c
 144 + 48 + 2a.b
= 48

 a.b = –72 Ans. (D)

 abc = 0
 ab  ac = 0
a  b  c  a = 2 = 2. 144.48 –  72 
2
 = 48 3 Ans. (C)

32. Column-  Column-

(A) In R2, if the magnitude of the projection vector of the vector (P) 1
 ˆi  ˆj on 3iˆ  ˆj is 3 and if  = 2 + 3  ,
then possible value(s) of || is (are)

(B) Let a and b be real numbers such that the function (Q) 2
 –3ax 2 – 2 , x  1
f(x)   is differentiable for all x  R.
 bx  a , x  1
2

Then possible value(s) of a is (are)

(C) Let  1 be a complex cube root of unity. (R) 3

If (3 – 3 + 22)4n + 3 + (2 + 3 – 32)4n + 3 + (–3 + 2 + 32)4n + 3 = 0,
then possible value(s) of n is (are)

(D) Let the harmonic mean of two positive real numbers a and b be 4. (S) 4
If q is a positive real number such that a, 5,q, b is an arithmetic
progression, then the value(s) of |q – a| is (are)
(T) 5
[JEE (Advanced) 2015, P-1 (2, –1)/ 88]

dkWye -  dkWye-

(A) ekuk fd R2 esa, ;fn lfn'k  ˆi  ˆj dk lfn'k 3iˆ  ˆj ij iz{ksi (P) 1
lfn'k (projection vector) dk ifjek.k (magnitude) 3 gks vkSj
;fn  = 2 + 3  gks] rc || ds laHko eku gS (gSa)

(B) ekuk fd okLrfod la[;k,a a vkSj b bl izdkj gS fd Qyu (Q) 2

 –3ax – 2 , x  1
2
f(x)   lHkh x  R ds fy, vodyuh; gSA
 bx  a , x  1
2

rc  ds laHko eku gS (gSa)

(C) ekuk fd  1, bdkbZ (unity) dk ,d lfEeJ ?kuewy gSA ;fn (R) 3
(3 – 3 + 2 ) + (2 + 3 – 3 )
2 4n + 3 2 4n + 3
+ (–3 + 2 + 3 ) 2 4n + 3
= 0,
rc n ds laHko eku gS (gSa)

(D) ekuk fd nks /kukRed okLrfod la[;k,a a vkSj b dk gjkRed ek/; 4 gSA (S) 4
;fn ,d /kukRed okLrfod la[;k q bl izdkj gS fd a, 5,q, b ,d
lekUrj Js.kh gSA rc |q – a| dk (ds) eku gS (gSa)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 101
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

[JEE (Advanced) 2015, P-1 (2, –1)/ 88]

Ans. (A)  P,Q ; (B)  P, Q ; (C)  P,Q,S,T ; (D)  Q, T

 3iˆ  ˆj 
Sol. (A)  ˆi  ˆj  .   3  3    2 3
 2 
2
3   2 3
 3 

  3 +  – 2 = ± 6    4 = 8, –4    = 2, –1
(A  P, Q)
(B) Continuous lrr~   –3a – 2 = b + a2
differentiable vodyuh;  –6a = b  6a = a2 + 3a + 2
 a2 – 3a + 2 = 0  a = 1,2
(B  P, Q)
2ab
(D) 4  ab = 2a + 2b ......(i)
ab
q = 10 – a and rFkk 2q = 5 + b
 20 – 2a = 5 + b  15 = 2a + b .......(ii)
From (i) and vkSj (ii) ls a (15 – 2a) = 2a + 2(15 – 2a)
5
 15a – 2a2 = –2a + 30  2a2 – 17a + 30 = 0  a = 6,
2

5
 q = 4,  |q – a| = 2, 5
2
(D  Q, T)
(C) Let a = 3 – 3 + 22
a = 3 – 32 + 2
a2 = 32 – 3 + 2
  Now a4n + 3 (1 + 4x + 3 + (2)4n + 3) = 0
 n should not be a multiple of 3
Hence P, Q, S, T

33. Column-  Column-

(A) In a triangle XYZ, let a, b and c be the lengths of the sides (P) 1
opposite to the angles X, Y and Z, respectively. If 2(a2 – b2) = c2
sin(X – Y)
and  = , then possible values of n for which
sinZ
cos(n) = 0 is (are)

(B) In a triangle XYZ, let a, b and c be the lengths of the sides (Q) 2
opposite to the angles X, Y and Z, respectively. If
a
1 + cos2X – 2cos2Y = 2sinXsinY, then possible value(s) of
b
is (are)
(C) In R2, let 3 ˆi  ˆj, ˆi  3ˆj and  ˆi  (1– ) ˆj be the position vectors (R) 3
of X, Y and Z with respect to the origin O, respectively. If the
distance of Z from the bisector of the acute angle of OX and OY

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 102
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

3
is , then possible value(s) of || is (are)
2

(D) Suppose that F() denotes the area of the region bounded by (S) 5
x = 0, x = 2, y2 = 4x and y = |x – 1| + |x – 2| + x, where
8
 {0, 1}. Then the value(s) of F() + 2 , when  = 0 and
3
 = 1, is (are)
(T) 6

[JEE (Advanced) 2015, P-1 (2, –1)/ 88]

dkWye -  dkWye-

(A) ekuk fd ,d f=kHkqt XYZ esa dks.kksa X, Y vkSj Z ds lkeus dh Hkqtkvksa dh (P) 1
yEckbZ;k¡ Øe'k% a, b vkSj c gSA ekukfd 2(a2 – b2) = c2 vkSj
sin(X – Y)
= gSA ;fn cos(n) = 0 rc n ds laHko eku gS (gSa)
sin Z

(B) ekuk fd ,d f=kHkqt XYZ esa dks.kksa X, Y vkSj Z ds lkeus dh Hkqtkvksa dh (Q) 2
yEckbZ;k¡ Øe'k% a, b vkSj c gSA ;fn 1 + cos2X – 2cos2Y = 2sinXsinY
a
rc ds laHko eku gS (gSa)
b

(C) ekuk fd R2 esa, ewy fcUnq O ds lkis{k 3 ˆi  ˆj, ˆi  3ˆj vkSj  ˆi  (1– ) ˆj (R) 3
Øe'k% X, Y vkSj Z ds fLFkfr lfn'k (position vectors) gSA ;fn OX vkSj OY
3
ds U;wu dks.k ds f}Hkktd ls Z dh nwjh gks] rks || dk (ds) laHko eku gS (gSa)
2

(D) ekuk fd F() ml {ks=k ds {ks=kQy dks n'kkZrk gS tks (S) 5

x = 0, x = 2, y2 = 4x vkSj y = |x – 1| + |x – 2| + x, ls f?kjk gS]
8
tgk¡  {0, 1} gSA  = 0 vkSj  = 1 ds fy, F() + 2 dk (ds)
3
eku gS (gSa)
(T) 6

Ans. (A)  P,R,S ; (B)  P ; (C)  P,Q ; (D)  S, T

Sol. (A)
X

a b

Y c Z
Given 2(a 2 – b 2 ) = c 2
 2(sin 2 x – sin 2 y) = sin 2 z
 2sin(x + y) sin(x – y) = sin 2 z
 2sin( – z) sin (x – y) = sin 2 z

sin z
z  sin(x – y) = ...(i)
2

also given,
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 103
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

sin x – y
1
 = 
sin z 2
Now, cos(n) = 0
 n 
 cos   = 0
 2 
 n = 1, 3, 5  (A  P,R,S)
X

c b

(B) Y a Z
1 + cos2X – 2cos2Y = 2sin X sin Y
2cos 2 X – 2cos2Y = 2sinX sinY
1 – sin 2 X – 1 + 2sin 2 Y = sin X sin Y
sin 2 X + sinX sinY = 2sin 2 Y
sin(sinX + sinY) = 2sin 2 Y sinX = ak, sinY = bk
a(a + b) = 2b 2
a 2 + ab – 2b 2 = 0
2
a a
b  b – 2 = 0
 
a
= –2, 1
b
a
= 1 (B  P)
b
Hence equation of acute angle bisector of OX and OY is y = x
Hence x – y = 0
 – 1–   3
Now, distance of  ˆi  1–   ˆj  z( , 1 –  ) from x – y is 
2 2

|2 – 1| = 3
2 – 1 = ±3
2 = 4, –2
 = 2, –1
| | = 2, 1 Ans. (P,Q)
 
y 1, 3

60°  
x 3,1
30°

(D)
For  = 1
 3x ; x 1

y = |x – 1| + |x – 2| + x =  1  x ; 1  x  2
3x  3 ; x2


Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 104
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2
1 1
A=
2 2 
(2  3)  1  (2  3)  1  2 xdx
0
8
A=5– 2
3
8
 F(1) +
2 =5
3
For  = 0, y = |–1| + |–2| = 3

2
8
A = 6 – 2 xdx 
0
 A=6–
3
2

8
 F(0) + 2 =6  (D  s, t)
3

Hindi. (A)
X

a b

Y c Z
fn;k x;k gS 2(a 2 – b 2 ) = c 2
 2(sin 2 x – sin 2 y) = sin 2 z
 2sin(x + y) sin(x – y) = sin 2 z
 2sin( – z) sin (x – y) = sin 2 z
sin z
z  sin(x – y) = ...(i)
2
;g Hkh Kkr gS,
sin x – y 1
 = 
sin z 2
vc , cos(n) = 0
 n 
 cos   = 0
 2 
 n = 1, 3, 5  (Q  P,R,S)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 105
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry
X
b
c

(B) Y a Z
1 + cos2x – 2cos2y = 2sin x sin y
2cos 2 x – 2cos2y = 2sinx siny
1 – sin 2 x – 1 + 2sin 2 y = sin x sin y
sin 2 x + sinx siny = 2sin 2 y
sin(sinx + siny) = 2sin 2 y sinx = ak, siny = bk
a(a + b) = 2b 2
a 2 + ab – 2b 2 = 0

1 + cos2x – 2cos2y = 2sin x sin y

2cos 2 x – 2cos2y = 2sinx siny
1 – sin 2 x – 1 + 2sin 2 y = sin x sin y
sin 2 x + sinx siny = 2sin 2 y
sin(sinx + siny) = 2sin 2 y sinx = ak, siny = bk
a(a + b) = 2b 2
a 2 + ab – 2b 2 = 0
2
a a
b  b – 2 = 0
 
a
= –2, 1
b
a
= 1 (B  P)
b
(C)
vr% OX rFkk OY ds e/; U;wu dks.k ds v/kZd dk lehdj.k y = x gSA
vr% x – y = 0
 – 1–   3
vc,  ˆi  1–  ˆj  z( , 1 –  ) dh x – y ls nwjh 
2 2
 
y 1, 3

60°  
x 3,1
30°

|2 – 1| = 3
2 – 1 = ±3
2 = 4, –2
 = 2, –1

| | = 2, 1 Ans. (P,Q)
(D)
 3x ; x 1

For  = 1 y = |x – 1| + |x – 2| + x =  1  x ; 1  x  2
3x  3 ; x2


Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 106
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2
1 1
2 2 
A = (2  3)  1  (2  3)  1  2 xdx
0
8
A=5– 2
3
 F(1) + = 5
ds fy, = 0, y = |–1| + |–2| = 3

2
8

A = 6 – 2 xdx
0
 A=6–
3
2

8
 F(0) + 2 =6  (D  s, t)
3

34. Suppose that p,q and r are three non-coplanar vectors in R 3 . Let the components of a
vector s along p,q and r be 4, 3 and 5, respectively. If the components of this vector s
along (–p  q  r ),(p  q  r ) and (–p – q  r ) are x, y and z, respectively, then the value of
2x + y + z is
ekuk fd R 3 esa] p,q vkSj r rhu vleryh; lfn'k gSA ekuk fd lfn'k ds ?kVd Øekxr lfn'kksa p,q rFkk r ds
vuqfn'k Øe'k% 4, 3 rFkk 5 gSA ;fn ds ?kVd Øekxr lfn'kksa (–p  q  r ),(p  q  r ) ,oa (–p – q  r ) ds
vuqfn'k Øe'k% x, y rFkk z gS] rc 2x + y + z dk eku gSA [JEE (Advanced) 2015, P-2 (4, 0) /
80]
Ans. BONUS
Sol. This question in seem to be wrong but examiner may think like this
;g iz'u xyr yx jgk gS ijUrq ijh{kd bl izdkj lksp ldrk gSA
Ans. BONUS
Sol. This question in seem to be wrong but examiner may think like this
;g iz'u xyr yx jgk gS ijUrq ijh{kd bl izdkj lksp ldrk gSA
S  4p  3q  5r
S  x(p  q  r )  y(p  q  r )  z(p  q  r )
–x + y – z = 4 .....(1)
x–y–z=3 .....(2)
x+y+z=5 .....(3)
add (1) and (2)
(1) vkSj (2) dks tksM+us ij
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 107
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

7
–2z = 7  z= 
2
2x = 8  x=4
y+z=1
2x + y + z = 2(4) + 1 = 9

35*. Consider a pyramid OPQRS located in the first octant (x  0, y  0, z  0) with O as origin, and OP and
OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with
OP = 3. The point S is directly above the mid point T of diagonal OQ such that TS = 3. Then


(A) the acute angle between OQ and OS is [JEE (Advanced) 2016, Paper-1, (4, –2)/62]
3

(B*) the equation of the plane containing the triangle OQS is x – y = 0

3
(C*) the length of the perpendicular from P to the plane containing the triangle OQS is
2

15
(D*) the perpendicular distance from O to the straight line containing RS is
2

fopkj dhft;s] ,d lwP;kdkj (pyramid) OPQRS tks çFke v””"Vka'kd (first octant) (x  0, y  0, z  0) esa fLFkr gS]
ftlesa O ewyfcUnq (origin) rFkk OP vkSj OR Øe’’'k% x- v{k vkSj y- v{k ij gSA bl lwP;kdkj dk vk/kkj (base)
OPQR ,d oxZ (square) gS ftlesa OP = 3 gSA fcUnq S d.kZ (diagonal) OQ ds e/;fcUnq T ds Bhd Åij bl
çdkj gS fd TS = 3 gS] rc


(A) OQ vkSj OS ds chp dk U;wudks.k (acute angle) gS
3

(B) f=kHkqt OQS dks varfo”Z"V (contain) djus okys lery dk lehdj.k x – y = 0 gS

3
(C) P ls f=kHkqt OQS dks varfo”Z"V djus okys lery ij yEc dh yEckbZ gS
2

15
(D) O ls RS dks varfoZ"V djrh gqbZ ljy js[kk dh yEcor~ nwjh gSA
2

Ans. (B,C,D)

R(0, 3, 0) Q(3,3,0)
3
2 3 3 
T  , ,0 
2 2 

O P(3,0,0)
Sol.

3 3  3 3
S   , , 3  OQ = 3iˆ  3ˆj  OS = î + ĵ + 3 k̂
2 2  2 2
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 108
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 1

2 2 1 1
cos = = 
1 1 3 3
2  1 2
2 4 2

n = OQ  OS = (iˆ  ˆj)  (iˆ  ˆj  2k)

ˆ = kˆ  2jˆ  kˆ  2iˆ  2iˆ  2jˆ

3
x–y=  x=y  (3, 0, 0) 
2

x 0 y 3 z0 3 3
RS    =  x= , y = –  + 3, z = 3
3 3 3 2 2

2 2

3 15
T distance ¼nwjh½  39 
2 2

(0, 0, 0,)

3 3 
 2 ,  2   3, 3 
 

2
9 2  3  27 2 9 1
 +  3  2   + 9 = 2  – 9 + 9  =
D= 2 =
4   27 3

36. Let P be the image of the point (3, 1, 7) with respect to the plane x – y + z = 3. Then the equation of the
x y z
plane passing through P and containing the straight line   is
1 2 1

ekuk fd fcUnq (3, 1, 7) dk] lery x – y + z = 3 ds lkis{k (with respect to), izfrfcEc (image) P gSA rc fcUnq P
x y z
ls xqtjus okys vkSj ljy js[kk   dks /kkj.k djus okys lery dk lehdj.k gS&
1 2 1

[JEE (Advanced) 2016, Paper-2, (3, –1)/62]

(A) x + y – 3z = 0 (B) 3x + z = 0 (C) x – 4y + 7z = 0 (D) 2x – y = 0

Ans. (C)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 109
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Q (3,1,7)

• x–y+z=3

Sol.

P (–1,5,3)

x–3 y –1 z–7 –2(6)

= = =  –4
1 –1 1 3

x =– 1, y = 5, z = 3 P (–1, 5, 3)

a(x + 1) + b(y – 5) + c (z – 3) = 0

a + 2b + c = 0 ...................(i)

a – 5b – 3c = 0

a b c
= =
–1 4 –7

–(x + 1) + 4 (y – 5) – 7 (z – 3) = 0

–x + 4y – 7z = 0

x – 4y + 7z = 0

1 ˆ ˆj
37*. Let û u1ˆi u2 ˆj u3kˆ be a unit vector in R3 and ŵ i 2kˆ . Given that there exists a vector
6
in R3 such that û ˆ uˆ
1 and w. 1. Which of the following statements(s) is (are) correct?

(A) There is exactly one choice for

(B) There are infinitely many choices for such [JEE (Advanced) 2016, Paper-2, (4, –2)/62]

(C) If û lies in the xy-plane then u1 u2

(D) If û lies in the xz-plane then 2 u1 u3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 110
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 ˆ
Ekkuk fd R3 esa û u1ˆi u2 ˆj u3kˆ ,d ek=kd lfn’'k (unit vector) gS vkSj ŵ i ˆj 2kˆ gSA fn;k gqvk gS
6

fd R3 esa lfn’'k dk vfLrRo bl izdkj gS fd û ˆ uˆ

1 vkSj w. 1 gSA fuEufyf[kr esa ls dkSu lk ¼ls½

dFku lgh gS ¼gSa½\

(A) bl izdkj ds ds fy, Bhd ,d (exactly one) p;u laHko gSA

(B) bl izdkj ds ds fy, vUkUr (infinitely) p;u laHko gSA

(C) ;fn û xy- lery ij gS rc u1 u2 gSA

(D) ;fn û xz-lery ij gS rc 2 u1 u3 gSA

Ans. (B,C)

Sol. | uˆ  v | 1

|v| sin  = 1  is angle between û & v (, û & v ds e/; dks.k gSA)

Also rFkk, ˆ ˆ  v)  1
w.(u

ˆ | | uˆ | | v | sin  cos  = 1
|w  is angle between ŵ & (uˆ  v) ( ŵ vkSj (uˆ  v) ds e/; dks.k gSA)

1.1(1) cos  = 1    0   uˆ  v  w
ˆ where tgk¡  > 0

ˆi ˆj kˆ
u1 u2 u3 =

6
 ˆi  ˆj  2kˆ  (u2vz – u3vy) î + (u3vx – u1vz) ĵ + (u1vy – u2vx) k̂ =

6
 ˆi  ˆj  2kˆ 
vx vy vz

(B) v is a vector such that (uˆ  v) is parallel to ŵ .

(B) v bl izdkj gS fd (uˆ  v) , ŵ ds lekUrj gSA

 
(C) u3 = 0  u2vz = & –u1vz =  | u2 |  | u1 |
6 6

(D) |u1| = 2|u3| (  u2 = 0)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 111
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

38. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

OP . OQ + OR . OS = OR . OP + OQ . OS = OQ . OR + OP . OS

Then the triangle PQR has S as its [JEE(Advanced) 2017, Paper-2,(3, –1)/61]

(A) centroid (B) orthocenter

(C) incentre (D) circumcenter

ekuk fd O ewyfcUnq (origin) gS ,oe~ PQR ,d LosfPNd f=kHkqt (arbitrary triangle) gSA fcUnq S bl izdkj gS fd

OP . OQ + OR . OS = OR . OP + OQ . OS = OQ . OR + OP . OS

rc fcUnq S f=kHkqt PQR dk gS

(A) dsUnzd(centroid) (B) yEcdsUnz (orthocenter)

(C) vUr%dsUnz (incentre) (D) ifjo`ÙkdsUnz (circumcenter)

Ans. (B)

Sol.

P(p)

•O

Q(q) R(r )

p.q + r.s = r.p + q.s = q.r + p.s

 p.  q – r  – s.  q – r  = 0  PS . QR = 0

Similarly blh izdkj PQ . SR = 0

 S is orthocentre of the triangle (S, f=kHkqt dk yEc dsUnz gksxk½

39. The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes
2x + y – 2z = 5 and 3x – 6y – 2z = 7, is [JEE(Advanced) 2017, Paper-2,(3, –1)/61]

leryksa 2x + y – 2z = 5 ,oe~ 3x – 6y – 2z = 7 ds yEcor~ vkSj fcUnq (1, 1, 1) ls xqtjus okys lery dk lehdj.k gS

(A) 14x + 2y – 15z = 1 (B) –14x + 2y + 15z = 3

(C) 14x – 2y + 15z = 27 (D) 14x + 2y + 15z = 31

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 112
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Ans. (D)

Sol. Let plane be

a(x – 1) + b(y – 1) + c(z – 1) = 0

ˆi ˆj kˆ
Now, direction ratio of its normal = 2 2 = î (–14) – ĵ (2) + k̂ (–15)
1
3 6 2

So, –14(x – 1) –2(y – 1) – 15(z – 1) = 0

14x + 2y + 15z = 31

Hindi. (D)

ekuk lery

a(x – 1) + b(y – 1) + c(z – 1) = 0

ˆi ˆj kˆ
vfHkyEc dh f}dksT;k = 2 2 = î (–14) – ĵ (2) + k̂ (–15)
1
3 6 2

vr%, –14(x – 1) –2(y – 1) – 15(z – 1) = 0

14x + 2y + 15z = 31

Comprehension (Q.40 & 41)

Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides QR, RP, PQ ,
respectively, of a triangle PQR.

vuqPNsn (Q.40 & 41)

ekuk fd O ewyfcUnq (origin) gS ,oe~ OX, OY, OZ Øe'k% f=kHkqt PQR dh Hkqtk;sa QR, RP, PQ , dh fn'kkvks es rhu
,dd lfn'k (unit vectors) gSA

40. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is
[JEE(Advanced) 2017, Paper-2,(3, 0)/61]

;fn f=kHkqt PQR ifjorhZ gS (If the triangle PQR varies), rc] cos(P + Q) + cos(Q + R) + cos(R + P) dk
U;wure eku (minimum value) gS Vector
3 3 5 5
(A) – (B) (C) (D) – 
2 2 3 3

Ans. (A)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 113
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. P Q

cos(P + Q) + cos(Q + R) + cos(R + P) = –cosR – cosP – cosQ

3
In any fdlh f=kHkqt esa  max of cosP + cosQ + cosR =
2

3
So minimum value of the given expression is 
2

3
vr% fn, x, O;atd dk U;wure eku 
2

41. | OX  OY| = [JEE(Advanced) 2017, Paper-2,(3, 0)/61]

(A) sin(P + Q) (B) sin(P + R) (C) sin(Q + R) (D) sin2R

Ans. (A) Vector

Sol.
P


OZ 
OY

Q 
R
OX

 
cosR = – OX. OY

 
 |cosR| = | OX. OY |

 
  OX  OY sinR= |sin( – (P + Q))| = |sin(P + Q)| = sin(P + Q)

42*. Let P1 : 2x + y – z = 3 and P2 : x + 2y + z = 2 be two planes. Then, which of the following statement(s) is
(are) TRUE? [JEE(Advanced) 2018, Paper-1,(4, –2),60] [Vector]
(A) The line of intersection of P1 and P2 has direction ratios 1, 2, – 1
3x  4 1  3y z
(B) The line = = is perpendicular to the line of intersection of P1 and P2
9 9 3
(C) The acute angle between P1 and P2 is 60º
(D) If P3 is the plane passing through the point (4, 2, –2) and perpendicular to the line of intersection of
2
P1 and P2, then the distance of the point (2, 1, 1) from the plane P3 is
3
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 114
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ekukfd P1 : 2x + y – z = 3 vkSj P2 : x + 2y + z = 2 nks lery (plane) gSA rc fuEufyf[kr esa ls dkSulk (ls)
dFku lR; gS(gSa) ?
(A) P1 vkSj P2 dh izfrPNsnu js[kk (line of intersection) ds fnd~-vuqikr (direction ratios) 1, 2, – 1 gSA
3x  4 1  3y z
(B) js[kk = = , P1 vkSj P2 dh izfrPNsnu js[kk ij yEcor~ (perpendicular) gSA
9 9 3

(C) P1 vkSj P2 ds chp dk U;wu dks.k (acute angle) 60º gSA

(D) ;fn lery P3 , fcUnq (4, 2, –2) ls xqtjrk gS rFkk P1 vkSj P2 dh izfrPNsnu js[kk ds yEcor~ gS] rc fcUnq
2
(2, 1, 1) dh lery P3 ls nwjh gSA
3

Ans. (CD)

Sol. Direction ratio of common line is n1 × n2

mHkfu"V js[kk ds fn~d vuqikr n1 × n2

ˆi ˆj kˆ
2 1 1 = ˆi(3)  ˆj(3)  k(3)
ˆ = 3 (iˆ  ˆj  k)
ˆ
1 2 1

x4/3 y  1/ 3 z
(B) = =
3 3 3

This is parallel to line of intersection

;g izfrPNsnu js[kk ds lekUrj gSA

xi x 2 2  2 1 3 1 
(C) cos  = = = =  =
| x1 || x 2 | 6 6 6 2 3

(D) P3 : x – y + z =  satisfy larq"V djrk gS (4, 2, –2)

4–2–2=  x–y+z=0

2  1 1 2
(2,1,1)   
3 3

43. Let a and b be two unit vectors such that a . b  0 . For some x, y  R, let c  xa  yb  (a  b) . If

| c | = 2 and the vector c is inclined at the same angle  to both a and b , then the value of 8 cos2 is

_______. [JEE(Advanced) 2018, Paper-1, (3, 0), 60] [Vector]

ekukfd a vkSj b nks ,sls bdkbZ lfn'k (unit vector) gS fd a . b  0 . fdUgh x, y  R ds fy, ekukfd

c  xa  yb  (a  b) .;fn | c | = 2 vkSj lfn'k c lfn'kksa a vkSj b nksuksa ds lkFk leku dks.k  cukrk gS] rc

8 cos2 dk eku gS _______.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 115
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Ans. (3)

Sol. c  xa  yb  a  b & a.b  0

a c  b  c  

c.a  c.b  2cos   x = y = 2cos

2
c  x2  y2  | a  b |2 = 2(4cos2) + 1 – 0

4 = 8cos2 + 1  8cos2 = 3

44. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is
perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the
z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length
of PR is _______ . [Vector] [JEE(Advanced) 2018, Paper-2,(3, 0)/60]
ekuk fd P izFke v"Bka'k (first octant) esa ,d fcanq gS] ftldk leery (plane) x + y = 3 esa izfrfcEc (imgae) Q
(vFkkZr js[kk[k.M PQ lery x + y = 3 ds yEcor gS vkSj PQ dk e/; fcanq leery x + y = 3 esa fLFkr gS) z-v{k
(axis) ij fLFkr gSA ekuk fd P dh x-v{k ls nwjh 5 gSA ;fn P dk xy-lery esa izfrfcEc R gS] rc PR yEckbZ gS
_______

Ans. (8)

Sol. P()

R(, , –)

x   y   z   2(    3)
  
1 1 0 2

x = 3 – , y = 3 – , z = 

Q(3 – , 3 – , ) lies on z –axis

Q(3 – , 3 – , ), z v+{k ij fLFkr gSA

  = 3,  = 3

P(3, 3, ) distance from x-axis is 5

P(3, 3, ) , x-v{k ls nwjh 5 gSA

9 + 2 = 25

2 = 16  =4

P(3, 3, 4)  PR = 8

R(3, 3, –4)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 116
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

45. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis and
 1 1 1
z-axis, respectively, where O(0, 0, 0) is the origin. Let S  , ,  be the centre of the cube and T be
2 2 2
the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p = SP , q =

SQ , r = SR and t = ST , then the value of |( p × q ) × ( r × t )| is ______ . [JEE(Advanced) 2018,

Paper-2,(3, 0), 60]

izFke v"Bka'k (first octant) esa ,d ,sls ?ku (cube) ij fopkj dhft;s] ftldh Hkqtkvksa (sides) OP, OQ vkSj OR dh
yEckbZ 1 gS vkSj tks Øe'k% x-v{k (axis), y-v{k vkSj z-v{k ds vuqfn'k (along) gSa] tgkW O(0, 0, 0) ewyfcUnq (origin)

gSA ekuk fd ?ku dk dsanz (centre) S  , ,  gS] vkSj 'kh"kZ (vertex) T ewyfcUnq O ds lEeq[k (opposite) okyk og
1 1 1
2 2 2
'kh"kZ gS fd fcanq S fod.kZ (diagonal) OT ij fLFkr gSA ;fn p = SP , q = SQ , r = SR vkSj t = ST rc
|( p × q ) × ( r × t )| dk eku gS ______A

Ans. (0.5)

(0, 1, 0)
Q

T(1, 1, 1)

Sol.

O(0, 0, 0) P(1, 0, 0)

R
(0, 0, 1)

 1 1 1
point fcUnq S  , , 
2 2 2

point fcUnq T(1, 1, 1)

ˆi  ˆj  kˆ
p  SP 
2

ˆi  ˆj  kˆ
q  SQ 
2

ˆi  ˆj  kˆ
r  SR 
2

ˆi  ˆj  kˆ
t  ST 
2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 117
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ˆi ˆj kˆ
1 1 i  ˆj
Now vc p q  1 1 1  = (2 ˆi  2 ˆj) 
4 4 2
1 1 1

ˆi ˆj kˆ
1 2iˆ  2jˆ ˆi  ˆj
r  t  1 1 1  = 
4 4 2
1 1 1

ˆi ˆj kˆ
1 kˆ 1
Now vc (p  q)  (r  t )  1 1 0    (p  q)  (r  t )   0.5
4 2 2
1 1 0

 
46. Let L1 and L2 denote the lines r = î + (– î + 2 ĵ + 2 k̂ ),  R and r = (2 î – ĵ + 2 k̂ ),  R

respectively. If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of
the following options describe(s) L3 ? [Vector] [T] {Vector[VT-SL]-T-305}

[JEE(Advanced) 2019, Paper-1,(4, –1)/62]

 
ekuk fd L1 vkSj L2 Øe'k% fuEu js[kk,a gS : r = î + (– î + 2 ĵ + 2 k̂ ),  R vkSj r = (2 î – ĵ + 2 k̂ ),  R
;fn L3 ,d js[kk gS tks L1 vkSj L2 nksuksa ds yEcor gS vkSj nksuksa dks dkVrh gS] rc fuEufyf[kr fodyiks a esa ls dkSu
lk (ls) L3 dks fu:fir djrk (djrs) gS (gSa) ?

 1
(A) r = (2 î + k̂ ) + t(2 î + 2 ĵ – k̂ ), t  R
3

 2
(B) r = (4 î + ĵ + k̂ ) + t(2 î + 2 ĵ – k̂ ), t  R
9

 2
(C) r = (2 î – ĵ + 2 k̂ ) + t(2 î + 2 ĵ – k̂ ), t  R
9


(D) r = t(2 î + 2 ĵ – k̂ ), t  R

Ans. (ABC)

Sol. Both given lines are skew lines.

So direction ratios of any line perpendicular to these lines are 6 î  6 ĵ  3k̂

<2, 2, –1>

Points at shortest distance between given lines are

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 118
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

A (1–, 2,2)

L1

L2

B (2, –, 2)

AB  line L1

AB  line L2

8 2 2
So A  , , 
9 9 9

 8 2 2 

Now equation of required line r   î  ĵ  k̂    2 î  2 ĵ  k̂ 
9 9 9 

Now by option B, C, D are correct.

Hindi. nh xbZ nksuksa js[kk,a fo"ke ryh; js[kk,a gSA

blfy, fdlh js[kk f}dvuqikr bu js[kkvksa ds yEcor gSA 6 î  6 ĵ  3k̂

<2, 2, –1>

nh xbZ js[kkvksa ds e/; y?kqre nwjh

A (1–, 2,2)

L1

L2

B (2, –, 2)

AB  js[kk L1

AB  js[kk L2

 
8 2 2
blfy, A  , , 
9 9 9

 8 2 2 

r   î  ĵ  k̂    2 î  2 ĵ  k̂ 
9 9 9 

47. Three lines are given by {Vector[VT-PL]-M-303}

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 119
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

r   î,   R

 
r  ( î  ĵ) , n R and r  ( î  ĵ  k̂ ),   R

Let the lines cut the plane x + y + z = 1 at he points A, B and C respectively. If the area of the triangle
ABC is  then the value of (6)2 equals ………. [JEE(Advanced) 2019, Paper-1,(4, –1)/62]

rhu js[kk,a Øe'k%


r   î,   R

 
r  ( î  ĵ) , n R and r  ( î  ĵ  k̂ ),   R

}kjk nh x;h gSaA ekuk fd js[kk,a lery (plane) x + y + z = 1 dks Øe'k% fcUnqvksa A, B vkSj C ij dkVrh gSaA ;fn
f=kHkqt ABC dk {ks=kQy  gS rc (6)2 dk eku cjkcj ……….

Ans. (0.75)

Sol. Put (, 0, 0) in j[kus ij x + y + z = 1esa  = 1  P(1, 0, 0)

1 1 
Put (, , 0) j[kus ij  2 = 1  Q  , ,0 
2 2 

1  1 1 1
Put (,,)j[kus ij =  R , , 
3 3 3 3

1 1  î  ĵ   2 î  ĵ  k̂ 
Area of triangle f=kHkqt PQR dk {ks=kQy = PQ  PR =    =
2 2  2   3 
   

1 3
î  ĵ  k̂  (6)2 = 0.75
12 12

48. Three lines


L1 : r = î ,  R

L2 : r = k̂ + µ ĵ , µ  R and

L3 : r = î  ĵ  k̂,   R .

are given. For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so that P,Q and R
are collinear. [Vector_M] [JEE(Advanced) 2019, Paper-2 ,(4, –1)/62]{[VT-CL]-M-
303}

rhu js[kk,a
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 120
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

L1 : r = î ,  R

L2 : r = k̂ + µ ĵ , µ  R rFkk

L3 : r = î  ĵ  k̂,   R .

nh x;ha gSA L2 ds fdl fcUnq ¼fdu fcUnqvks½ Q ds fy, ge L1 ij ,d fcUnq P, vkSj L3 ij ,d fcUnq R izkIr dj
ldrs gSaA rkfd P, Q vkSj R ljs[k (collinear) gks tk,¡ ?
1 1
(A) k̂  ĵ (B) k̂  ĵ (C) k̂ (D) k̂  ĵ
2 2

Ans. (AD)

Sol. P(, 0, 0)

Q(0, , 1)

R(1, 1, )

  1
PQ  kPR   
 1 1  

1 1
 1+ =    cannot take value of 1 and 0
 1 

  1 vkSj 0 eku ugha ys ldrk gSA

    
49. Let a = 2 î  ĵ  k̂ and b  î  2 ĵ  k̂ be two vectors. Consider a vector c  a  b , ,   R. If the
      
projection of c on the vector (a  b) is 3 2 , then the minimum value of (c  (a  b)) . c equal to
     
;fn a = 2 î  ĵ  k̂ rFkk b  î  2 ĵ  k̂ nks lfn'k fn;s gq, gS ekuk ,d lfn'k c bl izdkj gS fd c  a  b , ,
      
  R. ;fn lfn'k c dk lfn'k (a  b) ij iz{ksi 3 2 gS rks (c  (a  b)) . c dk U;wure eku gksxk&

Ans. (18.00) [JEE(Advanced) 2019, Paper-2 ,(4, –1)/62]

[Vector_M] {[VT-DP]-M-303}

Sol. c  (2 î  ĵ  k̂ ) + ( î  2 ĵ  k̂ )

 c  (2  )î  (  2) ĵ  (  )k̂
  
c . (a  b )
   = 3 2  9( + ) = 18  +=2
|ab|
         
(c  a  b) . c = (a  b  a  b) . (a  b) = 62 + 6 + 62 = 6[2 + (2 – ) + (2 – )2]

= 6(2 – 2 + 4) = min value 18

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 121
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)

Hkkx - II : JEE (MAIN) / AIEEE ¼fiNys o"kks±½ ds iz'u

1. Let a  ˆj – kˆ and c  ˆi – ˆj – kˆ . Then the vector b satisfying a  b  c  0 and a.b  3 is

ekuk a  ˆj – kˆ rFkk c  ˆi – ˆj – kˆ rks lfn'k b tks a  b  c  0 rFkk a.b  3 dks larq"V djrk gS] gksxk&
[AIEEE 2010 (4, –1), 144]
(1) 2iˆ – ˆj  2kˆ (2) ˆi – ˆj – 2kˆ (3) ˆi  ˆj – 2kˆ (4*) – ˆi  ˆj – 2kˆ
Ans. (4)
Sol. Since a  b  c  0
 a  (a  b)  a  c  0  (a . b)a  (a . a) b  a  c  0
Since  a  c = 2iˆ  ˆj  kˆ
 3(ˆj  k)
ˆ  2b  2iˆ  ˆj  kˆ  0  b  ˆi  ˆj  2kˆ
Hence correct option is (4)
vr% lgh fodYi (4) gSA

2. If the vectors a  ˆi – ˆj  2kˆ , b  2 ˆi  4ˆj  kˆ and c   ˆi  ˆj  µkˆ are mutually orthogonal, then ( , µ)
=
;fn lfn'k a  ˆi – ˆj  2kˆ , b  2 ˆi  4ˆj  kˆ rFkk c   ˆi  ˆj  µkˆ ijLij yEcdks.kh; gS] rks ( , µ) cjkcj gS&
[AIEEE 2010 (4, –1), 144]
(1) (2, – 3) (2) (–2, 3) (3) (3, – 2) (4*) (–3, 2)
Ans. (4) 
Sol.  a, b, c are mutually orthogonal ijLij yEcor~ gSA
 a . c 0   – 1 + 2 = 0 ........(i)
and rFkk
b . c 0   + 4 +  = 0 ........(ii)
solving (i) and (ii), we get  = – 3 and  = 2
(i) o (ii) dks gy djus ij  = – 3 rFkk  = 2
Hence correct option is (4)
vr% lgh fodYi (4) gSA

3. Statement -1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5.
Statement -2 : The plane x – y + z = 5 bisects the line segment joining A(3, 1,6) and B(1, 3, 4).
(1*) Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement
-1.
(2) Statement-1 is true, Statement-2 is false. [AIEEE 2009 (4, –1), 144]
(3) Statement -1 is false, Statement -2 is true.
(4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.
izzdFku-1 : lery x – y + z = 5 esa fcUnq B(1, 3, 4) dk niZ.k izfrfcEc fcUnq A(3, 1, 6) gSA
izzdFku-2 : lery x – y + z = 5, A(3, 1,6) rFkk B(1, 3, 4) dks feykus okys js[kk[k.M dks lef}Hkkftr djrk gSA
(1*) izzdFku-1 lR; gS] izdFku-2 lR; gS ; izdFku-2, izdFku-1 dh lgh O;k[;k ugha gSA
(2) izdFku-1 lR; gS] izdFku-2 feF;k gSA
(3) izdFku-1 feF;k gS] izdFku-2 lR; gSA
(4) izdFku-1 lR; gS] izdFku-2 lR; gS ; izdFku-2, izdFku-1 dh lgh O;k[;k gSA
Ans. (1)
Sol. Let image be ()

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 1
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 1 3 4  1 3  4  5 
= = = – 2 
1 1 1  3 
 1 3 4
 = = =2
1 1 1
  = 3,  = 1,  = 6
 A(3, 1, 6) statement 1 is true
Now midpoint of A(3, 1, 6) and B(1, 3, 4) is (2, 2, 5)
equation of plane is x – y + z = 5
coordinates of midpoint lies on the plane so plane bisects the line segment AB. But it is not correct
explanation of statement-1
Hence correct option is (1)
Hindi ekuk izfrfcEc () gS rks
 1 3 4  1 3  4  5 
= = = – 2 
1 1 1  3 
 1 3 4
 = = =2
1 1 1
  = 3,  = 1,  = 6
 A(3, 1, 6) dFku -1 lR; gSA
vc A(3, 1, 6) vkSj B(1, 3, 4) dk e/; fcUnq (2, 2, 5) gSA
lery dk lehdj.k x – y + z = 5
e/; fcUnq ds funsZ'kkad lery ij fLFkr gS blfy, lery js[kk[k.M AB dks lef}Hkkftr djrk gS ijUrq ;g dFku -
1 dk lgh Li"Vhdj.k ugha gSA
vr% lgh fodYi (1) gSA
4. A line AB in three-dimensional space makes angles 45º and 120º with the positive x-axis and the
positive
y-axis respectively. If AB makes an acute angle  with the positive z-axis, then  equal
f=kfoe lef"V esa js[kk AB, x-v{k dh /kukRed&fn'kk ds lkFk 45º dk dks.k rFkk y-v{k dh /kukRed&fn'kk ds lkFk
120º dk dks.k cukrh gSA ;fn AB, z-v{k dh /kukRed fn'kk ls U;wu dks.k  cukrh gS] rks  cjkcj gS&
[AIEEE 2010 (4, –1), 144]
(1) 45º (2*) 60º (3) 75º (4) 30º
Ans. (2) 
1 1
Sol. = , m=–
2 2
1 1
2 + m2 + n2 = 1  n2 =  n=±
4 2
1
cos  = ,  = 60º Hence correct option is (2) vr% lgh fodYi (2) gSA
2

1 ˆ and b  1 (2iˆ  3ˆj  6k)

ˆ , then the value of (2a  b) . [(a  b)  (a  2b)] is:
5. If a  (3iˆ  k)
10 7
1 ˆ rFkk b  1 (2iˆ  3ˆj  6k)
;fn b  (2iˆ  3ˆj  6k) ˆ , rks (2a  b) . [(a  b)  (a  2b)] cjkcj gS %
7 7
[AIEEE 2011, I, (4, –1), 120]
(1*) – 5 (2) –3 (3) 5 (4) 3

Sol. (1)
(2 a – b ) . [( a × b ) × ( a + 2 b )]
= – (2 a – b ) . [( a + 2 b ) × ( a × b )]

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 2
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

= – (2 a – b ) . [( a +2 b ) . b ) a – (( a + 2 b ) . a ) b ]
= – (2 a – b ) . [( a . b ) +2. b ) a – ( a . a + 2 b . a ) b )]
= – (2 a – b ) . [0 + 2 a – (0 + b )]
= – (2 a – b ) . (2 a – b )
= – (2 a – b )2 = – 4 a 2 + 4 a . b – b 2
= – 4 + 0 – 1 = – 5 Ans.

6. The vectors a and b are not perpendicular and c and d are two vectors satisfying : b  c  b  d and
a.d  0 . Then the vector d is equal to : [AIEEE 2011, I, (4, –1), 120]
a rFkk b yEcor ugha gS rFkk c rFkk d bl izdkj ds gS fd b  c  b  d rFkk a . d  0 gS] rks d cjkcj gS:
[AIEEE 2011, I, (4, –1), 120]
 b.c   a.c   b.c   a.c 
(1) b   (2) c   (3) b   (4*) c  
 a.d   a.b   a.b   a.b 
c b c b
       
Sol. (4)
a . b  0 , bc  bd , a . d = 0
(b  c) × a = (b  d) × a
(b.a) – (c.a) = (b . a) d – (d . a) b
a . c
d = c – 
 a. b 
b
 

7. If the vector p î + ĵ + k̂ , î + q ĵ + k̂ and î + ĵ + r k̂ (p  q  r  1) are coplanar, then the value of pqr

– (p+q+r) is-

;fn lfn'k p î + ĵ + k̂ , î + q ĵ + k̂ rFkk î + ĵ + r k̂ (p  q  r  1) leryh; gS] rks pqr – (p+q+r) dk eku gS

%
[AIEEE 2011, II, (4, –1), 120]
(1) 2 (2) 0 (3) –1 (4*) –2
Sol. (4)
p 1 1
1 q 1 =0
1 1 r
 p (qr – 1) + 1(1 – r) + 1(1 – q) = 0
 pqr – p + 1– r + 1 – q = 0
 pqr – (p + q + r) = – 2

8. Let a , b , c be three non-zero vectors which are pairwise non-collinear. If a + 3 b is collinear with c
and b + 2 c is collinear with a , then a + 3 b + 6 c is :
ekuk a , b , c rhu 'kwU;srj lfn'k gS tks ;qXer% lajs[k ugha gSA ;fn a + 3 b ,oa c lajs[kh; gS rFkk b + 2 c ,oa
a lajs[kh; gSa] rks a + 3 b + 6 c cjkcj gS % [AIEEE 2011, II, (4, –1), 120]
(1) a (2) c (3*) 0 (4) a + c

Sol. (3)
a  3b  c ........(1)
b  2c  µa .........(2)
(1) – 3(2) gives (1 + 3µ) a – ( + 6) c = 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 3
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

As a and c are non collinear

a rFkk c vlajs[kh; gSA vr%
 1 + 3µ = 0 and vkSj  + 6 = 0
From (1) ls a  3b  6c  0

y 1 z3  5 
9. If the angle between the line x = = and the plane x + 2y + 3z = 4 is cos–1   , then 
2   14 
equals:

y 1 z3  5 
;fn js[kk x = = rFkk lery x + 2y + 3z = 4 ds chp dk dks.k cos–1   gS] rks dk eku gS%
2   14 
[AIEEE 2011, I, (4, –1), 120]
2 3 2 5
(1*) (2) (3) (4)
3 2 5 3
Sol. (1)
x – 0 y –1 z – 3
  ....... (1)
1 2 
x + 2y + 3z = 4 ....... (2)
Angle between the line and plane is
a1a2  b1b2  c1c 2
cos (90° – ) =
a1  b12  c12 a22  b22  c 22
2

1  4  3 5  3
 sin=  ....... (3)
14  5   2
14  5   2
But given that angle between line and plane is
 5   3 
= cos–1 
 14 
= sin–1  
   14 
3
 sin  =
14
 from (3)
3 5  3
=
14 14  5   2
 9(5 + ) = 25 + 92 + 30 
 30= 20
2
= Ans.
3
x – 0 y –1 z – 3
Hindi   ....... (1)
1 2 
x + 2y + 3z = 4 ....... (2)
js[kk rFkk lery ds chp dk dks.k
a1a2  b1b2  c1c 2 1  4  3 5  3
cos (90° – ) =  sin=  ....... (3)
a b c
2
1
2
1
2
1 a b c
2
2
2
2
2
2 14  5   2
14  5   2
ijUrq fn;k gS] fd js[kk rFkk lery ds chp dk dks.k
 5   3 
= cos–1 
 14 
= sin–1  
   14 
3
 sin  =
14
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 4
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 from (3)
3 5  3
=
14 14  5   2
 9(5 + ) = 25 + 92 + 30 
 30= 20
2
= Ans.
3

10. Statement-1 : The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line :
x y 1 z  2
 
1 2 3
[AIEEE 2011, I, (4, –1), 120]
x y 1 z  2
Statement-2 : The line :   bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).
1 2 3
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(2*) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true.
x y 1 z  2
dFku-1 :   js[kk esa fcUnq B(1, 6, 3) dk niZ.k izfrfcac (mirror image) fcUnq A(1, 0, 7) gSA
1 2 3
x y 1 z  2
dFku-2 :   js[kk fcUnqvksa A(1, 0, 7) rFkk B(1, 6, 3) dks feykus okys js[kk[kaM dk lef}Hkktu
1 2 3
djrh gSA [AIEEE 2011, I, (4, –1), 120]
(1) dFku-1 lR; gS] dFku-2 lR; gSA dFku-2, dFku-1 dh lgh O;k[;k gSA
(2*) dFku-1 lR; gS] dFku-2 lR; gSA dFku-2, dFku-1 dh lgh O;k[;k ugha gSA
(3) dFku-1 lR; gS] dFku-2 vlR; gSA
(4) dFku-1 vlR; gS] dFku-2 lR; gSA
Sol. (2)

Mid- point of AB  M(1,3,5)

M lies on line
Direction ratios of AB is < 0, 6, – 4 >
Direction ratios of given line is < 1, 2, 3 >
As AB is perpendicular to line
 0.1 + 6.2 – 4.3 = 0
Hindi

AB dk e/; fcUnq  M(1,3,5)

M js[kk ij fLFkr gSA
AB ds fn~d vuqikr < 0, 6, – 4 >
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 5
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

nh xbZ js[kk ds fn~d vuqikr < 1, 2, 3 >

pawfd nh xbZ js[kk AB ds yEcor~ gSA
 0.1 + 6.2 – 4.3 = 0

11. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along a straight line
x = y = z is :
fcUnq (1, –5, 9) dh lery x – y + z = 5 ls nwjh tks ljy js[kk x = y = z ds vuqfn'k ekih x;h gS] gS %
[AIEEE 2011, II, (4, –1), 120]
(1*) 10 3 (2) 5 3 (3) 3 10 (4) 3 5
Sol. (1)
x –1 y5 z–9
Line through P(1, – 5, 9) parallel to x = y = z is = = = (say)
1 1 1
x –1 y5 z–9
fcUnq P(1, – 5, 9) ls xqtjus okyh rFkk x = y = z ds lekUrj js[kk = = = (ekuk) gSA
1 1 1
Q (x = 1 + , y = – 5 + , z = 9 + )
Given planefn;k x;k lery x – y + z = 5
 1++5–+9+=5
  = –10
 Q(–9, – 15, – 1)
 PQ = 300
= 300 = 10 3

x y2 z3
12. The length of the perpendicular drawn from the point (3, –1, 11) to the line = = is :
2 3 4
x y2 z3
fcUnq (3, –1, 11) ls js[kk = = ij Mkys x, yEc dh yEckbZ gS % [AIEEE 2011, II, (4, –1), 120]
2 3 4
(1) 29 (2) 33 (3*) 53 (4) 66
Sol. (3)
Let feet of perpendicular is (2, 3+ 2, 4+ 3)
ekuk yEcikn ds funsZ'kkad (2, 3+ 2, 4+ 3) gSA
 Dratio of the perpendicular line < 2– 3, 3+ 3, 4– 8 >
yEcor~ js[kk ds fn~d vuqikr < 2– 3, 3+ 3, 4– 8 >
and Dratio of the line < 2, 3, 4 >
vkSj nh xbZ js[kk ds fn~d vuqikr < 2, 3, 4 >
 2(2– 3) + 3 (3+ 3) + 4 (4– 8) = 0
 29 – 29 = 0
 = 1
 feet of perpendicular is (2, 5, 7)
yEcikn ds funsZ'kkad (2, 5, 7)
 length perpendicular is yEc dh yEckbZ 12  62  42 = 53

13. Let â and b̂ be two unit vectors. If the vectors c  aˆ  2bˆ and d  5aˆ  4bˆ are perpendicular to each
other, then the angle between â and b̂ is :
ekuk â rFkk b̂ nks ek=kd lfn'k gSAa ;fn lfn'k c  aˆ  2bˆ vkSj d  5aˆ  4bˆ ijLij yacor~ gSa] rks â rFkk b̂ ds
chp dk dks.k gS % [AIEEE-2012, (4, –1)/120]
   
(1) (2) (3*) (4)
6 2 3 4
Sol. Ans. (3)
c = â  2bˆ
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 6
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

d = 5aˆ – 4bˆ
c.d = 0
 ˆ
(aˆ  2b).(5a ˆ =0
ˆ – 4b)  5 + 6aˆ . bˆ – 8 = 0
1 
 â . bˆ =  =
2 3

14. A equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is :
lery x – 2y + 2z – 5 = 0 ds lekarj rFkk ewy fcanq ls ek=kd nwjh ij ,d lery dk lehdj.k gS&
[AIEEE 2012, (4, –1), 120]
(1*) x – 2y + 2z – 3 = 0 (2) x – 2y + 2z + 1 = 0 (3) x – 2y + 2z – 1 = 0 (4) x – 2y + 2z + 5 = 0
Sol. Ans. (1)

Equation of parallel plane x – 2y + 2z + d = 0

d
Now =1
1  22  22
2

d=3
So equation required plane x – 2y + 2z ± 3 = 0

Hindi Ans. (1)

lekUrj lery dk lehdj.k x – 2y + 2z + d = 0
d
vc =1
1  22  22
2

d=3
vr% vHkh"V lery dk lehdj.k gS x – 2y + 2z ± 3 = 0

x 1 y 1 z 1 x3 y k z
15. If the line = = and = = intersect, then k is equal to :
2 3 4 1 2 1
x 1 y 1 z 1 x3 y k z
;fn js[kk,¡ = = rFkk = = ijLij izfrPNsn djrh gSa] rks k cjkcj gS % :
2 3 4 1 2 1
[AIEEE 2012, (4, –1), 120]
2 9
(1) – 1 (2) (3*) (4) 0
9 2
Sol. Ans. (3)

x –1 y 1 z –1
= =
2 3 4
x–3 y–k z
= =
1 2 1
a (1, – 1, 1) ; r = a  b
b (2, 3, 4)
c (3, k, 0); r = c  µd
d (1, 2, 1)
These lines will intersect if lines are coplaner
a – c , b̂ & d are coplaner
 [a – c, b, d] = 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 7
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2 k  1 –1
2 3 4 =0
1 2 1
 2(–5) – (k + 1) (– 2) – 1 (1) = 0
 2(k + 1) = 11
9
 k=
2
x –1 y 1 z –1
Hindi = =
2 3 4
x–3 y–k z
= =
1 2 1
a (1, – 1, 1) ; r = a  b
b (2, 3, 4)
c (3, k, 0); r = c  µd
d (1, 2, 1)
ljy js[kk,sa izfrPNsn djsxh ;fn os leryh; gS
a – c , b̂ & d leryh; gSA
 [a – c, b, d] = 0
2 k  1 –1
2 3 4 =0
1 2 1
 2(–5) – (k + 1) (– 2) – 1 (1) = 0
 2(k + 1) = 11
9
 k=
2

16. Let ABCD be a parallelogram such that AB  q , AD  p and BAD be an acute angle. If r is the
vector that coincides with the altitude directed from the vertex B to the side AD, then is given by :
ekuk ABCD ,d ,slk lekarj prqHkqZt gS fd AB  q , AD  p rFkk BAD ,d U;wu dks.k gSA ;fn lfn'k r ] 'kh"kZ
B ls AD ij [khaps x, yac ds laikrh gS] rks fuEu }kjk iznÙk gS : [AIEEE-2012, (4, –1)/120]
3(p . q) p . q p . q 3(p . q)
(1) r  3q  p (2*) r  q    p (3) r  q    p (4) r  3q  p
(p . p) p . p p . p (p . p)
Sol. Ans. (2)

p . q p p . q
AX = = p
|p| |p| | p |2
BX = BA + AX
p . q
=– q + p
| p |2

17. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

[AIEEE - 2013, (4, –1),360]
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 8
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

nks lekUrj leryksa 2x + y + 2z = 8 rFkk 4x + 2y + 4z + 5 = 0 ds chp dh nwjh gS&

[AIEEE - 2013, (4, –1),360]
3 5 7 9
(1) (2) (3*) (4)
2 2 2 2
Sol. (3)
2x + y + 2z – 8 = 0 ...(P1)
5
2x + y + 2z + =0 ...(P2)
2
5
–8 –
2 7
Distance between P1 and P2 = 
2  1  22
2 2 2

Hindi. (3)
2x + y + 2z – 8 = 0 ...(P1)
5
2x + y + 2z + =0 ...(P2)
2
5
–8 –
2 7
P1 vkSj P2 ds chp dh nwjh = 
2  1  22
2 2 2

x–2 y–3 z–4 x –1 y–4 z–5

18. If the lines = = and = = are coplanar, then k can have
1 1 –k k 2 1
[AIEEE - 2013, (4, –1),360]
(1) any value (2) exactly one value (3*) exactly two values (4) exactly three values
x–2 y–3 z–4 x –1 y–4 z–5
;fn js[kk,a = = rFkk = = leryh; gS] rks k dk
1 1 –k k 2 1
(1) dksbZ Hkh eku laHko gSA (2) dsoy ,d eku laHko gSA (3) dsoy nks eku laHko gSA (4) dsoy rhu eku laHko gSA
[AIEEE - 2013, (4, –1),360]
Sol. (3)
[a – c, b, d] = 0
2 –1 3 – 4 4 – 5
1 1 –k =0
k 2 1
1 –1 –1
1 1 –k =0
k 2 1
 1(1 + 2k) + (1 + k2) – (2 – k) = 0
 k2 + 2k + k = 0
 k2 + 3k = 0
 k = 0, –3
Note : If 0 appears in the denominator, then the correct way of representing the equation of straight line is
x–2 y–3
 ; z=4
1 1
x–2 y–3
uksV : ;fn 0 gj esa vkrk gS] rc ljy js[kk dks fu:fir djusa dk lgh rjhdk  ; z = 4 gSA
1 1
19. If the vectors AB  3iˆ  4kˆ and AC  5iˆ – 2jˆ  4kˆ are the sides of a triangle ABC, then the length of
the median through A is [AIEEE - 2013, (4, –¼),360]
;fn lfn'k AB  3iˆ  4kˆ rFkk AC  5iˆ – 2jˆ  4kˆ ,d f=kHkqt ABC dh Hkqtk,¡ gS] rks A ls gksdj tkrh gqbZ
ekf/;dk dh yEckbZ gS& [AIEEE - 2013, (4, –¼),360]
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 9
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(1) 18 (2) 72 (3*) 33 (4) 45

Sol. (3)

AB  BC  CA = 0
 BC  AC – AB
AC – AB
 BM 
2
 AB  BM  MA = 0
AC – AB
 AB   AM
2
AB  AC
 AM   4iˆ – ˆj  4kˆ
2
 AM  33 

x 1 y  3 z  4
20. The image of the line   in the plane 2x – y + z + 3 = 0 is the line :
3 1 5
x 1 y  3 z  4
lery 2x – y + z + 3 = 0 esa js[kk   ds izfrfcEc okyh js[kk gS :
3 1 5
[JEE(Main) 2014, (4, – 1), 120]
x 3 y 5 z2 x 3 y 5 z2
(1)   (2)  
3 1 5 3 1 5
x 3 y 5 z2 x 3 y 5 z2
(3*)   (4)  
3 1 5 3 1 5
Sol. Ans. (3)
x 1 y3 z4 2(2  3  4  3)
= = =
2 1 1 4  1 1
x 1 y3 z4
 = = =–2
2 1 1
Image of point (1, 3, 4) on given line in the given plane is (–3, 5, 2)
js[kk ij fLFkr fcUnq (1, 3, 4) dk lery es izfrfcEc gS (–3, 5, 2)
Line is parallel to given plane js[kk fn;s lery ds lekUrj gS 3, 1, –5
So, image blfy, izfrfcac
x3 y5 z2
= =
3 1 5

21. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l 2 = m2 +
n2 is
[Vector & 3-D] [JEE(Main) 2014, (4, – 1), 120]
nks js[kk,sa ftuds fnd~ dksT;k] lehdj.kksa l + m + n = 0 rFkk l2 = m2 + n2 dks larq"V djrs gS] ds chp dk dksa.k gSµ
[Vector & 3-D] [JEE(Main) 2014, (4, – 1), 120]
   
(1) (2) (3*) (4)
6 2 3 4
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 10
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. Ans. (3)

mn  0 ...............(1)
2
 m2  n2 ...............(2)
 2
 m2 – (–  – m)2 = 0

 2m(m + ) = 0
m=0 or =–m
so direction ratios are blfy, f}d~ vuqikr gS – 1, 0, 1 vkSj and – 1, 1, 0
 a1a2  b1b2  c1c 2 
 cos=  
 a2  b2  c 2 a2  b2  c 2 
 1 1 1 2 2 2 
1 0  0 1
 cos   =
2 2 2

 
3

2
22. If a  b bc c  a   a b c  then  is equal to [JEE(Main) 2014, (4, – 1), 120]
   
2
;fn a  b b  c c  a   a b c  gS] rks cjkcj gSµ [JEE(Main) 2014, (4, – 1), 120]
(1) 0 (2*) 1 (3) 2 (4) 3
Sol. Ans. (2)
LHS = [ a × b b × c c × a ]
= [ p b × c c × a ] (where tgk¡ p = a × b )
= { p × ( b × c )}. ( c × a )
= {p × (b × c ) (c × a )
= {( p . c ) b – ( p . b ) c } . ( c × a )
= {[ a b c ] – [ a b c ]}.( c × a ) (As p = a × b )
= [ a b c ][ b c a ] – 0 ( [ a b b ] = 0)
= [ a b c ]2 ([ b c a ] = [ a b c ])
RHS =  [ a b c ]2
 = 1

x–2 y 1 z–2
23. The distance of the point (1,0,2) from the point of intersection of the line = = and the
3 4 12
plane x – y + z = 16, is [JEE(Main) 2015, (4, – ¼), 120]
(1) 2 14 (2) 8 (3) 3 21 (4) 13
x–2 y 1 z–2
js[kk = = rFkk lery x – y + z = 16 ds izfrPNsn fcUnq dh] fcUnq (1, 0, 2) ls nwjh gS&
3 4 12
[JEE(Main) 2015, (4, – ¼), 120]
(1) 2 14 (2) 8 (3) 3 21 (4) 13
Ans. (4)
Sol. Point of intersection
(3 + 2, 4 – 1, 12 + 2)
3 + 2 – 4 + 1 + 12 + 2 = 16
11 = 11  = 1
(5, 3, 14)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 11
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Distance = 16  9  144  169  13

Hindi. izfrPNsn fcUnq
(3 + 2, 4 – 1, 12 + 2)
3 + 2 – 4 + 1 + 12 + 2 = 16
11 = 11  = 1
(5, 3, 14)
nqjh = 16  9  144  169  13

24. The equation of the plane containing the line 2x – 5y + z = 3, x + y + 4z = 5 and parallel to the plane
x + 3y + 6z = 1, is [JEE(Main) 2015, (4, – ¼), 120]
(1) 2x + 6y + 12z = 13 (2) x + 3y + 6z = –7 (3) x + 3y + 6z = 7 (4) 2x + 6y + 12z = –
13
js[kk 2x – 5y + z = 3, x + y + 4z = 5 dks varfoZ"V djus okys lery] tks lery x + 3y + 6z = 1 ds lekUrj gS]
dk lehdj.k gS& [JEE(Main) 2015, (4, – ¼), 120]
(1) 2x + 6y + 12z = 13 (2) x + 3y + 6z = –7 (3) x + 3y + 6z = 7 (4) 2x + 6y + 12z = –
13
Ans. (3)
Sol. Equation of real plane
2x – 5y + z – 3 + (x + y + 4z – 5) = 0
x(2 + ) + y(– 5) + z(4 + 1) – 3 – 5 = 0
  2   5 4  1 55
    3 
1 3 6 2
3 + 6 =  – 5
2 = –11
11
=
2
7x 21y 49
 equation of plane   21z  0  7x + 21y + 42z – 49 = 0
2 2 2
 x + 3y + 6z – 7 = 0
Hindi. okLrfod lery dk lehdj.k
2x – 5y + z – 3 + (x + y + 4z – 5) = 0
x(2 + ) + y(– 5) + z(4 + 1) – 3 – 5 = 0
  2   5 4  1 55
    3 
1 3 6 2
3 + 6 =  – 5
2 = –11
11
=
2
7x 21y 49
 lery dk lehdj.k   21z  0  7x + 21y + 42z – 49 = 0
2 2 2
 x + 3y + 6z – 7 = 0

25. Let a, b and c be three non-zero vectors such that no two of them are collinear and
1
(a b)  c  | b || c | a . If is the angle between vectors b and c , then a value of sin is
3
[JEE(Main) 2015, (4, – ¼), 120]
2 2 – 2 2 –2 3
(1) (2) (3) (4)
3 3 3 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 12
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1
ekuk a, b rFkk c rhu 'kwU;srj ,sls lfn'k gS fd muesa ls dksbZ nks lajs[k ugha gS rFkk (a b)  c  | b || c | a
3
gSA ;fn lfn'kksa b vkSj c ds chp dk dks.k  gS] rks sin dk ,d eku gS&[JEE(Main) 2015, (4, – ¼), 120]
2 2 – 2 2 –2 3
(1) (2) (3) (4)
3 3 3 3
Ans. (1)

Sol.

a  b  c  31 | b || c | a
–c   a  b 

–  c.b  a   c.a  b  | b || c | a
1
3
1
  
 3 | b || c |  c.b  a   c.a  b
 
Since a & b are not collinear
1
c.b  |b| |c| = 0 & c.a = 0
3
1
cos + =0
3
1 8 2 2
cos = –  sin = =
3 3 3

Aliter : a  b c = a.c b – b.c a  31 | b || c | a

 a.c  b   3 | b || c | b.c  a
1
 
1
a.c = 0 & | b || c | b.c = 0
3
1
| b || c |  | b || c | cos = 0
3
1
cos = –
3
1 2 2
sin = 1– 
9 3

Hindi.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 13
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

a  b  c  31 | b || c | a
–c   a  b 

–  c.b  a   c.a  b  | b || c | a
1
3
1
 
 3 | b || c |  c.b  a   c.a  b
 
vr% a vkSj b lajs[kh; ugha gSA
1
c.b  |b| |c| = 0 & c.a = 0
3
1
cos + =0
3
1 8 2 2
cos = –  sin = =
3 3 3
oSdfYid : a  b c = a.c b – b.c a  31 | b || c | a
 a.c  b   3 | b || c | b.c  a
1
 
1
a.c = 0 & | b || c | b.c = 0
3
1
| b || c |  | b || c | cos = 0
3
1
cos = –
3
1 2 2
sin = 1– 
9 3

x–3 y2 z4

26. If the line,   lies in the plane, lx + my – z = 9, then l2 + m2 is equal to
2 –1 3
[JEE(Main) 2016, (4, – 1), 120]
x–3 y2 z4
;fn js[kk   , lery lx + my – z = 9 esa fLFkr gS] rks l2 + m2 cjkcj gS&
2 –1 3
(1) 18 (2) 5 (3) 2 (4) 26
Ans. (3)
Sol. (i) (3, – 2, –4) lies on the plane ¼lery ij fLFkr gS½
 3 – 2m + 4 = 9  3 – 2m = 5 ....... (i)

(ii) 2 – m – 3 = 0  2 – m = 3 ....... (ii)

from (i) and (ii) (i) rFkk (ii) ls
= 1 and ¼vkSj½ m = – 1

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 14
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

27. Let a, b and c be three unit vectors such that a  b  c =  2

3
(b  c) . If b is not parallel to c , then

the angle between a and b is [JEE(Main) 2016, (4, – 1), 120]

ekuk a, b rFkk c rhu ,sls ek=kd lfn’k gS fd a  b  c  =

3
(b  c) gSA ;fn b , c ds lekUrj ugha gS] rks a
2
rFkk b ds chp dk dks.k gS&
 2 5 3
(1) (2) (3) (4)
2 3 6 4
Ans. ( 3)
3
Sol. a × (b  c) = (b  c)
2
3 3
(a .c ) – (a . b)c = b + c
2 2
3 3
Hence ¼vr%½ a . c = and ¼vkSj½ a . b = –
2 2
3
a.b= –
2
3
cos = –
2
5
=
6

28. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is
[JEE(Main) 2016, (4, – 1), 120]
fcUnq (1, –5, 9) dh lery x – y + z = 5 ls og nwjh tks js[kk x = y = z dh fn’'kk esa ekih xbZ gS] gS&
10 20
(1) 10 3 (2) (3) (4) 3 10
3 3

Ans. (1)
P(1,–5, 9)

x= y =z

x– y+ z =5

Sol.
x 1 y  5 z  9
Equation of line PQ :   
1 1 1
x 1 y  5 z  9
js[kk PQ dk lehdj.k gS   
1 1 1
Q can be taken as ( + 1,  – 5,  + 9)
Q dks ( + 1,  – 5,  + 9) ekuk tk ldrk gSA
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 15
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

As Q lies on plane x – y + z = 5
tSlk fd Q lery x – y + z = 5 ij fLFkr gS

 ( + 1) – ( – 5) + ( + 9) = 5
 = –10  Q(–9, –15, –1)
(1  9)2  (5  15)2  (9  1)2 100  100  100 = 10 3
 Required distance PQ = =
(1  9)2  ( 5  15)2  (9  1) 2 100  100  100 = 10 3
 vHkh”V nwjh PQ = =

29. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,
x y z
= = is Q, then PQ is equal to : [JEE(Main) 2017, (4, – ¼), 120]
1 4 5
;fn fcanq P(1, –2, 3) dk lery 2x + 3y – 4z + 22 = 0 esa ;g izfrfcEc dh js[kk
x y z
= = ds lekarj gS Q gS] rks PQ cjkcj gS :
1 4 5
(1) 3 5 (2) 2 42 (3) 42 (4) 6 5
Ans. (2)
Sol. Let R be the point of intersection of plane and line passing through P and parallel to given line.
lery rFkk P ls xqtjus okyh js[kk vkSj nh xbZ js[kk ds lekUrj js[kk dk izfrPNsn fcUnq ekuk R gSA
So, vr% R is gSA (1 + , – 2 + 4, 3 + 5)
substituting co-ordinates of R in plane
lery esa R ds funsZ’kkad izfrLFkkfir djus ij
2 + 2 – 6 + 12 – 12 – 20 + 22 = 0  6 = 6   = 1
So vr%, R is (2,2,8)

Hence bl izdkj PR = 1  16  25 = 42

So blfy,, PQ = 2 42

30. The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, –1), having normal
x 1 x2 x4 x  2 y 1 z  7
perpendicular to both the lines = = and   , is
1 2 3 2 1 1
[JEE(Main) 2017, (4, – ¼), 120]
x 1 x2 x4
,d lery tks fcUnq (1, –1, –1) ls gksdj tkrk gS rFkk ftldk vfHkyEc nksuksa js[kkvksa = =
1 2 3
x  2 y 1 z  7
rFkk   ij yEc gS] dks fcUnq (1, 3, – 7) ls nqjh gS :
2 1 1
20 10 5 10
(1) (2) (3) (4)
74 83 83 74
Ans. (2)
Sol. Let the plane be ekuk lery
a(x – 1) + b(y + 1) + c(z + 1) = 0
a – 2b + 3c = 0
2a – b – c = 0
a b c
 
5 7 3
5(x – 1) + 7(y + 1) + 3(z + 1) = 0
5x + 7y + 3z + 5 = 0
P(1, 3, –7)
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 16
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

5  21  21  5 10
d= 
25  49  9 83

31. Let a  2iˆ  ˆj – 2kˆ and b  ˆi  ˆj . Let c be a vector such that c – a = 3, a  b  c = 3 and the angle

between c and a  b be 30º. Then a.c is equal to [JEE(Main) 2017, (4, – 1), 120]

ekuk a  2iˆ  ˆj – 2kˆ rFkk b  ˆi  ˆj gSA ekuk c ,d ,slk lfn’'k gS fd c – a = 3,  a  b   c = 3 rFkk c vkSj
a  b ds chp ds dks.k 30º gS] rks a.c cjkcj gS&
25 1
(1) (2) 2 (3) 5 (4)
8 8
Ans. (2)
Sol. a  2iˆ  j  2k,
ˆ b  ˆi  ˆj, | c  a | 3

| (a  b)  c | 3 , c ^ a  b =
6
Now vc | a  b | | c | sin300 = 3, | a  b | | c | = 6
 | a | b | c | sin = 6,  = a ^b
 2  1 
| a | = 3, |b| = 2  = cos–1  
3 2  4
6
|c| = . 2 =2
3 2
|c a| = 3
2
c
Squaring, we get oxZ djus ij | c | 2a.c  | a |  9
2 2
 a.c  2
2

32. If L1 is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L2 is the line of

intersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the
plane, containing the lines L1 and L2 , is : [JEE(Main) 2018, (4, –1), 120]

;fn leryksa 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 dh ifjPNsnh js[kk L1 gS rFkk leryksa x + 2y – z – 3 = 0, 3x

– y + 2z – 1 = 0 dh ifjPNsnh js[kk L2 gS ] rks ewy fcanq dh nwjh ml leery ls tks js[kkvksa L1 vkSj L2 dk varfoZ"V
djrk gS] gS %

1 1 1 1
(1) (2) (3) (4*)
2 2 2 4 2 3 2

Sol. (4)

2x  2y  3z  2  0
L1 : L1 : 
 x  y  z 1 0

Let a point on L1(0, 5, 4) and dr; s of L1 be a, b, c

ekuk fcUnq L1(0, 5, 4) gS vkSj L1 ds fnd~ vuqikr a, b, c gS

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 17
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

n ai + bi ck

L1

L2

2a1 + 2b1 + 3c1 = 0

a1 + b1 + c1 = 0

__________________

a1 b1 c1
 
1 1 0

so dr's of L2 be a2, b2, c2

dr's of L2 can be 3, –5, –7

so dr's of normal to the plane ca be a + b + oc = 0

3a – 5b –7c = 0

___________________

a b c
 
7 7 8

equation req. plane 7x – 7 (y –5) + 8 (z–4) = 0

7x –7y + 8z + 3 = 0

3 3 3 1
so req. distance =   
49  49  64 162 9 2 3 2

Hindi (4)

2x  2y  3z  2  0
L1 : L1 : 
 x  y  z 1 0

ekuk fcUnq L1(0, 5, 4) gS vkSj L1 ds fnd~ vuqikr a, b, c gS


n ai + bi ck

L1

L2

2a1 + 2b1 + 3c1 = 0

a1 + b1 + c1 = 0

__________________

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 18
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

a1 b1 c1
 
1 1 0

blfy, L2 ds fnd~ vuqikr a2, b2, c2

L2 ds fnd~ vuqikr gks ldrs gS 3, –5, –7

lery ds vfHkyEc ds fnd~ vuqikr a + b + oc = 0

3a – 5b –7c = 0

___________________

a b c
 
7 7 8

lery dk lehdj.k 7x – 7 (y –5) + 8 (z–4) = 0

7x –7y + 8z + 3 = 0

3 3 3 1
vHkh"V nwjh =   
49  49  64 162 9 2 3 2

33. Let u be a vector coplanar with the vectors a = 2 î + 3ˆj – k̂ and b = ĵ + k̂ . If u is perpendicular to a

2
and u . b = 24, then u is equal to : [JEE(Main) 2018, (4, –1), 120]

ekuk u ,d ,slk lfn'k gS tks lfn'kksa a = 2 î + 3ˆj – k̂ rFkk b = ĵ + k̂ ds lkFk leryh; gSA ;fn u , a ij
2
yacor~ gS rFkk u . b = 24 gS] rks u cjkcj gS %

(1) 256 (2) 84 (3*) 336 (4) 315

Sol. (3)

u = x î + y ĵ + z k̂

u . a = 0  2x + 3y – z = 0 ………(i)

u . b = 24  y + z = 24 ……..(ii)

x y z
[ u a b ] = 0  2 3 –1 = 0
0 1 1

 4x – 2y + 2z = 0

2x – y + z = 0 …………(iii)

(2) + (3)

2x + 2z = 24

x + z = 12 …………(iv)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 19
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Now vr% 24 – 2z + 3 (24 – z) – z = 0

96 = 6z

z = 16  x = – 4  y = 8

u = – 4 i + 8 j + 16 k̂

| u |2 = 16 + 64 + 256 = 336

34. The length of the projection of the line segment joining the points (5, – 1, 4) and (4,–1,3) on the plane ,

x + y + z = 7 is : [JEE(Main) 2018, (4, –1), 120]

fcanqvksa (5, – 1, 4) rFkk (4,–1,3) dks feykus okys js[kk[k.M dk lery x + y + z = 7 ij Mkys x, iz{ksi dh yEckbZ gS
%

1 2 2 2
(1) (2*) (3) (4)
3 3 3 3

Sol. (2)

A (5, –1, 4)

B (4,–1,3)

AB = 2

Direction ratio of AB < 1,0,1>

AB ds fnd~vuqikr < 1,0,1>

2
Angle between line AB and plane is   sin =
6

2 1
js[kk AB dk iz{ksi   sin =  cos =
6 3

2
Projection of AB on plane = AB cos =
3

35. If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :

;fn js[kk,¡ x = ay + b, z = cy + d rFkk x = a'z + b', y = c'z + d' yEcor gSa] rks&
[JEE(Main) 2019, Online (09-01-19),P-2 (4, – 1), 120]

(1) ab' + bc' + 1 = 0 (2) bb' + cc' + 1 = 0

(3) cc' + a + a' = 0 (4) aa' + c + c' = 0

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 20
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Vector 3D XII E,

Ans. (4)

x b y zd x  b' y  d' z

Sol. Given lines = = and = =
a 1 c a' c' 1

which are  to each other

 aa + c + c = 0

x b y zd x  b' y  d' z

Hindi. nh xbZ js[kk,sa = = rFkk = =
a 1 c a' c' 1

,d nwljs ds yEcor
 aa + c + c = 0

  
36. Let a  î  ĵ  2 k̂ , b  b1î  b2 ĵ  2 k̂ and c  5 î  ĵ  2 k̂ be three vectors such that the projection
      
vector of b on a is a . If a + b is perpendicular to c , then | b | is equal to : Vector 3D XII T,

[JEE(Main) 2019, Online (09-01-19),P-2 (4, – 1), 120]

    
ekuk a  î  ĵ  2 k̂ , b  b1î  b2 ĵ  2 k̂ vkSj c  5 î  ĵ  2 k̂ rhu lfn'k bl izdkj gS fd lfn'k b dk a ij
    
iz{ksi lfn'k a gSA ;fn a + b , lfn'k c ds yEcor gS rc | b | cjkcj gS&

(1) 22 (2) 4 (3) 32 (4) 6

Ans. (4)

Sol. ˆ ˆ | a | aˆ
(b . a)a

ˆ | a |
(b  a)

b1  b2  2
2
2

 b1 + b2 = 2 ......(1)

ab  c

 5(b1 + 1) + 1(b2 + 1) + 2 (2 2 ) = 0

 5b1 + b2 + 10 = 0 ......(2)

from (1) and (2) b1 = –3 and b2 = 5 (lehdj.k (1) o (2) ls)

 | b|= 9  25  2 = 6

37. A tetrahedron has vertices P(1,2,1), Q(2,1,3), R(–1,1,2) and O(0,0,0) the angle between the faces OPQ
and PQR is :

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 21
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

,d prq"Qyd (tetrahedron) ds 'kh"kZ P(1,2,1), Q(2,1,3), R(–1,1,2) rFkk O(0,0,0) gSA Qyd OPQ rFkk PQR ds
chp dk dks.k gS&[JEE(Main) 2019, Online (12-01-19),P-1 (4, – 1), 120]

 19  7  17   9 
(1) cos1   (2) cos–1  (3) cos1   (4) cos1  
 35   31  31   35 

Ans. (1)

Sol.

A(1,2,1) B(2,1,3)

C(–1,1,2)

î ĵ k̂
Vector perpendicular to face OAB = 1 2 1 = 5 î – ĵ – 3k̂
2 1 3

î ĵ k̂
Qyd OAB ds yEcor~ lfn'k = 1 2 1 = 5 î – ĵ – 3k̂
2 1 3

î ĵ k̂
Vector perpendicular to face ABC = 2 – 1 = î – 5 ĵ – 2k̂
1
1 –1 2

î ĵ k̂
Qyd ABC ds yEcor~ lfn'k = 2 – 1 = î – 5 ĵ – 2k̂
1
1 –1 2

559 19
Angle between two faces cos = =
35 35 35

559 19
nksuksa Qydksa ds e/; dks.k cos = =
35 35 35

 19 
 = cos–1  
 35 

38. Let S be the set of all real values of such that a plane passing through the points (–2 , 1, 1),
(1, –2, 1) and (1, 1, –2) also passes through the point (–1, –1, 1). Then S is equal to -

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 22
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

;fn  ds mu lHkh okLrfod ekuksa] ftuds fy, fcUnqvksa (–2 , 1, 1), (1, –2, 1) rFkk (1, 1, –2) ls gksdj tkus okyk
,d lery] fcUnq (–1, –1, 1) ls Hkh gksdj tkrk gS] dk leqPp; S gS] rks S cjkcj gS&

[JEE(Main) 2019, Online (12-01-19),P-2 (4, – 1), 120]

(1) {1, –1} (2) { 3 } (3) { 3,– 3 } (4) {3, –3}

Ans. (3)

 2  1 2 0
Sol. 2   1 2
0 0  – (2 + 1).{(1 – 2)2 – 4} = 0  2 – 1 = ±2
2 2  2  1

 2 = 3  =  3

39. The magnitude of the projection of the vector 2 î  3 ĵ  k̂ on the vector perpendicular to the plane
containing the vectors î  ĵ  k̂ and î  2 ĵ  3k̂ , is :

lfn'k 2î  3 ĵ  k̂ ds lfn'kksa î  ĵ  k̂ rFkk î  2 ĵ  3k̂ dks varfoZ"V djus okys lery ds yacorhZ; lfn'k ij iz{ksi
dk ifjek.k gS&

3 3
(1) 3 6 (2) 6 (3) (4)
2 2

[JEE(Main) 2019, Online (08-04-19),P-1 (4, – 1), 120][ Vector & 3-D]

Ans. (3)

Sol. Normal vector to the plane containing î  ĵ  k̂ and î  2 ĵ  3k̂ is


n = î  ĵ  k̂  × î  2 ĵ  3k̂ 

n = î – 2 ĵ  k̂


projection of 2î  3 ĵ  k̂  on n

=
2î  3 ĵ  k̂ . î – 2 ĵ  k̂ 
1 4  1

3 3
= =
6 2

Hindi. î  ĵ  k̂ rFkk î  2 ĵ  3k̂ dks lekfgr djus okys lery ds yEcor lfn'k&

n = î  ĵ  k̂  × î  2 ĵ  3k̂ 

n = î – 2 ĵ  k̂

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 23
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

2î  3 ĵ  k̂  dk n ij iz{ksi

=
2î  3 ĵ  k̂ . î – 2 ĵ  k̂ 
1 4  1

3 3
= =
6 2

40. If the volume of parallelepiped formed by the vectors î   ĵ  k̂, ĵ  k̂ and  î + k̂ is minimum , then  is
equal to :

;fn lfn'k î   ĵ  k̂, ĵ  k̂ rFkk  î + k̂ }kjk cuk;s x;s lekUrj "kV~Qyd dk vk;ru U;wure gS] rks cjkcj gS

1 1
(1) 3 (2) – 3 (3) (4) –
3 3

Ans. (3) [JEE(Main) 2019, Online (12-04-19),P-1 (4, – 1), 120] [Vector]

1  1
Sol. V = a b c  = 0 1  = 1–  (–2) + 1.(0–) = 3 – + 1
 0 1

1
Whose minimum value occur at  =
3

x  2 y 1 2
41. The vertices B and C of a ABC lie on the line,   such that BC = 5 units. Then the area
3 0 4
(in sq. units) of this triangle, given that the point A(1, –1, 2), is :

x  2 y 1 2
ABC ds 'kh"kZ B rFkk C js[kk   ij fLFkr gS rFkk BC = 5 bdkbZ gSA ;fn fn;k gS fd fcUnq
3 0 4

A(1, –1, 2) gS] rks bl f=kHkqt dk {ks=kQy (oxZ bdkbZ;ksa esa) gS&

(1) 5 17 (2) 6 (3) 34 (4) 2 34

Ans. (3) [JEE(Main) 2019, Online (09-04-19),P-2 (4, – 1), 120] [Vector & 3-D]

Sol.

A a (1, –1, 2)

5 units
B C
d (–2, 1, 0) u (3, 0, 4)
direction cosine of line


 a – d u 
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 24
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 
3 4 
 3i – 2j  2k   i  k 
5 5 

8 6 6
= – i j k
5 5 5

136
=
25

1 136
Area of triangle = 5 = 34 units2
2 25

42. Let A(3, 0, –1), B(2, 10, 6) and C(1,2,1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos(  GOA) (O being the origin) is equal to :

ekuk ,d f=kHkqt ds 'kh"kZ fcUnq A(3, 0, –1), B(2, 10, 6) rFkk C(1,2,1) gSa rFkk AC dk e/;fcUnq M gSA ;fn G, BM
dks 2 : 1 ds vuqikr esa foHkkftr djrk gS] rks cos(  GOA) (O ewyfcUnq gSSa) cjkcj gS %
1 1 1 1
(1) (2) (3) (4)
6 10 2 15 15 30

Ans. (3) [JEE(Main) 2019, Online (10-04-19),P-1 (4, – 1), 120] [Vector and 3D]

Sol. A (3,0–1,), B (2,10,6) and C (1,2,1)

C (1,2,1)

m (2, 4,2)
m

B(2,10,6)
A(3, 0,–1)

0(0, 0,0)

G is centroid of  from given information .

OA . OG = OA OG cos  2î  4 ĵ  2k̂  . 3î – k̂  = 2 6. 10 cos

 6 – 2 = 2 3. 5.2 cos

1
cos =
15

x –1 y 1 z
43. A perpendicular is drawn from a point on the line   to the plane x + y + z = 3 such that
2 –1 1
the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 25
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x –1 y 1 z
js[kk   ds ,d fcUnq ls lery x + y + z = 3 ij ,d yEc bl izdkj Mkyk x;k fd bldk yEcikn
2 –1 1
Q, lery x – y + z = 3 ij Hkh fLFkr gSA rks Q ds funsZ'kkad gS&

(1) (4, 0, –1) (2) (2, 0, 1) (3) (–1, 0, 4) (4) (1, 0, 2)

Ans. (2) [JEE(Main) 2019, Online (10-04-19),P-2 (4, – 1), 120] [3D]

x 1 y 1 z
Sol.   
2 1 1

P(2 + 1, – – 1, )

foot of perpendicular

x  (2  1) y  (  1) z   (2  1    1    3)
  
1 1 1 3

x  (2  1) y    1 z   (2  3) (2  3) 4  6

    x = 2 + 1 –  
1 1 1 3 3 3

(2  3) 3  3  2  3 5
 y = – – 1 –   
3 3 3

(2  3)   3
 z=– 
3 3

 4  6  5   3 
 point P is  , , 
 3 3 3 

It lies on x – y + z = 3

4  6 5   3
   3  10 + 9 = 9 = 0
3 3 3

 point P becomes (2, 0, 1)  (4) option is correct

44. A vector a  ˆi  2jˆ  kˆ ( R) lies in the plane of the vectors, b  ˆi  ˆj and c  ˆi  ˆj  4kˆ .

If a bisects the angle between b and c , then: [JEE(Main) 2020, Online (07-01-20),P-1 (4, – 1), 120]

,d lfn'k a  ˆi  2jˆ  kˆ ( R) ml lery esa] ftlesa nksuksa lfn'k b  ˆi  ˆj rFkk c  ˆi  ˆj  4kˆ fLFkr gSa]
;fn a lfn'kksa b vkSj c ds chp ds dks.k dks lef}Hkkftr djrk gS] rks %

(1) a . k̂ + 4 = 0 (2) a . k̂ + 2 = 0 (3) a . î + 1 = 0 (4) a . î + 3 = 0

Ans. (2)

Sol. angle bisector can be dks.k v}Zd gks ldrk gS a  (bˆ  c)

ˆ or ;k a  (bˆ  c)
ˆ

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 26
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

 ˆi  ˆj ˆi  ˆj  4kˆ    ˆ   ˆ
a      =  3i  3ˆj  ˆi  ˆj  4kˆ  = 4i  2jˆ  4kˆ 
 2 3 2  3 2 3 2

Compare with ls rqyuk djus ij a  ˆi  2jˆ  kˆ

2
2 3 2
3 2

a  4iˆ  2jˆ  4kˆ

 ˆi  ˆj ˆi  ˆj  4kˆ 
Not in option so now consider ekuk fd a     
 2 3 2 

 
a =
3 2
3iˆ  3ˆj  ˆi  ˆj  4kˆ 

=
3 2
 2iˆ  4ˆj  4kˆ 
Compare withrqyuk djus ij a  ˆi  2jˆ  kˆ

4 3 2
2
3 2 2

a  î  2 ĵ  2k̂

a .k̂  2  0

–2 + 2 = 0

45. Let a, band c be three unit vectros such that a  b  c  0 . If   a. b  b. c  c.a and
d  a  b  b  c  c  a , then the ordered pair, , d is equal to:  
ekuk a, b rFkk c rhu ek=kd (unit) lfn'k bl izdkj gSa fd a  b  c  0 . ;fn
  a. b  b. c  c.a rFkk d  a  b  b  c  c  a , rks Øfer ;qXe , d cjkcj gS&  
3   3  3   3 
(1)  ,3a  c  (2)   ,3c  b  (3)  ,3b  c  (4)   ,3a  b 
2   2  2   2 

Ans. (4) [JEE(Main) 2020, Online (07-01-20),P-2 (4, –1), 120]

  
Sol. | a  b  c |2 = 0

  
3 + 2 (a.b  b.c  c.a)  0

   3
(a.b  b.c  c.a) =
2

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 27
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

3
=
2
     
  
  
d  ab  b  a  b   a  b a 
     
= ab  ab  ab
  
d = 3 (a  b )

46. If the distance between the plane, 23x –10y –2z +48 = 0 and the plane containing the lines

x 1 y  3 z 1 x  3 y  2 z 1
  and     R is equal to k
, then k is equal to ___________
2 4 3 2 6  633

;fn lery 23x –10y –2z +48 = 0rFkk js[kkvkssa

x 1 y  3 z 1 x  3 y  2 z 1
  vkSj     R dks varfoZ"V djus okys lery ds chp dh nwjh
k
gS]
2 4 3 2 6  633
rks k cjkcj gS___________A [JEE(Main) 2020, Online (09-01-20),P-2 (4, 0), 120]

Ans. 3

Sol. Lines must be intersecting

 (2s – 1, 4s + 3, 3s –1) = (2t – 3, 6t – 2 , t + 1)

1 1
2s–1 = 2t – 3, 4s + 3 = 6t–2 , 3s – 1 = t + 1 t= , s =  ,  = –7
2 2

distance of plane contains given lines from given plane is same as distance between point (–3, –2,1)
from given plane.

| –69  20 – 2  48 | 3 k
Required distance equal to = = k=3
529  100  4 633 633

Sol. js[kk,sa izfrPNsnh gksxh

 (2s – 1, 4s + 3, 3s –1) = (2t – 3, 6t – 2 , t + 1)

1 1
2s–1 = 2t – 3, 4s + 3 = 6t–2 , 3s – 1 = t + 1 t= , s =  ,  = –7
2 2

js[kkvksa dk lekfgr djus okys lery dh fn;s gq;s lery ls nwjh = fcanq (–3, –2,1) ls fn;s x;s lery dh nwjh

| –69  20 – 2  48 | 3 k
vHkh"V nwjh = = = k=3
529  100  4 633 633

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 28
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

SUBJECTIVE QUESTIONS
fo"k;kRed iz'u ¼SUBJECTIVE QUESTIONS½
1. Using Vectors prove that
lfn'k fof/k ls n'kkZb;s fd
(i) cos(A – B) = cosA cosB + sinA sinB (ii) sin(A + B) = sinA cosB + cosA sinB

Sol. (i)
r1r2 cos(A – B) = r1r2 (cos Aiˆ  sin Aj).(cosBi
ˆ ˆ  sinBj)
ˆ
cos(A – B) = cosA cosB + sinA sinB

(ii)
p  q  r1 r2 (cos Aiˆ  sin A ˆj) × (cosBiˆ  sinB ˆj)
ˆ = k̂r r (cos A sinB  sin A cosB)
r1 r2 sin(A  B)(k) 1 2

sin(A + B) = sinA cosB + cosA sinB

2. Using vectors, prove that the altitudes of a triangle are concurrent.

lfn'k fof/k ls fl} djks fd ,d f=kHkqt ds 'kh"kZ yEc laxkeh gksrs gSA
Sol. Let AD  BC and BE  AC. Let AD and BE meet at O. Join CO and produce it to meet AB at F.
Take O as the origin. Let a, b, c be the position vectors of A, B and C respectively.
 BC = c  b and CA = a  c
Now AD  BC  OA  BC  a.(b  c) = 0
and BE  CA  OB  CA  b.(a  c)  0
These imply
a.b  a.c and a . b  b . c  a.c b.c  ( b  a ). c = 0
 OC  AB  CF  AB
  Altitudes of a  are concurrent.
Hindi ekuk AD  BC rFkk BE  AC ekuk AD rFkk BE fcUnq O ij feykrk gSA CO dks feykus ij rFkk bldks vkxs
c<kus ij ;g AB dks F ij feysxhA
ekuk O ewy fcUnq gS A, B, C ds fLFkfr lfn'k Øe'k% a, b, c gSaA
 BC = c  b rFkk CA = a  c

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 1
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

vc AD  BC  OA  BC  a.(b  c) = 0
rFkk BE  CA  OB  CA  b.(a  c)  0
bldk eryc gS fd
a.b  a.c rFkk a . b  b . c  a.c b.c  ( b  a ). c = 0
 OC  AB  CF  AB
vr% f=kHkqt ds 'kh"kZ yEc laxkeh gksxsA

3. Prove that the direction cosines of a line equally inclined to three mutually perpendicular lines having
 2  3 m1  m2  m3 n1  n2  n3
D.C.’s as 1, m1, n1 ; 2, m2, n2 ; 3, m3, n3 are 1 , , .
3 3 3
fl) dhft, fd rhu ijLij yEcor~ js[kkvksa ftudh fnDdksT;k,sa 1, m1, n1 ; 2, m2, n2 ; 3, m3, n3 gS] ls cjkcj >qdh
  m1  m2  m3 n1  n2  n3
gqbZ js[kk dh fnDdksT;k,sa 1 2 3
, , gSA
3 3 3
ˆi  m ˆj  n kˆ , OB = ˆ ˆ ˆ ˆi  m ˆj  n kˆ be mutually perpendicular
Sol. Let OA = 1 1 1 2 i  m2 j  n2k and OC = 3 3 3

vectors. Let OP = ˆi  mjˆ  nkˆ be equally inclined to OA , OB and OC .

Then OP = OA + OB + OC =  1ˆi   m1ˆj   n1kˆ
2
OP = ( 1 )2  ( m1 )2  ( n1)2 = 3 + 2  (12 + m1m2 + n1n2) = 3
  m1  m2  m3 n1  n2  n3
 OP = 3  = 1 2 3
î + ĵ + k̂
3 3 3
Hindi ekukfd OA = 1ˆi  m1ˆj  n1kˆ , OB = 2 ˆi  m2 ˆj  n2kˆ vkSj OC = 3 ˆi  m3 ˆj  n3kˆ rhu ijLij yEcor lfn'k gSaA
ekukfd OP = ˆi  mjˆ  nkˆ , OA , OB rFkk OC ls leku dks.k cukrk gS] rks
OP = OA + OB + OC =  1ˆi   m1ˆj   n1kˆ
2
OP = ( 1 )2  ( m1 )2  ( n1)2 = 3 + 2  (12 + m1m2 + n1n2) = 3

  m1  m2  m3 n1  n2  n3
 OP = 3  = 1 2 3
î + ĵ + k̂
3 3 3

4. If A (a) , B (b) , C (c) , D (d) are the position vector of cyclic quadrilateral then find the value of
| a b  b  d  d a | | b c  c  d  db |
+ . (It is given that no angle of cyclic quadrilateral ABCD is
[(b  a).(d  a)] [(b  c).(d  c)]
right angle)
| a b  b  d  d a |
;fn A (a) , B (b) , C (c) , D (d) pØh; prqHkqZt ds fLFkfr lfn'k gks] rc +
[(b  a).(d  a)]
| b  c  c  d  d b |
dk eku Kkr dhft,A ¼;g fn;k x;k gS fd pØh; prqHkqZt ABCD dk dksbZ Hkh dks.k ledks.k
[(b  c).(d  c)]
ugha gSA½
Ans. 0

Sol.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 2
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

(d)D C(c )

(a) A B(b)
Now vc | (b  c).(d  a) | = AB.AD.cos A..............(1)
| a  b  b  d  d  a | = | (b  a)  (d  a) |
| (b  c)  (d  a) | = AB.AD.sin A.............(2)
(b  c).(d  c) = BC.CD.cosC
| (b  c)  (d  c) | = BC.CD sinC
Now putting the values dk eku j[kus ij
tanA + tanC [ Now (A + C = 180)]
tanA + tan ( – A)
=0

5. Prove that the volume of tetrahedron bounded by the planes,

3
r.(mjˆ  nk)
ˆ  0 , r .(nkˆ  ˆi)  0 , r .( ˆi  m ˆj)  0 , r.( ˆ  p is 2p .
ˆi  m ˆj  n k)
3 mn
ˆ ˆ ˆ ˆ ˆ ˆ
fl) dhft, fd fn;s x;s leryks r.(mj  nk)  0 , r .(nk  i)  0 , r .( i  m j)  0 ,
ˆi  m ˆj  n k)
ˆ  p }kjk f?kjs leprq"Qyd dk vk;ru 2p3
r.( gSA
3 mn
Sol. Vectors along the edges passing through the origin.
(mjˆ  nk) ˆ = mniˆ  n ˆj  mkˆ
ˆ  ( ˆi  nk)

(nkˆ  ˆi)  ( ˆi  mj)

ˆ = mniˆ  n ˆj  mkˆ

( ˆi  mj) ˆ = mniˆ  n ˆj  mkˆ

ˆ  (mjˆ  nk)

Let the equation of one edge be r  (mniˆ  n ˆj  mk)

ˆ

For its point of intersection with the plane r.( ˆi  mjˆ  nk)
ˆ  p , we have
p
 (mniˆ  n ˆj  mk)
ˆ . ( ˆi  mjˆ  nk)
ˆ =p   =
mn
mn n  m
mn n m = 42m2n2
mn n m
1 p3 2 p3
 volume of the tetrahedron = 3
. 42m2n2 =
6 ( mn) 3 mn
Hindi ewy fcUnq ls xqtjus okys dksjksa ds vuqfn'k lfn'k
(mjˆ  nk) ˆ = mniˆ  n ˆj  mkˆ
ˆ  ( ˆi  nk)

(nkˆ  ˆi)  ( ˆi  mj)

ˆ = mniˆ  n ˆj  mkˆ

( ˆi  mj) ˆ = mniˆ  n ˆj  mkˆ

ˆ  (mjˆ  nk)

ekukfd ,d dksj dk lehdj.k r  (mniˆ  n ˆj  mk)ˆ

bldk lery r.( ˆi  mjˆ  nk)

ˆ  p ds lkFk izfrPNsn fcUnq ds fy,

(mniˆ  n ˆj  mk)
ˆ . ( ˆi  mjˆ  nk)
ˆ =p

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 3
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

mn n  m
p
 = mn n m = 42m2n2
mn
mn n m
1 p3 2 p3
 prq"Qyd dk vk;ru = . 42
m n
2 2
=
6 ( mn)3 3 mn

6. In a  ABC, let M be the mid point of segment AB and let D be the foot of the bisector of  C. Then
ar( CDM) 1 ab 1 A B A B
prove that = = tan cot .
ar( ABC) 2 ab 2 2 2
fdlh  ABC esa Hkqtk AB dk e/; fcUnq M gS vkSj D,  C ds v/kZd dk ikn gSA rc fl) dhft,
CDM dk {ks0 1 a  b 1 A B A B
= = tan cot
 ABC dk {ks0 2 a  b 2 2 2
Sol.

aa  bb
CD 
ab
ab
CM 
2
1
 ar ( CDM)= | CD  CM |
2
1 1 1
= × (aa  bb)  (a  b) = |(a – b) ( a  b )|
4 (a  b) 4 (a  b)
1 ab 1 1 ab
= . . | ab | = ar (ABC)
2 ab 2 2 ab
ar( CDM) 1 ab 1 A B A B
 = = tan cot
ar( ABC) 2 ab 2 2 2
Hindi.

aa  bb
CD 
ab
ab
CM 
2
1
  CDM dk {ks=kQy = | CD  CM |
2
1 1 1
= × (aa  bb)  (a  b) = |(a – b) ( a  b )|
4 (a  b) 4 (a  b)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 4
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 ab 1 1 ab
= . . | ab | = ar (ABC)
2 ab 2 2 ab
ar( CDM) 1 ab 1 A B A B
 = = tan cot
ar( ABC) 2 ab 2 2 2

7. Let ABC be a triangle.Points M, N and P are taken on the sides AB, BC and CA respectively such that
AM BN CP
= = =  . Prove that the vectors AN , BP and CM form a triangle. Also find  for which
AB BC CA
the area of the triangle formed by these vectors is the least.
1
Ans.  =
2
ekuk ABC ,d f=kHkqt gSA Hkqtk AB, BC vkSj CA ij Øe'k% fcUnq M, N vkSj P bl izdkj ls gS fd
AM BN CP
= = =  fl) dhft, fd lfn'k AN , BP rFkk CM ,d f=kHkqt cukrs gSA  dk og eku Kkr
AB BC CA
dhft, ftlds fy, bu lfn'kksa }kjk fufeZr f=kHkqt dk {ks=kQy U;wure gksA
Sol.

Let a, b, c be the position vectors of A, B, C respectively.

OM = b + (1 – ) a , ON = c  (1  )b , OP = a  (1  )c
 AN  c  (1  )b  a
BP  a  (1  )c  b and CM  b  (1  )a  c
Clearly AN  BP  CM = 0
 AN, BP, CM form a triangle
Area of triangle formed by the vectors AN , BP, CM is
1 1
= | (c  (1  ) b  a)  (a  (1  ) c  b) | = |2 –  + 1| (ABC)
2 2
is least if |2 –  + 1| is least
1
Which is when  =
2
Hindi.

ekukfd A, B, C ds fLFkfr lfn'k Øe'k% a, b o c gSA

OM = b + (1 – ) a , ON = c  (1  )b , OP = a  (1  )c
 AN  c  (1  )b  a

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 5
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

BP  a  (1  )c  b rFkk CM  b  (1  )a  c

Li"Vr;k AN  BP  CM = 0
 AN, BP, CM ,d f=kHkqt cukrs gSaA
lfn'kksa AN, BP, CM ls fufeZr f=kHkqt dk {ks=kQy gSA
1 1
= | (c  (1  ) b  a)  (a  (1  ) c  b) | = |2 –  + 1| (ABC)
2 2
;g U;wure gksxk ;fn |2 –  + 1| U;wure gksA
1
tks fd lEHko gS ;fn  = gksA
2
8. In any triangle, show that the perpendicular bisectors of the sides are concurrent.
fdlh f=kHkqt esa fl) dhft, fd Hkqtkvksa ds yEc v)Zd laxkeh gSA
Sol. Let ABC be the triangle and D, E and F are respectively middle points of sides BC, CA and
AB. Let the perpendicular bisectors of BC and CA meet at O. Join OF. We are required to a
prove that OF is  to AB. Let the position vectors of A, B, C with O as origin of reference
be a , b and c respectively.
1 1 1
  OD = ( b + c ), OE = ( c + a ) and OF = (a +b )
2 2 2
Also BC = c – b , CA = a – c and AB = b – a
Since OD  BC
1
  ( b + c ) . ( c – b ) = 0  b2 = c2 ............(i)
2
1
Similarly OE  CA   ( c + a ) . ( a – c ) = 0  a2 = c2 ...........(ii)
2
from (i) and (ii) we have a2 – b2 = 0
1
  (a +b ) . (b –a ) = 0  ( b + a ) . ( b – a ) = 0  OF  AB .
2
Sol. ekuk ABC ,d f=kHkqt gS rFkk D, E rFkk F Øe'k% Hkqtkvksa BC, CA rFkk AB ds e/; fcUnq gSA ekuk BC vkSj
CA dk yEc v)Zd fcUnq O ij feyrs gSA OF dks feykb;sA gesa a dh vko';drk gS rc fl) djuk gS fd OF 
AB ekuk A, B, C ds O ds lkis{k fLFkfr lfn'k Øe'k% a , b vkSj c gSA
1 1 1
  OD = ( b + c ), OE = ( c + a ) rFkk OF = (a +b )
2 2 2
rFkk BC = c – b , CA = a – c vkSj AB = b – a
pwafd OD  BC
1
  ( b + c ) . ( c – b ) = 0  b2 = c2 ............(i)
2
1
blhizdkj OE  CA   ( c + a ) . ( a – c ) = 0  a2 = c2 ...........(ii)
2
lehdj.k (i) vkSj (ii) ls a2 – b2 = 0
1
  (a +b ) . (b –a ) = 0  ( b + a ) . ( b – a ) = 0  OF  AB .
2

9. Let ABC be an acute-angled triangle AD be the bisector of BAC with D on BC and BE be the altitude
from B on AC. Show that CED > 45º.
fdlh U;wudks.k f=kHkqt ABC esa ekuk BAC dk v/kZd AD bl izdkj gS fd fcUnq D Hkqtk BC ij fLFkr gS vkSj 'kh"kZ
B ls Hkqtk AC ij Mkyk x;k yEc BE gSA iznf'kZRk dhft, fd CED > 45º.
Sol. Let AB = c and AC = b

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 6
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

c  c  b b
p.v. of D is
c   b
Also (a – c) . b = 0
c  c  b b
. c
Now cos CED = c   b
c  c  b b
. c
c   b
2
c c c c
 cos = =
c  c  b b c ( c  )2 | c |2  ( b )2 | b |2
cos2  sin2 C 1
 ( b)2 | b |2 cos2  = ( c)2 | c |2 sin2   = ×
sin2  sin2 B tan2 C
cos2 C
 cot2  = ... (i)
sin2 B
Since A,B,C are all acute.
cos2 C
Hence to prove that <1
sin2 B
 
or cos C < sin B Now since B + C >  B> –C
2 2
or sin B > cos C from equation (i) cot < 1  > 45°
Hence Proved.
Hindi. ekuk AB = c vkSj AC = b
c  c  b b
fcUnq D dk fLFkfr lfn'k
c   b
rFkk (a – c) . b = 0
c  c  b b
. c
 cos CED = c   b
c  c  b b
. c
c   b
2
c c c c
 cos = =
c  c  b b c ( c  )2 | c |2  ( b )2 | b |2
 ( b)2 | b |2 cos2  = ( c)2 | c |2 sin2 
cos2  sin2 C 1 cos2 C
 = ×  cot2  = ... (i)
sin 
2
sin2 B tan2 C sin2 B
pw¡fd dks.k A, B, C lHkh U;wudks.k gSA
cos2 C
vr% ;g fl) djus ds fy;s fd <1 ;k cos C < sin B
sin2 B
 
 B+C>  B> –C
2 2
;k sin B > cos C  lehdj.k (i) ls cot < 1  > 45°
vr% fl) gqvkA

10. In a quadrilateral ABCD, it is given that AB || CD and the diagonals AC and BD are perpendicular to
each other. Show that
(a) AD. BC  AB.CD (b) AD + BC  AB + CD

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 7
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

fdlh prqHkqZt ABCD esa Hkqtk,¡ AB vkSj CD ijLij lekUrj gS vkSj fod.kZ AC vkSj BD ijLij yEcor~ gSA
iznf'kZr dhft, fd
(a) AD. BC  AB.CD (b) AD + BC  AB + CD
Sol.

a. b = 0
Also µ a – b | | a – b  µ = .
(a) | b – a | | µ a – b |  |( b – a )| |( b –µ a )|
 (2 b2 + a2)(µ2a2 + b2)  (b2 + a2)(2 b2 + µ 2a2)
 2 µ2 a2 b2 + a2b2  2 a2 b2 + µ 2 b2 a2
 (2 – 1) (µ2 – 1)  0
 (2 – 1)2  0 which is true.
(b) (AD + BC)2  (AB + CD)2
 AD2 + OD2 + OC2 + OB2 + 2AD.BC  OA2 + OB2 + OD2 + OC2 + 2AB. CD
which follows from (a).

Hindi.

a. b = 0
Also µ a – b | | a – b  µ = .
(a) | b – a | | µ a – b |  |( b – a )| |( b –µ a )|
 (2 b2 + a2)(µ2a2 + b2)  (b2 + a2)(2 b2 + µ 2a2)
 2 µ2 a2 b2 + a2b2  2 a2 b2 + µ 2 b2 a2
 (2 – 1) (µ2 – 1)  0
 (2 b2 + a2)(µ2a2 + b2)  (b2 + a2)(2 b2 + µ 2a2)
 2 µ2 a2 b2 + a2b2  2 a2 b2 + µ 2 b2 a2
 (2 – 1) (µ2 – 1)  0
 (2 – 1)2  0 tks lR; gSA
(b) (AD + BC)2  (AB + CD)2
 AD2 + OD2 + OC2 + OB2 + 2AD.BC  OA2 + OB2 + OD2 + OC2 + 2AB. CD

11. A, B, C, D are four points in space. using vector methods, prove that
AC2 + BD2 + AD2 + BC2  AB2 + CD2 what is the implication of the sign of equality.
lef"V esa A, B, C, D pkj fcUnq gS lfn'k fof/k ls fl) dhft, fd AC2 + BD2 + AD2 + BC2  AB2 + CD2 vlfedk
ds fpUg dk vFkZ le>kb;sA

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 8
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. Let the position vector of A, B, C, D be a , b , c and d respectively then

        
AC2+BD2 + AD2 + BC2 =  c  a  .  c  a  + d  b . d  b + d  a . d  a + c  b . c  b = | c |2 
+ | a | –2 a . c + | d | + | b | – 2 d . b + | d | + | a | – 2 a . d + | c | + | b | – 2b . c
2 2 2 2 2 2 2

= | a |2 + | b |2 – 2 a . b + | c |2 + | d |2 – 2 c . d + | a |2 + | b |2 + | c |2 + | d |2
+ 2 a . b + 2c . d – 2a . c – 2b . d – 2a . d – 2b . c


= ab . ab   +  c  d .  c  d +  a  b  c  d  2

= AB2 + CD2 + a  b  c  d . a  b  c  d  AB + CD
2 2

 AC2 + BD2 + AD2 + BC2  AB2 + CD2

for the sign of equality to hold, a  b  c  d = 0 or a c  db
 AC and BD are collinear, the four points A, B, C, D are collinear
Hindi. ekuk A, B, C, D ds fLFkfr lfn'k
a , b , c vkSj d gS rc

        
AC2+BD2 + AD2 + BC2 =  c  a  .  c  a  + d  b . d  b + d  a . d  a + c  b . c  b = | c |2 
+ | a | –2 a . c + | d |2 + | b |2 – 2 d . b + | d |2 + | a | – 2 a . d + | c | + | b |2 – 2b . c
2 2 2

= | a |2 + | b |2 – 2 a . b + | c |2 + | d |2 – 2 c . d + | a |2 + | b |2 + | c |2 + | d |2
+ 2 a . b + 2c . d – 2a . c – 2b . d – 2a . d – 2b . c


= ab . ab   +  c  d .  c  d +  a  b  c  d  2

= AB2 + CD2 + a  b  c  d . a  b  c  d  AB + CD
2 2

 AC2 + BD2 + AD2 + BC2  AB2 + CD2

lfedk ds fpUg ds fy,, a  b  c  d = 0 ;k a c  db
 AC vkSj BD lajs[kh; gS rc pkj fcUnq A, B, C, D lajs[kh; gSA

12. The direction cosines of a variable line in two near by positions are l, m, n; l + l, m + m, n + n. Show
that the small angle  between the two position is given by ()2 = (l)2 + (m)2 + (n)2.
nks fudVre fLFkfr;ksa esa ,d pj js[kk dh fnd~dksT;k,a l, m, n; l + l, m + m, n + n gaAS iznf'kZr dhft, fd bu
nks fLFkfr;ksa ds e/; y?kq dks.k  dks ()2 = (l)2 + (m)2 + (n)2 ds }kjk fn;k tkrk gSA
Sol. As (l, m, n) and (l + l, m + m, n + n) are the direction cosines of lines so we have
l2 + m2 + n2 = 1 ... (1)
and (l + l) + (m + m) + (n + n) = 1
2 2 2
.... (2)
Subtracting (1) from (2) we have
2(ll + mm + nn) + [(l)2 + (m)2 + (n)2] = 0
or 2(ll) = – (l)2 ... (3)
Also as  is the angle between these lines so we have
cos = l (l + l) + m (m + m) + n(n + n)
= (l2 + m2 + n2) + (ll + mm + nn)
2
()
2
1 1 1  1 
= 1– [(  )2 ], from (1) and (3) or ( )2   1 – cos= 2sin2     2    =
2 2  2   2  2
1  1
 sin      or (l)2 + (m)2 + (n)2 = ()2
2  2
Hindi. js[kkvksa dh fnd~dksT;k,¡ (l, m, n) rFkk (l + l, m + m, n + n) gaSA vr%
l2 + m2 + n2 = 1 ... (1)
rFkk (l + l)2 + (m + m)2 + (n + n)2 = 1 .... (2)
lehdj.k (2) esa ls lehdj.k (1) dks ?kVkus ij
2(ll + mm + nn) + [(l)2 + (m)2 + (n)2] = 0
;k 2(ll) = – (l)2 .... (3)
bu nks js[kkvksa ds e/; dks.k  gSSA vr%
Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 9
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

cos = l (l + l) + m (m + m) + n (n + n)

= (l2 + m2 + n2) + (ll + mm + nn)
1
= 1– [(  )2 ] lehdj.k (1) rFkk (3) lss
2
2
()
2
1 1  1 
;k ( )2   1 – cos= 2sin2     2    =
2  2   2  2
1  1
 sin      ;k (l)2 + (m)2 + (n)2 = ()2 Tkks fd (a) dk vuqlaj.k djrs gq,
2  2
gSA
 a2  b2  c 2 
13. In a ABC, prove that distance between centroid and circumcentre is R2   
 9 
where R is the circumradius and a, b, c denotes the sides of ABC.
 a2  b2  c 2 
ABC esa fl) dhft, fd dsUnzd vkSj ifjdsUnz ds e/; nwjh R2   
 9 
tgka R ifjf=kT;k gS rFkk ABC dh Hkqtk,a a, b, c gSA
Sol. Let the circum centre be (O)
ekuk (O) ifjdsUnz gSA
abc 
G=  
 3
 
1
OG = | a  b  c |
3
1
= | a |2  | b |2  | c |2 2(a.b  b.c  c.a)
3
| a |2 | b |2 | c |2 = R
1
3R2  2R2 (cos2A  cos2B  cos2C)
3
1
3R2  2R2 (1  2sin2 A  1  2sin2 B  1  2sin2 C
3
1
9R2  (a2  b2  c 2 )
3
 a2  b2  c 2 
R2   
 9 

14. Prove that the square of the perpendicular distance of a point P (p, q, r) from a line through A(a, b, c)
and whose direction cosines are , m, n is  {(q – b) n – (r – c) m}2.
fl) dhft, fd fcUnq A (a,b,c) ls xqtjus okyh js[kk ftldh fnddksT;k,¡ , m, n gSa dh fcUnq P(p,q,r) ls yEcor
nwjh dk oxZ  {(q – b) n – (r – c) m}2 gS
Sol.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 10
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

ˆi ˆj kˆ
Consider AP × ( ˆi  mjˆ  nk)
ˆ = p a qb r c
m n

=  (n (q – b) – m (r – c)) î
AP  ( ˆi  mjˆ  nk)
ˆ
2
 MP2 = =  {n (q – b) – m (r – c)}2

Hindi.
ˆi ˆj kˆ
ekukfd AP × ( ˆi  mjˆ  nk)
ˆ = p a qb r c
m n

=  (n (q – b) – m (r – c)) î
AP  ( ˆi  mjˆ  nk)
ˆ
2
 MP2 = =  {(n (q – b) – m (r – c)}2

15. (i) Let 1 & 2 be two skew lines. If P, Q are two distinct points on 1 and R, S are two
distinct points on 2 , then prove that PR can not be parallel to QS.
(ii) A line with direction cosines proportional to (2, 7 – 5) is drawn to intersect the lines
x–5 y–7 z2 x3 y–3 z–6
  and   . Find the coordinate of the points of
3 –1 1 –3 2 4
intersection and the length intercepted on it. Also find the equation of intersecting straight line.
(i) ekukfd 1 & 2 nks fo"keryh; js[kk,¡ (skew lines) gaSA ;fn 1 ij fLFkr nks fHkUu fcUnq P, Q gaS
rFkk 2 ij fLFkr nks fHkUu fcUnq R, S gS] rks fl) dhft, PR, QS ds lekUrj ugha gks ldrh gSA
x–5 y–7 z2
(ii) ,d js[kk ftldh fnd~dksT;k,¡ (2, 7 – 5) ds lekuqikrh gaS] js[kkvksa   ,oa
3 –1 1
x3 y–3 z–6
  dks izfrPNsn djrh gSA izfrPNsnu fcUnqvksa ds funsZ'kkad rFkk bl ij cus vUr%[kf.Mr
–3 2 4
Hkkx dh yEckbZ Kkr dhft,A lkFk gh izfrPNsnu ljy js[kk dh lehdj.k Hkh Kkr dhft,A
x – 2 y –8 z3
Ans. (2, 8, – 3) & (0, 1, 2) ; 78 ;  
2 7 –5

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 11
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

Sol. (i) Let equation of the line 1 be r = a  b and equation of the line 2 be r = c  d ,
where a – c , b and d are non-coplanar.
Let the position vectors of points P and Q be a  1 b and a  2 b respectively.
Let the position vectors of points R and S be c  1 d and c  2 d respectively.
Then the lines PR and QS are parallel if and only if c  a  1d  1b  k (c  a  2 d  2b)
i.e. (1 – k) (c  a) + (1 – k2) d – (1 – k2) b = o
 1 – k = 0, 1 – k2 = 0, 1 – k2 = 0
i.e. 1 = 2 and 1 = 2 which is not possible  PR can not be parallel to QS.
(ii) The given equation
x–5 y–7 z2 x3 y–3 z–6
  ... (i) and   ... (ii)
3 –1 1 –3 2 4
any point P on (i) is (3r1 + 5, – r1 + 7, r1 – 2) and any point Q on (ii) is
(– 3r2 – 3, 2r2 + 3, 4r2 + 6)
the direction ratios of PQ are
(3r1 + 3r2 + 8, – r1 – 2r2 + 4, r1 – 4r2 – 8) ... (iii)
suppose the line with d.r’s 2, 7, – 5 will be proportional to the d.r.’s given by (iii)
3r1  3r2  8 – r1 – 2r2  4 r1 – 4r2 – 8
    ... (iv)
2 7 –5
Solving (iv), we get r1 = r2 = – 1
So point of intersection are P(2, 8, – 3) and Q(0, 1, 2)
and intercepted length = PQ = (2 – 0)2  (8 – 1)2  (– 3 – 2)2  78
x –2 y –8 z3
and equation of PQ is  
2 7 –5
Hindi (i) ekuk fd js[kk 1 dk lehdj.k r = a  b gS rFkk js[kk 2 dk lehdj.k r = c  d gSA
tgk¡ a – c , b rFkk d vleryh; gaSA
ekuk fd fcUnqvksa P rFkk Q ds fLFkfr lfn'k Øe'k% a  1 b rFkk a  2 b gSaA
ekuk fd fcUnqvksa R rFkk S ds fLFkfr lfn'k Øe'k% c  1 d rFkk c  2 d gSaA
rks js[kk,¡ PR rFkk QS lekUrj gSa ;fn vkSj dsoy ;fn c  a  1d  1b  k (c  a  2d  2b)
i.e. (1 – k) (c  a) + (1 – k2) d – (1 – k2) b = o
 1 – k = 0, 1 – k2 = 0, 1 – k2 = 0
i.e. 1 = 2 rFkk 1 = 2 tks fd laHko ugha gSA
 PR, QS ds lekUrj ugha gks ldrh gSA
x–5 y–7 z2 x3 y–3 z–6
(ii) nh xbZ lehdj.k   ... (i) rFkk   ... (ii)
3 –1 1 –3 2 4
dksbZ fcUnq P js[kk (i) ij (3r1 + 5, – r1 + 7, r1 – 2) rFkk js[kk (ii) ij dksbZ fcUnq
Q (– 3r2 – 3, 2r2 + 3, 4r2 + 6) gSA
PQ ds fnd~vuqikr fuEu gS&
(3r1 + 3r2 + 8, – r1 – 2r2 + 4, r1 – 4r2 – 8) ... (iii)
ekuk fd (iii) esa fn;s x;s fnd~vuqikr js[kk ftlds fnd~vuqikr 2, 7, – 5 gS ] ds lekuqikrh gksxsaA 
3r1  3r2  8 – r1 – 2r2  4 r1 – 4r2 – 8
     ... (iv)
2 7 –5
(iv) dks gy djus ij izkIr gksrk gS fd r1 = r2 = – 1
vr% izfrPNsnu fcUnq P(2, 8, – 3) rFkk Q(0, 1, 2) gaSA
rFkk vUr%[kf.Mr yEckbZ = PQ = (2 – 0)2  (8 – 1)2  (– 3 – 2)2  78

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 12
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

x –2 y –8 z3
rFkk PQ dk lehdj.k   gSA
2 7 –5

16. The base of the pyramid AOBC is an equilateral triangle OBC with each side equal to 4 2 . ' O ' is the
origin of reference, AO is perpendicular to the plane of  OBC and | AO | = 2 . Then find the cosine of
the angle between the skew straight lines one passing through A and the mid point of OB and the other
passing through 'O' and the mid point of BC.
,d fijkfeM AOBC dk vk/kkj ,d leckgq f=kHkqt OBC gS] ftldh izR;sd Hkqtk 4 2 ds cjkcj gSA ' O ' dks ewy
fcUnq ekuk x;k gSA AO, OBC ds lery ds yEcor~ gS vkSj | AO | = 2 gSA nks fo"keryh; js[kkvkas ds chp ds
dks.k dh dksT;k ¼cosine) Kkr dhft, tcfd ,d js[kk A vkSj OB ds e/; fcUnq ls xqtjrh gS rFkk nwljh js[kk 'O'
rFkk BC ds e/; fcUnq ls xqtjrh gSA
1
Ans.
2
Sol.

AD = 2 2 ˆi  2kˆ
OE  3 2 ˆi  6 ˆj
12
cos  =
12 24
1
cos  =
2

17. If D, E, F be three point on BC, CA, AB respectively of a ABC. Such that the line AD, BE, CF are
BD CE AF
concurrent then find the value of . . .
CD AE BF
;fn ABC ds BC, CA rFkk AB ij Øe'k% rhu fcUnq D, E, F bl izdkj gS fd js[kk,sa AD, BE, CF laxkeh gS rc
BD CE AF
. . . dk eku Kkr dhft, &
CD AE BF
Ans. 1

Sol. xa + yb + zc + wp
x+y+z+w=0
From the two equation nks lehdj.kksa ls
xa  yb zc  wp
=
xy zw
xa  yb
Position vector of F dk fLFkfr lfn'k =
xy
y
F divide AB in the ratio 
x

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 13
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

y
F, AB dks ds vuqikr esa foHkkftr djrk gSA
x
AF y
=
FB x
BD z CE x AF BD CE
Similarly blh izdkj = & = ; . . =1
CD y AE z BF CD AE

18. Without expanding the determinant, Prove that

na1  b1 na2  b2 na3  b3 a1 a2 a3
nb1  c1 nb2  c 2 nb3  c 3 = (n + 1)
3
b1 b2 b3
nc1  a1 nc 2  a2 nc 3  a3 c1 c 2 c3
lkjf.kd dk foLrkj fd, fcuk fl) dhft,
na1  b1 na2  b2 na3  b3 a1 a2 a3
nb1  c1 nb2  c 2 nb3  c 3 = (n + 1) b1 b2
3
b3
nc1  a1 nc 2  a2 nc 3  a3 c1 c 2 c3
Sol. Let's consider ekuk fd

a  a1ˆi  a2 ˆj  a3kˆ

b  b1ˆi  b2 ˆj  b3kˆ

c  c1ˆi  c 2 ˆj  c 3kˆ
we have to prove gesa fl) djuk gS
 a 

 

      

na  b na  c n c  a   n3  1

b c

     
   
 na  b  .   nb  c    n c  a  
     
 (na  b).[n2 (b  c)  n(b  a)  n(c  c)  c  a] [ c  c = 0]
 na.(n (b c)  n(b  a)  c  a]  b.(n (b  c)  n(b  a)  (c  a))
2 2

n3 [a b c]  [b c a]
(n + 1) [a
3
b c]

19. (i) OABC is a regular tetrahedron D is circumcentre of OAB and E is mid point of edge AC.
Prove that DE is equal to half the edge of tetrahedron.
(ii) If V be the volume of a tetrahedron and V  be the volume of the tetrahedron formed by the
centroids and V = k V  then find the value of k.
Ans. (ii) 27
(i) OABC ,d leprq"Qyd gS] OAB dk ifjdsUnz D gS vkSj fdukjs (Edge) AC dk e/; fcUnq E gSA fl)
dhft, fd DE, leprq"Qyd dh fdukj (Edge) dh vk/kh gSA
(ii) ;fn V leprq"Qyd dk vk;ru gS vkSj V  dsUnzdkas ds }kjk cuk;s x;s leprq"Qyd dk vk;ru gS rFkk
V = kV  - rks k dk eku Kkr dhft, A

Sol. (i)

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 14
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

OA  OB ab
OD  = ( OAB is an equilateral )
3 3
ac
OE 
2
a  3c  2b
 DE 
6
2
 a  3c  2b 
| DE | = 
  ( | a | = | b | = | c | and 
 6 
1 6a2 12a2 4a2 1 a
  DE = a2  9a2  4a2     DE = 9a2  DE =
6 2 2 2 6 2
1
(ii) V= [a b c]
6
ab bc c a abc
The centroid are , , ,
3 3 3 3
1 c  a c b c  1 1
 V =   = [a b c] = V  k = 27
6  3 3 3  6  27 27
Hindi (i)

OA  OB ab
OD  = ( OAB ,d leckgq f=kHkqt gSA)
3 3
ac
OE 
2
a  3c  2b
 DE 
6
2
 a  3c  2b 
| DE | =   ( | a | = | b | = | c | rFkk 
 6 
1 6a2 12a2 4a2 1 a
  DE = a2  9a2  4a2     DE = 9a2  DE =
6 2 2 2 6 2
1
(ii) V= [a b c]
6
ab bc c a abc
dsUnzd , , , gSA
3 3 3 3

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 15
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

1 c  a c b c 1 1
 V =   = [a b c] = V  k = 27
6  3 3 3  6  27 27

20. Given that u  ˆi  2jˆ  3kˆ v  2iˆ  ˆj  4kˆ , w  2iˆ  j  3kˆ and (u.R  10) i + (v.R  20) j + (w.R  20) k̂ =
0 then find R
Ans. R  10iˆ
fn;k x;k gS u  ˆi  2jˆ  3kˆ v  2iˆ  ˆj  4kˆ , w  2iˆ  j  3kˆ vkSj (u.R  10) i + (v.R  20) j + (w.R  20) k̂ =
0 rc R dks Kkr dhft,A
Sol. (u.R)iˆ + (v.R)jˆ + (w.R)kˆ = 10iˆ  20jˆ  20kˆ
R  xiˆ  yjˆ  zkˆ
u.R = x – 2y + 3z
v.R = 2x + y + 4z
w.R = x + 3y + 3z
(x – 2y + 3z) î + (2x + y + 4z) ĵ + (x + 3y + 3z) k̂ = 10iˆ  20jˆ  20kˆ
 x – 2y + 3z = 10 ..................(1)
2x + y + 4z = 20 ..................(2)
2x + y + 3z = 20 ..................(3)
(x, y, z) = (10, 2, 5)
R  10iˆ

21. AB , AC and AD are three adjacent edges of a parallelopiped . The diagonal of the parallelopiped
passing through A and directed away from it is vector a . The vector area of the faces containing
vertices A , B , C and A , B , D are b and c respectively i.e. AB  AC = b and AD  AB  c . If
|a|
projection of each edge AB and AC on diagonal vector a is , then find the vectors AB, AC and
3
AD in terms of a, b, c and | a | .
AB , AC vkSj AD ,d ?kukHk dh rhu vklUu Hkqtk;sa gSA ?kukHk ds 'kh"kZ fcUnq A ls xqtjus okyk vkSj blls nwj dh
vksj tkrh fn'kk esa fod.kZ lfn'k a gSA Qydsa ftuds 'kh"kZ A , B , C rFkk A , B , D gaS ds {ks=kQy Øe'k% lfn'k b

rFkk c gSA vFkkZr~ AB  AC = b vkSj AD  AB  c ;fn Hkqtk AB vkSj AC izR;sd dk] fod.kZ a ij iz{ksi
|a| gS]
3
  
rks a , b , c rFkk a ds inksa esa lfn'k AB , AC rFkk AD dks Kkr dhft,A
ca 1 a  (b  c) 3(b  a) 1 a  (b  c) 3(c  a)
Ans. AB  AD  3 2
; AC = a + + ; AD = a + –
|a| 3 |a| 2
|a| 2
3 | a |2 | a |2
Sol.

a  AP  AB  AC  AD ......(i)
AB  AC  b
AD  AB  c

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 16
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

a |a|
AB. 
|a| 3 | a |2
i.e. AB.a 
3
a |a| | a |2
AC.  i.e. AC . a 
|a| 3 3
 (AB  AC)  a  b  a
| a |2 | a |2
 AC  AB  b  a
3 3
ba
 AC  AB  3 ..........(ii)
| a |2
| a |2  AB . a  AC . a  AD . a
| a |2
  AD.a
3
ca
(AD  AB)  a  c  a  AB  AD  3 ........(iii)
| a |2
1 a  (b  c)
from (i) , (ii) and (iii), we get AB  a
3 | a |2
Now from (ii) and (iii), we can easily find AC and AD
1 a  (b  c) 3 (b  a) 1 a  (b  c) 3 (c  a)
AC = a + + ; AD = a + –
3 | a |2 | a |2 3 | a |2 | a |2
Hindi

a  AP  AB  AC  AD ......(i)
AB  AC  b
AD  AB  c

a |a|
AB. 
|a| 3 | a |2
vFkkZr~ AB.a 
3
a |a| | a |2
AC.  vFkkZr~ AC . a 
|a| 3 3
 (AB  AC)  a  b  a
| a |2 | a |2
 AC  AB  b  a
3 3
ba
 AC  AB  3 ..........(ii)
| a |2
| a |2  AB . a  AC . a  AD . a

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 17
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Vector & Three Dimensional Geometry

| a |2
  AD.a
3
ca
(AD  AB)  a  c  a  AB  AD  3 ........(iii)
| a |2
1 a  (b  c)
lehdj.k (i) , (ii) vkSj (iii) ls AB  a
3 | a |2
lehdj.k (ii) vkSj (iii) ls AC and AD
1 a  (b  c) 3 (b  a) 1 a  (b  c) 3 (c  a)
AC = a + + ; AD = a + –
3 | a |2 | a |2 3 | a |2 | a |2

22. Prove that if the equation r  a  b and r  c  d are consistent a.d  0,b.c  0,d  b  0 then

b.c  a.d  0
fl) dhft, lehdj.k r  a  b vkSj r  c  d ds fudk; laxr gS  a.d  0,b.c  0,d  b  0  rc b.c  a.d  0
Sol. r  a  b  d  (r  a)  d  b
db
   
 a.d r – d.r a = d  b  r 
a.d
......(i)

Now vc r  c = d  b  (r  c)  b  d
db
   
 b.c r – b.r c = b  d  r
–b.c
.....(ii)

from (i) and (ii) we get a.d = –b.c ((i) vkSj (ii) ls a.d = –b.c )

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
ADVQE- 18
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029