VECTOR

1. INTRODUCTION :
Vectors constitute one of the several Mathematical systems which can be usefully employed to provide
mathematical handling for certain types of problems in Geometry, Mechanics and other branches of Applied
Mathematics.
Vectors facilitate mathematical study of such physical quantities as possess Direction in addition to Magnitude.
Velocity of a particle, for example, is one such quantity.

2. Physical quantities are broadly divided in two categories viz (a) Vector Quantities & (b) Scalar quantities.

(a) Vector quantities :

Any quantity, such as velocity, momentum, or force, that has both magnitude and direction and for
which vector addition is defined and meaningful; is treated as vector quantities.
Note :
Quantities having magnitude and direction but not obeying the vector law of addition will not be
treated as vectors.
For example, the rotations of a rigid body through finite angles have both magnitude & direction but do
not satisfy the law of vector addition therefore not a vector.

(b) Scalar quantities :

A quantity, such as mass, length, time, density or energy, that has size or magnitude but does not
involve the concept of direction is called scalar quantity.

3. MATHEMATICAL DESCRIPTION OF VECTOR & SCALAR :

To understand vectors mathematically we will first understand directed line segment.

Directed line segment :

Any given portion of a given straight line where the two end points are distinguished as Initial and Terminal
is called a Directed Line Segment.

The directed line segment with initial point A and terminal point B is denoted by the symbol AB .
The two end points of a directed line segment are not interchangeable and the directed line segments.
 
AB and BA must be thought of as different.

(a) Vector :
A directed line segment is called vector. Every directed line segment have three essential characteristics.

 
(i) Length : The length of AB will be denoted by the symbol AB
 
Clearly, we have AB  BA
(ii) Support : The line of unlimited length of which a directed line segment is a part is called its line
of support or simply the Support.

1
  
(iii) Sense : The sense of AB is from A to B and that of BA from B to A so that the sense of a
directed line segment is from its initial to the terminal point.
(b) Scalar :
Any real number is a scalar.

4. CLASSIFICATION OF VECTORS :
There is a very important classification of vector in which vectors are divided into two categories.
(a) Free vectors : If a vector can be translated anywhere in space without changing its magnitude &
direction, then such a vector is called free vector. In other words, the initial point of free vector can be
taken anywhere in space keeping its magnitude & direction same.
c

b
c 
a b

(b) Localised vectors : For a vector of given magnitude and

direction, if its initial point is fixed in space, then such a vector

r F
is called localised vector. O
P   rF
Examples : Torque, Moment of Inertia etc.

Unless & until stated, vectors are treated as free vectors.

5. EQUALITY OF TWO VECTORS :

Two vectors are said to be equal if they have
(a) the same length,
(b) the same or parallel supports and
(c) the same sense.
Note : Components of two equal vectors taken in any arbitrary direction are equal. i.e If
 
ˆ b  b ˆi  b ˆj  b kˆ , where ˆi, ˆj & k̂ are the unit vectors taken along co-ordinate axes,
a  a 1ˆi  a 2 ˆj  a 3 k, 1 2 3
 
then a  b  a 1  b1 , a 2  b2 , a 3  b3

 
Illustration 1 :
 ˆ b  ˆi  3 ˆj  2kˆ and c  2ˆi  ˆj  3kˆ . r  a  b  c , then
ˆ a  2 ˆi  ˆj  k,
Let r  3 ˆi  2 ˆj  5k,
find      .
Solution : 3 ˆi  2 ˆj  5 kˆ  (2 ˆi  ˆj  k)
ˆ  (iˆ  3 ˆj  2k)
ˆ  (2 ˆi  ˆj  3k)
ˆ

= (2     2  )iˆ  (  3    )jˆ  (  2   3  )kˆ

Equating components of equal vectors
2    2 = 3 ........(i)
  3    = 2 ........(ii)
  2   3  = –5 ........(iii)
on solving (i), (ii) & (iii)
we get   3,   1,   2
So       6 Ans.

2
 Do yourself - 1 :
 
(i) If a = 2 î +  ˆj – 7 k̂ and b =  î + 3 ˆj – 7 k̂ are two equal vectors, then find 2 + 2.
 
(ii) If a , b are two vectors then which of the following statements is/are correct -
       
(A) a = – b  | a |= | b | (B) | a |= | b |  a = ± b
       
(C) | a |= | b |  a = b (D) | a |= | b |  a = ±2 b

6. LEFT AND RIGHT - HANDED ORIENTATION (CONFIGURATIONS) :

z z

x x

y y
Right - Handed Triplet Left - Handed Triplet
For each hand take the directions Ox, Oy and Oz as shown in the figure. Thus we get two rectangular
coordinate systems. Can they be made congruent ? They cannot be, because the two hands have different
orientations. Therefore these two systems are different.
A rectangular coordinate system which can be made congruent with the system formed with the help of right
hand (or left hand) is called a right handed (or left handed) rectangular coordinates system.
Thus we have the following condition to identify these two systems using sense of rotation :
(a) If the rotation from Ox to Oy is in the anticlockwise direction and Oz is directed upwards (see right
hand), then the system is right handed.
(b) If the rotation from Ox to Oy is clockwise and Oz is directed upward (see left hand) then the system
is left handed.
Here after we shall use the right-handed rectangular Cartesian coordinate system (or Ortho-normal system).

7. ALGEBRA OF VECTORS :
It is possible to develop an Algebra of Vectors which proves useful in the study of Geometry, Mechanics and
other branches of Applied Mathematics.
(a) Addition of two vectors :
The vectors have magnitude as well as direction, therefore their addition is different than addition of
real numbers.
  
Let a and b be two vectors in a plane, which are represented by AB and CD . Their addition can be
performed in the following two ways :
(i) Triangle law of addition of vectors : If two vectors can be represented in magnitude and
direction by the two sides of a triangle, taken in order, then their sum will be represented by the
third side in reverse order.

Let O be the fixed point in the plane of vectors. Draw a line segment OE F D
 
from O, equal and parallel to AB , which represents the vector a . a+b
  b b
Now from E, draw a line segment EF , equal and parallel to CD ,
  O
which represents the vector b . Line segment OF obtained by a EC
A B
  a
joining O and F represents the sum of vectors a and b .
  
i.e. OE  EF  OF
  
or a  b  OF
This method of addition of two vectors is called Triangle law of addition of vectors.

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 (ii) Parallelogram law of addition of vectors : If two vectors be represented in magnitude and
direction by the two adjacent sides of a parallelogram then their sum will be represented by the
diagonal through the co-initial point.
  Q R
Let a and b be vectors drawn from point O denoted by line segments
  +
b
OP and O Q . Now complete the parallelogram OPRQ. Then the b a

vector represented by the diagonal OR will represent the sum of P

O
  a
the vectors a and b .
  
i.e. OP  O Q  OR

or   
a  b  OR
This method of addition of two vectors is called Parallelogram law of addition of vectors.
(iii) Properties of vector addition :
         
(1) a  b  b  a (commutative) (2) (a  b)  c  a  (b  c) (associativity)
         
(3) a  0  a  0  a (additive identity) (4) a  (a )  0  ( a )  a (additive inverse)

(b) Polygon law of vector Addition (Addition of more than two vectors): D d C
Addition of more than two vectors is found to be by repetition of triangle
e c
law.

a+b+c+d
c
b+ B
   
Suppose we have to find the sum of five vectors a, b, c, d and e . If a+
E
a+
b+
    c+ a+b b
d+
these vectors be represented by line segment OA, AB, BC, CD and e O
a
A

 
DE respectively, then their sum will be denoted by OE . This is the
vector represented by rest (last) side of the polygon OABCDE in reverse
order. We can also make it clear this way :
By triangle's law
     
OA  AB  OB or a  b  OB
      
OB  c  OC or (a  b)  c  OC
       
OC  d  OD or (a  b  c)  d  O D
        
OD  e  OE or (a  b  c  d)  e  OE

Here, we see that OE is represented by the line segment joining the initial point O of the first vector
 
a and the final point of the last vector e .
In order to find the sum of more that two vectors by this method, a polygon is formed. Therefore this
method is known as the polygon law of addition.
Note : If the initial point of the first vector and the final point of the last vector are the same, then the
sum of the vectors will be a null vector. B
(c) Subtraction of Vectors :
  b
a+
Earlier we have described the vector – b whose length is equal to vector b but b
   O
a
A
direction is opposite. Subtraction of vector a and b is defined as addition of a a
 –
and (– b ). It is written as follows : b –b

    C
a  b  a  ( b)

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 Geometrical representation :
   
In the given diagram, a and b are represented by OA and AB . We extend the line AB in opposite
 
direction upto C, where AB = AC. The line segment AC will represent the vector b . By joining
    
the points O and C, the vector represented by OC is a  ( b) . i.e. denotes the vector a  b.
Note :
    
(i) a  a  a  ( a )  0
 
(ii) a  b  b  a
Hence subtraction of vectors does not obey the commutative law.
     
(iii) a  (b  c)  (a  b)  c
i.e. subtraction of vectors does not obey the associative law.
(d) Multiplication of vector by scalars :
  
If a is a vector & m is a scalar, then m( a ) is a vector parallel to a whose modulus is |m| times that
  
of a . This multiplication is called SCALAR MULTIPLICATION. If a & b are vectors & 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

Illustration 2 : ABCD is a parallelogram whose diagonals meet at P. If O is a fixed point, then

   
OA + OB + OC + OD equals :-
   
(A) OP (B) 2 OP (C) 3 OP (D) 4 OP
Solution : Since, P bisects both the diagonal AC and BD, so
          
 OA + OC = 2 OP and OB + OD = 2 OP  OA + OB + OC + OD = 4 OP Ans. [D]
  
Illustration 3 : A, B, P, Q, R are five points in any plane. If forces AP , AQ , AR acts on point A and force
  
PB , QB , RB acts on point B then resultant is :-
   
(A) 3 AB (B) 3 BA (C) 3 PQ (D) 3 PR
Solution : From figure
   A P
AP + PB = AB
  
AQ + QB = AB
   Q
AR + RB = AB
      
So ( AP + AQ + AR ) + ( PB + QB + RB ) = 3 AB
 B R
so required resultant = 3 AB . Ans. [A]
Illustration 4 : Prove that the line joining the middle points of two sides of a triangle is parallel to the third side
and is of half its length.
Solution : Let the middle points of side AB and AC of a ABC be D and E respectively.
    A
BA  2DA and AC  2 AE
Now in ABC, by triangle law of addition D E
  
BA  AC  BC B C

     1 

2DA  2 AE  BC  DA  AE  BC
2
 1 
DE  BC
2
Hence, line DE is parallel to third side BC of triangle and half of it.

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 Do yourself - 2 :
  
(i) If a, b, c be the vectors represented by the sides of a triangle taken in order, then prove that
  
abc=0
   
(ii) If PO  OQ  QO  OR , then prove that the points P, Q and R are collinear..
 
(iii) For any two vectors a and b prove that
           
(a) | a + b |  | a | +| b | (b) | a – b |  | a | +| b | (c) | a + b |  | a | –| b |
     
Note : In general for any non-zero vectors a, b & c one may note that although a  b  c  0 but it will not
always represent the three sides of a triangle.

8. COLLINEAR VECTORS : a
Two vectors are said to be collinear if their supports are parallel disregards b
to their direction. Collinear vectors are also called Parallel vectors. If
they have the same direction they are named as like vectors otherwise c
unlike vectors.
Note :
 
(i) Symbolically two non zero vectors a & b are collinear if and only if, a  Kb , where K  R

  a1 a 2 a 3
(ii) If a  a 1ˆi  a 2 ˆj  a 3 kˆ and b  b1ˆi  b2ˆj  b3 kˆ are two collinear vectors then b  b  b .
1 2 3
   
(iii) If a & b are two non-zero, non-collinear vectors such that xa  yb  0  x = y = 0

D A
9. CO-INITIAL VECTORS :
d a
Vectors having same initial point are called Co-initial Vectors.

C c O B
b
 
Illustration 5 : If a and b are non-collinear vectors, then find the value of x for which vectors :
     
  (x  2 )a  b and   (3  2 x)a – 2 b are collinear..

Solution : Since the vectors  and  are collinear..

 there exist scalar  such that  =  
     
 (x – 2) a + b =  {(3 + 2x) a – 2 b }  (x – 2 –  (3 + 2x)) a + (1 + 2) b = 0
 x – 2 –  (3 + 2x) = 0 and 1 + 2 = 0
1
x – 2 – (3 + 2x) = 0 and  = –
2
1 1
 x–2+ (3 + 2x) = 0  4x – 1 = 0  x= Ans.
2 4
Illustration 6 : If A  (2 î + 3 ˆj ), B  (p î + 9 ˆj ) and C  ( î – ˆj ) are collinear, then the value of p is :-
(A) 1/2 (B) 3/2 (C) 7/2 (D) 5/2
 ˆ  ˆ
Solution : AB = (p – 2) î + 6 j , AC = – î – 4 j
  p 2 6
Now A, B, C are collinear  AB || AC  =  p = 7/2 Ans. [C]
–1 –4

6
  
Illustration 7 : The value of  when a = 2 î – 3 ˆj + k̂ and b = 8 î +  ˆj + 4 k̂ are parallel is :-
(A) 4 (B) – 6 (C) – 12 (D) 1

  2 3 1
Solution : Since a & b are parallel  =– =   = –12 Ans. [C]
8  4
E D
Do yourself - 3 :
(i) In the given figure, which vectors are (provided vertices d
x c
A, B, C, D, E, F are all fixed) : F C
(a) Parallel (b) Equal z
y
(c) Coinitial (d) Parallel but not equal. b
    A a B
(ii) If a  2 ˆi  3 ˆj  4 kˆ and b  8 ˆi  12ˆj  16kˆ such that a =  b , then  equals to
     
(iii) If 3 a  2 b  5 c and 8 a  7 b  4 c , then which statement is/are true :
   
(A) | a| | b| (B) | c| | b|
   
(C) a, b and c are collinear vectors. (D) | a| | b|
10. COPLANAR VECTORS :
A given number of vectors are called coplanar if their supports are all parallel to the same plane. Note that
“TWO VECTORS ARE ALWAYS COPLANAR ”.
Note : Coplanar vectors may have any directions or magnitude.

11. REPRESENTATION OF A VECTOR IN SPACE IN TERMS OF 3

ORTHONORMAL TRIAD OF UNIT VECTORS :
Let P(x, y, z) be a point in space with reference to OX, OY and OZ as C A'
the coordinate axes, then OA = x, OB = y and OC = z
P
B' (x,y,z)
Let ˆi, ˆj, kˆ be unit vectors along OX, OY and OZ respectively, then
  


ˆ OB  yj, ˆ OC  zkˆ r
OA  xi,
       
OP  OC '  C 'P = OB  OA  OC [ C ' P  OC ] O B
  
= OA  OB  OC = xiˆ  yjˆ  zkˆ A
C'
 
If OP  r

r  xiˆ  yjˆ  zkˆ

| r |  OP  x2  y2  z2

B
12. POSITION VECTOR :
Let O be a fixed origin, then the position vector of a point P is the vector b
   A
OP . If a & b are position vectors of two point A and B, then
a
  
AB  b – a = pv of B – pv of A.

13. ZERO VECTOR OR NULL VECTOR :

A vector of zero magnitude i.e. which has the same initial & terminal point is called a zero vector. It is denoted

by O . It can have any arbitrary direction and any line as its line of support.

14. UNIT VECTOR :

 
A vector of unit magnitude in direction of a vector a is called unit vector along a and is denoted by

a 
â symbolically â   (provided | a|  0 )
| a|
15. SECTION FORMULA :
7
   
If a & b are the position vectors of two points A & B then the p.v. of a point C (r ) which divides AB in the
ratio m : n is given by : A(a ) m C( c ) n B( b )
(a) Internal Division :

  m b  na
OC  r 
m n
 
ab
Note : Position vector of mid point of AB = O
2
(b) External division : m
A( a ) B( b ) n C( r )

  mb  na
OC  r 
m n

Illustration 8 : Prove that the medians of a triangle are concurrent.

Solution : Let ABC be a triangle and position vectors of three vertices A, B and C with respect to the origin
 
O be a, b and c respectively..
     
 OA  a, OB  b, OC  c A
O

Again, let D be the middle point of the side BC, E


 b  c G
so the position vector of point D is OD 
2 B C
D
[ Position vector of the middle point of any line =1/2
(Sum of position vectors of end point of line)]
Now take a point G, which divides the median AD in the ratio 2 : 1.

   1  
 1.OA  2.OD 1 .a  2 . b  c   
 
Position vector of point G is OG  2 abc
 
1 2 1 2 3
Similarly, the position vector of the middle points of the other two medians, which divide the
  
a bc
medians in the ratio 2 : 1 will comes out to the same , which is the position vector of G..
3
Hence, the medians of the triangles meet in G i.e. are concurrent.
Illustration 9 : If the middle points of sides BC, CA & AB of triangle ABC are respectively D, E, F then position
vector of centroid of triangle DEF, when position vector of A,B,C are respectively i + j, j + k, k + i is
-
1 2
(A) (i + j + k) (B) (i + j + k) (C) 2(i + j + k) (D) (i + j + k)
3 3

i j kj ik
Solution : The position vector of points D, E, F are respectively + k, i + and +j
2 2 2

1 i  j k  j i k  2
So, position vector of centroid of DEF = ki   j = [i + j + k]. Ans. [D]
3  2 2 2  3

8
 Do yourself - 4 :
   
(i) Find the position vectors of the points which divide the join of the points 2 a – 3 b and 3 a – 2 b
internally and externally in the ratio 2 : 3,
(ii) ABCD is a parallelogram and P is the point of intersection of its diagonals. If O is the origin of
    
reference, show that OA  OB  OC  OD  4OP

(iii) Find the unit vector in the direction of 3 î – 6 ˆj + 2 k̂ .

16. VECTOR EQUATION OF A LINE :

B(b)
R(r)
R(r)

A(a) A(a)

O r=a+ b
O r=a+µ(b–a)
     
Parametric vector equation of a line passing through two points A(a ) & B(b) is given by,, r = a + t(b – a)
 
where t is a parameter. If the line passes through the point A(a ) & is parallel to the vector b then its equation
  
is r = a + tb .
Note :
       
(i) Equations of the bisectors of the angles between the lines r  a  b & r  a  c is, r = a + t(bˆ + c) ˆ
  ˆ .
& r = a + p(cˆ – b)
(ii) In a plane, two lines are either intersecting or parallel.
(iii) Two non parallel nor intersecting lines are called skew lines.

Illustration 10 : In a triangle ABC, D and E are points on BC and AC respectively, such that BD = 2DC
and AE = 3EC. Let P be the point of intersection of AD and BE. Find BP/PE using vector
methods. (JEE-1993)
Solution : Let the position vectors of A and B be a and b respectively. Equations of AD and BE are
   
r  a  t(b / 3  a ) ........ (i)
    C(0)


r  b  s(a / 4  b) ........ (ii)

If they intersect at P we must have identical values of r. 1 1
E D
1
Comparing the coefficients of a and b in (i) and (ii), we get
P
3 k 2
s t
1–t= , =1–s
4 3 

A(a) B(b)
9 8
solving we get t = ,s .
11 11
 
2a  3b
Putting for t or s in (i) or (ii), we get the point P as .
11

a   
k.  b
Let P divide BE in the ratio k : 1, then P is 4 2a  3 b .

k 1 11
 8
Comparing a and b , we get 11k = 8(k + 1) and 11 = 3(k + 1)  k=
3
and this satisfies the 2nd relation also. Hence the required ratio is 8 : 3. Ans.

9
Illustration 11 : Find whether the given lines are coplanar or not
 ˆ ˆ
r  i  j  10kˆ  (2 ˆi  3 ˆj  8 k)
ˆ


r  4 ˆi  3ˆj  kˆ  (iˆ  4 ˆj  7k)
ˆ


Solution : L1 : r  (2   1)iˆ  (1  3  )jˆ  (8   10)kˆ

L2 : r  (4  )iˆ  (4   3)jˆ  (7   1)kˆ
The given lines are not parallel. For coplanarity, the lines must intersect.
 (2   1)iˆ  (1  3  )jˆ  (8   10 )kˆ  (4  )iˆ  (4   3 )jˆ  (7   1)kˆ

2  1  4   .........(i)

1  3   4  3 .........(ii)

8   10  7   1 .........(iii)
Solving (i) & (ii),  = 2, µ = 1 and  = 2, µ = 1 satisfies equation (iii)
Given lines are intersecting & hence coplanar. Ans.
17. TEST OF COLLINEARITY OF THREE POINTS :
b
(a) 3 points A B C will be collinear if AB =  BC,
   b+c
(b) 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
   c
such that ; xa  yb  zc  0 , where x + y + z = 0
(c) Collinearly can also be checked by first finding the equation of line through two points and satisfying
the third point.
 
Illustration 12 : ˆ b  2 ˆi  3 ˆj  4kˆ & 7ˆj  10kˆ are
Prove that the points with position vectors a  ˆi  2 ˆj  3k,
collinear.

Solution : If we find, three scalars , m & n such that a  mb  nc  0 where   m  n  0
then points are collinear.
ˆ  m(2 ˆi  3 ˆj  4k)
(iˆ  2 ˆj  3k) ˆ  n(7ˆj  10k)
ˆ 0

 (  2m)iˆ  (2   3m  7n)jˆ  (3   4m  10n)kˆ  0

  + 2m = 0, –2 + 3m – 7n = 0, 3 – 4m + 10n = 0
Solving, we get = 2, m = –1, n = –1
since  + m + n = 0
Hence, the points are collinear.
Aliter :
  
   
AB  b  a  2 ˆi  3 ˆj  4kˆ  ˆi  2 ˆj  3kˆ  ˆi  5 ˆj  7kˆ

  
    
BC  c  b  7ˆj  10kˆ  2 ˆi  3 ˆj  4kˆ  2 ˆi  10 ˆj  14 kˆ  2 ˆi  5 ˆj  7kˆ 
 
 AB  2BC
 
Hence a, b & c are collinear..

Do yourself - 5 :
(i) The position vectors of the points P, Q, R are î +2 ˆj +3 k̂ , –2 î +3 ˆj +5 k̂ and 7 î – k̂ respectively..
Prove that P, Q and R are collinear.

10
18. SCALAR PRODUCT OF TWO VECTORS (DOT PRODUCT) :
  
Definition : Let a and b be two non zero vectors inclined at an angle . Then the scalar product of a with
      
b is denoted by a . b and is defined as a . b = | a || b | cos  ; 0    .

b
b a 
 
a

Geometrical Interpretation of Scalar product :

   
| OA| = | a |, | OB| = | b |
B
   
Now a . b = | a || b | cos  M

= | a | (OB cos ) b
  
= (magnitude of a ) (Projection of b on a )

   
Again, a . b = | a || b | cos  O a N A
 
= | b | (| a | cos  )
  
= (Magnitude of b ) (Projection of a on b )
   
(a) a . b | a || b | co s (0    )
   
Note that if  is acute then a . b  0 & if  is obtuse then a . b  0
       
(b) (i) a . a | a | 2  a 2 (ii) a . b  b . a (commutative)
      
(c) a . (b  c)  a . b  a . c (distributive)
     
(d) a . b  0  a  b ; (a , b  0 )

(e) ˆ .k
ˆi . ˆi  ˆj. ˆj  k ˆ  1 ; ˆi . ˆj  ˆj. k
ˆ k
ˆ . ˆi  0
A

  
(f) Projection of a on b  a . b . (Provided b  0 ) a
| b|
b B
O
Note : Projection of a on b

  a.  
 b
(i) The vector component of a along b    2  b and perpendicular
 b 
  a


   a . b  
to b  a    2  b [by triangle law of vector Addition]
 b  b
 

 
 a.b
(ii) The angle  between a & b is given by cos     0    
| a || b |
   
(iii) If a  a 1 ˆi  a 2 ˆj  a 3 kˆ & b  b1 ˆi  b2 ˆj  b3 kˆ , then a. b  a 1 b1  a 2 b2  a 3 b 3

  2 2 2
2 2 2
a  a1  a 2  a 3 , b  b1  b2  b 3
   
(iv) Maximum value of a . b | a| | b|
   
(v) Minimum values of a . b  | a| | b|
     ˆ ˆ
(vi) Any vector a can be written as, a  (a . ˆi)iˆ  (a . ˆj)jˆ  (a . k)k

11
 (g) Vector equation of angle bisector :
A
A vector in the direction of the bisector of the angle C' C
  a b
  a b
a
  
b   
| a| | b|
between the two vectors a & b is    . | a| | b| a
| a| | b|
Hence bisector of the angle between the two vectors –b b
  B'
O B
ˆ , where   R  . Bisector of the
a & b is (aˆ  b)
 
exterior angle between a & b is (aˆ – b) ˆ ,   R

   
Illustration 13 : a, b, c, d are the position vectors of four coplanar points A, B, C and D respectively..
       
If (a  d).(b  c)  0  (b  d).(c  a ) , then for the ABC, D is :-
(A) incentre (B) orthocentre (C) circumcentre (D) centroid
        
 
Solution : (a  d) . (b  c) = 0  (a  d)  (b  c)  AD  BC
     
Similarly (b  d).(c  a ) = 0 BD  AC
 D is the orthocentre of ABC. Ans. [B]

Illustration 14 : The vector c , directed along the internal bisector of the angle between the vector 7 î – 4 ˆj –

4 k̂ and –2 î – ˆj + 2 k̂ with | c | = 5 6 is -

5 5 5
(A) ( î – 7 ˆj + 2 k̂ ) (B) (5 î + 5 ˆj + 2 k̂ ) (C) ( î + 7 ˆj + 2 k̂ ) (D) none of these
3 3 3

Solution : Let a = 7 î – 4 ˆj – 4 k̂

and b = – 2 î – ˆj + 2 k̂
  C
internal bisector divides the BC in the ratio of AB : AC , | AC|
  b c
AB = 9, AC = 3 D
| AB|
A B
 ˆ  3(7ˆi – 4 ˆj – 4k)
 9(–2 ˆi – ˆj  2k) ˆ  ˆ ˆ ˆ a
AD =   = i  7 j  2k
 9  3  4

  
 AD 5
c = ±   5 6 = ± ( î – 7 ˆj + 2k) Ans.[A]
 AD  3

         
Illustration 15 : If modulii of vectors a, b, c are 3, 4 and 5 respectively and a and b + c , b and c + a , c and
    
a + b are perpendicular to each other, then modulus of a + b + c is -

(A) 5 2 (B) 2 5 (C) 50 (D) 20

      
Solution :  a  (b + c )  a .b + a .c = 0
      
Similarly b  ( c + a )  b.c + b.a = 0
      
and c  ( a + b )  c .a + c .b = 0
     
 a .b + b.c + c .a = 0
           
Now | a + b + c |2 = | a |2 + | b |2 + | c |2 + 2( a . b + b . c + c . a ) = 9 + 16 + 25 = 50
  
 |a + b + c | =5 2 Ans. [A]

12
Illustration 16 : If pth, qth, rth terms of a G.P. are the positive numbers a, b, c then angle between the vectors
log a 2 ˆi  log b2 ˆj  log c 2 kˆ and (q  r)iˆ  (r  p)jˆ  (p  q)kˆ is :-

   1 
(A) (B) (C) sin 1   (D) none of these
3 2 2 2 2 
 a b c 
Solution : Let x0 be first term and x the common ratio of the G.P.
 a = x 0 x p – 1, b = x 0 x q – 1, c = x 0 x r – 1 log a = log x0 + (p – 1) log x;
log b = log x0 + (q – 1) log x; log c = log x 0 + (r – 1) log x
 
If a  log a 2 i  log b2 j  log c 2 k and b  (q  r ) i  (r  p) j  (p  q ) k

    
 a . b  2 (log a ) (q  r )  2  (log x0  (p – 1)log x)(q  r)  0  a, b   Ans.
2
Illustration 17 : Find the distance of the point B( î + 2 ˆj + 3 k̂) from the line which is passing through

A(4 î + 2 ˆj + 2 k̂) and which is parallel to the vector C  2 ˆi  3 ˆj  6 kˆ . (Roorkee 1993)

Solution : AB = 3 2  1 2  10 B(1,2,3)

 ˆ ˆ ˆ
ˆ (2 i  3 j  6 k)
AM  AB.iˆ  (3ˆi  k).
7
=–6+6=0
BM2 = AB2 – AM2 A(4i+2j+2k) M
C
So, BM = AB = 10 Ans.
Illustration 18 : Prove that the medians to the base of an isosceles triangle is perpendicular to the base.
Solution : The triangle being isosceles, we have
C(c)
AB = AC ........... (i)

 b  c
Now AP  where P is mid-point of BC. P
2
 
Also BC  c  b

  b  c   1 2 B(b) A(0)
 AP . BC  .(c  b)  (c  b2 )
2 2

1
 (AC 2  AB2 )  0 {by (i)}
2
 Median AP is perpendicular to base BC.

Do yourself - 6 :
 
(i) Find the angle between two vectors a & b with magnitude 2 and 1 respectively and such that
 
a .b = 3 .
     
(ii) Find the value of ( a + 3 b ) . (2 a – b ) if a = î + ˆj +2 k̂ , b = 3 î +2 ˆj – k̂ .

(iii) The scalar product of the vector î + ˆj + k̂ with a unit vector along the sum of the vectors

2 î + 4 ˆj – 5 k̂ and  î +2 j + 3 k̂ is equal to 1, find  .
 
(iv) Find the projection of the vector a  4 ˆi  2 ˆj  kˆ on the vector b  3 ˆi  6 ˆj  2kˆ . Also find component
  
of a along b and perpendicular to b .

(v) Find the unit vectors along the angle bisectors between the vectors a  ˆi  2 ˆj  2kˆ and

b  3 ˆi  6 ˆj  2kˆ .

13
19. LINEAR COMBINATIONS :
      
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 .
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, b i.e. there exist some unique x, y  R such that xa  yb  r .

   
Illustration 19 : Find a vector c in the plane of a  2 ˆi  ˆj  kˆ and b  ˆi  ˆj  kˆ such that c is perpendicular to
  ˆ  1
b and c.(2 ˆi  3 ˆj  k)
   
Solution : Any vector in the plane of a & b can be written as xa  yb
  
let c  xa  yb [by fundamental theorem in plane]
Now, given that
   
c.b  0  (xa  yb).b  0
 
xa.b  yb2  0
 x(–2 + 1 – 1) + y(3) = 0
–2x + 3y = 0 ...........(i)
 
ˆ  1
Also (xa  yb).(2 ˆi  3 ˆj  k)
 
 ˆ  yb.(2 ˆi  3 ˆj  k)
xa.(2 ˆi  3 ˆj  k) ˆ  1
 x(–4 + 3 + 1) + y(2 + 3 – 1) = –1
1
y
4
3y 3
x 
2 8
 3 ˆ  1 (ˆi  ˆj  k)
ˆ
Hence the required vector c   (2 ˆi  ˆj  k)
8 4
1 ˆ  1 [4 ˆi  5 ˆj  k]
= [6 ˆi  3 ˆj  3 kˆ  2 ˆi  2 ˆj  2k] ˆ Ans.
8 8

Do yourself - 7 :
  
(i) Find a vector r in the plane of p  ˆi  ˆj and q  ˆj  kˆ such that r is perpendicular to p and

r.q  2
20. VECTOR PRODUCT OF TWO VECTORS (CROSS PRODUCT) :
 
(a) If a & b are two vectors &  is the angle between them, then C = A×B B
   
a  b  a b sin nˆ , where n̂ is the unit vector perpendicular to 

     A
both a & b such that a , b & n forms a right handed screw
–C = B ×A
system.
Sign convention :
 
Right handed screw system : a , b and n̂ form a right handed

system it means that if we rotate vector a towards the direction

of b . through the angle , then n̂ advances in the same direction
as a right handed screw would, if turned in the same way.

14
     
  2 2 2   2 a.a a.b
(b) Lagranges Identity : For any two vectors a & b ; (a  b)  a b  (a . b)     
a . b b. b
  
(c) Formulation of vector product in terms of scalar product : The vector product a  b is the vector c ,
such that
           
(i) c  a 2 b 2  (a . b)2 (ii) c. a  0 ; c. b  0 and (iii) a , b , c form a right handed system
       
(d) (i) a  b  0  a & b are parallel (collinear) (a  0 , b  0 ) i.e. a  Kb , where K is a scalar
   
(ii) a  b  b  a (not commutative)
     
(iii) (ma)  b  a  (mb)  m (a  b) where m is a scalar..
      
(iv) a  (b  c )  (a  b)  (a  c ) (distributive over addition)
(v) ˆ k
ˆi  ˆi  ˆj  ˆj  k ˆ 0 i
j
(vi) ˆ ˆj  k
ˆi  ˆj  k, ˆ  ˆi, k
ˆ  ˆi  ˆj

k ˆi ˆj ˆ
k
   
(e) ˆ & b  b ˆi  b ˆj  b k
If a  a 1 ˆi  a 2 ˆj  a 3 k ˆ
1 2 3 , then a  b  a 1 a2 a3
b1 b 2 b 3
 
(f) Geometrically | a  b| = area of the parallelogram whose two adjacent sides
 
are represented by a & b . a× b b
 
  a  b a
(g) (i) Unit vector perpendicular to the plane of a & b is n̂    
| a  b|
  
 r(a  b)
(ii) A vector of magnitude ‘r’ & perpendicular to the plane of a & b is   
| a  b|
 
  | a  b|
(iii) If  is the angle between a & b , then s in    
| a || b|
(h) Vector area :
  
(i) If a, b & c are the pv’s of 3 points A, B & C then the vector area of triangle
1       
ABC  a  b  b  c  c  a .
2 
     
(ii) The points A, B & C are collinear if a  b  b  c  c  a  0
  1  
(iii) Area of any quadrilateral whose diagonal vectors are d1 & d2 is given by d1  d 2 .
2

Illustration 20 : Find the vectors of magnitude 5 which are perpendicular to the vectors a  2 ˆi  ˆj  3kˆ and

b  ˆi  2 ˆj  kˆ .
  
 ab
Solution : Unit vectors perpendicular to a & b    
| a  b|
ˆi ˆj kˆ
 
 a b  2 1 3 = 5 ˆi  5 ˆj  5kˆ
1 2 1

ˆ
(5 ˆi  5 ˆj  5k)
 Unit vectors  
5 3
5 3 ˆ ˆ ˆ
Hence the required vectors are  (i  j  k) Ans.
3

15
        
Illustration 21 : If a, b, c a re three non zero vectors such that a  b  c and b  c  a , prove that a, b, c are
  
mutually at right angles and | b|  1 and | c| | a| .
Solution :      
a  b  c and a  b  c
       
 c  a, c  b and a  b, a  c
     
 a  b, b  c and c  a
  
 a, b, c are mutually perpendicular vectors.
Again,      
a  b  c and b  c  a
     
 | a  b| | c| and | b  c| | a|
           
 | a| | b| sin | c| and | b| | c| sin | a|
2 2
 a  b and b  c 
     
 | a| | b|  | c| and | b| | c| | a|
  
 | b| 2 | c|  | c|

 | b| 2  1

 | b|  1
  
putting in | a| | b| | c|
 
 | a|  | c|
Illustration 22 : Show that the area of the triangle formed by joining the extremities of an oblique side of a
trapezium to the midpoint of opposite side is half that of the trapezium.

Solution : Let ABCD be the trapezium and E be the midpoint of BC. Let A be the initial point and let b be
 
the position vector of B and d that of D. Since DC is parallel to AB, tb is a vector along DC,
 
so that the position vector of c is d  tb .
     D (d) C
b  d  tb d  (1  t )b
 the position vector of E is 
2 2
E
 
1 d  (1  t)b  1  
Area of AED =  d = (1  t)| b  d|
2 2 4 (b)
A(0) B
Area of the trapezium = Area (ACD) + Area (ABC).
1    1   
= | b  (d  tb)|  | (d  tb) d|
2 2
1   t   1  
 | b  d|  | b  d|  (1  t)| b  d| = 2AED
2 2 2
     
Illustration 23 :

Let a & b be two non-collinear unit vectors. If u  a   a . b  b & v   a  b  , then v is -
[JEE 99]
          
(A) u (B) u  u. a (C) u  u . b 
(D) u  u . a  b 
   
Solution : u.a  a.a  (a.b)(a.b)
 2  2  
= 1 | a| | b| cos2  (where  is the angle between a and b )
= 1 – cos2 = sin2
  
| v| | a  b|  sin 
 
| u|  u.u
    
 a.a  2(a.b)2  (a.b)2 | b| 2 = 1  (a.b)2 = sin 
  
 | v| | u| also u.b  0
    
Hence, | v|  | u| | u| | u. b| Ans. (A, C)

16
 Do yourself - 8 :
             
(i) If a × b = c × d and a × c = b × d , then show that ( a – d ) is parallel to ( b – c ) when a  d
 
and b  c .
   
(ii) Find a × b , if a = 2 î + k̂ and b = î + ˆj + k̂ .
 
(iii) For any two vectors u & v , prove that [JEE 98]
   2 2 2 2 2       2
(a) (u. v)2  u  v  u v (b) (1  u )(1  v )  (1  u . v)2  u  v  (u  v)

21. SHORTEST DISTANCE BETWEEN TWO LINES :

If two lines in space intersect at a point, then obviously the A
shortest distance between them is zero. Lines which do not a L p
intersect & are also not parallel are called skew lines. In
other words the lines which are not coplanar are skew
lines. For Skew lines the direction of the shortest distance p× q
vector would be perpendicular to both the lines. The
magnitude of the shortest distance vector would be equal to

M q
that of the projection of AB along the direction of the line of b
B
  
shortest distance, LM is parallel to p  q
  
i.e. LM = | Projection of AB on LM|

      
   AB . (p  q) (b  a).(p  q)
= | Projection of AB on p  q|      
p q | p  q|
 
(a) The two lines directed along p & q will intersect only if shortest distance = 0 i.e.
           
 
(b  a). (p  q)  0 i.e. (b  a ) lies in the plane containing p & q   b  a p q   0
 
     
(b) If two lines are given by r1  a 1  K1 b & r2  a 2  K 2 b i.e. they
a2
  
b  (a 2  a 1 ) d
are parallel then, d  
| b| b
a1
Illustration 24 : Find the shortest distance between the lines
 ˆ and r  (iˆ – ˆj  2k)  µ(2 ˆi  4 ˆj – 5k)
ˆ
r  (4 ˆi – ˆj)  (iˆ  2ˆj – 3k)
     
Solution : We known, the shortest distance between the lines r  a 1  b1 and r  a 2  b2 is given by
   
(a 2  a 1 ).(b1  b2 )
d=  
| b1  b 2 |
    
Comparing the given equation with the equations r  a 1  b1 and r  a 2  b2 respectively,,
   
we have a 1  4 ˆi  ˆj, a 2  ˆi  ˆj  2kˆ , b1  ˆi  2 ˆj – 3kˆ and b2  2 ˆi  4 ˆj – 5 kˆ

ˆi ˆj kˆ
   
Now, a 2  a 1  –3ˆi  0 ˆj  2kˆ and b1  b2 = 1 2 –3  2 ˆi  ˆj  0 kˆ
2 4 –5

     
 ˆ ˆi – ˆj  0k)
(a 2 – a 1 ).(b1  b2 )  (–3 ˆi  0 ˆj  2k).(2 ˆ = –6 and | b  b |  4  1  0  5
1 2

   
(a 2 – a 1 ).(b1  b2 ) –6 6
 Shortest distance d =     . Ans.
| b1  b 2 | 5 5

17
 Do yourself - 9 :
(i) Find the shortest distance between the lines :
 ˆ & r  (2 ˆi  4 ˆj  5k)
ˆ  (2 ˆi  3 ˆj  4 k)
r1  (iˆ  2 ˆj  3k) ˆ  (3 ˆi  4 ˆj  5k)
ˆ .
2

22. SCALAR TRIPLE PRODUCT / BOX PRODUCT / MIXED PRODUCT :

  
(i) The scalar triple product of three vectors a, b & c
is def ined as :
      ccos  c
 a  b  . c | a || b || c | s in  co s  where  is
   b
the angle between a & b &  is the angle a
  
between a  b & c . It is also defined as
  
[a b c] , spelled as box product.
(ii) Scalar triple product geometrically represents the volume of the parallelopiped whose three coterminous
   
edges are represented by a, b & c i.e. V  [a b c ]
     
(iii) In a scalar triple product the position of dot & cross can be interchanged i.e. a . (b  c )  (a  b). c OR
        
[a b c]  [b c a]  [c a b]
        
(iv) a . (b  c)   a . (c  b) i.e. [a b c ] =  [a c b]

a1 a2 a3
     
If a  a 1 ˆi  a 2 ˆj  a 3 kˆ ; b  b1 ˆi  b2 ˆj  b3 kˆ & c  c1 ˆi  c 2 ˆj  c 3 kˆ then [a b c] = b1 b2 b3
c1 c2 c3
           
In general , if a  a 1 l  a 2 m  a 3 n ; b  b1 l  b2 m  b3 n ; & c  c1 l  c2 m  c 3 n

a1 a2 a3  
      
then [a b c ]  b1 b2 b 3 [ l m n] ; where l , m & n are non coplanar vectors.
c1 c2 c3
        
(v) If a , b , c are coplanar  [a b c] = 0  a , b , c are linearly dependent.
  
(vi) Scalar product of three vectors, two of which are equal or parallel is 0 i.e. [a b c]  0
        
Note : If a, b, c are non- coplanar then [a b c]  0 for right handed system & [a b c]  0 for
left handed system.
               
(vii) [i j k]  1 (viii) [K a b c ]  K[a b c]
(ix) [(a  b) c d]  [a c d]  [b c d]
  
(viii) The Volume of the tetrahedron OABC with O as origin & the pv’s of A, B and C being a, b & c are

 
given by V  1 [a b c] O
c
6
The position vector of the centroid of a tetrahedron if the pv’s of its a b C

    1    
angular vertices are a , b , c & d are given by (a  b  c  d)
4 A B
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.
              
(ix) [a  b b  c c  a]  0 & [a  b b  c c  a ]  2 [a b c ]

18
          
Illustration 25 : For any three vectors a, b, c prove that [a  b b  c c  a]  2[a b c]
     
Solution : We have [a  b b  c c  a]
                 
= {(a  b)  (b  c)}.(c  a )  {a  b  a  c  b  b  b  c}.(c  a ) { b  b  0}
                         
= {a  b  a  c  b  c}.(c  a ) = (a  b).c  (a  c).c  (b  c).c  (a  b).a  (a  c).a  (b  c).a
     
= [a b c] + 0 + 0 + 0 + 0 + [b c a] { [a c c]  0,[b c c]  0,[a b a]  0,[a c a]  0}
  
= [a b c]  [a b c]  2[a b c] . Ans.
 
Illustration 26 : If a, b are non-zero and non-collinear vectors then show
    ˆ ˆ
a  b  [a b ˆi]iˆ  [a b ˆj]jˆ  [a b k]k
 
Solution : Let a  b  xiˆ  yjˆ  zkˆ
  ˆ ˆi
(a  b). ˆi  (xiˆ  yjˆ  zk).
 
(a  b). ˆi = x
   
also (a  b). ˆj = y & (a  b). kˆ = z
    ˆ ˆ
 a  b  [a b ˆi]iˆ  [a b ˆj]jˆ  [a b k]k Ans.
Do yourself - 10 :
     
(i) If a , b , c are three non coplanar mutually perpendicular unit vectors then find [ a , b , c ].
            
(ii) If r be a vector perpendicular to a + b + c , where [ a b c ] = z and r = ( b × c ) + m( c × a ) + n( a ×

b ), then find l + m + n.
(iii) Find the volume of the parallelepiped whose coterminous edges are represented by
 
ˆ b  ˆi  2 ˆj  kˆ and c  3 ˆi  ˆj  2kˆ
a  2 ˆi  3 ˆj  4 k,
 
(iv) Examine whether the vectors a  2 ˆi  3 ˆj  2k, ˆ b  ˆi  ˆj  2kˆ and c  3 ˆi  2 ˆj  4 kˆ form a left handed
or right handed system.
  
(v) For three vectors u, v, w which of the following expressions is not equal to any of the remaining
three ? [JEE 98]
           
(A) u.(v  w) (B) (v  w). u (C) v. (u  w) (D) (u  v).w

23. VECTOR TRIPLE PRODUCT :

     
Let a, b & c be any three vectors, then the expression a  (b  c ) is a vector & is called a vector triple
product.
  
Geometrical interpretation of a × (b × c)
c
  
Consider the expression a  (b  c) which itself is a vector, since it (a× b)× c a× b
     
is a cross product of two vectors a & (b  c) . Now a  (b  c) is
   b
vector perpendicular to the plane containing a & (b  c) but a
   
(b  c) is a vector perpendicular to the plane b & c , therefore
  
a  (b  c)
     
is vector lies in the plane of b & c and perpendicular to a . Hence we can express a  (b  c) in terms
      
of b & c i.e. a  (b  c ) = xb  yc where x & y are scalars.
                 
(a) a  (b  c )  (a . c )b  (a . b)c (b) (a  b)  c  (a . c)b  (b . c )a
     
(c) (a  b)  c  a  (b  c )

19
      2
Illustration 27 : Prove that  a  b b  c c  a   abc
   
       
Solution : We have, a  b b  c c  a  = a  b  b  c .  c  a 
   
 
       
  
= d  b  c .c  a  (where, d  (a  b) )
               
=  d.c  b   d.b  c.  c  a  =   a  b  .c  b    a  b  .b  c.  c  a 
     
 abc  b  0.  c  a  ( [abb]  0 )
              2
 
  abc  b.  c  a   abc   bca   abc  abc   abc 
         
           
Illustration 28 : Show that (b  c).(a  d)  (c  a). (b  d)  (a  b).(c  d)  0
        
Solution : Let b  c  u, c  a  v, c  d  w
                 
 L.H.S = u.(a  d)  v.(b  d)  (a  b).w = (u  a).d  (v  b).d  a . (b  w)
           
 [(b  c)  a].d  [(c  a)  b].d  a.[b  (c  d)]
              
 [(b. a)c – (c. a)b].d  [(c. b)a – (a . b)c].d  a.[(b. d)c – (b. c)d]
                      
 {(a . b)(c. d)}– {(c. a)(b. d)} {(c. b)(a . d)}– {(a . b)(c. d)} {(a . c)(b.d)}– {(a . d)(b. c)}  0 = R.H.S.

Do yourself - 11 :
        
(i) If a = 2 î – 4 ˆj + 7 k̂ , b = 3 î + 5 ˆj – 9 k̂ and c = î + ˆj + k̂ then find [ a b c ] and also a × ( b × c ).

24. LINEAR INDEPENDENCE AND DEPENDENCE OF VECTORS :

  
(a) If x1 , x2 ,.......xn are n non zero vectors, & k1 , k2 ,.......k n are n scalars & if the linear combination
  
k 1 x 1  k 2 x 2  ... .k n x n  0  k1  0 , k 2  0 ... .k n  0 then we say that v ectors
  
x 1 , x 2 , .... .x n are linearly independent vectors.
  
(b) If x1 , x2 ,.......x n are not linearly independent then they are said to be linearly dependent
     
vectors. i.e. if k 1 x 1  k 2 x 2  .... ... k n x n = 0 & if there exists at least one k r  0 then x 1 , x 2 , . ..x n
are said to be linearly dependent.
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 r  xa  yb  zc .
Note :
  
(i) If a  3 ˆi  2 ˆj  5kˆ then a is expressed as a linear combination of vectors ˆi, ˆj, kˆ . Also, a , ˆi, ˆj, kˆ
form a linearly dependent set of vectors. In general, every set of four vectors is a linearly dependent
system.
     
(ii) If a, b, c are three non-zero, non-coplanar vectors then xa  yb  zc  0  x = y = z = 0

(iii) ˆi, ˆj, kˆ are linearly independent set of vectors. For K ˆi  K ˆj  K kˆ  0  K  0  K  K

1 2 3 1 2 3

     
(iv) Two vectors a & b are linearly dependent  a is a parallel to b i.e. a  b  0  linear
     
dependence of a & b . Conversely if a  b  0 then a & b are linearly independent.
     
(v) 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.

20
         
Illustration 29 : Show that points with position vectors a  2 b  3 c, –2 a  3 b – c, 4 a – 7 b  7c are collinear. It is
  
given that vectors a, b, c are non-coplanar..
Solution : The three points are collinear, if we can find 1, 2 and 3, such that
        
1 (a – 2b  3c)  2 (–2a  3b – c)  3 (4 a – 7b  7c)  0
with 1 + 2 + 3 = 0
  
equating the coefficients a, b and c separately to zero, we get
1 – 22 + 43 = 0, –21 + 32 – 73 = 0 and 31 – 2 + 73 = 0
on solving we get, 1 = – 2, 2 = 1, 3 = 1
So that 1 + 2 + 3 = 0
Hence the given vectors are collinear.

 
Illustration 30 : ˆ b  4 ˆi  3ˆj  4kˆ and c  ˆi  ˆj  kˆ are linearly dependent vectors and | c|
If a  ˆi  ˆj  k, 
= 3
then -
(A)  =1 ,  = –1 (B)  =1 ,  = ±1 (C)  =–1 ,  = ±1 (D)  = ±1 ,  = 1
     
Solution : If a, b, c are linearly dependent vectors, then c should be a linear combination of a and b .
  
Let c  pa  qb

i.e. ˆi  ˆj  kˆ  p(iˆ  ˆj  k)

ˆ  q(4 ˆi  3 ˆj  4k)
ˆ

Equating coefficients of ˆi, ˆj, kˆ we get 1 = p + 4q,  = p + 3q,  = p + 4q

from first and third,  = 1.

Now | c| = 3
 1 + 2 + 2 = 3
 1 + 2 + 1 = 3 {Using  = 1}
 =±1
Hence,  = ±1,  = 1. Ans. (D

25. COPLANARITY OF FOUR POINTS :

   
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

26. RECIPROCAL SYSTEM OF VECTORS :

           
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 systems are called Reciprocal System of vectors.
     
b  c c  a a  b
Note : a '     ; b '     ; c '    
[a b c] [a b c] [a b c]

27. PROPERTIES OF RECIPROCAL SYSTEM OF VECTORS :

         
(a) a . b '  a . c '  b. c '  c. a '  c. b '  0
     
(b) The scalar triple product [a b c] formed by three non-coplanar vectors a, b, c is the reciprocal of the
scalar triple product formed from reciprocal system.

21
    
Illustration 31 : Find a set of vectors reciprocal to the vectors a, b and a  b .
     
Solution : Let the given vectors be denoted by a, b and c where c  a  b
         
 [a b c ]  (a  b).c  (a  b).(a  b)  (a  b)2
  
and let the reciprocal system of vectors be a ' b ' and c '
             
 bc b  (a  b)  ca (a  b)  a  ab a b
 a '        2 ; b'        2 ; c'        2
[a b c] | a  b| [a b c] | a  b| [a b c] | a  b|
      
 a ', b ', c ' are required reciprocal system of vectors for a, b and a  b . Ans.
     
 bc  ca  ab      
Illustration 32 : If a '     , b '     , c '     then shown that ; a  a ' b  b ' c  c '  0
[a b c] [a b c] [a b c]
  
where a, b, c are non-coplanar vectors.
  
  a  (b  c)
Solution : Here a a'  
[a b c]
  
  (a.c)b  (a.b)c
a a'  
[a b c]
     
  (b.a )c  (b.c)a   (c.b)a  (c.a)b
Similarly b  b'  
  & c c'  
[a b c] [a b c]
              
(a.c)b  (a.b)c  (b.a )c  (b.c)a  (c.b)a  (c.a)b    
a  a ' b  b ' c  c ' =  [ a . b  b. a etc.]
[a b c]
= 0. Ans.
Do yourself - 12 :
     
 bc  ca  a b        
(i) If p     , q     , r     , then find the value of ( a + b ). p + ( b + c ) . q + ( c + a )
[b c a] [c a b] [a b c ]

. r.
(ii) Find the set of vectors reciprocal to the set 2 ˆi  3 ˆj  k, ˆ ˆi  ˆj  2kˆ and ˆi  2 ˆj  2kˆ . Hence prove that
     
a . a ' b. b ' c. c '  3 .
     
(iii) If a, b and c are non zero, non coplanar vectors determine whether the vectors r1  2a  3 b  c ,
       
r2  3a  5 b  2c and r3  4a  5 b  c are linearly independent or dependent.

28. APPLICATION OF VECTORS :

 L
(a) Work done against a constant force F over a
  
displacement s is defined as W  F . s
 w
a
(b) The tangential velocity V of a body moving in a
    A
circle is given by V  w  r where r is the pv of r p v
the point P.
    O
(c) The moment of F about ‘O’ is defined as M  r  F
r
 
where r is the pv of P wrt ‘O’. The direction of M F
 
is along to the normal to the plane OPN such that r' P
  
r, F & M form a right handed system.
   N
(d) Moment of couple = (r1  r2 )  F where
 
r1 & r2 are pv’s of the point of the
  Q
application of the forces F &  F .

22
 Do yourself - 13 :

(i) Force F  4 ˆi  ˆj  3kˆ acting on a particle displaces it from the point ˆi  2 ˆj  3kˆ to the point

5 ˆi  3 ˆj  4 kˆ . Find the work done by the force F .

Miscellaneous Illustrations :
Illustration 33 : Forces of magnitudes 5, 4, 3 units act on a particle in the directions 2 ˆi – 2 ˆj  kˆ , ˆi  2 ˆj  2kˆ

and –2ˆi  ˆj  2kˆ respectively, and the particle gets displaced from the point A whose position

vector is 6 ˆi  2 ˆj  3kˆ , to the point B whose position vector is 9 ˆi  7ˆj  5 kˆ . Find the work
done.
      3
5 ˆ ; F  4 (iˆ  2 ˆj  2k) ˆ
ˆ and F3  (–2 ˆi  ˆj – 2k)
Solution : If the forces are F1 , F2 , F3 then F1  (2 ˆi  2 ˆj  k)
3
2
3 3
    1 ˆ
and hence the sum force F  F1  F2  F3  (8 ˆi  ˆj  7k)
3
  
Displacement vector AB  OB – OA = 9 ˆi  7ˆj  5 kˆ – (6 ˆi  2 ˆj  3k) ˆ = 3 ˆi  5 ˆj  2kˆ

Work done =
1 ˆ ˆ
(8 i  j  7k) ˆ = 1 (2 4  5  1 4 ) = 4 3 units.
ˆ . (3 ˆi  5 ˆj  2k) Ans.
3 3 3
 
Illustration 34 : Let u and v be unit vectors. If w is a vector such that w + ( w  ×  ) =  , then prove that
u v
   1  
|( u × v ) . w |  and that the equality holds if and only if u is perpendicular to v .
2
   
Solution : w  (w  u)  v ......(i)
     2 2 2  
 w  u  v – w  (w  u)  v  w – 2 v.w
   
 2 v.w  1  w 2 – (u  w)2 ......(ii)

also taking dot product of (i) with v , we get
     
w.v  (w  u).v  v.v
      
 v.(w  u)  1 – w.v .......(iii)  v.v | v| 2  1

   1  
Now ; v.(w  u)  1 – (1  w 2  (u  w)2 ) (using (ii) and (iii))
2
 
1 w 2 (u  w)2
 –  ( 0  cos2  1)
2 2 2
1
= (1 – w2 + w2 sin2 ) ......(iv)
2
as we know ; 0  w2 cos2   w2
1 1  w 2 cos 2  1  w 2
  
2 2 2
1  w2 cos 2  1
  ......(v)
2 2
from (iv) and (v)
   1
| v.(w  u)| 
2

Equality holds only when cos2 = 0  =
2
       
i.e., u  w  u. w  0  w  (w  u)  v
       
 u·w  u. (w  u)  u. v (taking dot with u )
     
 0 + 0 = u·v  u·v = 0  uv
23
Illustration 35 : A point A(x1, y1) with abscissa x1 = 1 and a point B(x2, y2) with ordinate y2 = 11 are given in a
rectangular cartesian system of co-ordinates OXY on the part of the curve y = x 2 – 2x + 3
 
which lies in the first quadrant. Find the scalar product of OA and OB .
Solution : Since (x1, y1) and (x2, y2) lies on y = x2 – 2x + 3.
 y1 = x12 – 2x1 + 3
y1 = 12 – 2(1) + 3 (as ; x1 = 1)
y1 = 2
so the co-ordinates of A(1, 2)
Also, y2 = x22 – 2x2 + 3
2
11 = x2 – 2x2 + 3  x2 = 4, x2  –2 (as B lie in 1st quadrant)
 co-ordinates of B (4, 11).
 
Hence, OA  ˆi  2ˆj and OB  4 ˆi  11ˆj
 
 OA . OB = 4 + 22 = 26.

Illustration 36 : If ‘a’ is real constant and A, B, C are variable angles and a 2  4 tan A + a tan B + a 2  4 tan C =
6a
then find the least value of tan2 A + tan2 B + tan2 C
Solution : The given relation can be re-written as ;
ˆ = 6a
ˆ . (tan Aiˆ  tan Bjˆ  tan Ck)
( a 2  4 ˆi  aˆj  a 2  4 k)

 (a 2  4 )  a 2  (a 2  4 ). tan 2 A  tan 2 B  tan 2 C . cos  = 6a

(as, a.b = |a| |b| cos )

 3 a . tan 2 A  tan 2 B  tan 2 C cos   6a

 tan2A + tan2B + tan2C = 12 sec2  ......(i)
also, 12 sec2  12 (as, sec2   1) ......(ii)
from (i) and (ii),
tan2 A + tan2 B + tan2 C  12
 least value of tan2 A + tan2 B + tan2 C = 12.

Illustration 37 : Prove that the right bisectors of the sides of a triangle are concurrent.
Solution : Let the right bisectors of sides BC and CA meet at O and taking O as origin, let the position
  
vectors of A, B and C be taken as a, b, c respectively. Hence the mid-points D, E, F are
     
bc c a a b A (a)
, ,
2 2 2
 
  bc   F E
 OD  BC,  .(c  b)  0 O
2
i.e. b2 = c2 B(b)
D C(c)
   
ca  
Again since OE  CA,  .(a  c)  0
2
or a2 = c2  a2 = b2 = c2 ......... (i)
 
  a b  
Now we have to prove that OF is also  to AB which will be true if .(b  a )  0
2
i.e. b2 = a2
which is true by (i)

24
  
Illustration 38 : A, B, C and D are f our point s such that AB  m(2 ˆi  6 ˆj  2k), ˆ BC  (iˆ  2 ˆj) and

ˆ .
CD  n(6 ˆi  15 ˆj  3k)
 
Find the conditions on the scalars m and n so that CD intersects AB at some point E. Also
find the area of the triangle BCE.
(Roorkee 1995)
 
Solution : ˆ BC  (iˆ  2 ˆj)
AB  m(2 ˆi  6 ˆj  2k),

ˆ
CD  n(6 ˆi  15 ˆj  3k) B
    D
If AB and CD intersect at E, then EB  p AB, CE  qCD
where both p and q are positive quantities less than 1 E
    C
Now we know that EB  BC  CE  EE  0 A

  

 p AB  BC  q CD  0 {by (i)}

or pm(2 î – 6 ˆj + 2 k̂ ) + ( î – 2 ˆj ) + q.n(–6 î + 15 ˆj – 3 k̂ ) = 0

Since î , ˆj , k̂ are non-coplanar, the above relation implies that if x î + y ˆj + z k̂ = 0, then x =

0, y = 0 and z = 0
 2mp + 1 – 6qn = 0, – 6pm – 2 + 15qn = 0
2pm – 3qn = 0
Solving these for pm and qn, we get
1 1 1 1
pm = , qn =  p= , q
2 3 2m 3n
1 1 1 1
 0  1, 0  1 or m ,n
2m 3n 2 3

ˆi ˆj kˆ

1   1 qCD  1
Again area of BCE = EC  EB =  p AB = pqnm 6 15 3
2 2 2
2 6 2

1 1
Put pm  , qn 
2 3
1 1 1 ˆ  1 .6 6  1 6
= . . | 1 2 ˆi  6 ˆj  6 k|
2 2 3 12 2

  
Illustration 39 : a , b , c are three non-coplanar unit vectors such that angle between any two is  If
      
a × b + b × c =  a + m b + n c , then determine , m, n in terms of . (JEE-1997)
2 2 2
Solution : a = b = c = 1, [abc]  0
     
a . b = b . c = c . a = cos ......... (i)
  
Multiply both sides of given relation scalarly by a , b and c , we get

0 + [ a b c ] = .1 + (m + n) cos ......... (ii)
0 = m + (n + ) cos ......... (iii)

[ a b c ] + 0 = ( + m)cos + n ......... (iv)
Adding, we get

2[ a b c ] = ( + m + n) + 2( + m + n) cos

or 2[ a b c ] = ( + m + n) (1 + 2cos) ......... (v)

25
 
[a b c]  
From (ii), (m + n) =
cos 

 
    [a b c]   
Putting in (v), we get 2[a b c]     1  2 cos  
 cos  

  1  2 cos    1 
or [a b c]2   =  1  (1  2 cos  )
 cos    cos  


[a b c]
  n {as above}
(1  2 co s  )(1  co s  )

 2 [a b c]co s 
and m = –(n + ) cos =
(1  2 cos  )(1  co s  )

Thus the values of , m, n depend on [ a b c ]

Hence we now find the value of scalar [ a b c ] in terms of .
  
a.a a.b a.c 1 cos  cos 
   
Now [ a b c ]2 = b.a b.b b.c = cos  1 cos  (Apply C1 + C2 + C3)
     cos  cos  1
c.a c.b c.c

1 cos  cos 
= (1 + 2 cos) 1 1 cos  (Apply R2 – R1 and R3 – R1)
1 cos  1

 [ a b c ]2 = (1 + 2 cos)(1 – cos)2


[a b c]
  1  2 co s 
1  cos 
1 2 cos 
Putting in the value of , m, n we have    n, m  Ans.
(1  2 cos  ) (1  2 cos  )

      
Illustration 40 : Vectors x, y and z each of magnitude 2 , make angles of 60° with each other. If x  (y  z)  a ,
            
y  (z  x)  b and x  y  c , then find x, y and z in terms of a, b and c . (Roorkee 1997)
    
Solution : x.y  2 2 cos 60   1  y. z  z. x ...... (i)
2 2 2
Also x = y = z = 2
      
Again a  (x.z)y  (x.y)z  y  z {by (i)}
     
 a  y  z, b  z  x ........(ii)
           
Now a  c  (y  z)  (x  y)  y  (x  y)  z  (x  y)
           
= [(y . y)x – (y . x)y] – [(z . y)x – (z . x)y] = (2x – y) – (x – y) {by (i)}
  
or a× c=x
  
Similarly, b× c= y
     
Now z = y – a or z= b+ x {by (ii)}
      
 z = (b × c – a) or b + (a × c) Ans.

26
               
Illustration 41 : If x × y = a , y × z = b , x . b =  , x . y = 1 and y . z = 1, then find x, y and z in terms
 
of a, b and  (Roorkee 1998)
  
Solution : x× y= a ........ (i)
  
y× z= b ........ (ii)
     
Also x . b = , x . y = 1, y . z = 1 ........ (iii)
We have to make use of the relations given above.
From (i)
    
x . (x × y) = x . a
  
 x.a =0  [x x y] = 0
     
Similarly y. a = 0, y . b = 0, z . b = 0 ........ (iv)

Multiplying (i) vectorially by b ,
          
b × (x × y) = b × a or (b . y)x – (b . x)y = b × a
 
    (a  b)
y
or 0 –  y = –(a × b)  ........ (v)

by using relations is (iii) and (iv).

Again multiplying (i) vectorially by y ,
          
(x × y) × y = a × y or (x . y)y – (y . y)x = a × y
    
y – a × y = | y| 2 x {by (iii)}

 1   
 x   2 [y  a  y]
| y|

 
 ab
where y {by (v)}

 
Hence x is known in terms of a, b and .

Again multiplying (ii) vectorially by y , we get
               
(y × z) × y = b × y or | y| 2 z – (y . z)y = b × y or | y| 2 z = b × y + y {by (iii)}

1   
or z   2 [b  y  y]
| y|

where y is given by (v) ......... (vi)
    
Results (v) and (vi) give the values of x, y and z in terms of a, b and . Ans.

27
 ANSWERS FOR DO YOURSELF

1: (i) 7 (ii) A
                   
3: (i) (a) a, d ; b, x, z ; c, y (b) b, x ; a, d ; c, y (c) a, y, z (d) b, z; x, z
1
(ii) (iii) A,B,C
4
 
12a  13b  3ˆ 6ˆ 2 ˆ
4: (i) , 5 b (iii) i j k
5 7 7 7
 2 2 ˆ ˆ ˆ ˆ
6: (i) (ii) – 15 (iii) 1 (iv) , ˆ and 190 i  110 j  45 k
(3 i  6 ˆj  2 k)
6 7 49 49
2 ˆ 32 ˆ 8 ˆ 16 ˆ 4 ˆ 20 ˆ
(v)  i j k & i j k
21 21 21 21 21 21
 2 ˆ ˆ
7. (i) r
3
i  j  2 kˆ 
8: (ii) ˆi  ˆj  2kˆ (iii) –3 î + 6 ˆj + 6 k̂

1
9: (i)
6

10 : (i) ±1 (ii) 0 (iii) 7 (iv) Right handed system (v) C

11 : (i) 62, 92 ˆi  102 ˆj  32kˆ
ˆ ˆ
2 ˆi  kˆ 8 ˆi  3ˆj  7k 7ˆi  3 ˆj  5k
12 : (i) 3 (ii) , , (iii) linearly dependent.
3 3 3
13 : (i) 20 units

28
 3-DIMENSIONAL GEOMETRY
A. DISTANCE BETWEEN TWO POINTS
Let P and Q be two given points in space. Let the co-ordinates of the points P and Q be (x1, y1 z1) and (x2,
y2, z2) with respect to a set OX, OY, OZ of rectangular axes. The position vectors of the points P and Q are

given by OP = x1 î + y1 ĵ + z1 k̂ and OQ = x2 î + y2 ĵ + z2 k̂
  
Now we have PQ  OQ  OP . = (x2 î + y2 ĵ + z2 k̂ ) – (x1 î + y1 ĵ + z1 k̂ )

= (x2 – x1) î – (y2 – y1) ĵ – (z2 – z1) k̂ .

 2 2 2
 PQ = | PQ |  (x 2  x1)  (y2  y1)  (z2  z1)
Distance (d) between two points (x1, y1, z1) and (x2, y2, z2) is

d= ( x 2  x1 )2  ( y 2  y1 )2  ( z 2  z1 )2

B. SECTION FORMULA

m2 x1  m1x 2 m2 y1  m1y 2 m2 z1  m1z 2

x= ; y= ; z=
m1  m2 m1  m2 m1  m2

(for external division take –ve sign)

To determine the co-ordinates of a point R which divides the joining of two points P(x1, y1, z1) and Q(x2, y2,
z2) internally in the ratio m1 : m2. Let OX, OY, OZ be a set of rectangular axes.
The position vectors of the two given points P(x1, y1, z1) and Q(x2, y2, z2) are given by

OP = x1 î + y1 ĵ + z1 k̂ ....(1) and OQ = x2 î + y2 ĵ + z2 k̂ ....(2)

m1 m2
P Q
R
(x1,y1,z1) (x2,y2,z2)

Also if the co-ordinates of the point R are (x, y, z), then OR = x î + y ĵ + z k̂ . .....(3)

Now the point R divides the join of P and Q in the ratio m1 : m2, so that

m1OQ  m2 OP
Hence m 2 PR  m1RQ or m2 (OR  OP ) = m1 (OQ  OR ) or OR 
m1  m2

(m1x 2  m2 x1) ˆi  (m1y 2  m2 y1) ˆj  (m1z2  m2z1)kˆ

or x î + y ĵ + z k̂ = [Using (1), (2) and (3)]
(m1  m2 )

m1x 2  m2 x1 m1y 2  m2 y1 m1z 2  m2 z1

Comparing the coefficients of î , ĵ , k̂ we get x = ,y= ,z=
m1  m2 m1  m2 m1  m2

Remark : The middle point of the segment PQ is obtained by putting m 1 = m 2. Hence the

1 1 1 
co-ordinates of the middle point of PQ are  ( x1  x 2 ), ( y1  y 2 ), ( z1  z 2 ) 
2 2 2 

29
 CENTROID OF A TRIANGLE :
Let ABC be a triangle. Let the co-ordinates of the vertices A, B and C be (x1, y1, z1), (x2, y2, z2) and (x3,
y3, z3) respectively. Let AD be a median of the ABC. Thus D is the mid point of BC.

 x2  x3 y 2  y3 z2  z3 
 The co-ordinates of D are  , , 
 2 2 2 

Now if G is the centroid of ABC, then G divides AD in the ratio 2 : 1. Let the co-ordinates of G be (x,

 x  x3 
2.  2   1.x1
A
 2  x  x 2  x3
y, z). Then x  , or x  1 .
2 1 3 2

1 1 1
G
Similarly y = (y1 + y2 + y3), z = (z + z2 + z3).
2 2 1 B D C

CENTROID OF A TETRAHEDRON :

Let ABCD be a tetrahedron, the co-ordinates of whose vertices are (xr, yr, zr), r = 1, 2, 3, 4.
Let G1 be the centroid of the face ABC of the tetrahedron. Then the co-ordinates of G1 are

 x1  x 2  x 3 y1  y 2  y 3 z1  z 2  z 3 
 , , 
 3 3 3 

The fourth vertex D of the tetrahedron does not lie in the plane of ABC. We know from statics that the
centroid of the tetrahedron divides the line DG1 in the ratio 3 : 1. Let G be the centroid of the tetrahedron and
if (x, y, z) are its co-ordinates, then
x1  x 2  x 3
3.  1.x 4 x  x 2  x3  x 4
x 3 or x  1 .
3 1 4

1 1
Similarly y  (y1 + y2 + y3 + y4), z  (z1 + z2 + z3 + z4).
4 4

Ex.1 P is a variable point and the co-ordinates of two points A and B are (–2, 2, 3) and (13, –3, 13) respectively.
Find the locus of P if 3PA = 2PB.
Sol. Let the co-ordinates of P be (x, y, z).

 PA = ( x  2 )2  ( y  2 )2  ( z  3 ) 2 ....(1) and PB

= ( x  13 ) 2  ( y  3 ) 2  ( z  13 ) 2 ....(2)

Now it is given that 3PA = 2PB i.e., 9PA2 = 4PB2. ....(3)

Putting the values of PA and PB from (1) and (2) in (3), we get
9{(x + 2)2 + (y – 2)2 + (z – 3)2} = 4 {(x – 13)2 + (y + 3)2 + (z – 13)2}
or 9 {x2 + y2 + z2 + 4x – 4y – 6z + 17} = 4{x2 + y2 + z2 – 26x + 6y – 26z + 347}
or 5x2 + 5y2 + 5z2 + 140x – 60 y + 50 z – 1235 = 0 or x2 + y2 + z2 + 28x – 12y + 10z – 247 = 0
This is the required locus of P.

30
Ex.2 Find the ratio in which the xy-plane divides the join of (–3, 4, –8) and (5, –6, 4). Also find the point of
intersection of the line with the plane.
Sol. Let the xy-plane (i.e., z = 0 plane) divide the line joining the points (–3,4, –8) and (5, –6, 4) in the ratio :
1, in the point R. Therefore, the co-ordinates of the point R are

 5  3 6  4 4  8 
 , ,  ....(1)
  1  1  1 
But on xy-plane, the z co-ordinate of R is zero
 (4– 8) / ( + 1) = 0, or  = 2. Hence : 1 = 2 : 1. Thus the required ratio is 2 : 1.
Again putting  = 2 in (1), the co-ordinates of the point R become (7/3, –8/3, 0).

Ex.3 ABCD is a square of side length ‘a’. Its side AB slides between x and y-axes in first quadrant. Find the
locus of the foot of perpendicular dropped from the point E on the diagonal AC, where E is the midpoint of
the side AD.
Sol. Let vertex A slides on y-axis and vertex B slides on x-axis coordinates of the point A are
(0, a sin ) and that of C are (a cos  + a sin , a cos )
a a a 3a
In AEF, AF = cos 45   and FC = AC – AF = 2a  
2 2 2 2 2 2 2
a 3a
 AF : FC = : =1:3
2 2 2 2
 Let the coordinates of the point F are (x, y)
3  0  1(a cos   a sin ) a(sin   cos )
 x= 
4 4
4x 3a sin   a cos  4y
 = sin  + cos  ....(1)and y =  = 3sin + cos  ...(2)
a 4 a
2( y  x ) 6 x  2y
Form (1) and (2), sin  = and cos  =
a a
a2
 (y – x)2 + (3x – y)2 = is the locus of the point F..
4

C. DIRECTION COSINES OF A LINE

If are the angles which a given 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 (briefly written as d.c.’s) of the line.
These d.c.’s are usually denote by , m, n.
Let AB be a given line. Draw a line OP parallel to the line AB and passing through the origin O. Measure
angles , then cos , cos , cos  are the d.c.’s of the line AB. It can be easily seen that , m, n, are
the direction cosines of a line if and only if  î + m ĵ + n k̂ is a unit vector in the direction of that line.
Clearly OP(i.e. the line through O and parallel to BA) makes angle 180° – , 180°– , 180° –  with OX, OY
and OZ respectively. Hence d.c.’s of the line BA are cos (180° – ), cos (180º – ), cos (180° – ) i.e., are
–cos , –cos , – cos .
If the length of a line OP through the origin O be r, then the co-ordinates of P are (r, mr, nr) where , m, n
are the d c.’s of OP.
If , m, n are direction cosines of any line AB, then they will satisfy 2 + m2 + n2 = 1.
DIRECTION RATIOS :
If the direction cosines , m, n of a given line be proportional to any three numbers a, b, c respectively, then
the numbers a, b, c are called direction ratios (briefly written as d.r.’s of the given line.

31
 RELATION BETWEEN DIRECTION COSINES AND DIRECTION RATIOS :
Let a, b, c be the direction ratios of a line whose d.c.’s are , m, n. From the definition of d.r.’s. we have /
a = m/b = n/c = k (say). Then  = ka, m = kb, n = kc. But 2 + m2 + n2 = 1.

1
 k2 (a2 + b2 + c2) = 1, or k2 = 1/(a2 + b2 + c2) or k = ± .
(a  b 2  c 2 )
2

a b c
Taking the positive value of k, we get  = ,m= ,n=
2 2 2 2 2 2
(a  b  c ) (a  b  c ) (a 2  b 2  c 2 )

Again taking the negative value of k,

a b c
we get = , m= ,n= .
2 2
(a  b  c ) 2 2
(a  b  c )2 2
(a  b 2  c 2 )
2

Remark. Direction cosines of a line are unique. But the direction ratios of a line are by no means unique.
If a, b, c are direction ratios of a line, then ka, kb, kc are also direction ratios of that line where k is any non-

zero real number. Moreover if a, b, c are direction ratios of a line, then a î + b ĵ + c k̂ is a vector parallel to

that line.

Ex.4 Find the direction cosines  + m + n of the two lines which are connected by the relation  + m + n = 0 and
mn – 2n –2m = 0.
Sol. The given relations are  + m + n = 0 or  = –m – n ....(1) and mn – 2n – 2m = 0 ...(2)
Putting the value of  from (1) in the relation (2), we get
mn – 2n (–m –n) – 2(–m – n) m = 0 or 2m2 + 5mn + 2n2 = 0 or (2m + n) (m + 2n) = 0.

m 1  m  n m
   and –2. From (1), we have   –1 ...(3)
n 2 n n n

m 1  1 1 m n  n
Now when = – , (3) given = –1=– .  = and =
n 2 n 2 2 1 2 1 2

 m n ( 2  m2  n2 ) 1
i.e.      The d.c.’s of one line are
1 1 2 2 2
{1  1  ( 2) } 2 6

1 1 2
, , .
6 6 6

m 
Again when = –2, (3) given = 2 – 1 = 1.
2 n

 m n ( 2  m2  n2 ) 1 1 2 1
i.e.     The d.c.’s of the other line are , , .
1 2 1 2
{1  ( 2)  1 }2 2 6  6 6 6

32
 To find the projection of the line joining two points P(x1, y1, z1) and Q(x2, y2, z2) on the another
line whose d.c.’s are , m, n.

Let O be the origin. Then OP = x1 î + y1 ĵ + z1 k̂ and OQ = x2 î + y2 ĵ + z2 k̂ .

 PQ  OQ  OP = (x2 – x1) î + (y2 – y1) ĵ + (z2 – z1) k̂ .

Now the unit vector along the line whose d.c.’s are ,m,n =  î + m ĵ + n k̂ .

 projection of PQ on the line whose d.c.’s are , m, n

= [(x2 – x1) î + (y2 – y1) ĵ + (z2 – z1) k̂ ] . ( î + m ĵ + n k̂ ) = (x2 – x1) + m (y2 – y1) + n(z2 – z1).

a1a 2  b1b2  c 1c 2
The angle  between these two lines is given by cos  =
(a12  b12  c 12 ) (a 22  b22  c 22 )

If  1, m1, n1 and  2 , m2, n2 are two sets of real numbers, then

( 12 + m12 + n12) ( 22 + m22 + n22) – ( 1 2 + m1m2 + n1n2)2
= (m1n2 – m2n1)2 + (n1 2 – n2 1)2 + ( 1m2 –  2m1)2
Now, we have
sin2  = 1 – cos2 = 1 – (12 + m1m2 + n1n2)2 = (12 + m12 + n12) (22 + m22 + n22) – (12 + m1m2 + n1n2)2

2 2 2
m n  n  m
= (m1n2 – m2n1)2 + (n1 2 – n2 1)2 + ( 1m2 –  2m1)2 = m1 n1  1 n1  1 m1
2 2 2 2 2 2

Condition for perpendicularity  12 + m1m2 + n1n2 = 0.

a1 b1 c 1
Condition for parallelism  1 = 2, m1 = m2, n1 = n2.   
a2 b 2 c 2

Ex.5 Show that the lines whose d.c.’s are given by  + m + n = 0 and 2mn + 3n – 5m = 0 are at right angles.
Sol. From the first relation, we have  = –m – n. ....(1)
Putting this value of  in the second relation, we have
2mn + 3 (–m –n) n – 5 (–m –n) m = 0 or 5m2 + 4mn – 3n2 = 0 or 5(m/n)2 + 4(m/n) – 3 = 0 ....(2)
Let 1, m1, n1 and 2, m2, n2 be the d,c,’s of the two lines. Then the roots of (2) are m1/n1 and m2/n2.

m1 m2 3 mm nn
 product of the roots = .   or 1 2  1 2 . ....(3)
n1 n2 5 3 5

Again from (1), n = –  – m and putting this value of n in the second given relation, we have
2m (– – m) + 3(– – m) – 5m = 0 or 3(/m)2 + 10 (/m) + 2 = 0.

1  2 2  mm 1 2 m1m2 n1n2

 .  or 1 2  1 2 From (3) and (4) we have  = k (say)
m1 m2 3 2 3 2 3 5

 12 + m1m2 + n1n2 = (2 + 3 – 5) k = 0 . k = 0. The lines are at right angles.

33
 Remarks :
(a) Any three numbers a, b, c proportional to the direction cosines are called the direction ratios i.e.
 m n 1
   same sign either +ve or –ve should be taken throughout.
a b c a  b2  c 2
2

Note that d.r’s of a line joining x1, y1, z1 and x2, y2, z2 are proportional to x2 – x1, y2 – y1 and z2 – z1
(b) If  is the angle between the two lines whose d.c’s are  1 , m1, n1 and  2 , m2, n2
cos  =  1 2 + m1m2 + n1n2

Hence if lines are perpendicular then  1 2 + m1m2 + n1n2 = 0.

 1 m1 n1
if lines are parallel then  
 2 m2 n2

1 m1 n1

Note that if three lines are coplanar then 2 m2 n2 = 0
 3 m3 n3

(c) Projection of the join of two points on a line with d.c’s , m, n are
 (x2 – x1) + m(y2 – y1) + n(z2 – z1)
(d) If 1, m1, n1 and 2, m2, n2 are the d.c.’s of two concurrent lines, show that the d.c.’s of two lines
bisecting the angles between them are proportional to 1 ± 2, m1 ± m2, n1 ± n2.

D. AREA OF A TRIANGLE

Show that the area of a triangle whose vertices are the origin and the points A(x1, y1, z1) and
1
B(x2, y2, z2) is (y1z2  y 2 z1)2  (z1x2  z2 x1)2  (x1y 2  x 2 y1)2 .
2
The direction ratios of OA are x1, y1, z1 and those of OB are x2, y2, z2.

Also OA = (x1  0)2  (y1  0)2  (z1  0)2  (x12  y12  z12 )

and OB = (x 2  0)2  (y 2  0)2  (z2  0)2  (x 22  y 22  z 22 ) .

x1 y1 z1
, ,
 the d.c.’ s of OA are
( x12  y12  z12 ) ( x12  y12  z12 ) ( x12  y12  z12 )

x2 y2 z2
, ,
and the d.c.’s of OB are
( x22  y 22  z22 ) ( x 22  y 22  z 22 ) ( x 22  y 22  z 22 )
Hence if  is the angle between the line OA and OB, then

{( y1z 2  y 2 z 2 )2 } {( y1z 2  y 2 z1 )2 }

sin  = 
( x12  y 12  z12 ) ( x 22  y 22  z 22 ) OA.OB

1
Hence the area of OAB = . OA . OB sin  [ AOB = ]
2

1 {(y1z 2  y 2z 2 )2 } 1
= . OA. OB. = {( y1z 2  y 2 z 2 )2 } .
2 OA.OB 2

34
Ex.6 Find the area of the triangle whose vertices are A(1, 2, 3), B(2, –1, 1)and C(1, 2, –4).
Sol. Let x, y , z be the areas of the projections of the area  of triangle ABC on the yz, zx and
xy-planes respectively. We have

1 y1 z1 1 1 2 3 1 21 1 x1 z1 1 1 1 3 1 7
x = y2 z2 1  1 1 1  ; y = x 2 z2 1  2 1 1 
2 y z 1 2 2 4 1 2 2 x z 1 2 1 4 1 2
3 3 3 3

1 x1 y1 1 1 1 2 1 7 10
z = x2 y2 1  2 1 1  0  the required area  = [2x  2y  z2 ] = sq. units.
2 x y 1 2 1 2 1 2
3 3

Ex.7 A plane is passing through a point P(a, –2a, 2a), a  0, at right angle to OP, where O is the origin to meet
the axes in A, B and C. Find the area of the triangle ABC.

Sol. OP = a 2  4a 2  4a 2 = |3a|.

Equation of plane passing through P(a, –2a, 2a) is

A(x – a) + B(y + 2a) + C(z – 2a) = 0.
 the direction cosines of the normal OP to the
plane ABC are proportional to
a – 0, –2a – 0, 2a – 0 i.e. a, –2a, 2a.
 equation of plane ABC is
a(x – a) – 2a(y + 2a) + 2a(z – 2a) = 0
or ax – 2ay + 2az = 9a2 ....(1)
Now projection of area of triangle ABC on ZX, XY and YZ
planes are the triangles AOC, AOB and BOC respectively.
 (Area ABC)2 = (Area AOC)2 + (Area AOB)2 + (Area BOC)2

2 2 2
1  1  1 
=  . AO . OC    . AO . BO    . BO . OC 
2  2  2 

2 2 2
1  9   9  9 9   1 812 a 4  1
= 4  9a. 2 a    9a . 2 a    2 a. 2 a   , 1  1  
       4 4  4

95 4 35 2 243 2
 (Area ABC) = 3 a  Area of ABC = 3 a 
2 a .
4 2 8

35
E. PLANE

(i) General equation of degree one in x, y, z i.e. ax + by + cz + d = 0 represents a plane.

(ii) Equation of a plane passing through (x1, y1, z1) is a(x – x1) + b (y – y1) + c(z – z1) = 0
where a, b, c are the direction ratios of the normal to the plane.

x y z
(iii) Equation of a plane if its intercepts on the co-ordinate axes are x1, y1, z1 is  
x1 y1 z1 = 1.

(iv) Equation of a plane if the length of the perpendicular from the origin on the plane is ‘p’ and d.c’s of the
perpendiculars as  , m, n is x + my + nz = p
(v) Parallel and perpendicular planes :
Two planes a1 x + b1 y + c1z + d1 = 0 and a2 x + b2 y + c2 z + d2 = 0 are

a1 b1 c 1 a1 b1 c1 d1
Perpendicular if a1a2 + b1b2 + c1c2 = 0, parallel if   and Coincident if   
a2 b 2 c 2 a 2 b 2 c 2 d2

(vi) Angle between a plane and a line is the complement of the angle between the normal to the plane and the

   
Line : r  a  b  b.n
line. If   then cos (90 – ) = sin  =   .
Plane : r .n  d  | b|.|n|

where  is the angle between the line and normal to the plane.

ax1  by1  cz1  d

(vii) Length of the ar from a point (x1, y1, z1) to a plane ax + by + cz + d = 0 is p =
a2  b 2  c 2

d1  d2
(viii) Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is
a  b2  c 2
2

(ix) Planes bisecting the angle between two planes a1x + b1y + c1z + d1 = 0 and a2 x + b2y + c2 z + d2 = 0 is

a1x  b1y  c 1z  d1 a 2 x  b 2 y  c 2 z  d2
given by  of these two bisecting planes, one bisects
a12  b12  c 12 a 22  b 22  c 22

the acute and the other obtuse angle between the given planes.

(x) Equation of a plane through the intersection of two planes P1 and P2 is given by P1 + P2 = 0

36
Ex.8 Reduce the equation of the plane x + 2y – 2z – 9 = 0 to the normal form and hence find the length of the
perpendicular drawn form the origin to the given plane.
Sol. The equation of the given plane is x + 2y – 2z – 9 = 0
Bringing the constant term to the R.H.S., the equation becomes x + 2y – 2z = 9 ...(1)
[Note that in the equation (1) the constant term 9 is positive. If it were negative, we would have changed
the sign throughout to make it positive.]
Now the square root of the sum of the squares of the coefficients of x, y, z in (1)

= (1)2  (2) 2  (  2) 2  9 3.

1 2 2
Dividing both sides of (1) by 3, we have x  y  z  3.
2 3 3
....(2)
The equation (2) of the plane is in the normal form x + my + nz = p.

1 2 2
Hence the d.c.’s , m, n of the normal to the plane are , ,  and the length p of the perpendicular
2 3 3
from the origin to the plane is 3.
Ex.9 Find the equation to the plane through the three points (0, –1, –1), (4, 5, 1) and (3, 9, 4).
Sol. The equation of any plane passing through the point (0, –1, –1) is given by
a(x – 0) + b{y – (–1)} + c{z – (–1)} = 0 or ax + b(y + 1) + c (z + 1) = 0 ....(1)
If the plane (1) passes through the point (4, 5, 1), we have 4a + 6b + 2c = 0 ....(2)
If the plane (1) passes through the point (3, 9, 4), we have 3a + 10b + 5c = 0 ....(3)

a b c
Now solving the equations (2) and (3), we have   =  (say).
30  20 6  20 40  18
 a = 10, b = –14, c = 22.
Putting these value of a, b, c in (1), the equation of the required plane is given by
[10x – 14(y + 1) + 22(z + 1)] = 0 or 10x – 14(y + 1) + 22(z + 1) = 0 or 5x – 7y + 11z + 4 = 0.

Ex.10 Find the equation of the plane through (1, 0, –2) and perpendicular to each of the planes
2x + y – z – 2 = 0 and x – y – z – 3 = 0.
Sol. The equation of any plane through the point (1, 0, –2) is
a (x – 1) + b (y – 0) + c(z + 2) = 0. ...(1)
If the plane (1) is perpendicular to the planes 2x + y – z – 2 = 0 and x – y – z – 3 = 0, we have
a (2) + b(1) + c(–1) = 0 i.e., 2a + b – c = 0, ...(2)
and a(1) + b(–1) + c(–1) = 0 i.e., a – b – c = 0. ...(3)

3 1
Adding the equation (2) and (3), we have c = a. Subtracting (3) from (2), we have b = – a.
2 2
Putting the values of b and c in (1), the equation of the required plane is given by

1 3
a (x – 1) – ay + a (z + 2) = 0 or 2x – 2 – y + 3z + 6 = 0 or 2x – y + 3z + 4 = 0.
2 2

37
Ex.11 Find the equation of the plane passing through the line of intersection of the planes
2x – 7y + 4z = 3, 3x – 5y + 4z + 11 = 0, and the point (–2, 1, 3)
Sol. The equation of any plane through the line of intersection of the given plane is
(2x – 7y + 4z – 3) +  (3x – 5y + 4z + 11) = 0. ....(1)
If the plane (1) passes through the point (–2, 1, 3), then substituting the co-ordinates of this point in the
equation (1), we have
{2(–2) – 7(1) + 4(3) – 3} + {3 (–2) – 5(1) + 4(3) + 1} = 0 or (–2) + (12) = 0 or  = 1/6.
Putting this value of  in (1), the equation of the required plane is
(2x – 7y + 4z – 3) + (1/6) (3x – 5y + 4z + 11) = 0 or 15x – 47y + 28z = 7.
Ex.12 A variable plane is at a constant distance 3p from the origin and meets the axes in A, B and C. Prove that
the locus of the centroid of the triangle ABC is x–2 + y–2 + z–2 = p–2.
Sol. Let the equation of the variable plane be x/a + y/b + z/c = 1. .....(1)
It is given that the length of the perpendicular from the origin to the plane (1) is 3p.

1 1 1 1 1
 3p = or    , .....(2)
2
(1/ a  1/ b  1/ c ) 2 2
9p 2 a2 b2 c 2

The plane (1) meets the coordinate axes in the points A, B and C whose co-ordinates are respectively given
by (a, 0, 0), (0, b, 0) and (0, 0, c). Let (x, y, z) be the co-ordinates of the centroid of the triangle ABC. Then
x = (a + 0 + 0)/3, y = (0 + b + 0)/3, z = (0 + 0 + c)/3

1 1 1
i.e., x= a, y = b, z = c.  a = 3x, b = 3y, c = 3z. .....(3)
3 3 3

The locus of the centroid of the triangle ABC is obtained by eliminating a, b, c between the equation (2) and
(3). Putting the value of a, b, c from (3) in (2), the required locus is given by

1 1 1 1
   or x–2 + y–2 + z–2 = p–2.
9p 2 9x 2 9y 2 9z2

Ex.13 Show that the origin lies in the acute angles between the planes x + 2y + 2z – 9 = 0 and
4x – 3y + 12z + 13 = 0. Find the planes bisecting the angles between them and point out the one which
bisects the acute angle.
Sol. In order that the constant terms are positive, the equations of the given planes may be written as
–x – 2y – 2z + 9 = 0 ...(1) and 4x – 3y + 12z + 13 = 0.
We have a1a2 + b1b2 + c1c2 = (–1).4 + (–2). (–3) + (–2).(12) = – 4 + 6 – 24 = – 22 = negative.
Hence the origin lies in the acute angle between the planes (1) and (2)
The equation of the plane bisecting the angle between the given planes (1) and (2) when contains the origin

x  2y  2 z  9 4 x  3 y  12 z  13
is 
(1  4  4 ) (16  9  144 )

or 13 (–x – 2y – 2z + 9) = 3(4x – 3y + 12z + 13) or 25x + 17y + 62z – 78 = 0 ...(3)

38
 We have proved above that origin lies in the acute angle between the planes and so the equation (3) is the
equation of the bisector plane which bisects the acute angle between the given planes.
The equation of the other bisector plane (i.e., the plane bisecting the obtuse angle) is

x  2y  2z  9 4x  3y  12z  13
  or x + 35y – 10z – 156 = 0 ....(4)
(1  4  4) (16  9  144)

the equation (3) and (4) given the planes bisecting the angle between the given planes and the equation (3)
is the bisector of the acute angle.
Ex.14 The mirror image of the point (a, b, c) about coordinate planes xy, xz and yz are A, B and C. Find the
orthocentre of the triangle ABC.
Sol. Let the point P be (a, b, c)  A  (a, b, –c), B  (a, –b, c) and C  (–a, b, c)
Let the orthocentre of ABC be H  (x, y, z)
 (x – a) (2a) + (y – b) (–2b) + (z + c) 0 = 0  ax – by = a2 – b2 ...(1)
Similarly, by – cz = b2 – c2 ...(2)

x  a y b z  c A(a, b, –c)
Also 0 2b 2c = 0 (As A, B, C and H are coplanar)
2a 0 2c

H(x)
 bcx + acy + abz = abc ...(3)

for solving (1), (2) and (3),

B(a,–b, c) C(–a, b, c)

a b 0 a 2  b2 b 0
2 2
D = 0 b c = a b + b c + a c , D1 = b  c
2 2 2 2 2 2 b c = a2 (b2 + c2) – b2c2
bc ac ab abc ac ab

 Similarly D2 = b2(c2 + a2) – a2c2 and D3 = c2(a2 + b2) – a2b2

 a 2 (b 2  c 2 )  b2c 2 b2 (c 2  a2 )  a2c 2 c 2 (a2  b2 )  a2b 2 

 Orthocentre is H   2 2 2 2 2 2
, 2 2 2 2 2 2
, 2 2 2 2

2 2 .
 a b  b c  c a a b  b c  c a a b  b c  c a 

F. STRAIGHT LINE

(i) Equation of a line through A(x1, y1, z1) and having direction cosines  , m , n are

x  x1 y  y1 z  z1
  and the lines through (x1, y1, z1) and (x2, y2, z2)
 m n

x  x1 y  y1 z  z1
 
x 2  x1 y 2  y1 z 2  z1

(ii) Intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 together represent
the unsymmetrical form of the straight line.

39
 x  x1 y  y1 z  z1
(iii) General equation of the plane containing the line   is
 m n

A(x – x1) + B(y – y1) + c(z – z1) = 0 where A  + bm + cn = 0.

(iv) Line of Greatest Slope
AB is the line of intersection of G-plane and H
is the h orizontal plane. Line of greatest slope
on a given plane, drawn through a given point
on the plane, is the line through the point ‘P’
perpendicular to the line of intersection of the
given plane with any horizontal plane.

x  2 y 1 z  2
Ex.15 Show that the distance of the point of intersection of the line   and the plane x
3 4 12
– y + z = 5 from the point (–1, –5, –10) is 13.

x  2 y 1 z  2
Sol. The equation of the given line are   = r (say). ....(1)
3 4 12
The co-ordinates of any point on the line (1) are (3r + 2, 4r - 1, 12 r + 2). If this point lies on the plane x – y
+ z = 5, we have 3r + 2 – (4r – 1) + 12r + 2 = 5, or 11r = 0, or r = 0.
Putting this value of r, the co-ordinates of the point of intersection of the line (1) and the given plane are (2,
–1, 2).
 The required distance = distance between the points (2, –1, 2) and (–1, –5, –10)

= (2  1)2  ( 1  5)2  (2  10)2 = (9  16  144)  (169)  13

Ex.16 Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 =
0. Find also the co-ordinates of the point on the line which is at the same distance from the foot of the
perpendicular as the origin is.
Sol. The equation of the plane is 3x + 4y – 6z + 1 = 0. ....(1)
The direction ratios of the normal to the plane (1) are 3, 4, –6. Hence the line normal to the plane (1) has
d.r.’s 3, 4, –6, so that the equations of the line through (0, 0, 0) and perpendicular to the plane (1) are
x/3 = y/4 = z/–6 = r (say) ....(2)
The co-ordinates of any point P on (2) are (3r, 4r, – 6r) ....(3)
If this point lies on the plane (1), then 3(3r) + r(4r) – 6(–6r) + 1 = 0, or r = –1/61.
Putting the value of r in (3), the co-ordinates of the foot of the perpendicular P are (–3/61, –4/61, 6/61).
Now let Q be the point on the line which is at the same distance from the foot of the perpendicular as the
origin. Let (x1, y1, z1) be the co-ordinates of the point Q. Clearly P is the middle point of OQ. Hence we have

x1  0 3 y  0 4 z1  0 6
 , 1  , 
2 61 2 61 2 61
or x1 = 6/61, y1 = –8/61, z1 = 12/61.
 The co-ordinates of Q are (–6/61, –8/61, 12/61).

40
Ex.17 Find in symmetrical form the equations of the line 3x + 2y – z – 4 = 0 & 4x + y – 2z + 3 = 0
and find its direction cosines.
Sol. The equations of the given line in general form are 3x + 2y – z – 4 = 0 & 4x + y – 2z + 3 = 0 ..(1)

Let , m, n be the d.c.’s of the line. Since the line is common to both the planes, it is perpendicular to the

normals to both the planes. Hence we have 3 + 2m – n = 0, 4 + m – 2n = 0.

 m n  m n (  2  m 2  n2 ) 1
Solving these, we get   or    
4  1 4  6 3  8 3 2 5 (9  4  25) 38)

3 2 5
the d.c.’s of the line are – , ,

(38 ) (38 ) (38 ) .

Now to find the co-ordinates of a point on the line given by (1), let us find the point where it meets the plane
z = 0. Putting z = 0 i the equations given by (1), we have 3x + 2y – 4 = 0, 4x + y + 3 = 0.

x y 1
Solving these, we get   , or x = –2, y = 5.
6  4 16  9 3  8

x2 y5 z0

Therefore the equation of the given line in symmetrical form is   .
3 2 5

Ex.18 Find the equation of the plane through the line 3x – 4y + 5z = 10, 2x + 2y – 3z = 4
and parallel to the line x = 2y = 3z.
Sol. The equation of the given line are 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 ...(1)
The equation of any plane through the line (1) is (3x – 4y + 5z – 10) +  (2x + 2y – 3z – 4) = 0
or (3 + 2)x + (–4 +2) y + (5 – 3) z – 10 – 4 = 0. ...(2)

x y z
The plane (1) will be parallel to the line x = 2y = 3z i.e.   if
6 3 2

4
(3 + 2) . 6 + (–4 + 2). 3 + (5 – 3).2 = 0 or (12 + 6 – 6) + 18 – 12 + 10 = 0 or  = – .
3

Putting this value of  in (2), the required equation of the plane is given by

 8  8 16
 3   x    4   y  (5  4 )z  10  0 or x – 20y + 27z = 14.
 3  3 3

41
 x 1 y  2 z  2
Ex.19 Find the equation of a plane passing through the line   and making an angle of 30°
1 1 2

with the plane x + y + z = 5.

Sol. The equation of the required plane is (x – y + 1) +  (2y + z – 6) = 0  x + (2 – 1) y + z + 1 – 6 = 0
Since it makes an angle of 30° with x +y + z = 5

| 1  ( 2  1)   | 3
   |6| = 3 52  4  2  42 = 52 – 4 + 2
3. 1  2  (2  1)2 2

 2— – 4 + 2 = 0   = (2 ± 2 )  (x – y + 1) + (2 ± 2 ) (2y + x – 6) = 0 are two required planes.

Ex.20 Prove that the lines 3x + 2y + z – 5 = 0 = x + y – 2z – 3 and 2x – y – z = 0 = 7x + 10y – 8z – 15 are
perpendicular.
Sol. Let 1, m1, n1 be the d.c.’s of the first line. Then 31 + 2m1 + n1 = 0, 1 + m1 – 2n1 = 0. Solving, we get

1 m n  m n
 1  1 or 1  1  1 .
4  1 1  6 3  2 5 7 1

Again let 2, m2,n2 be the d.c.’s of the second line, then 22 – m2 – n2 = 0, 72 + 10m2 – 8n2 = 0.

2 m2 n2  m n
Solving,   or 2  2  2 .
8  10 7  16 20  7 2 1 3

Hence the d.c.’s of the two given lines are proportional to –5, 7, 1 and 2, 1, 3. We have
–5.2 + 7.1 + 1.3 = 0  the given lines are perpendicular.
Ex.21 Find the equation of the plane which contains the two parallel lines

x 1 y  2 z x  3 y  4 z 1
  and   .
3 2 1 3 2 1

Sol. The equation of the two parallel lines are

(x + 1)/3 = (y – 2)/2 = (z – 0)/1 ....(1) and (x – 3)/3 = (y + 4)/2 = (z – 1)/1. ....(2)
The equation of any plane through the line (1) is
a(x + 1) + b (y – 2) + cz = 0, ....(3) where 3a + 2b + c = 0. ....(4)
The line (2) will also lie on the plane (3) if the point (3, –4, 1) lying on the line (2) also lies on the plane (3),
and for this we have a (3 + 1) + b (–4 – 2) + c. 1 = 0 or 4a – 6b + c = 0. ....(5)

a b c
Solving (4) and (5), we get   .
8 1  26

Putting these proportionate values of a, b, c in (3), the required equation of the plane is
8(x + 1) + 1.(y – 2) – 26z = 0, or 8x + y – 26 + 6 = 0.

42
 x 1 y  3 z  2
Ex.22 Find 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.

Sol. The equation of the given line are (x – 1)/2 = (y – 3)/4 = (z – 2)/3 = r, (say). ...(1)

Any point Q on the line (1) is (2r + 1, 4r + 3, 3r + 2).

Now P is the point (3, 8, 2) and hence d.r.’s of PQ are

2r + 1 – 3, 4r + 3 – 8, 3r + 2 – 2 i.e. 2r – 2, 4r – 5, 3r.

It is required to find the distance PQ measured parallel to the plane 3x + 2y – 2z + 17 = 0 ...(2)

Now PQ is parallel to the plane (2) and hence PQ will be perpendicular to the normal to the plane (2). Hence

we have (2r – 2) (3) + (4r – 5) (2) + (2r) (–2) = 0 or 8r – 16 = 0, or r = 2.

Putting the value of r, the point Q is (5, 11, 8) = [(3  5 )2  (8  11)2  ( 2  8 )2 ]  ( 4  9  36 )  7 .

Ex.23 Find the projection of the line 3x – y + 2z = 1, x + 2y – z = 2 on the plane 3x + 2y + z = 0.

Sol. The equations of the given line are 3x – y + 2z = 1, x + 2y – z = 2. ....(1)

The equation of the given plane is 3x + 2y + z = 0. ....(2)

The equation of any plane through the line (1) is (3x – y + 2z – 1) + (x + 2y – z – 2) = 0

or (3 + ) x + (–1 + 2) y + (2 – ) z – 1 – 2 = 0 ....(3)

3
The plane (3) will be perpendicular to the plane (2), if 3(3 + )+ 2(–1 + 2) + 1 (2 – ) = 0 or  = – .
3

Putting this value of in (3), the equation of the plane through the line (1) and perpendicular to the plane (2)

 3  3
is given by  3   x + (–1 –3) y +  2   z – 1 + 3 = 0 or 3x – 8y + 7z + 4 = 0. ....(4)
 2  2

 The projection of the given line (1) on the given plane (2), is given by the equations (2) and (4) together.
4 2
x
y
Note : The symmetrical form of the projection given above by equations (2) and (4) is 5  5 z.
 11 9 5

43
 x 1 y 1 z  3
Ex.24 Find the image of the line   in the plane x + 2y + z = 12
2 1 4
Sol. Any point on the given line is 2r + 1, –r – 1, 4r + 3. If this point lies on the planes,

5
then 2r + 1 – 2r – 2 + 4r + 3 = 12  r = .
2

 7 
Hence the point of intersection of the given line and that of the plane is  6, ,13  .
 2 
Also a point on the line is (1, –1, 3).

 1  1   3
Let () be its image in the given plane. In such a case   
1 2 1
  =  + 1,  = 2 – 1,  =  + 3. Now the midpoint of the image and the point (1, –1, 3) lies on the plane

   10  8 7 14 
i.e. 1  ,   1, 3   lies in the plane = . Hence the image of (1, –1, 3) is  , , .
 2 2 3 3 3 3 

7 7
y y
x6 2  z  13 x  6 2 z  13

Hence the equation of the required line is
10  35 25 or 4   7  10 .
3 6 3
Ex.25 Find the foot and hence the length of the perpendicular from the point (5, 7, 3) to the line
(x – 15)/3 = (y – 29)/8 = (z – 5)/(–5). Find the equations of the perpendicular. Also find the equation of the
plane in which the perpendicular and the given straight line lie.
Sol. Let the given point (5, 7, 3) be P.
The equations of the given line are (x – 15)/3 = (y – 29)/8 = (z – 5)/(–5)= r (say). ...(1)
Let N be the foot of the perpendicular from the point P to the line (1). The co-ordinates of N may be taken
as (3r + 15, 8r + 29, – 5r + 5). ...(2)
 the direction ratios of the perpendicular PN are
3r + 15 – 5, 8r + 29 – 7, – 5r + 5 – 3, i.e. are 3r + 10, 8r + 22, –5r + 2. ...(3)
Since the line (1) and the line PN are perpendicular to each other, therefore
3 (3r + 10) + 8 (8r + 22) – 5(–5r + 2) = 0 or 98r + 196 = 0 or r = –2
Putting this value of r in (2) and (3), the foot of the perpendicular N is (9, 13, 15) and the direction ratios of
the perpendicular PN are 4, 6, 12 or 2, 3, 6.
 the equations of the perpendicular PN are (x – 5)/2 = (y – 7)/3 = (z – 3)/6. ...(4)
Length of the perpendicular PN

= the distance between P(5, 7, 3) and N(9, 13, 15) = (9  5) 2  (13  7 ) 2  (15  3 ) 2 = 14.

Lastly the equation of the plane containing the given line (1) and the perpendicular (4) is given by

x  15 y  29 z  5
3 8 5 =0
2 3 6

or (x – 15) (48 + 15) – (y – 29) (18 + 10) + (z – 5) (9 – 16) = 0 or 9x – 4y – z =– 14 = 0.

44
Ex.26 Show that the planes 2x – 3y – 7z = 0, 3x – 14y – 13z = 0, 8x – 31y – 33z = 0 pass through the one line
find its equations.

2 3 7 0
Sol. The rectangular array of coefficient is 3 14 13 0 .
8 31 33 0

2 3 7 2 1 1
We have, 4 = 3 14 13  3 11 4 (by C2 + C1, C2 + 3C1)
8 31 33 8 23 9

0 0 1
= 5 7 4 = –1(70 – 70) = 0, (by C1 + 2C2, C2 – C2)
10 14 9

since 4 = 0, therefore, the three planes either intersect in a line or form a triangular prism.
2 3 0
Now 3 = 3 14 0 = 0 Similarly 2 = 0 and 1 = 0,
8 31 0
Hence the three planes intersect in a common line.
Clearly the three planes pass through (0, 0, 0) and hence the common line of intersection will pass through
(0, 0, 0). The equations of the common line are given by any of the two given planes. Therefore the
equations of the common line are given by 2x – 3y – 7z = 0 and 3x – 14y – 13z = 0.

x y z x y z
 the symmetric form of the line is given by   or   .
39  98  21  26  28  9  59 5  19
Ex.27 For what values of k do the planes x – y + z + 1 = 0, kx + 3y + 2z – 3 = 0, 3x + ky + z – 2 = 0
(i) intersect in a point ; (ii) intersect in a line ; (iii) form a triangular prism ?

1 1 1 1
Sol. The rectangular array of coefficients is k 3 2 3
3 k 1 2

Now we calculate the following determinants

1 1 1 0 1 0
4 = k 3 2  k  3 3 5 (adding 2nd column to 1st and
3 k 1 3  k k k 1

3rd)

0 1 0
= (k + 3) 1 3 5 = (k + 3) (k + 1 – 5) = (k + 3) (k – 4).
1 k k 1

1 1 1 0 1 0
2 = k 3 3  k  3 3 0 = (k + 3) (k – 2), (adding 2nd column to 1st and 3rd)
3 k 2 3k k k2

1 1 1 0 1 0
2 = k 2 3  k  2 2 5 (adding (–1) times 2nd column to 1st and 3rd)
3 1 2 2 1 3

= –{(k – 2) (–3) + 10} = 3k – 16,

45
 1 1 1 0 1 1
and 1 = 3 2 3  0 2 3 = –5 (k – 2) (adding 3rd column to 1st)
k 1 2 k  2 1 2

(i) The given planes will intersect in a point if 4  0 and so we must have k  –3 and k  4. Thus the
given planes will intersect in a point for all real values of k other than –3 and 4.
(ii) If k = –3, we have 4 = 0, 3 = 0 but 2  0. Hence the given planes will form a triangular prism if
k = –3.
(iii) If k = 4, we have 4 = 0 but 3  0. Hence the given planes will form a triangular prism if k = 4.
We observe that for no value of k the given planes will have a common line of intersection.

Ex.28 Find the equation of the line passing through (1, 1, 1) and perpendicular to the line of intersection of the
planes x + 2y – 4z = 0 and 2x – y + 2z = 0.
Sol. Equation of the plane through the lines x + 2y – 4z = 0 and 2x – y + 2z = 0 is
x + 2y – 4z +  (2x – y + 2z) = 0 ...(1)
If (1, 1, 1) lies on this plane, then –1 + 3 = 0

1
= , so that the plane becomes 3x + 6y – 12z + 2x – y + 2z = 0  x + y – 2z = 0
3

....(2)
11
Also (1) will be perpendicular to (2) if 1 + 2 + 2 –  – 2(–4 + 2) =   = .
3
 Equation of plane perpendicular to (2) is 5x – y + 2z = 0. ...(3)
Therefore the equation of line through (1, 1, 1) and perpendicular to the given line is parallel to the normal

x 1 y 1 z 1
to the plane (3). Hence the required line is  
5 1 2

Alternate :

x y z
Solving the equation of planes x + 2y – 4z = 0 and 2x – y + 2z = 0, we get   ...(1)
0  10  5

Any point P on the line (1) can be written as (0, –10, –5).
Direction ratios of the line joining P and Q(1, 1, 1) is (1, 1, + 10, 1 + 5).
Line PQ is perpendicular to line (1)  0(1) – 10(1 + 10) – 5(1 + 5) = 0

 15  3  6 3
0 – 10 – 100 – 5 – 25x = 0 or 125 + 15 = 0  =   P =  0, , 
125 25  5 5

 1 2 x 1 y 1 z 1
Direction ratios of PQ =   1, ,  . Hence equations of lien are   .
 5 5  5 1 2

46
 x3 y8 z3 x3 y7 z6
Ex.29 Find the shortest distance (S.D.) between the lines   ,   .
3 1 1 3 2 4

Find also its equations and the points in which it meets the given lines.
Sol. The equations of the given lines are (x – 3)/3 = (y – 8)/–1 = (z – 3)/1 = r1 (say) ...(1)
and (x + 3)/(–3) = (y + 7)/2 = (z – 6)/4 = r2 (say) ...(2)
Any point on line (1) is (3r1 + 3, –r1 + 8, r1 + 3), say P. ...(3)
any point on line (2) is (–3r2 – 3, 2r2 – 7, 4r2 + 6), say Q. ...(4)
The d.r.’s of the line PQ are (–3r2 – 3) – (3r1 + 3), (2r2 – 7) – (–r1 + 8), (4r2 + 6) – (r1 + 3)
or –3r2 – 3r1– 6, 2r2 + r1 – 15, 4r2 – r1 + 3. ...(5)
Let the line PQ be the lines of S.D., so that PQ is perpendicular to both the given lines (1) and (2), and so
we have 3 (–3r2 – 3r1 – 6) – 1 (2r2 + r1 – 15) + 1. (4r2 – r1 + 3) = 0
and –3(–3r2 – 3r1 – 6) + 2. (2R2 + r1 – 15) + 4 (4r2 – r1 + 3) = 0
or –7r2 – 11r1 = 0 and 11r2 + 7r1 = 0. Solving these equations, we get r1 = r2 = 0.
Substituting the values of r1 and r2 in (3), (4) and (5), we have P(3, 8, 3), Q(–3, –7, 6)
And the d.r.’s of PQ (the line of S.D.) are –6, –15, 3 or –2, –5, 1.

The length of S.D. = the distance between the points P and Q = (3  3)2  ( 7  8)2  (6  3)2  3 30 .

Now the line PQ of shortest distance is the line passing through P(3, 8, 3) and having d.r.’s –2, –5, 1 and

x3 y8 z3 x3 y8 z3

hence its equations are given by   or   .
2 5 1 2 5 1

47
 [VECTORS]
    
Q.1 If a&b are non col linear v ectors such that , p  ( x  4 y )a  ( 2 x  y  1) b &
    
q  ( y  2 x  2) a  ( 2 x  3y  1) b , find x & y such that 3p  2q .

       
Q.2 (a) Show that the points a  2 b  3 c ; 2 a  3 b  4 c &  7 b  10 c are collinear .
(b) Prove that the points A = (1,2,3), B (3,4,7), C (3,2,5) are collinear & find the ratio in which B
divides AC.

Q.3 Find out whether the following pairs of lines are parallel, non-parallel & intersecting, or non-parallel &
non-intersecting.
  
r1  i  j  2 k   3 i  2 j  4 k
 
(i) 
r2  2 i  j  3 k    6 i  4 j  8 k
 
  
r1  i  j  3 k   i  j  k
 
(ii) 
r2  2 i  4 j  6 k   2 i  j  3 k
 
  
r1  i  k   i  3 j  4 k
 
(iii) 
r2  2 i  3 j   4 i  j  k
 
Q.4 Let OACB be paralelogram with O at the origin & OC a diagonal. Let D be the mid point of OA. Using
vector method prove that BD & CO intersect in the same ratio. Determine this ratio.

AE AF
Q.5 In a ABC, points E and F divide sides AC and AB respectively so that = 4 and = 1.
EC FB
Suppose D is a point on side BC. Let G be the intersection of EF and AD and suppose D is situated so

AG 3 BD a
that = . If the ratio = , where a and b are in their lowest form, find the value of (a + b).
GD 2 DC b

 
Q.6 Let u be a vector on rectangular coordinate system with sloping angle 60°. Suppose that u  î is
  
geometric mean of u and u  2î where î is the unit vector along x-axis then find the value of u .

Q.7 (a) Find a unit vector â which makes an angle (/4) with axis of z & is such that â  î  ˆj is a unit vector..

  2   2
 a b   ab 
(b) Prove that          
 a 2 b2   | a | | b | 
   
 
  3 ab  
(c) If a and b are any two unit vectors, then find the range range of + 2 ab .
2

48
  
Q.8 Let V1 and V2 are two vectors such that
 
V1 = 2(sin  + cos ) î  ĵ and V2 = sin  î + cos  ˆj where  and  satisfy the relation
2(sin  + cos )sin  = 3 – cos , find the value of (3 tan2 + 4 tan2).

Q.9 Given three points on the xy plane on O(0, 0), A(1, 0) and B(–1, 0). Point P is moving on the plane

  
satisfying the condition P A ·P B + 3 O A ·O B = 0 
If the maximum and minimum values of P A P B are M and m respectively then find the value of
M2 + m2.

Q.10 (a) Find the minimum area of the triangle whose vertices are A(–1, 1, 2); B(1, 2, 3) and C(t, 1, 1)
where t is a real number.
   
(b) Let OA = a ; OB = 100 a  2 b and OC = b where O, A, and C are non collinear points.
s. Let

P denotes the area of the parallelogram with OA and OC as adjacent sides and Q denotes the
area of the quadrilateral OABC. If Q = P. Find the value of .

Q.11 (a) Determine vector of magnitude 9 which is perpendicular to both the vectors:

4î  ĵ  3k̂ &  2î  ˆj  2k̂

(b) A triangle has vertices (1, 1, 1) ; (2, 2, 2), (1, 1, y) and has the area equal to csc  4 sq. units.
Find the value of y.

  
Q.12 Let a  3 i  2 j  4 k ; b  2 i  k

  and c  4 i  2 j  3 k .

  
If the equation x a  y b  z c =  x i  y j  z k has a non trivial solution, then find the sum of all
 
distinct possible values of 

Q.13 The position vectors of the points A, B, C are respectively (1, 1, 1) ; (1, 1, 2); (0, 2, 1). Find a unit vector
parallel to the plane determined by ABC & perpendicular to the vector (1, 0, 1) .


Q.14 The vector OP = î  2ˆj  2k̂ turns through a right angle, passing through the positive x-axis on the way..
Find the vector in its new position.

         
Q.15 If a and b are two vectors such that | a | 1 , | b | 4 , a ·b  2 . If c  ( 2a  b)  3b then find the angle
 
between b and c .

Q.16 The pv's of the four angular points of a tetrahedron are: A j  2 k ; B 3 i  k ; C 4 i  3 j  6 k

     
 
& D 2 i  3 j  2 k . Find :
(i) the perpendicular distance from A to the line BC.
(ii) the volume of the tetrahedron ABCD.
(iii) the perpendicular distance from D to the plane ABC.
(iv) the shortest distance between the lines AB & CD.

49
   
Q.17 If a  a1î  a 2 ĵ  a 3k̂ ; b  b1î  b 2 ĵ  b3k̂ and c  c1î  c 2ˆj  c 3k̂ then show that the value of the
  
a ·î a · ĵ a ·k̂
     
scalar triple product [ na  b nb  c nc  a ] is (n3 + 1) b ·î b ·ˆj b ·k̂
  
c ·î c ·ˆj c ·k̂

Q.18

          
Let r  a  b sin x  b  c cos y  2  c  a  , where a , b, c are non-zero and non-coplanar vectors.

    20
If r is orthogonal to a  b  c , then find the minimum value of 2 (x2 + y2).


            
Q.19 If A , B & C are vectors such that | B |  | C | , Prove that: A  B x A  C x Bx C.B  C  0 .
             
Q.20 Find the scalars  &  if a x (b x c)  (a . b) b  ( 4  2  sin ) b  ( 2  1) c & ( c . c) a  c while b & c
are non zero non collinear vectors.

  
Q.21 Let a   i  2 j  3 k , b  i  2  j  2 k and c  2 i   j  k . Find the value(s) of , if any, such that
 
a  b   b  c  c  a  = 0. Find the vector product when  = 0.
   
Q.22 Suppose V1  î  ˆj  2k̂ , V2  î  2 ĵ  k̂ and V3   2î  2 ĵ  k̂ are three vectors. Let V be
   
a vector such that it can be expressed as a linear combination of V1 and V2 . Also V ·V3 = 0
 
and the projection of the vector V on î  ĵ  k̂ is 6 3 . Find the vector V .

 
Q.23 Let A( t ) = f1 (t )i  f2 (t )j and B( t )  g1 ( t ) i  g 2 ( t ) j , t  [0, 1], where f1, f2, g1, g2 are continuous
  
functions. If A( t ) and B( t ) are nonzero vectors for all t and A(0) = 2 i  3j ,
    
A(1) = 6 i  2j , B(0) = 3i  2j and B(1) = 2i  6 j , then show that A(t ) and B(t ) are parallel for
some t.

  
Q.24 Let A  î  2ˆj  3k̂ , B  2î  ˆj  k̂ , C  ˆj  k̂
      
If the v ector B  C can be expressed as a linear combination B  C = x A  y B  z C
where x, y, z are scalars, then find the value of (100x + 10y + 8z).

  1    2   1      2
Q.25 Let a   0  b  1  c   1
; ; . Find the numbers , ,  such that  a   b   c    5 .
  3  0  1   6 

50
   
Q.26 Let a 3 dimensional vector V satisfies the condition, 2V  V  ( î  2 ĵ) = 2î  k̂ .

If 3 V = m where m  N, then find m.

 
Q.27 If x , y are two non-zero and non-collinear vectors satisfying
   
[(a – 2)2 + (b – 3) + c] x + [(a – 2)2 + (b – 3) + c] y + [(a – 2)2 + (b – 3) + c] (x  y) = 0
where , ,  are three distinct real numbers, then find the value of (a2 + b2 + c2).

    
Q.28 Let a , b, c be three vectors of magnitude 1,
1
, 2 respectively, satisfying a b c  = 1.
2

   13
     
If 2a  b  c · a  c   a  c   b =  k
, then find the value of k.

  
Q.29 Let a  î  ˆj  2k̂ ; b  3î  4 ĵ  k̂ , c  î  ĵ  k̂ ,
 
   
then find the value of a  2b ·   c  2a   b  2c   
.


51
 [THREE DIMENSION GEOMETRY]

Q.1 Find the angle between the two straight lines whose direction cosines l, m, n are given by
2l + 2m – n = 0 and mn + nl + lm = 0.

Q.2 The plane denoted by 1 : 4 x  7 y  4 z  81  0 is rotated through a right angle about its line of

intersection with the plane  2 : 5 x  3 y  10 z  25 . If the plane in its new position be denoted by , and

the distance of this plane from the origin is k where k  N, then find k.

Q.3 Find the equations of the straight line passing through the point (1, 2, 3) to intersect the straight line
x + 1 = 2 (y – 2) = z + 4 and parallel to the plane x + 5y + 4z = 0.

x 3 y 3 z
Q.4 Find the equations of the two lines through the origin which intersect the line   at an
2 1 1

angle of .
3

Q.5 Let  : x + y – z – 4 = 0 be the equation of a plane and A be the point with position vector î  2 ĵ  3k̂ .
L is a line which passes through the point (1, 2, 3) with direction ratios 3, – 1 and 4.
If the distance of the point A from the line L measured parallel to the plane  is d1 and the distance
of the point A from the plane  measured parallel to the line L is d2 , then find the value of

d12  d 2 2 .

Q.6 Find the equation to the line passing through the point (1, –2, –3) and parallel to the line
2x + 3y – 3z + 2 = 0 = 3x – 4y + 2z – 4.

Q.7 Let P = (1, 0, – 1) ; Q = (1, 1, 1) and R = (2, 1, 3) are three points.

(a) Find the area of the triangle having P, Q and R as its vertices.
(b) Give the equation of the plane through P, Q and R in the form ax + by + cz = 1.
(c) Where does the plane in part (b) intersect the y-axis.
(d) Give parametric equations for the line through R that is perpendicular to the plane in part (b).

Q.8 Find the point where the line of intersection of the planes x – 2y + z = l and x + 2y – 2z = 5, intersects
the plane 2x + 2y + z + 6 = 0.

Q.9 Feet of the perpendicular drawn from the point P (2, 3, –5) on the axes of coordinates are A, B and C. Find
the equation of the plane passing through their feet and the area of ABC.

Q.10 Find the equations to the line which can be drawn from the point (2, –1, 3) and perpendicular to the lines

x 1 y  2 z  3 x4 y z3
  and   .
2 3 4 4 5 3

52
 x 1 y  2 z
Q.11 Find the equation of the plane containing the straight line   and perpendicular to the
2 3 5
plane x – y + z + 2 = 0.

x 1 y  p z2 x y7 z7

Q.12 Find the value of p so that the lines   and   are in the same
3 2 1 1 3 2
plane. For this value of p, find the coordinates of their point of intersection and the equation of the plane
containing them.

2   1
Q.13 
Let L be the line given by r =   2 +   0  and let P be the point (2, – 1, 1). Also suppose that E be
  1   1 
the plane containing three non collinear points A = (0, 1, 1); B(1, 2, 2) and C = (1, 0, 1)
Find
(a) Distance between the point P and the line L.
(b) Equation of the plane E.
(c) Equation the plane F containing the line L and the point P.
(d) Acute between the plane E and F.
(e) Volume of the parallelopiped by A, B, C and the point D(– 3, 0, 1).

x 1 y  2 z  3
Q.14 Find the equation of the line which is reflection of the line   in the plane
9 1 3
3x – 3y + 10z = 26.

x 1 y z x 3 y z  2
Q.15 Find the equation of the plane containing the line   and parallel to the line   .
2 3 2 2 5 4
Find also the S.D. between the two lines.

x 1 y z  1 x 1 y z 1
Q.16 If the distance between the lines = = and = = when they are nearest to
1 2 1 1 1 1

m
each other is where m and n are relatively prime positive integers then find the value of (n – 5m).
n

Q.17 Let P be the plane containing the line y + z = 2, x = 0 and parallel to the line x – z = 2, y = 0.
If the distance of the plane P from origin is d, then find the value of 3d2.

x 1 y  3 z  4
Q.18 Let image of the line   in the plane 2x – y + z + 3 = 0 be L.
3 5 2
A plane 7x + py + qz + r = 0 (p, q, r  R) is such that it contains the line L and perpendicular to the plane
2x – y + z + 3 = 0. Find the value of (p + 3q + r).

53
 VECTOR

SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER)

     

1. If ABCDEF is a regular hexagon and if AB  AC  AD  AE  AF   AD , then  is -
(A) 0 (B) 1 (C) 2 (D) 3

   
2. If a  b is along the angle bisector of a & b then -

   
(A) a & b are perpendicular (B) a  b

   
(C) angle between a & b is 60° (D) a  b


3. Given the points A (2, 3, 4) , B (3, 2, 5) , C (1, 1, 2) & D (3, 2, 4) . The projection of the vector AB on

the vector CD is -

22 21 47
(A) (B)  (C)  (D) –47
3 4 7

 
4. ˆ are the adjacent sides of a parallelogram ABCD then the
The vectors AB  3 ˆi  2 ˆj  2kˆ and BC   î  2k
angle between the diagonals is -

1
 1  1
 49   1   3 
(A) cos   (B)  – cos   (C) cos 1   (D) cos 1  
 85   85  2 2   10 
 

5. ˆ ˆi – 3ˆj – 5kˆ and aˆi  3 ˆj  kˆ

The values of a, for which the points A, B, C with position vectors 2ˆi – ˆj + k,


respectively are the vertices of a right angled triangle with C  are -
2
(A) –2 and 1 (B) 2 and –1 (C) 2 and 1 (D) –2 and –1

     
a.a a.b a.c
     
6.

If a  ˆi  ˆj  k, ˆ c  ˆi  2 ˆj  kˆ , then the value of b. a
ˆ b  ˆi  ˆj  k, b. b b.c 
     
c.a c.b c.c

(A) 2 (B) 4 (C) 16 (D) 64

7. The area of the triangle whose vertices are A (1, –1, 2) ; B (2, 1, –1) ; C (3, –1, 2) is -
(A) 13 (B) 2 13 (C) 13 (D) none

         
8. Let a  ˆi  ˆj & b  2 ˆi  kˆ . The point of intersection of the lines r x a  b x a & r x b  a x b is -

(A)  ˆi  ˆj  kˆ (B) 3 ˆi  ˆj  kˆ (C) 3 ˆi  ˆj  kˆ (D) ˆi  ˆj  kˆ

54
       2        
9. If a, b, c are non-coplanar vectors and  is a real number then  (a  b)  b  c    a b  c b  for -
   
(A) exactly two values of  (B) exactly three values of 
(C) no value of  (D) exactly one value of 

10. Volume of the tetrahedron whose vertices are represented by the position vectors , A (0, 1, 2) ; B (3, 0, 1) ;
C (4, 3, 6) & D (2, 3, 2) is -
(A) 3 (B) 6 (C) 36 (D) none

11. The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where
A  (3, –2, 1) ; B  (3, 1, 5); C  (4, 0, 3) and D  (1, 0, 0) is -

2 5 3 3 2
(A) (B) (C) (D)
29 29 29 29

12. 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) none

        1   
13. Let a , b and c be non-zero vectors such that a and b are non-collinear & satisfies (a  b)  c  | b| | c| a .
3
 
If is the angle between the vectors b and c then sin equals -

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

   ˆ
14. The value of ˆi  (r  ˆi)  ˆj  (r  ˆj)  kˆ  (r  k) is -
   
(A) r (B) 2 r (C) 3 r (D) 4 r

     
15. A, B, C, D be four points in a space and if, | AB  CD  BC  AD  CA  BD| =  (area of triangle
ABC)then the value of  is -
(A) 4 (B) 2 (C) 1 (D) none of these

16. If the v olume of the parallelopiped whose conterminous edges are represented by
ˆ 3ˆj – k,
–12ˆi  k, ˆ 2ˆi  ˆj – 15kˆ is 546, then equals-
(A) 3 (B) 2 (C) –3 (D) –2
 
17. Let a = 2 î + 3 ˆj – k̂ and b = î – 2 ˆj + 3 k̂ . Then the value of  for which the vector
  
c =  î + ˆj + (2 – 1) k̂ is parallel to the plane containing a and b , is-
(A) 1 (B) 0 (C) –1 (D) 2
     
18. If a + 5 b = c and a – 7 b = 2 c , then-
   
(A) a and c are like but b and c are unlike vectors
   
(B) a and b are unlike vectors and so also a and c
   
(C) b and c are like but a and b are unlike vectors
   
(D) a and c are unlike vectors and so also b and c
    
   
19. If a , b , c are three non-coplanar and p , q , r are reciprocal vectors to a , b and c respectively,, then
  
  
( a + m b + n c ).( p + m q + n r ) is equal to : (where , m, n are scalars)
(A) 2 + m2 + n2 (B) m + mn + n (C) 0 (D) none of these

55
  
20. If x & y are two non collinear vectors and a, b, c represent the sides of a ABC satisfying
   
(a  b)x  (b  c)y  (c  a)(x  y)  0 then ABC is -
(A) an acute angle triangle (B) an obtuse angle triangle
(C) a right angle triangle (D) a scalene triangle

         
21. If A , B and C are three non-coplanar vectors then ( A + B + C ).[( A + B ) × ( A + C )] equals -
        
(A) 0 (B) [ A B C ] (C) 2[ A B C ] (D) –[ A B C ]

SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS)

22. ABCD is a parallelogram. E and F be the middle points of the sides AB and BC, then -
(A) DE trisect AC (B) DF trisect AC
(C) DE divide AC in ratio 2 : 3 (D) DF divide AC in ratio 3 : 2

      
23. a , b , c are mutually perpendicular vectors of equal magnitude then angle between a  b  c and a is -

1 FG IJ F1I F1I
(A) cos 1 3 H K (B) cos 1 GH 3 JK (C)  – cos 1 GH 3 JK (D) tan1 2

              
24. If (a  b)  c  a  (b  c) , where a, b and c are any three vectors such that a . b  0, b. c  0 then a and c
are -
(A) perpendicular (B) parallel
(C) non collinear (D) linearly dependent


     b  c
25.

e
If a , b & c are non coplanar unit vectors such that a  b  c =
2
j
, then the angle between -

  3      3   
(A) a & b is (B) a & b is (C) a & c is (D) a & c is
4 4 4 4

     
26. If a, b, c, d, e, f are position v ectors of 6 points A, B, C, D, E & F respectiv ely such that
      
3 a  4 b  6 c  d  4 e  3 f  x , then -
 
(A) AB is parallel to CD
(B) line AB, CD and EF are concurrent

x
(C) is position vector of the point dividing CD in ratio 1 : 6
7
(D) A, B, C, D, E & F are coplanar

27. Read the following statement carefully and identify the true statement -
(a) Two lines parallel to a third line are parallel.
(b) Two lines perpendicular to a third line are parallel.
(c) Two lines parallel to a plane are parallel.
(d) Two lines perpendicular to a plane are parallel.
(e) Two lines either intersect or are parallel.
(A) a & b (B) a & d (C) d & e (D) a

56
 1 ˆ ˆ is -
28. The vector (2 i  2 ˆj  k)
3

(A) unit vector (B) makes an angle /3 with vector 2 ˆi  4 ˆj  3kˆ

(C) parallel to the vector ˆi  ˆj  (1 / 2)kˆ (D) perpendicular to the vector 3 ˆi  2 ˆj  2kˆ

 
29. If a vector r of magnitude 3 6 is collinear with the bisector of the angle between the vectors a  7 i  4 j  4 k
 
& b   2 i  j  2 k , then r =

13 ˆi  ˆj  10kˆ
(A) i  7 j  2 k (B) i  7 j  2 k (C) (D) i  7 j  2 k
5

30. A parallelopiped is formed by planes drawn through the points (1, 2, 3) and (9, 8, 5) parallel to the coordinate
planes then which of the following is the length of an edge of this rectangular parallelopiped -
(A) 2 (B) 4 (C) 6 (D) 8

31. If A (a ) ; B (b) ; C (c) and D (d) are four points such that a = –2 î + 4 ˆj + 3 k̂ ; b = 2 î – 8 ˆj ; c = î – 3 ˆj + 5 k̂ ;

d = 4 î + ˆj – 7 k̂ , d is the shortest distance between the lines AB and CD, then

  
[AB CD BD]
(A) d = 0, hence AB and CD intersect (B) d =  
AB  CD
  
23 [AB CD AC]
(C) AB and CD are skew lines and d = (D) d =  
13 AB  CD

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10
Ans. D B C D C C A C C B
Que. 11 12 13 14 15 16 17 18 19 20
Ans. B C D B A C B A A A
Que. 21 22 23 24 25 26 27 28 29 30
Ans. D A,B B,D B,D A,D B,C B,D A,C,D A,C A,C,D
Que. 31
Ans. B,C,D
57
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS)

1.
 ˆ b  ˆi  2 ˆj  kˆ and c  ˆi  ˆj  2kˆ be three vectors. A vector in the plane of b and c whose
Let a  2 ˆi  ˆj  k,

projection on a is magnitude 2 / 3 is -

(A) 2 ˆi  3 ˆj  3kˆ (B) 2 ˆi  3 ˆj  3kˆ (C) 2 ˆi  5 ˆj  kˆ (D) 2 ˆi  ˆj  5 kˆ

              
2. 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  1 r1  2 r2   3 r3 , then -
(A) 1  7 (B) 1   3  3 (C) 1  2   3  4 (D)  3  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 CB
3
   
such that BN  CB then, for the point of intersection X of AB & MN which of the
following holds good ?
   1     
(A) XB  1 AB (B) AX  AB (C) XN  3 MN (D) XM  3 XN
3 2 4

4. Vector A has components A1, A2, A3 along the three axes. If the co-ordinates system is rotated by 90°
about z-axis, then the new components along the axes are -
(A) A1,  A2, A3 (B)  A1,  A2, A3 (C) A2,  A1, A3 (D)  A2,  A1, A3

  
5. Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the
            
equation p  ((x  q)  p)  q  ((x  r )  q)  r  ((x  p)  r )  0 . Then x is given by -

1    1    1    1   
(A) (p  q  2 r) (B) (p  q  r ) (C) (p  q  r ) (D) (2p  q  r )
2 3 2 3
1 ˆ ˆ 1 (4 ˆi  3k),
ˆ ˆj is -
6. A vector which makes equal angles with the vectors (i  2 ˆj  2k),
3 5

(A) 5 ˆi  ˆj  5kˆ (B) 5 ˆi  ˆj  5kˆ (C) 5 ˆi  ˆj  5 kˆ (D) 5 ˆi  ˆj  5kˆ

     
7.     
The triple product d  a · a  b  c  d  simplifies to -
 
              
(A) (b. d)[d a c] (B) (b.c)[a b d] (C) (b.a )[a b d] (D) none
  
8. If the vectors a, b, c are non-coplanar and  ,m,n are distinct real numbers, then
  
(a  mb  nc)
    
(b  mc  na) (c  ma  nb) = 0 implies -
 
(A) m + mn+ n= 0 (B) + m + n= 0 (C) 2 + m2+ n2= 0 (D) 3 + m3 + n3= 0
    
9. If unit vectors ˆi & ˆj are at right angles to each other and p  3 ˆi  4 ˆj , q  5 ˆi , 4 r  p  q and
  
2 s  p  q , then -
     
(A) r  k s = r  k s for all real k (B) r is perpendicular to s
       
(C) r  s is perpendicular to r  s (D) r  s  p  q

58
10. ˆ kˆ  ˆi taken two at a time form three planes, The three unit vectors drawn
The three vectors ˆi  ˆj, ˆj  k,
perpendicular to these planes form a parallelopiped of volume :
1 3 3 4
(A) (B) 4 (C) (D)
3 4 3 3
11. If a, b, c are different real numbers and a ˆi  b ˆj  c kˆ ; b ˆi  c ˆj  a kˆ & c ˆi  a ˆj  b kˆ are position vectors
of three non-collinear points A, B & C then -
a bc ˆ ˆ ˆ
(A) centroid of triangle ABC is
3

i jk 
(B) ˆi  ˆj  kˆ is equally inclined to the three vectors
(C) perpendicular from the origin to the plane of triangle ABC meet at centroid
(D) triangle ABC is an equilateral triangle.

12. Identify the statement (s) which is/are incorrect ?

      2
  
(A) a  a  a  b   a  b a 
         
(B) If a, b, c are non coplanar vectors and v. a  v. b  v. c  0 then v must be a null vector
       

(C) If a and b lie in a plane normal to the plane containing the vectors c and d then a  b  c  d =0   
           
(D) If a, b, c and a ', b ', c ' are reciprocal system of vectors then a . b ' b. c ' c. a '  3
  
13. Given a parallelogram OACB. The lengths of the vectors OA , OB & AB are a, b & c respectively..
 
The scalar product of the vectors OC & OB is -

a 2  3 b2  c2 3a 2  b 2  c 2 3a 2  b 2  c 2 a 2  3b 2  c 2
(A) (B) (C) (D)
2 2 2 2
           
14. Consider ABC with A  (a) , B  (b) and C = (c) . If b .  a  c  = b. b + a . c ; b – a = 3; c – b = 4, then the
 
angle between the medians AM and BD is -
 1   1   1   1 
(A) – cos–1   (B)  – cos–1   (C) cos–1   (D) cos–1  
 5 13  13 5   5 13   13 5 

    
15. If the non zero vectors a & b are perpendicular to each other then the solution of the equation, r  a  b is -
  1     1     
(A) r  xa    a  b
a.a
  
(B) r  xb    a  b
b. b
 
(C) r  x a  b  (D) none of these
   
16. a , b , c be three non coplanar vectors and r be any arbitrary vector, then
           
( a × b ) × ( r × c ) + ( b × c ) × ( r × a ) + ( c × a ) × ( r × b ) is equal to-
           
(A) [ a b c ] r (B) 2 [ a b c ] r (C) 3[ a b c ] r (D) none of these
         
17. a and b are mutually perpendicular unit vectors. r is a vector satisfying r . a = 0, r . b = 1 and [ r a b ] =

1, then r is -
             
(A) a + ( a × b ) (B) b + ( a × b ) (C) a + b ( a × b ) (D) a – b + ( a × b )’

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10
Ans. A,C B,C B,C C C B,C A B A,B,C D
Que. 11 12 13 14 15 16 17
Ans. A,B,C,D A,C,D D A A B B
59
TRUE / FALSE

1. There exists infinitely many vectors of given magnitude which are perpendicular to a given plane.
2. There exists infinitely many vectors of given magnitude which are perpendicular to a given line.
           
3. Given that  a b c  =  b c d  = 0 & r =  a  b  then r . a  r . b  r.c  r.d  0 .

 ˆ   (iˆ  ˆj  k)
ˆ .
4. The point (1, 2, 3) lies on the line r  (2 ˆi  3 ˆj  4k)

5. The area of a parallelogram whose two adjacent edges are two diagonals of a given parallelogram is double
the area of given parallelogram.
     
   A.(B  C) B.(A  C)
6. If A, B, C are three non-coplanar vectors, then         0 [JEE 1985]
(C  A).B C.(A  B)
     
7. [2a  3 b 3a  4 b 4 a  5 b]  0

MATCH THE COLUMN

Following question contains statements given in two columns, which have to be matched. The statements in
Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given
statement in Column-I can have correct matching with ONE statement in Column-II.

1. Column-I Column-II

(A) ABC is a triangle. If P is a point inside the ABC (p) centroid

such that areas of the triangle PBC, PCA and PAB,
all are equal, then with respect to the ABC, P is its
  
(B) If a, b, c are the position vectors of the three (q) orthocentre
non-collinear points A, B and C respectively such
   
that the vector V  PA  PB  PC is a null vector,,
then with respect to the ABC, P is its
(C) If P is a point inside the ABC such that the (r) incentre
   
vector R  (BC)(PA)  (CA)(PB)  (AB)(PC) is a
null vector, then with respect to the ABC, P is its
(D) If P is a point in the plane of the triangle ABC (s) circumcentre
   
such that the scalar product PA. CB and PB. AC
vanishes, then with respect to the ABC, P is its

2. Let a, b, c be vectors then -

Column-I Column-II
         
(A) [a + b, b + c , c + a ] (p) | b |2 [ a c b ]
         
(B) [( a × b ) × ( a × c )]. b (q) ( a . b )[ a b c ]
        
(C) [a × b, b × c , c × a ] (r) 2[ a b c ]
       
(D) b . {( a × b ) × ( c × b )} (s) [ a b c ]2
60
3. Column-I Column-II
     
(A) Let a  ˆi  ˆj & b  2 ˆi  kˆ . If the point of intersection of the lines r  a  b  a (p) 0
 
& r  b  a  b is 'P', then 2 (O P) (where O is the origin) is
   
(B) ˆ b  2 ˆi  ˆj  kˆ and c  3 ˆi  2 ˆj  kˆ and a  (b  c)
If a  ˆi  2 ˆj  3k, is equal to (q) 5
  
xa  yb  zc , then x + y + z is equal to
(C) The number of values of x for which the angle between the vectors (r) 7
  1
a  x9 ˆi  (x 3  1)jˆ  2kˆ & b  (x  1)iˆ  xjˆ  kˆ is obtuse
3

2
(D) Let P1  2x – y + z = 7 & P2  x + y + z = 2. If P be a point that lies on (s) 11
P1, P2 and XOY plane, Q be the point that lies on P1, P2 and YOZ plane
and R be the point that lies on P1, P2 & XOZ plane,
then [Area of triangle PQR]
(where [.] is greatest integer function)

ASSERTION & REASON

These questions contains, Statement I (assertion) and Statement II (reason).

(A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
(B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I
(C) Statement-I is true, Statement-II is false
(D) Statement-I is false, Statement-II is true
1. Statement-I : The volume of a parallelopiped whose co-terminous edges are the three face diagonals of
a given parallelopiped is double the volume of given parallelopied.
Because
        
Statement-II : For any vectors a, b, c we have [a  b b  c c  a]  2[a b c]

(A) A (B) B (C) C (D) D

  
2. Statement-I : Let A(a) & B(b) be two points in space. Let P(r ) be a variable point which moves in space
 
such that PA.PB  0 , such a variable point traces a three-dimensional figure whose volume is given by
 2  2    
6
 
a  b  2a . b . a  b

Because
Statement-II : Diameter of sphere subtends acute angle at any point inside the sphere & its volume is given
4 3
by r , where 'r' is the radius of sphere.
3
(A) A (B) B (C) C (D) D
      
3. Statement-I : Let a, b, c be there non-coplanar vectors. Let p 1 be perpendicular to plane of a & b, p 2
       
perpendicular to plane b & c, p 3 perpendicular to plane of c & a then p 1 , p 2 & p 3 are non-coplanar..

Because
      
Statement-II : [a  b b  c c  a]  [a b c]2
(A) A (B) B (C) C (D) D

61
           
4. Statement-I : If r  a  b & r  p  µd be two lines such that b  td & a  p  s b where , µ t
& s be non-zero scalars then the two lines have unique point of intersection.
Because
Statement-II : Two non-parallel coplanar lines have unique point of intersection.
(A) A (B) B (C) C (D) D
 ˆ  ˆ  ˆ ˆ     
5. Statement-I : If a  i, b  j and c  i  j , then a and b are linearly independent but a, b and c are linearly
dependent.
Because
     
Statement-II : If a and b are linearly dependent and c is any vector, then a, b and c are linearly depen-
dent.
(A) A (B) B (C) C (D) D

COMPREHENSION BASED QUESTIONS

Comprehension # 1 :
F'
   G F
Three forces ƒ 1 , ƒ 2 & ƒ 3 of magnitude 2, 4 and 6 units respectively act ƒ1
D'
along three face diagonals of a cube as shown in figure. Let P1 be a E
D
parallelopiped whose three co-terminus edges be three vectors ƒ2
C
   O
ƒ 1 , ƒ 2 & ƒ 3 . Let the joining of mid-points of each pair of opposite edges ƒ3

of parallelopiped P1 meet in point X. A B B'

On the basis of above information, answer the following questions :

1. The magnitude of the resultant of the three forces is -

(A) 5 (B) 10 (C) 15 (D) none of these
2. The volume of the parallelopiped P1 is -

(A) 4 8 2 (B) 9 6 2 (C) 2 4 2 (D) 5 0 2

3. (OX) is equal to -
(A) 5 (B) 1.5 (C) 2 (D) 2.5

Comprehension # 2 :
 
Consider three vectors p  ˆi  ˆj  kˆ , q  2 ˆi  4 ˆj  kˆ and r  ˆi  ˆj  3kˆ and let s be a unit vector..

On the basis of above information, answer the following questions :

  
1. p , q and r are -
(A) linearly dependent
(B) can form the sides of a possible triangle
  
(C) such that the vector  q  r  is orthogonal to p
(D) such that each one of these can be expressed as a linear combination of the other two
     
2. If  p  q   r = up  vq  wr , then (u+v+w) equals to -

(A) 8 (B) 2 (C) –2 (D) 4

        
3. The magnitude of the vector  p.s  q  r  +  q.s   r  p  +  r.s   p  q  is -

(A) 4 (B) 8 (C) –2 (D) 2

62
Comprehension # 3 :
Three points A(1, 1, 4), B(0, 0, 5) & C(2, –1, 0) forms a plane. P is a point lying on the line r  ˆi  3 ˆj   (iˆ  ˆj  k)
ˆ .

2 6
The perpendicular distance of point P from plane ABC is .
3
   
'Q' is a point inside the tetrahedron PABC such that resultant of vectors AQ , BQ , CQ & PQ is a null
vector.
On the basis of above information, answer the following questions :

1. Co-ordinates of point 'P' is -

(A) (2, 4, 1) (B) (1, 3, 0) (C) (4, 6, 3) (D) (7, 9, 6)

2. Volume of tetrahedron PABC is -

4 81 2 81 81 6 81
(A) (B) (C) (D)
9 9 9 9

3. Co-ordinates of point 'Q' is -

5 5 5 5 5 5
(A)  4 , 1, 2  (B) (5, 1, 5) (C)  2 , 1, 4  (D)  4 , 5, 2 
     

ANSWER KEY
 True / False
1. F 2. T 3. F 4. T 5. T 6. T 7. T
 Match the Column
1. (A) (p), (B) (p), (C) (r), (D) (q) 2. (A) (r), (B) (q), (C) (s), (D) (p)
3. (A) (s), (B) (r), (C) (p), (D) (p)
 Assertion & Reason
1. A 2. C 3. A 4. D 5. B
 Comprehension Based Questions
Comprehension # 1 : 1. B 2. C 3. A Comprehension # 2 : 1. C 2. B 3. A
Comprehension # 3 : 1. A 2. B 3. A

63
1. The sides of parallelogram are 2 ˆi  4 ˆj  5kˆ and ˆi  2 ˆj  3kˆ . Find the unit vectors, parallel to their diagonals.

  

2. If G is the centroid of a triangle ABC, then prove that GA  GB  GC  0

3. Find out whether the following pairs of lines are parallel, non-parallel & intersecting, or non-parallel &
non-intersecting.
 ˆ ˆ
 r1  i  j  3kˆ  (iˆ  ˆj  k)ˆ
(a) r1  ˆi  ˆj  2kˆ  (3 ˆi  2 ˆj  4k)
ˆ (b) 
 r2  2 ˆi  4 ˆj  6kˆ  µ(2 ˆi  ˆj  3k)
ˆ
r2  2 ˆi  ˆj  3kˆ  µ(6 ˆi  4 ˆj  8 k)
ˆ

r1  ˆi  kˆ  (iˆ  3 ˆj  4k)ˆ
(c)  ˆ
r2  2 ˆi  3 ˆj  µ(4 ˆi  ˆj  k)
       
4. (a) Show that the points a  2b  3c;2a  3b  4c &  7b  10c are collinear..
(b) Prove that the points A = (1, 2, 3), B(3, 4, 7), C(–3, –2, –5) are collinear & find the ratio in which
B divides AC.
 
5. Points X & Y are taken on the sides QR & RS, respectively of a parallelogram PQRS, so that QX  4 XR &
   
RY  4 YS . The line XY cuts the line PR at Z. Prove that PZ   21  PR .
 25 
6. Using vectors prove that the altitudes of a triangle are concurrent.

7. Using vectors show that the mid-point of the hypotenuse of a right angled triangle is equidistant from its
vertices.

8. Using vectors show that a parallelogram whose diagonals are equal is a rectangle.

9. Using vectors show that a quadrilateral whose diagonals bisect each other at right angles is a rhombus.

10. Two medians of a triangle are equal, then using vector show that the triangle is isosceles.


11. 'O' is the origin of vectors and A is a fixed point on the circle of radius 'a' with centre O. The vector OA is

   
denoted by a . A variable point 'P' lies on the tangent at A & OP  r . Show that a.r | a| 2 . Hence if P (x,
y) & A (x1, y1) deduce the equation of tangent at A to this circle.

 
12. Let u be a vector on rectangular coordinate system with sloping angle 60°. Suppose that u  ˆi is geometric

mean of u and u  2 ˆi where î is the unit vector along x-axis then u has the value equal to a  b
where a, b  N . Find the value (a + b)3 + (a – b)3.

   
13. a, b, c and d are the position vectors of the points A  (x, y, z); B  (y, –2z, 3x) ; C  (2z, 3x, –y) and

        
D  (1, –1, 2) respectively. If a  2 3;  a ^ b    a ^ c  ;  a ^ d   and  a ^ ˆj  is obtuse, then find x, y, z.
2

   
14. If r and s are nonzero constant vectors and the scalar b is chosen such that r  bs is minimum, then
2  2 2
show that the value of bs  r  bs is equal to r .

64
15. (a) Find a unit vector â which makes an angle (/4) with axis of z & is such that â+i+j is a unit
vector.

 2    2
 a b  ab
(b) Prove that   2   2      
 a b   a b 

    
 
16. 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  = 3 ,  d a  = ,  d b =
  

^

 
then prove that d c = cos–1 (cos – cos)

17. Given three points on the xy plane O(0, 0), A(1, 0) and B(–1, 0). Point P is moving on the plane satisfying
     
the condition  PA. PB  + 3  OA. OB  = 0. If the maximum and minimum values of PA PB are M and
m respectively then find the values of M2 + m2.

       
   
18. If O is origin of reference, point A( a ); B( b ); C( c ); D( a + b ); E( b + c ); F( c + a ); G( a + b + c ) where
 

a = a1 î + a2 ˆj + a3 k̂ ; b = b1 î + b2 ˆj + b3 k̂ and c = c1 î + c2 ˆj + c3 k̂ , then prove that these points are
vertices of a cube having length of its edge equal to unity provided the matrix.

a1 a2 a3 
 
 b1 b2 b 3  is orthogonal. Also find the length XY such that X is the point of intersection of CM and GP;
 c1 c2 c 3 

Y is the point of intersection of OQ and DN where P, Q, M, N are respectively the midpoint of sides CF,
BD, GF and OB

           
19. Let A = 2 i + k , B = i + j + k , and C = 4 i – 3 j + 7 k Determine a vector R , satisfying
     
R × B = C × B and R · A = 0

   
20. If a , b, c, d are position vectors of the vertices of a cyclic quadrilateral ABCD prove that :

           
a xb  b xd  d xa bxc  cxd  d xb
         0
(b  a ) . (d  a ) (b  c) . (d  c)


 1 3 ˆ   2
  
  
21. Let a = 3 i – j and b = î + j and x = a + (q – 3) b , y = – p a + q b . If x  y , then express p
2 2
as a function of q, say p = f(q), (p  0 and q 0) and find the intervals of monotonicity of f(q).
 
22.
 ˆ b  ˆi  2ˆj  2kˆ & c  ˆi  2 ˆj  kˆ , find a unit vectors normal to the vectors a  b and b  c .
If a  ˆi  ˆj  k,

[REE 2000]

23. Prove that a  b =

      

 b.  a  a  b  

65
    
24. If a, b, c are non-coplanar v ectors and d is a unit v ector, then f ind the v alue of ,
            
| (a . d)(b  c)  (b. d)(c  a)  (c. d)(a  b)| independent of d .
[REE 99]

   
25. Find the vector r which is perpendicular to a  ˆi  2 ˆj  5kˆ and b  2 ˆi  3 ˆj  kˆ and r.(2 ˆi  ˆj  k)
ˆ 8  0 .

26. Two vertices of a triangle are at ˆi  3 ˆj and 2 ˆi  5 ˆj and its orthocentre is at ˆi  2 ˆj . Find the position vector
of third vertex.
[REE 2001]

27. Find the point R in which the line AB cuts the plane CDE where
    
a = i + 2j + k , b = 2i + j + 2k c =  4j + 4k , d = 2i  2j + 2k & e = 4i + j + 2k.

          
28. Solve for x : x × a + ( x . b ) a = c , where a and c are non zero non collinear and a . b  0

ANSWER KEY
3 ˆ 6 ˆ 2 ˆ 1 ˆ 2 ˆ 8 ˆ
1. i  j  k, i j k 3. (a) parallel (b) the lines intersect at the point p.v.  2 ˆi  2 ˆj
7 7 7 69 69 69
(c) lines are skew 4. (b) Externally in ratio 1 : 3 11. xx 1 + yy1 = a2 12. 28
1 1 1 11
13. x = 2, y = –2, z = –2 15. (a) i  j k 17. 34 18. 19. – î – 8 ˆj + 2 k̂
2 2 2 3
q(q 2  3 )  
21. p = ; decreasing in q (–1, 1), q  0 22.  î 24. [a b c] 25. r  13ˆi  11ˆj  7kˆ
4

5 ˆ 17 ˆ  1    a.c    
26.
7
i
7
j   kˆ where   R 27. p.v. of R = r = 3i + 3k 
28.  
a.b 
(b  c)  
 a  (b  a) 
a2 

66
1. The position vectors of the points A, B, C are respectively (1, 1, 1) ; (1, 1, 2) ; (0, 2, 1). Find a unit vector
parallel to the plane determined by ABC & perpendicular to the vector (1, 0, 1).

  
2. If a = a1 î + a2 ˆj + a3 k̂ ; b = b1 î + b2 ˆj + b3 k̂ and c = c1 î + c2 ˆj + c3 k̂ then show that the value of the
  
a .iˆ a .ˆj a .kˆ
  
      
b .ˆj b .kˆ
scalar triple product  n a  b n b  c n c  a  is (n3 + 1) b .iˆ
  
c .iˆ c .ˆj c .kˆ

3.
    

 
Given that a, b, p, q are four vectors such that a + b = µ p , b . q = 0 & b

 

= 1, where µ is a scalar

then prove that  a . q  p  p . q  a

       
= p.q

4. ABCD is a tetrahedron with pv's of its angular point as A(–5, 22, 5); B(1, 2, 3); C(4, 3, 2) and
D(–1, 2, –3). If the area of the triangle AEF where the quadrilaterals ABDE and ABCF are parallelograms
is S then find the values of S.

5. Given four points P1, P2, P3 and P4 on the coordinate plane with origin O which satisfy the condition
  3 
OPn  1 + OPn  1 = OPn , n = 2, 3
2
(a) If P1, P2 lie on the curve xy = 1, then prove that P3 does not lie on this curve.
(b) If P1, P2, P3 lie on the circle x2 + y2 = 1, then prove that P4 lies on this circle.


6. Find a vector v which is coplanar with the vectors î + ˆj – 2 k̂ and î – 2 ˆj + k̂ and is orthogonal to the

vector 2 ˆi  ˆj  kˆ . It is given that the projection of v along the vector ˆi  ˆj  kˆ is equal to 6 3 .

     
     p 2 b  (b. a )a  p (b x a )
7. If px  (x x a)  b ; (p  0 ) prove that x   .
p (p 2  a 2 )

          
8.  
Solve the following equation for the vector p ; p x a  p . b c  b x c where a , b, c are non zero non

 
    a b c   

coplanar vectors and a is neither perpendicular to b nor to c , hence show that  p x a     c  is

 a.c 
 
 
perpendicular to b  c .

         
9. Solve the simultaneous vector equations for the vectors x and y . x  c  y  a and y  c  x  b where

c is a non zero vector..

67
 (a 1  a )2 (a 1  b)2 (a 1  c)2
 
10. Let (b1  a )2 (b1  b)2 (b1  c)2 = 0 and if the vectors  = î + a ˆj + a2 k̂ ;  = î + b ˆj + b2 k̂ ;
(c1  a )2 (c1  b)2 (c1  c)2

  
 = î + c ˆj + c2 k̂ are non coplanar, show that the vectors 1 = î + a1 ˆj + a12 k̂ ; 1 = î + b1 ˆj + b12 k̂


and  1 = î + c1 ˆj + c12 k̂ are coplanar..


11. The vector OP = î + 2 ˆj + 2 k̂ turns through a right angle, passing through the positive x-axis on the way..
Find the vector in its new position.

             
12. If x  y  a, y  z  b, x.b  , x.y  1 and y.z  1 , then find x, y & z in terms of a, b &  . [REE 98]

13. Find the value of  such that a, b, c are all non-zero and
ˆ  (iˆ  ˆj  3k)c
(4 ˆi  5 ˆj)a  (3 ˆi  3 ˆj  k)b ˆ   (aˆi  bjˆ  ck)
ˆ [REE 2001]

A NS W ER KE Y
1
1.  (i  5 j  k) 4. 110 5. 9 6. ˆ
9 (ˆj  k)
3 3
      
         
    
 a b c  b. b c 
     b. c b  a  (c.a )c  b  c b  (c.b)c  a  c
8. p      
a c xb        9. x   , y  
  
a.c a.b   a.b    
a.b 

1  c2 1  c2

       
a b  a b  a b  a b
a   b
4 ˆ 1 ˆ 1 ˆ ab  
11. i j k 12. x   
; y  ; z  13.   2  29
2 2 2   2    2
a b a b
    
    
68
 3-DIMENSIONAL GEOMETRY

SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER)

1. The plane XOZ divides the join of (1, –1, 5) and (2, 3, 4) in the ratio  : 1, then is -
(A) –3 (B) –1/3 (C) 3 (D) 1/3

2. Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of
triangles ABC, ACD and ADB be 3, 4 and 5 sq. units respectively. Then the area of the triangle BCD, is -

5 5
(A) 5 2 (B) 5 (C) (D)
2 2

3. Which one of the following statement is INCORRECT ?

        
(A) If n . a = 0, n . b = 0 and n . c = 0 for some non zero vector n , then [a b c] = 0
(B) there exist a vector having direction angles  = 30° and  = 45°
(C) locus of point in space for which x = 3 and y = 4 is a line parallel to the z-axis whose distance from the
z-axis is 5
  
(D) In a regular tetrahedron OABC where 'O' is the origin, the vector OA + OB + OC is perpendicular to the
plane ABC.

4. Consider the following 5 statements

(I) There exists a plane containing the points (1, 2, 3) and (2, 3, 4) and perpendicular to the vector

V1 = î + ˆj – k̂
(II) There exist no plane containing the point (1, 0, 0); (0, 1, 0); (0, 0, 1) and (1, 1, 1)
   
(III) If a plane with normal vector N is perpendicular to a vector V then N · V = 0
(IV) If two planes are perpendicular then every line in one plane is perpendicular to every line on the other
plane
(v) Let P1 and P2 are two perpendicular planes. If a third plane P3 is perpendicular to P1 then it must be
either parallel or perpendicular or at an angle of 45° to P2.
Choose the correct alternative.
(A) exactly one is false (B) exactly 2 are false (C) exactly 3 are false (D) exactly four are false

 
5. Let L1 be the line r1 = 2 î + ˆj – k̂ + ( î + 2 k̂ ) and let L2 be the line r2 = 3 î + ˆj + µ( î + ˆj – k̂ ).

Let be the plane which contains the line L1 and is parallel to L2. The distance of the plane from the origin
is -

(A) 2/ 7 (B) 1/7 (C) 6 (D) none of these

 
6. The intercept made by the plane r . n = q on the x-axis is -


q ˆi. n  q
(A) 
(B) (C) (î. n ) q (D) 
ˆi. n q | n|

69
7. If from the point P(f, g, h) perpendiculars PL, PM be drawn to yz and zx planes then the equation to the
plane OLM is -

x y z x y z
(A) + + =0 (B) + – =0
f g h f g h

x y z x y z
(C) – + =0 (D) – + + =0
f g h f g h

8. The line which contains all points (x, y, z) which are of the form (x, y, z) = (2, –2, 5) + (1, –3, 2) intersects
the plane 2x – 3y + 4z = 163 at P and intersects the YZ plane at Q. If the distance PQ is a b , where
a, b N and a > 3 then (a + b) equals -
(A) 23 (B) 95 (C) 27 (D) none of these

9. A plane passes through the point P(4, 0, 0) and Q(0, 0, 4) and is parallel to the y-axis. The distance of the
plane from the origin is -

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


10. The distance between the parallel planes given by the equations, r . (2 î – 2 ˆj + k̂ ) + 3 = 0 and

r . (4 î – 4 ˆj + 2 k̂ ) + 5 = 0 is -
(A) 1/2 (B) 1/3 (C) 1/4 (D) 1/6

11. If the plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(k) with x-axis, then k is equal to -

3 2 2
(A) (B) (C) (D) 1
2 7 3

12. A variable plane forms a tetrahedron of constant volume 64K3 with the coordinate planes and the origin, then
locus of the centroid of the tetrahedron is -
(A) x3 + y3 + z3 = 6K2 (B) xyz = 6k3 (C) x2 + y2 + z2 = 4K2 (D) x–2 + y–2 + z–2 = 4k–2

  
13. The expression in the vector form for the point r1 of intersection of the plane r · n = d and the perpendicular
  
line r = r0 + t n where t is a parameter given by -
   
   d  r0 ·n      r0 ·n  
(A) r1 = r0 +   2  n (B) r1 = r0 –   2  n
 n   n 

   
   r ·n  d      r0 ·n  
(C) r1 = r0 –  0   n (D) r1 = r0 +    n
 n   n 

x 1 y 3 z 2
14. The equation of the plane containing the line   and the point (0, 7, –7) is -
3 2 1
(A) x + y + z = 1 (B) x + y + z = 2 (C) x + y + z = 0 (D) none of these

70
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS)

   
15. Consider the plane r . n1  d1 and r . n 2  d2 , then which of the follwoing are true -
 
(A) they are perpendicular if n1 . n 2  0

 
 n .n 
(B) angle between them is cos 1   1 2 
|
 1n | | n |
2 

  d   d
(C) normal form of the equation of plane are r . n1   1 & r . n 2   2
| n1 | | n2 |
(D) none of these

 ˆ and r  ˆi  2 ˆj  kˆ  µ(iˆ  ˆj  3k)

16. The equation of the plane which contains the lines r  ˆi  2 ˆj  kˆ  (iˆ  2 ˆj  k) ˆ

must be -
 ˆ 0
(A) r.(7ˆi  4 ˆj  k) (B) 7(x – 1) – 4(y – 2) – (z + 1) = 0

 ˆ 0  ˆ 0
(C) r.(iˆ  2ˆj  k) (D) r.(iˆ  ˆj  3k)

     
17. The plane containing the lines r  a  ta ' and r  a ' sa -
 
(A) must be parallel to a  a ' (B) must be the perpendicular to a  a '
      
(C) must be [r, a, a '] = 0 (D) (r  a).(a  a ')  0

18. The points A(5, –1, 1), B(7, –4, 7), C(1, –6, 10) and D(–1, –3, 4) are the vertices of a -
(A) parallelogram (B) rectangle (C) rhombus (D) square

19. If P1, P2, P3 denotes the perpendicular distances of the plane 2x – 3y + 4z + 2 = 0 from the parallel planes
2x – 3y + 4z + 6 = 0, 4x – 6y + 8z + 3 = 0 and 2x – 3y + 4z – 6 = 0 respectively, then -
(A) P1 + 8P2 – P3 = 0 (B) P3 = 16P2
(C) 8P2 = P1 (D) P1 + 2P2 + 3P3 = 29

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10
Ans. D A B D A A B A D D
Que. 11 12 13 14 15 16 17 18 19
Ans. B B A C A,B A,B B,C,D A,C A,B,C,D

71
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS)


1. If the line r = 2 î – ˆj + 3 k̂ + ( î + ˆj + 2 k̂ ) makes angles , , with xy, yz and zx planes respectively
then which one of the following are not possible ?
(A) sin2+ sin2+ sin2= 2 and cos2+ cos2+ cos2= 1
(B) tan2+ tan2+ tan2= 7 and cot2+ cot2+ cot2= 5/3
(C) sin2+ sin2+ sin2= 1 and cos2+ cos2+ cos2= 2
(D) sec2+ sec2+ sec2= 10 and cosec2+ cosec2+ cosec2= 14/3

2. A plane meets the coordinate axes in A, B, C such that the centroid of the triangle ABC is the point (1, r, r2).
The plane passes through the point (4, –8, 15) if r is equal to -
(A) –3 (B) 3 (C) 5 (D) –5

3. Indicate the correct order statements -

x 4 y6 z6 x 1 y2 z 3

(A) The lines = = and = = are orthogonal
3 1 1 1 2 2
(B) The planes 3x – 2y – 4z = 3 and the plane x – y – z = 3 are orthogonal.
(C) The function f(x) = n(e–2 + ex) is monotonic increasing x  R.
(D) If g is the inverse of the function, f(x) = n(e–2 + ex) then g(x) = n(ex – e–2)

x 1 y 1
4. The coordinates of a point on the line = = z at a distance 4 1 4 from the point (1, –1, 0) are-
2 3

(A) (9, –13, 4) (B) ( 8 1 4 +1, –12 1 4 –1, 4 1 4 )

(C) (–7, 11, –4) (D) (– 8 1 4 +1, 12 1 4 –1, –4 1 4 )

x 9 y4 z 5
5. Let 6x + 4y – 5z = 4, x – 5y + 2z = 12 and = = be two lines then-
2 1 1

 5
(A) the angle between them must be (B) the angle between them must be cos–1
3 6
(C) the plane containing them must be x + y – z = 0 (D) they are non-coplanar

x 1 y 1 z 3 x y z 1
6. The lines = = and = = are -
2 1  1 2 1
(A) coplanar for all  (B) coplanar for  = 19/3

 1 2 4 1 1 
(C) if coplanar then intersect at   ,  ,   (D) intersect at  ,  ,  1 
 5 5 5 2 2 

7. If two pairs of opposite edges of a tetrahedron are perpendicular then -

(A) the third is also perpendicular (B) the third pair is inclined at 60°
(C) the third pair is inclined at 45° (D) (B), (C) are false

72
8. The equation of a plane bisecting the angle between the plane 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0
is -
(A) 5x – y – 4z – 45 = 0 (B) 5x – y – 4z – 3 = 0
(C) 23x – 13y + 32z + 45 = 0 (D) 23x – 13y + 32z + 5 = 0


9. A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors î , î + ˆj and

the plane determined by the vectors î – ˆj , î – k̂ ,. The possible angle between a and î – 2 ˆj + 2 k̂ is -
(A) /3 (B) /4 (C) /6 (D) 3/4

10. If 1, m1, n1 and 2, m2, n2 are DCs of the two lines inclined to each other at an angle , then the DCs of the
bisector of the angle between these lines are-

1   2 m1  m 2 n n 2 1   2 m1  m 2 n1  n 2
(A) , , 1 (B) , ,
2 sin  / 2 2 sin  / 2 2 sin  / 2 2 cos  / 2 2 cos  / 2 2 cos  / 2

1   2 m1  m 2 n  n2 1   2 m1  m 2 n1  n 2
(C) , , 1 (D) , ,
2 sin  / 2 2 sin  / 2 2 sin  / 2 2 cos  / 2 2 cos  / 2 2 cos  / 2

x2 y3 z6

11. Points that lie on the lines bisecting the angle between the lines   and
2 3 6

x2 y3 z6

  are -
3 6 2
(A) (7, 12, 14) (B) (0, –3, 14) (C) (1, 0, 10) (D) (–3, –6, –2)

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10
Ans. A,B,D B,C C,D A,C A,C B,C A,D B,C B,D B,C
Que. 11
Ans. A,B,C,D
73
TRUE / FALSE

1. If the plane xbc + yac + zab = abc cuts x, y & z-axis in A, B & C respectively then area of ABC is

a 2 b2  b 2 c 2  c 2 a 2 .
 
      –1
 b·n 
2. The angle between the line r = a +  b and plane r · n = d is – cos    
2  b n 

3. The perpendicular distance of the plane r · n̂ = d, from the origin is d where d > 0

4. If A(1, 2, –1), B(2, 6, 2) and C(, –2, –4) are collinear then value of is 0.

5. The projection of line segment on the axes of reference are 3, 4 and 12 respectively. The length of such a line
segment is 13

MATCH THE COLUMN

Following questions contains statements given in two columns, which have to be matched. The statements in
Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given
statement in Column-I can have correct matching with ONE statement in Column-II.

1. Match the following pair of planes with their lines of intersections :

Column-I Column-II

x 2 y  2007 z  2004
(A) x +y=0=y+z (p) = =
0 1 1

x 2 y z 1
(B) x = 2, y = 3 (q) = =
0 1 1
(C) x = 2, y + z = 3 (r) x = –y = z

x2 y3 z
(D) x = 2, x + y + z = 3 (s) = =
0 0 1

2. Consider three planes

P1  2x + y + z = 1
P2  x – y + z = 2
P3  x – y + 3z = 5
The three planes intersects each other at point P on XOY plane and at point Q on YOZ plane. O is the origin.
Column-I Column-II
(A) The value of  is (p) 1
(B) The length of projection of PQ on x-axis is
(C) If the co-ordinates of point R situated at a minimum (q) 2
distance from point 'O' on the line PQ are (a, b, c),
then value of 7a + 14b + 14c is (r) 4

a
(D) If the area of POQ is , then value of a – b is (s) 3
b

74
 Following question contains statements given in two columns, which have to be matched. The statements in
Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given
statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II.

3. Consider the following four pairs of lines in column–I and match them with one or more entries in column–II
Column-I Column-II
(A) L1 : x=1 + t, y=t, z=2–5t (p) non coplanar lines
L2 : r = (2, 1,–3) +  (2,2,–10)
x 1 y3 z2
(B) L1 : = = (q) lines lie in a unique plane
2 2 1
x 2 y6 z 2
L2 :  
1 1 3
(C) L1 : x = – 6t, y=1 + 9t, z=–3t (r) infinite planes containing both
the lines
L2 : x=1 +2s, y=4–3s, z=s
x y 1 z2
(D) L1 : = = (s) lines are not intersecting
1 2 3
x3 y2 z 1
L2 : = =
4 3 2

ASSERTION & REASON

These questions contains, Statement I (assertion) and Statement II (reason).
(A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
(B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I.
(C) Statement-I is true, Statement-II is false.
(D) Statement-I is false, Statement-II is true.
  
1. Statement - I : If a plane contains point A( a ) and is parallel to vectors b and c , then its vector equation
     
is r = a +  b + µ c , where  & µ are parameters and b c .
Because
Statement - II : If three vectors are co-planar, then any one can be expressed as the linear combination
of other two.
(A) A (B) B (C) C (D) D

2. Statement - I : If ax + by + cz = a 2  b 2  c2 be a plane and (x1, y1, z1) and (x2, y2, z2) be two points
on this plane then a(x1 – x2) + b(y1 – y2) + c(z1 – z2) = 0.
Because
Statement - II : If two vectors p 1 î + p 2 ˆj + p 3 k̂ and q 1 î + q 2 ˆj + p 3 k̂ are orthogonal then
p1q1 + p2q2 + p3q3 = 0.
(A) A (B) B (C) C (D) D

x  x1 y  y1 z  z1 x  x2 y  y2 z  z2
3. Statement - I : If the lines = = and = = are coplanar then
a1 b1 c1 a2 b2 c2

x1 y1 z1 x2 y2 z2
a1 b1 c1 = a 1 b1 c1
a2 b2 c2 a2 b2 c2
Because
Statement - II : If the two lines are coplanar then shortest distance between them is zero.

(A) A (B) B (C) C (D) D

75
4. Statement - I : ABCDA1B1C1D1 is a cube of edge 1 unit. P and Q are the mid points of the edges B1A1,
8
and B1C1 respectively. Then the distance of the vertex D from the plane PBQ is .
3
Because
Statement - II : Perpendicular distance of point (x1, y1, z1) from the plane ax + by + cz + d = 0 is given by

ax1  by1  cz1  d

.
a 2  b2  c 2
(A) A (B) B (C) C (D) D

5. Statement - I : If 2a + 3b + 6c = 14, where a, b & c  R, then the minimum value of a2 + b2 + c2 is 4.

Because
1
Statement - II : The perpendicular distance of the plane px + qy + rz = 1 from origin is .
p  q2  r 2
2

(A) A (B) B (C) C (D) D

6. Consider following two planes
   
P1  [r  p a b]  0
   
P2  [r  p c d]  0
    
such that | (a  b)  (c  d)|  0 & let x be any vector in space.
       
Statement-I : x.{(a  b)  (c  d)}  0  x.{t1 a  t 2 b}  0,  t1 , t 2  R
Because
       
Statement-II : x.{t1 a  t2 b}  0  t 1 , t 2  R  x.{(a  b)  (c  d)}  0 .
(A) A (B) B (C) C (D) D
 ˆ  0 and P : (r  (2ˆi  ˆj  k)).{(i
ˆ  (iˆ  2k)} ˆ ˆ  2k)
ˆ  (2 ˆi  ˆj  3k)}
ˆ 0
7. Consider planes P1 : (r  ˆi).{(iˆ  ˆj  k) 2
 ˆ
and line L : r  5 ˆi  (iˆ  ˆj  k)
Statement-I : P1 & P2 are parallel planes.
Because
Statement-II : L is parallel to both P1 & P2.
(A) A (B) B (C) C (D) D

COMPREHENSION BASED QUESTIONS

Comprehension # 1 :
Given four points A(2, 1, 0); B(1, 0, 1); C(3, 0, 1) and D(0, 0, 2). The point D lies on a line L orthogonal to the
plane determined by the point A, B, and C
On the basis of above information, answer the following questions :

1. Equation of the plane ABC is -

(A) x + y + z – 3 = 0 (B) y + z – 1 = 0 (C) x + z – 1 = 0 (D) 2y + z – 1 = 0
2. Equation of the line L is -
 
(A) r = 2 k̂ + ( î + k̂ ) (B) r = 2 k̂ + (2 ˆj + k̂ )

(C) r = 2 k̂ + ( ˆj + k̂ ) (D) none of these

3. Perpendicular distance of D from the plane ABC, is -

1 1
(A) 2 (B) (C) 2 (D)
2 2
76
Comprehension # 2 :
x  x1 y  y1 z  z1
If a line passes through P (x1, y1, z1) and having Dr’s a, b, c, then the equation of line is
= =
a b c
and equation of plane perpendicular to it and passing through P is a(x – x1) + b(y – y1) + c(z – z1) = 0.
Further equation of plane through the intersection of the two planes
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is
(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0
On the basis of above information, answer the following questions :

1. The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line
x y z 3
= = is -
2 3 4
1 1 1 2
(A) 21 (B) 29 (C) 13 (D)
5 5 5 5
x 3 y 1 z 2
2. The equation of the plane through (0, 2, 4) and containing the line = = is -
3 4 2
(A) x – 2y + 4z – 12 = 0 (B) 5x + y + 9z – 38 = 0
(C) 10x – 12y – 9z + 60 = 0 (D) 7x + 5y – 3z + 2 = 0
3. The plane x – y – z = 2 is rotated through 90° about its line of intersection with the plane x + 2y + z = 2. Then
equation of this plane in new position is -
(A) 5x + 4y + z – 10 = 0 (B) 4x + 5y + 3z = 0 (C) 2x + y + 2z = 9 (D) 3x + 4y – 5z = 9
Comprehension # 3 :
Consider a triangular pyramid ABCD the position vectors of whose angular point are A(3, 0, 1);
B(–1, 4, 1); C(5, 2, 3) and D(0, –5, 4). Let G be the point of intersection of the medians of the triangle
BCD.
On the basis of above information, answer the following questions :

1. The length of the vector AG is-
51 51 59
(A) 17 (B) (C) (D)
3 9 4
2. Area of the triangle ABC in sq. units is-
(A) 24 (B) 8 6 (C) 4 6 (D) none of these

3. The length of the perpendicular from the vertex D on the opposite face is -
14 2 3
(A) (B) (C) (D) none of these
6 6 6

4. Equation of the plane ABC is -

(A) x + y + 2z = 5 (B) x – y – 2z = 1 (C) 2x + y – 2z = 4 (D) x + y – 2z = 1

A NS W ER KE Y
 True / False
1. F 2. T 3. T 4. T 5. F
 Match the Column
1. (A)  (r); (B)  (s); (C)  (p); (D)  (q) 2. (A)  (r); (B)  (p); (C)  (q); (D)  (s)
3. (A)  (r); (B)  (q); (C)  (q,s); (D)  (p,s)
 Assertion & Reason
1. C 2. A 3. C 4. D 5. A 6. D 7. B
 Comprehension Based Questions
Comprehension # 1 : 1. B 2. C 3. D Comprehension # 2 : 1. B 2. C 3. A
Comprehension # 3 : 1. B 2. C 3. A 4. D
77
1. Find the angle between the two straight lines whose direction cosines  , m, n are given by
2+ 2m – n = 0 and mn + n+ m = 0.

2. A variable plane is at a constant distance p from the origin and meets the coordinate axes in points A, B and
C respectively. Through these points, planes are drawn parallel to the coordinates planes. Find the locus of
their point of intersection.

3. P is any point on the plane x + my + nz = p. A point Q taken on the line OP (where O is the origin) such that
OP.OQ = p2. Show that the locus of Q is p(x + my + nz) = x2 + y2 + z2.

4. The plane x + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle . Prove

that the equation to the plane in new position is x + my ±z 2  m 2 tan = 0

x 3 y3 z
5. Find the equations of the two lines through the origin which intersect the line = = at an
2 1 1

angle of
3

x 1 y2 z3
6. Find the equation of the line which is reflection of the line = = in the plane
9 1 3
3x – 3y + 10z = 26

7. Find the point where the line of intersection of the planes x – 2y + z = 1 and x + 2y – 2z = 5, intersects the
plane 2x + 2y + z + 6 = 0

8. Find the foot and hence the length of the perpendicular from the point (5, 7, 3) to the line
x  15 y29 5 z
= = . Also find the equation of the plane in which the perpendicular and the given straight
3 8 5
line lie.

x 1 y2 z
9. Find the equation of the plane containing the straight line = = are perpendicular to the
2 3 5
plane x – y + z + 2 = 0

x 1 y z x 3 y z 2
10. Find the equation of the plane containing the line = = and parallel to the line = = .
2 3 2 2 5 4
Find also the S.D. between two lines.

ANSWER KEY
1 1 1 1 x y z x y z
1.  = 90° 2. 2
+ 2 + 2 = 2 5. = = or = =
x y z p 1 2 1 1 1 2
x 4 y 1 z 7
6. = = 7. (1, – 2, –4) 8. (9, 13, 15) ; 14; 9x – 4y – z = 14
9 1 3

9. 2x + 3y +z + 4 = 0 10. x – 2y + 2z – 1 = 0; 2 units

78
1. Through a point P(f, g, h), a plane is drawn at right angles to OP where 'O' is the origin, to meet the coordinate

r5
axes in A, B, C. Prove that the area of the triangle ABC is where OP = r..
2fgh

2. Find the equations to the line which can be drawn from the point (2, –1, 3) perpendicular to the lines

x 1 y2 z 3 x 4 y z3
= = and = = at right angles.
2 3 4 4 5 3

3. The position vectors of the four angular points of a tetrahedron OABC are (0, 0, 0); (0, 0, 2); (0, 4, 0) and
(6, 0, 0) respectively. A point P inside the tetrahedron is at the same distance 'r' from the four plane faces of
the tetrahedron. Find the values of 'r'.

x 6 y  10 z 14
4. The line = = is the hypotenuse of an isosceles right angled triangle whose opposite
5 3 8
vertex is (7, 2, 4). Find the equation of the remaining sides.

5. If two straight lines having direction cosines , m, n satisfy a+ bm + cn = 0 and fmn + gn+ hm = 0 are
f g h
perpendicular, then show that + + = 0.
a b c

6. Find the equations to the line of greatest slope through the point (7, 2, –1) in the plane x – 2y + 3z = 0
assuming that the axes are so placed that the plane 2x + 3y – 4z = 0 is horizontal.

7. Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of
triangles ABC, ACD and ADB be denoted by x, y and z sq. units respectively. Find the area of the
triangle BCD.

ANSWER KEY
x 2 y1 z3 2 x 7 y2 z 4 x 7 y2 z4
2. = = 3. 4. = = ; = =
11 1 0 2 3 3 6 2 2 3 6
x 7 y2 z 1
6. = = 7. (x 2  y 2  z2 )
22 5 4

79
 VECTOR  
8. The vectors AB = 3 î + 4 k̂ and AC =5 î –
1. If a, b, c are three non zero vectors out of which
2 ˆj +4 k̂ are the sides of a triangle ABC. The
two are not collinear. If a + 2b and c ; b + 3c and
a are collinear then a + 2b + 6c is- length of the median through A is-
[AIEEE-2002] [AIEEE-2003]
(A) Parallel to c (B) Parallel to a (A) 2 8 8 (B) 18 (C) 72 (D) 33
(C) Parallel to b (D) 0
  
        9. a , b , c are three v ectors, such that
2. If [ a b c ] = 4 then [ a × b b × c c × a ] =
      
[AIEEE-2002] a + b + c = 0 , | a | = 1, | b | = 2, | c | = 3,
(A) 4 (B) 2 (C) 8 (D) 16      
then a . b + b . c + c . a is equal to-
[AIEEE-2003]
3. If c = 2(a × b) + 3µ(b × a) ; a × b 0, c.(a ×
b)=0 then- [AIEEE-2002] (A) 1 (B) 0 (C) –7 (D) 7
(A)  = 3µ (B) 2 = 3µ 10. Consider point A, B, C and D with postion vectors
(C)  + µ = 0 (D) None of these
7 î – 4 ˆj + 7 k̂ , î – 6 ˆj + 10 k̂ , – î – 3 ˆj + 4 k̂
 
4. If a = 2 î + ˆj + 2 k̂ , b = 5 î – 3 ˆj + k̂ , then
  and 5 î – ˆj + 5 k̂ respectively. Then ABCD is a-
orthogonal projection of a on b is-
[AIEEE-2003]
[AIEEE-2002]
(A) parallelogram but not a rhombus
ˆ
9(5 ˆi  3 ˆj  k)
(A) 3 î – 3 ˆj + k̂ (B) (B) square
35
(C) rhombus
ˆ
(5 ˆi  3 ˆj  k)
(C) (D) 9(5 î – 3 ˆj + k̂ ) (D) None of these
35
5. A unit vector perpendicular to the plane of   
11. If u , v and w are three non-coplanar vectors,
        
a = 2 î – 6 ˆj – 3 k̂ , b = 4 î + 3 ˆj – k̂ is- then ( u + v – w ).( u – v ) × ( v – w ) equals-
[AIEEE-2002] [AIEEE-2003]
  
4 ˆi  3ˆj  kˆ (A) 3 u .( v × w ) (B) 0
2 ˆi – 6 ˆj  3kˆ
(A) (B)      
26 7 (C) u .( v × w ) (D) u .( w × v )
3ˆi – 2ˆj  6 kˆ 2 ˆi – 3ˆj – 6 kˆ   
(C) (D) 12. Let a , b and c be three non-zero vectors such
7 7 that no two of these are collinear. If the vector
       
6. Let u = î + ˆj , v = î – ˆj and w = î + 2 ˆj + a + 2 b is collinear with c and b + 3 c is
 
3 k̂ . If n̂ is a unit vector such that u . n̂ = 0 and collinear with a (being some non-zero scalar)
    
v . n̂ =0, then | w . n̂ | is equal to- then a + 2 b + 6 c equals- [AIEEE-2004]
[AIEEE-2003]   
(A)  a (B)  b (C)  c (D) 0
(A) 3 (B) 0 (C) 1 (D) 2

7. A particle acted on by constant forces 4 î + ˆj – 13. A particle is acted upon by constant forces 4 î

+ ˆj – 3 k̂ and 3 î + ˆj – k̂ which displace it from

3 k̂ and 3 î + ˆj – k̂ is displaced from the point
a point î + 2 ˆj + 3 k̂ to the point 5 î + 4 ˆj + k̂ .
î + 2 ˆj + 3 k̂ to the point 5 î + 4 ˆj + k̂ . The total
Then work done in standard units by the forces
work done by the forces is- [AIEEE-2003]
is given by- [AIEEE-2004]
(A) 50 units (B) 20 units
(A) 40 (B) 30 (C) 25 (D) 15
(C) 30 units (D) 40 units

80
      
14. If a , b , c are non-coplanar vectors and  is a 20. If a , b , c are non-coplanar vectors and is a
         
real number, then the vectors a + 2 b + 3 c , real number then [( a + b )2 b  c ] = [ a b + c
   
 b + 4 c and (2 – 1) c are non-coplanar for- d ] for- [AIEEE-2005]
[AIEEE-2004] (A) exactly one value of 
(A) all values of  (B) no value of 
(B) all except one value of  (C) exactly three values of 
(C) all except two values of  (D) exactly two values of 
(D) no value of 
  
      21. Let a = î – k̂ , b = x î + ˆj + (1 – x) k̂ and c =
15. Let u , v , w be such that | u | = 1, | v | = 2, | w |
    
= 3. If the projection of v along u is equal to y î + x ˆj + (1 + x – y) k̂ . Then [ a , b , c ] depends
    on- [AIEEE-2005]
projection of w along u and v and w are
   (A) only y (B) only x
perpendicular to each other then | u – v + w |
equals- [AIEEE-2004] (C) both x and y (D) neither x nor y
     
(A) 2 (B) 7 (C) 1 4 (D) 14 22. If ( a × b ) × c = a × ( b × c ), where a , b and

   c are any three vectors such that a . b  0,

16. Let a , b and c be non-zero vectors such that
b . c 0, then a and c are- [AIEEE-2006]
   1   
(a × b) × c = | b || c | a . If is the acute (A) inclined at an angle of /6 between them
3
(B) perpendicular
 
angle between the vectors b and c , then sin (C) parallel
equals- [AIEEE-2004] (D) inclined at an angle of /3 between them

23. ABC is a triangle, right angled at A. The resultant

1 2 2 2 2  
(A) (B) (C) (D) of the forces acting along AB , AC with
3 3 3 3

17. If C is the mid point of AB and P is any point 1 1

magnitudes and respectively is the
outside AB, then- [AIEEE-2005] AB AC
   
(A) PA + PB = 2 PC force along AD , where D is the foot of the
   perpendicular from A onto BC. the magnitude of
(B) PA + PB = PC the resultant is- [AIEEE-2006]
   
(C) PA + PB + 2 PC = 0 (AB)(AC) 1
1
    (A) (B) +
(D) PA + PB + PC = 0 AB  AC AB AC

  AB2  AC 2
18. For any vector a , the value of ( a × î )2 + 1
(C) (D)
  AD (AB)2 (AC)2
( a × ˆj )2 + ( a × k̂ )2 is equal to- [AIEEE-2005]
   
(A) 3 a 2 (B) a 2 (C) 2 a 2 (D) 4 a 2 24. The values of a, for which the points A, B, C with
19. Let a, b and c be distinct non-negative numbers. position vectors 2 î – ˆj + k̂ , î – 3 ˆj – 5 k̂ and

If the vectors a î + a ˆj + c k̂ , î + k̂ and c î + a î – 3 ˆj + k̂ respectively are the vertices of a

c ˆj + b k̂ lie in a plane, then c is- [AIEEE-2005] 

right-angled triangle with C = are-
(A) the Geometric Mean of a and b 2
(B) the Arithmetic Mean of a and b [AIEEE-2006]
(C) equal to zero (A) –2 and –1 (B) –2 and 1
(D) the Harmonic Mean of a and b (C) 2 and –1 (D) 2 and 1

81
  
25. If û and v̂ are unit vectors and  is the acute 32. The vectors a and b are not perpendicular and
     
angle between them, then 2 û × 3 v̂ is a unit c and d are two vectors satisfying: b  c  b  d
vector for- [AIEEE-2007]  
and a.d  0 . Then the vector d is equal to :
(A) Exactly two values of 
(B) More than two values of  [AIEEE-2011]
 
(C) No value of    b.c     a.c  
(D) Exactly one value of  (A) b      c (B) c      b
 a.b  a.b
26. Let a = î + ˆj + k̂ , b = î – ˆj + 2 k̂ and c =
 
  b.c     a.c  
x î + (x – 2) ˆj – k̂ . If the vector c lies in the (C) b      c (D) c      b
 a.b   a.b 
plane of a and b , then x equals-
[AIEEE-2007]  1  ˆ ˆ  1
(A) 0 (B) 1 (C) –4 (D) –2
33. If a 
7

3 i  k and b  2 ˆi  3 ˆj  6 kˆ , 
10
      
27. The vector a =  î + 2 ˆj +  k̂ , lies in the plane then the value of  2a  b  .  a  b    a  2 b   is:-
 
the vectors b = î + ˆj and c = ˆj + k̂ and [AIEEE-2011]
  (A) 5 (B) 3 (C) – 5 (D) – 3
bisect the angle between b and c . Then which
one of the following gives possible values of  34. ˆ ˆi  qjˆ  kˆ and ˆi  ˆj  rkˆ
If the vectors pˆi  ˆj  k,
and ? [AIEEE-2008]
(p  q  r  1) are coplanar, then the value of
(A)  = 2,  = 2 (B)  = 1,  = 2 pqr – (p + q + r) is :- [AIEEE-2011]
(C)  = 2,  = –1 (D)  = 1,  = 1
(A) –2 (B) 2 (C) 0 (D) –1
  
28. The non-zero vectors a , b and c are related
  
    35. Let a, b, c be three non-zero vectors which are
a = 8 b and c = –7 b . Then the angle between  
  pairwise non-collinear. If a  3b is collinear with
a and c is- [AIEEE-2008]    
c and b  2 c is colliner with a , then
(A) 0 (B) /4 (C) /2 (D)    
a  3b  6c is : [AIEEE-2011]
       
29. If u, v, w are non-coplanar vectors and p, q are (A) a  c (B) a (C) c (D) 0
real numbers, then the equality 36. Let â and b̂ be two unit vectors. If the vectors
         
[3 u pv p w] – [pv w qu] – [2 w qv qu] = 0 
c  aˆ  2bˆ and d  5aˆ  4 bˆ are perpendicular
holds for :- [AIEEE-2009]
to each other, then the angle between â and b̂
(A) More than two but not all values of (p, q) is : [AIEEE-2012]
(B) All values of (p, q)
   
(C) Exactly one value of (p, q) (A) (B) (C) (D)
4 6 2 3
(D) Exactly two values of (p, q)
  37. Let ABCD be a parallelogram such that
30. Let a  ˆj  kˆ and c  ˆi  ˆj  kˆ . Then the vector    
       AB  q, AD  p and BAD be an acute angle.
b satisfying a  b  c  0 and a . b  3 is : 
If r is the vector that coincides with the altitude
[AIEEE-2010] directed from the vertex B to the side AD, then

(A) ˆi  ˆj  2kˆ (B) 2 ˆi  ˆj  2kˆ r is given by : [AIEEE-2012]
 
 3 p . q 

(C) ˆi  ˆj  2kˆ (D) ˆi  ˆj  2kˆ (A) r   3q    p
(p . p )
    3  p . q  
31. If the vectors a  ˆi  ˆj  2kˆ , b  2ˆi  4 ˆj  kˆ and (B) r  3q    p
(p . p)

c  ˆi  ˆj  kˆ are mutually orthogonal, then    p . q  
(C) r   q      p
(, ) = [AIEEE-2010] p . p 
(A) (–3, 2) (B) (2, –3)    p . q  

(D) r  q      p
(C) (–2, 3) (D) (3, –2) p . p 

82
 
38. If the v ectors AB  3 ˆi  4 kˆ and  
 44. Let a,b and c be three units vectors such that
ˆ
AC  5 ˆi  2ˆj  4 k are the sides of a triangle    
ABC, then the length of the median through A a  b  c  0 . If = ⃗ · ⃗ + ⃗ · ⃗ + ⃗ · ⃗ and
is : [JEE (Main)-2013] ⃗ = ⃗ × ⃗ + ⃗ × ⃗ + ⃗ × ⃗, , t hen the
(A) 18 (B) 72 (C) 33 (D) 45 ordered pair, ( , ⃗) is equal to :
   [JEE (Main)-2020]
39. Let a  î  ĵ , b  î  ˆj  k̂ and c be a vector
       3    3  
such that a  c  b  0 and a · c = 4, then (A)   ,3a  b  (B)   ,3c  b 
 2   2 

| c |2 is equal to [JEE (Main)-2019]
3   3  
17 (C)  ,3b  c  (D)  ,3a  c 
(A) 8 (B) 2  2 
2
19
(C) 9 (D) 45. Let the volume of a parallelepiped whose
2
coterminous edges are given by
40. Let ⃗ = ̂+ ̂+ , ⃗ = ̂+ ̂+3
 
a  2î  1ˆj  3k̂, b  4î  (3   2 ) ĵ  6k̂ ⃗ = 2 ̂+ ̂+ be 1 cu. unit. If  be an angle
   , then
and c  3î  6ˆj  ( 3  1)k̂ be three vectors between the edges u and w cos can
 be [JEE (Main)-2020]
  
such that b  2a and a is perpendicular to c .
7 5
Then a possible value of (1, 2, 3) is : (A) (B)
[JEE (Main)-2019] 6 3 7
(A) (1, 5, 1) (B) (1, 3, 1)
7 5
(C) (D)
1   1  6 6 3 3
(C)  , 4,  2  (D)   , 4, 0
 2   2 
      46. Let ⃗ = ̂ − 2 ̂ + ⃗ = ̂− ̂+ be
41. Let   (  2)a  b and   ( 4  2)a  3b 
  two vectors. If c is a v ector such that
be two given vectors where vectors a and b are 
⃗ × ⃗ = ⃗ × ⃗ and ⃗ ∙ ⃗ = 0 , then c.b is
non-collinear. Then value of  for which vectors equal to [JEE (Main)-2020]
 
 and  are collinear, is 1
[JEE (Main)-2019] (A) (B) –1
2
(A) – 3 (B) – 4
(C) 4 (D) 3 1 3
(C)  (D) 
  2 2
42. Let a  î  2ˆj  4k̂ , b  î  ˆj  4k̂ and
 If the vectors, ⃗ = ( + 1) ̂ + ̂+
c  2î  4ˆj  (2  1) k̂ be coplanar vectors. 47. ,
 
Then the non-zero vector a  c is ⃗= ̂ + ( + 1) ̂ + and
[JEE (Main)-2019]
⃗= ̂+ ̂ + ( + 1) ( ∈ ) are coplanar
(A)  10 î  5 ĵ (B)  14 î  5 ĵ
2 2
and 3( ⃗ ∙ ⃗) − |⃗ × ⃗| = 0 , then the value of
(C)  14 î  5ˆj (D)  10 î  5ˆj  is ______. [JEE (Main)-2020]

43. Let 3 î  ˆj , î  3 ĵ and  î  (1  ) ĵ  

48. Let a,b and c be three v ectors such that
respectively be the position vectors of the points
A, B and C w.r.t. the origin O. If the distance of | ⃗| = √3, | ⃗ | = 5, ⃗ · ⃗ = 10 and the angle
C from the bisector of the acute angle between
   
3 between bandc is . If a is perpendicular to
OA and OB is , then the sum of all positive 3
2
    
values of  is: [JEE (Main)-2019] the vector b  c , then a  (b c) is equal to
(A) 4 (B) 3
(C) 2 (D) 1 _____. [JEE (Main)-2020]

83
 3-DIMENSIONAL GEOMETRY 57. A line makes the same angle , with each of the
x and z axis. If the angle , which it makes with
x 1 y2 z 3 y-axis, is such that sin2=3sin2, then cos2
49. If the line = =
3 2k 2 equals- [AIEEE-2004]
x 1 y 5 z 6 (A) 2/3 (B) 1/5 (C) 3/5 (D) 2/5
and = = are perpendicular to
3k 1 5 58. Distance between two parallel planes 2x + y +
each other then k = [AIEEE-2002] 2z = 8 and 4x + 2y + 4z + 5 = 0 is-
5 7 –7 –1 0 [AIEEE-2004]
(A) (B) (C) (D)
7 5 10 7 (A) 3/2 (B) 5/2 (C) 7/2 (D) 9/2
50. The angle between the lines, whose direction
59. A line with direction cosines proportional to 2,
ratios are 1, 1, 2 and 3 –1, – 3 –1, 4, is- 1, 2 meets each of the lines x = y + a = z and
[AIEEE-2002] x + a = 2y = 2z. The coordinates of each of the
(A) 45° (B) 30° (C) 60° (D) 90° points on intersection are given by-
[AIEEE-2004]
51. The acute angle between the planes 2x – y + z
(A) (3a, 3a, 3a), (a, a, a)
= 6 and x + y + 2z = 3 is- [AIEEE-2002]
(B) (3a, 2a, 3a), (a, a, a)
(A) 30° (B) 45° (C) 60° (D) 75°
(C) (3a, 2a, 3a), (a, a, 2a)
52. The shortest distance from the plane 12x + 4y (D) (2a, 3a, 3a), (2a, a, a)
+ 3z = 327 to the sphere x2 + y2 + z2 + 4x – 2y –
60. If the straight lines x = 1 + s, y = –3 – s,
6z = 155 is- [AIEEE-2003]
t
z = 1 + s and x = , y = 1 + t, z = 2 – t, with
4 2
(A) 39 (B) 26 (C) 11 (D) 13 parameters s and t respectively are coplanar then
13
equals- [AIEEE-2004]
x 2 y3 z4
53. The lines = = and 1
1 1 k (A) –2 (B) –1 (C) – (D) 0
2
x 1 y  4 z  5 61. The intersection of the spheres x2 + y2 + z2 + 7x
= = are coplanar if-[AIEEE-2003]
k 2 1 – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8
(A) k = 3 or –3 (B) k = 0 or –1 is the same as the intersection of one of the
(C) k = 1 or –1 (D) k = 0 or –3 sphere and the plane- [AIEEE-2004]
(A) x – y – z = 1 (B) x – 2y – z = 1
54. The radius of the circle in which the sphere
(C) x – y – 2z = 1 (D) 2x – y – z = 1
x2 + y2 + z2 + 2x – 2y – 4z – 19 = 0 is cut by the
plane x + 2y + 2z + 7 = 0 is- [AIEEE-2003] 62. if the angle between the line
(A) 4 (B) 1 (C) 2 (D) 3
x 1 y 1 z  2
= = and the plane 2x – y +  z
55. A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1 2 2
1), B(2, 1, 3) and C(–1, 1, 2). Then the angle 1
between the faces OAB and ABC will be- + 4 = 0 is such that sin= the value of is-
3
[AIEEE-2003] [AIEEE-2005]
 19  5 –3 3 –4
(A) 90° (B) cos–1   (A) (B) (C) (D)
 35  3 5 4 3
 17  63. The angle between the lines 2x = 3y = –z and
(C) cos–1   (D) 30° 6x = –y = –4z is- [AIEEE-2005]
 31 
56. The two lines x = ay + b, z = cy + d and x = a'y (A) 0° (B) 90° (C) 45° (D) 30°
+ b', z = c'y + d' will be perpendicular, if and only 64. If the plane 2ax – 3ay + 4az + 6 = 0 passes
if- [AIEEE-2003] through the midpoint of the line joining the
(A) aa' + cc' + 1 = 0 centres of the spheres x2 + y2 + z2 + 6x – 8y –
2z = 13 and x2 + y2 + z2 – 10x + 4y – 2z = 8 then
(B) aa' + bb' + cc' + 1= 0
a equals- [AIEEE-2005]
(C) aa' + bb' + cc' = 0 (A) –1 (B) 1 (C) –2 (D) 2
(D) (a + a') (b + b') + (c + c') = 0
84
65. The distance between the line
x 1 y2 z 3
 73. If the straight lines = = and
r = 2 î – 2 ˆj + 3 k̂ + ( î – ˆj + 4 k̂ ) and the plane k 2 3

r .( î + 5 ˆj + k̂ ) = 5 is- [AIEEE-2005] x 2 y3 z 1
= = intersect at a point, then
10 10 3 10 3 k 2
(A) (B) (C) (D)
9 3 3 10 3 the integer k is equal to- [AIEEE-2008]
(A) –5 (B) 5 (C) 2 (D) –2
66. The plane x + 2y – z = 4 cuts the sphere x2 + y2
+ z2 – x + z – 2 = 0 in a circle of radius- x 2 y 1 z  2
74. Let the line   lie in the
[AIEEE-2005] 3 5 2

(A) 3 (B) 1 (C) 2 (D) 2 plane x + 3y – z +  = 0. Then (, ) equals:

[AIEEE-2009]
67. The two lines x = ay + b, z = cy + d; and x = a'y
(A) (5, – 15) (B) (–5, 5)
+ b; z = c'y + d' are perpendicular to each other
if- [AIEEE-2006] (C) (6, –17)(D) (–6, 7)

a c 75. The projections of a vector on the three

(A) aa' + cc' = 1 (B) + = –1
a' c' coordinate axis are 6, –3, 2 respectively. The
a c direction cosines of the vector are :-
(C) + =1 (D) aa' + cc' = –1 [AIEEE-2009]
a' c'
6 3 2 6 3 2
(A) , , (B) , ,
68. The image of the point (–1, 3, 4) in the plane 7 7 7 7 7 7
x – 2y = 0 is - [AIEEE-2006]
6 3 2
(C) 6, –3, 2 (D) , ,
 17 19  5 5 5
(A) (15, 11, 4) (B)   ,  ,1 
 3 3  76. A line AB in three-dimensional space makes
(C) (8, 4, 4)(D) None of these angle 45º and 120º with the positive x-axis and
thepositive y-axis respectively. If AB makes an
69. Let L be the line of intersection of the planes 2x
+ 3y + z = 1 and x + 3y + 2z = 2. If L makes an acute angle  with the positive z-axis, then 
angle with the positive x-axis, then cosequals equals :- [AIEEE-2010]
- [AIEEE-2007] (A) 30° (B) 45° (C) 60° (D) 75°

(A) 1/ 3 (B) 1/2 (C) 1 (D) 1/ 2 77. Statement–I : The point A(3, 1, 6) is the mirror
 image of the point B(1, 3, 4) in the plane x – y
70. If a line makes an angle of with the positive + z = 5. [AIEEE-2010]
4
directions of each of x-axis and y-axis, then the Statement–II : The plane x – y + z = 5 bisects
angle that the line makes with the positive the line segment joining A(3, 1, 6) and B(1, 3, 4).
direction of the z-axis is- [AIEEE-2007] (A) Statement–I is true, Statement–II is true;
(A) /6 (B) /3 (C) /4 (D) /2 Statement–II is a correct explanation for
Statement–I.
71. If (2, 3, 5) is one end of a diameter of the sphere (B) Statement–I is true, Statement–II is true;
x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then the Statement–II is not a correct explanation
coordinates of the other end of the diameter are- for Statement–I.
[AIEEE-2007] (C) Statement–I is true, Statement–II is false.
(A) (4, 9, –3) (B) (4, –3, 3) (D) Statement–I is false, Statement–II is true.
(C) (4, 3, 5) (D) (4, 3, –3) y 1 z  3
78. If the angle between the line x = 
72. The line passing through the points (5, 1, a) and 2 
(3, b, 1) crosses the yz-plane at the point  5 
and the plane x + 2y + 3z = 4 is cos–1  ,
 17 13   14 
 0, ,  . Then- [AIEEE-2008]  
 2 2  then  equals - [AIEEE-2011]
(A) a = 2, b = 8 (B) a = 4, b = 6 2 5 2 3
(A) (B) (C) (D)
(C) a = 6, b = 4 (D) a = 8, b = 2 5 3 3 2

85
79. Statement-I: The point A(1, 0, 7) is the mirror x2 y3 z4
image of the point B(1, 6, 3) in the line : 85. If the lines   and
1 1 k
x y 1 z  2 x 1 y  4 z  5
    are coplanar, then k can
1 2 3 k 2 1
have : [JEE-MAIN 2013]
x y 1 z  2 (A) any value
Statement-II: The line :  
1 2 3 (B) exactly one value
(C) exactly two values
bisects the line segment joining A (1, 0, 7) and (D) exactly three values.
B(1, 6, 3). [AIEEE-2011]
(A) Statement-I is true, Statement-II is false. x 1 y  3 z  4
86. The image of the line   in the
(B) Statement-I is false, Statement-II is true 3 1 5
(C) Statement-I is true, Statement-II is true; plane 2x – y + z + 3 = 0 is the line :
Statement-II is a correct explanation for [JEE-MAIN 2014]
Statement-I
x 3 y5 z 2
(D) Statement-I is true, Statement-II is true; (A)  
Statement-II is not a correct explanation 3 1 5
for Statement-I. x 3 y5 z 2
(B)  
3 1 5
80. The length of the perpendicular drawn from the
x 3 y5 z2
x y 2 z3 (C)  
point (3, –1, 11) to the line   is 3 1 5
2 3 4
x 3 y5 z 2
: [AIEEE-2011] (D)  
3 1 5
(A) 66 (B) 29        2
87. If [a  b b  c c  a]  [a b c] then  is equal to :
(C) 33 (D) 53
[JEE-MAIN 2014]
81. The distance of the point (1, –5, 9) from the plane (A) 2 (B) 3 (C) 0 (D) 1
x – y + z = 5 measured along a straight line x
= y = z is: [AIEEE-2011] 88. The angle between the lines whose direction co-
sines satisfy the equations  + m + n = 0 and
(A) 3 5 (B) 10 3 2 2
 2 = m + n is : [JEE-MAIN 2014]
(C) 5 3 (D) 3 10
   
82. An equation of a plane parallel to the plane (A) (B) (C) (D)
3 4 6 2
x – 2y + 2z – 5 = 0 and at a unit distance from   
the origin is : [AIEEE-2012] 89. Let a, b and c be three non-zero vectors such
(A) x – 2y + 2z + 5 = 0 that no two of them are collinear and
(B) x – 2y + 2z – 3 = 0   
(C) x – 2y + 2z + 1 = 0  a  b   c  13 b c 
a . If  is the angle vectors b
(D) x – 2y + 2z – 1 = 0

x 1 y 1 z 1 and c , then a value of sin is: [JEE-MAIN 2015]
83. If the lines   and
2 3 4
2 2 3 2 2  2
x3 yk z (A) (B) (C) (D)
  intersect, then k is equal to:- 3 3 3 3
1 2 1
90. The equation of the plane containing the line 2x
[AIEEE-2012]
– 5y + z = 3 ; x + y + 4z = 5, and parallel to the
(A) 0 (B) – 1
plane, x + 3y + 6z = 1, is [JEE-MAIN 2015]
2 9 (A) x + 3y + 6z = 7
(C) (D) (B) 2x + 6y + 12z = – 13
9 2
(C) 2x + 6y + 12z = 13
84. Distance between two parallel planes (D) x + 3y + 6z = – 7
2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is:- 91. The distance of the point (1, 0, 2) from the point
[JEE-MAIN 2013]
x  2 y 1 z  2
3 5 of intersection of the line  
(A) (B) 3 4 12
2 2 and the plane x – y + z = 16, is
7 9 [JEE-MAIN 2015]
(C) (D)
2 2 (A) 3 21 (B) 12 (C) 2 14 (D) 8

86
 99. The length of the projection of the line segment
x 3 y 2 z 4
92. If the line ,   lies in the plane, joining the points (5, –1, 4) and (4, –1, 3) on the
2 1 3 plane, x + y + z = 7 is
2 2
lx + my – z = 9, then l + m is equal to : [JEE-MAIN 2018]
[JEE-MAIN 2016]
(A) 18 (B) 5 (C) 2 (D) 26 1 2 2 2
  (A) (B) (C) (D)
 3 3 3 3
93. Let a, b and c be three vectors such that
   3    
 
a  bc 
2
 
b  c . If b is not parallel to c , 100. The plane through the intersection of the planes
x + y + z = 1 and 2x + 3y – z + 4 = 0 and
 
then the angle between a and b is : parallel to y-axis also passes through the point
[JEE-MAIN 2019]
[JEE-MAIN 2016]
(A) (– 3, 0, – 1) (B) (3, 2, 1)
 2 5 3 (C) (3, 3, – 1) (D) (– 3, 1, 1)
(A) (B) (C) (D)
2 3 6 4
94. The distance of the point (1, –5, 9) from the plane 101. If the lines x = ay + b, z = cy + d and x = a'z +
x – y + z = 5 measured along the line b', y = c'z + d' are perpendicular, then
x = y = z is : [JEE-MAIN 2016] [JEE-MAIN 2019]
(A) bb' + cc' + 1 = 0 (B) aa' + c + c' = 0
10 20
(A) 10 3 (B) (C) (D) 3 10 (C) ab' + bc' + 1 = 0 (D) cc' + a + a' = 0
3 3
95. If the image of the point P (1, – 2, 3) in the plane, 102. The equation of the plane containing the straight
2x + 3y – 4z + 22 = 0 measured parallel to the
x y z
x y z line   and perpendicular to the plane
line, = = is Q, then PQ is equal to 2 3 4
1 4 5
[JEE-MAIN 2017] x y z
containing the straight lines   and
(A) 3 5 (B) 2 42 (C) 42 (D) 6 5 3 4 2
96. The distance of the point (1, 3, – 7) from the x y z
plane passing through the point (1, – 1, – 1)   is
having normal perpendicular to both the lines
4 2 3
[JEE-MAIN 2019]
x 1 y2 z4 x2 y 1 (A) 5x + 2y – 4z = 0 (B) 3x + 2y – 3z = 0
= = and =
1 2 3 2 1 (C) x – 2y + z = 0 (D) x + 2y – 2z = 0
z7
= is [JEE-MAIN 2017] 103. Let A be a point on the line
1 
20 10 5 10 r  (1  3 ) î  (  1)ˆj  ( 2  5 ) k̂ and B (3,
(A) (B) (C) (D) 2, 6) be a point in the space. Then the value of 
74 83 83 74
 for which the vector AB is parallel to the plane
 
97. Let a = 2î  ˆj  2 k̂ and b = î  ˆj . Let c be a x – 4y + 3z = 1 is [JEE-MAIN 2019]

     1 1
vector such that | c  a | = 3, (a  b)  c = 3 (A) (B)
8 4
  
and the angle between c and a  b be 30°. 1 
  (C) (D)
Then a ·c is equal to [JEE-MAIN 2017] 2 4
25 1 104. The plane passing through the point (4, – 1, 2)
(A) (B) 2 (C) 5 (D)
8 8
x  2 y  2 z 1
 and parallel to the lines  
98. Let u be a vector coplanar with the vectors 3 1 2
  
a  2î  3 ĵ  k̂ and b  ĵ  k̂ . If u is x 2 y 3 z  4
and   also passes
    2 1 2 3
perpendicular to a and u . b  24 , then u
through the point [JEE-MAIN 2019]
is equal to: [JEE-MAIN 2018] (A) (– 1, – 1, – 1) (B) (1, 1, – 1)
(A) 256 (B) 84 (C) 336 (D) 315 (C) (1, 1, 1) (D) (– 1, – 1, 1)

87
105. On which of the following lines lies the point of 111. Let P be a plane passing through the points
x  4 y 5 z 3 (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any
intersection of the line,   point (2, 1, 6). Then the image of R in the plane
2 2 1 P is : [JEE-MAIN 2020]
and the plane, x + y + z = 2?
[JEE-MAIN 2019] (A) (6, 5, –2) (B) (4, 3, 2)
(C) (3, 4, –2) (D) (6, 5, 2)
x  3 4  y z 1
(A)  
3 3 2
112. A vector ⃗ = + 2 ̂+ ( , ∈ ) lies in
x 4 y 5 z 5 the plane of the vectros
(B)  
1 1 1 
⃗ = ̂+ ̂ ⃗ = ̂ − ̂ + 4 . If a bisects
x 1 y  3 z  4  
(C)   the angle between band c , then:
1 2 5 [JEE-MAIN 2020]
x 2 y 3 z 3  
(D)   (A) a. ˆi  1  0 (B) a. ˆi  3  0
2 2 3  
(C) a.kˆ  4  0 (D) a.kˆ  2  0
106. The plane which bisects the line segment joining
the points (– 3, – 3, 4) and (3, 7, 6) at right
angles, passes through which one of the 113. If the foot of the perpendicular drawn from the
following points? [JEE-MAIN 2019] point (1, 0, 3) on a line passing through (,7,1)
(A) (– 2, 3, 5) (B) (2, 1, 3)  5 7 17 
(C) (4, – 1, 7) (D) (4, 1, – 2) is  , ,  , then a is equal to_____
3 3 3 
107. The plane containing the line
[JEE-MAIN 2020]
x - 3 y  2 z 1
  and also containing its
2 1 3 114. The shortest distance between the lines
projection on the plane 2x + 3y – z = 5,
contains which one of the following points? x3 y8 z3 x3 y7 z6
  and  
[JEE-MAIN 2019] 3 1 1 3 2 4
(A) (2, 2, 0) (B) (–2, 2, 2) is [JEE-MAIN 2020]
(C) (0, –2, 2) (D) (2, 0, –2)
7
(A) 30 (B) 3 30
108. The direction ratios of normal to the plane through 2
the points (0, –1, 0) and (0, 0, 1) and making an
(C) 3 (D) 2 30

angle with the plane
4 115. The projection of the line segment joining the
y – z + 5 = 0 are: [JEE-MAIN 2019] points (1, -1, 3) and (2, -4, 11) on the line joining
the points (-1, 2, 3) and (3, -2, 10) is _______.
(A) 2, –1, 1 (B) 2, 2,– 2
[JEE-MAIN 2020]
(C) 2 , 1, –1 (D) 2 3 , 1, –1
116. If the distance between the plane,
x  3 y 1 z  6 23x – 10y – 2z + 48 = 0 and the plane containing
109. Two lines   and
1 3 1 x 1 y  3 z 1
the lines   and
x 5 y2 z3 2 4 3
  intersect at the point R.
7 6 4 x  3 y  2 z 1
The reflection of R in the xy-plane has   (   R) is equal to
2 6 
coordinates [JEE-MAIN 2019]
(A) (2, – 4, –7) (B) (2, 4, 7) k
(C) (2, –4, 7) (D) (–2, 4, 7) , then k is equal to ______
633
110. If the points (2, , ) lies on the plane which [JEE-MAIN 2020]
passes through the points (3, 4, 2) and (7, 0, 6)
and is perpendicular to the plane 2x – 5y = 15,
then 2 – 3 is equal to: [JEE-MAIN 2019]
(A) 12 (B) 7
(C) 5 (D) 17
88
 ANSWER KEY

1. D 2. D 3. B 4. B 5. C

6. A 7. D 8. D 9. C 10. D

11. C 12. D 13. A 14. C 15. C

16. D 17. A 18. C 19. A 20. B

21. D 22. C 23. C 24. D 25. D

26. D 27. D 28. D 29. C 30. A

31. A 32. B 33. C 34. A 35. D

36. D 37. C 38. C 39. D 40. D

41. B 42. D 43. D 44. A 45. A

46. C 47. 1.00 48. 30 49. D 50. C

51. C 52. D 53. D 54. D 55. B

56. A 57. C 58. C 59. B 60. A

61. D 62. A 63. B 64. C 65. B

66. B 67. D 68. D 69. A 70. D

71. A 72. C 73. A 74. D 75. A

76. C 77. B 78. C 79. D 80. D

81. B 82. B 83. D 84. C 85. C

86. A 87. D 88. A 89. C 90. A

91. B 92. C 93. C 94. A 95. B

96. B 97. B 98. C 99. B 100. B

101. B 102. C 103. D 104. C 105. C

106. D 107. D 108. C 109. A 110. B

111. A 112. D 113. 4.00 114. B 115. 8.00

116. 3

89
 VECTOR 4. Show by vector methods, that the angular
1. Select the correct alternative :
bisectors of a triangle are concurrent and find
  
(i) If the vectors a, b & c form the sides an expression for the position vector of the point
BC, CA & AB respectively of a triangle of concurrency in terms of the position vectors
ABC, then
of the vertices. [JEE 2001]
     
(A) a . b  b. c  c. a  0
  
      5. Find 3-dimensional vectors v1 , v2 , v3 satisfying
(B) a  b  b  c  c  a      
      v1 . v1  4 , v1 . v2  2 , v1 . v3  6 ,
(C) a . b  b. c  c. a      
v2 . v2  2 , v2 . v3  5 , v3 . v3  29 .
     
(D) a  b  b  c  c  a  0 [JEE 2001]
     
B(t)  g1 (t)iˆ  g2 (t)jˆ ,
(ii) Let the vectors a, b, c & d be such that 6. Let A(t)  f1 (t)iˆ  f2 (t)jˆ and t  [0,1] ,
     where f 1, f 2, g1, g2 are continuous functions. If
   
a  b  c  d  0 . Let P1& P2 be planes  
A(t) and B(t) are non-zero vectors for all t and
determined by the pairs of v ectors
  
    A(0)  2ˆi  3 ˆj , A(1)  6ˆi  2ˆj , B(0 )  3ˆi  2 ˆj and
a, b & c, d respectively. Then the angle
  
between P1 and P2 is : B(1)  2ˆi  6ˆj , then show that A(t) and B(t) b

(A) 0 (B) /4 (C) /3 (D) /2 are parallel for some t. [JEE 2001]
    
(iii) If a, b & c are unit coplanar vectors, 7. (i) If a and b are two unit vectors such that
         
then the scalar triple product 2ab 2bc 2ca  a  2b and 5 a  4 b are perpendicular to
 
[JEE 2000] each other then the angle between
 
(A) 0 (B) 1 (C)  3 (D) 3 a and b is - [JEE 2002]
2. Let ABC and PQR be any two triangles in the (A) 45° (B) 60°
same plane. Assume that the perpendicular from
1  1  1  2 
the points A, B, C to the sides QR, RP, PQ (C) cos   (D) cos  
3 7
respectively are concurrent. Using vector
       
methods or otherwise, prov e that the (ii) Let V  2 i  j  k and W  i  3 k . If U
perpendiculars from P, Q, R to BC, CA, AB is a unit vector, then the maximum value
  
respectively are also concurrent. [JEE 2000] of the scalar triple product [U V W] is -

3. ˆ bˆ and cˆ are unit vectors, then

(i) If a, [JEE 2002]

2 2
(A) –1 (B) 1 0  6
2
aˆ  bˆ  bˆ  cˆ  cˆ  aˆ does not exceed
(C) 59 (D) 60
(A) 4 (B) 9 (C) 8 (D) 6
   8. Let v be the volume of the parallelopiped formed
(ii) Let a  ˆi  kˆ , b  xiˆ  ˆj  (1  x)k and 
by the v ectors ˆ
a  a 1 ˆi  a 2 ˆj  a 3 k,
   
c  yiˆ  xjˆ  (1  x  y)kˆ . Then [a, b, c]  
b  b1 ˆi  b2 ˆj  b3 kˆ , c  c1 ˆi  c 2 ˆj  c 3 k.
ˆ
depends on [JEE 2001]
(A) only x If ar, br, cr, where r = 1, 2, 3, are non-negative
3
(B) only y
real numbers and   a r  b r  c r   3L , show
(C) neither x nor y r 1
3
(D) both x and y that V  L . [JEE 2002]

90
9. The value of a for which the volume of 14. Incident ray is along the unit vector v and the
parallelopiped formed by the vectors ˆi  ajˆ  kˆ reflected ray is along the unit vector w  . The
normal is along unit vector â outwards. Express
and ˆj  akˆ and aˆi  kˆ as coterminous edge is
minimum is - [JEE 2003]  in terms of â and v .
w [JEE 2005]
(A) –3 (B) 3  
15. (i) Let ˆ b  ˆi  ˆj  kˆ
a  ˆi  2 ˆj  k, A and c  ˆi  ˆj  kˆ .
(C) 1 / 3 (D) none of these  
vector in the plane of a and b whose

10.
  
If u , v , w are three non-coplanar unit vectors projection on c has the magnitude equal
  1
and ,  ,  are the angles between u and v , to is - [JEE 2006]
    3
v and w , w and u respectiv ely and
   (A) 4 ˆi  ˆj  4kˆ (B) 3 ˆi  ˆj  3kˆ
x , y , z are unit vectors along the bisectors of
(C) 2 ˆi  ˆj  2kˆ (D) 4 ˆi  ˆj  4kˆ
the angles , ,  respectively. Prove that

     
[x  y y  z z  x] 
1   2
[u v w] sec 2

sec 2

sec 2
 (ii) Let A be v ector parallel to line of
16 2 2 2
intersection of planes P1 and P2 through
[JEE 2003]
  origin. P1 is parallel to the vectors 2ˆj  3kˆ
11. (i) If for vectors a and b ,
 and 4 ˆj  3kˆ and P2 is parallel to ˆj  kˆ and
    ˆ a  ˆi  ˆj  kˆ
a . b  1, a  b  ˆj  k, then vector b is – 
3 ˆi  3 ˆj , then the angle between vector A
(A) ˆi  ˆj  kˆ (B) 2 ˆj  kˆ
and 2 ˆi  ˆj  2kˆ is - [JEE 2006]
(C) î (D) 2 î    3
(A) (B) (C) (D)
2 4 6 4
(ii) A giv en unit vector is orthogonal to
16. The number of distinct real values of , for which
5 ˆi  2 ˆj  6 kˆ and coplanar with ˆi  ˆj  kˆ and
the v ectors 2 ˆi  ˆj  kˆ , ˆi  2 ˆj  kˆ and
2 ˆi  ˆj  kˆ then the vector is - [JEE 2004] ˆi  ˆj  2 kˆ are coplanar, is :- [JEE 2007]
(A) zero (B) one
3 ˆj  kˆ 6 ˆi  5kˆ (C) two (D) three
(A) (B)
10 61 
   
17. Let a, b, c be unit vectors such that a  b  c  0 .
2 ˆi  5 kˆ Which one of the following is correct ?
2 ˆi  ˆj  2kˆ
(C) (D) [JEE 2007]
29 3       
(A) a  b  b  c  c  a  0
      
    (B) a  b  b  c  c  a  0
12. a, b, c and d are four distinct vectors satisfying
          
    (C) a  b  b  c  a  c  0
the conditions a  b  c  d & a  c  b  d , then
     
        (D) a  b, b  c, c  a are mutually perpendicular
prove that a . b  c. d  a . c  b. d [JEE 2004]
    
18. Let the vectors PQ , QR , RS , ST , TU and
   
13. If a , b , c are three non-zero, non-coplanar vec-
 
UP represent the sides of a regular hexagon.
  b . a    b . a     
tors and b1  b   2 a , b2  b   2 a , Statement-I : PQ  (RS  ST)  0 . because
| a| | a|
     
    Statement-I : PQ  RS  0 and PQ  ST  0 .

 
 c .a  b . c  
 
 c .a  b1 . c 
c1  c   2 a +  b1 , c2  c   2 a   b1 , [JEE 2007]
| a| | c| 2 | a| | b1| 2
    (A) Statement-I is True, Statement-II is True ;
   
  c.a  b .c    c.a  b.c  Statement-II is a correct explanation for
c3  c   2 a + 1 2 b1 , c 4  c   2 a –  2 b1
| a| | c| | a| | b| Statement-I.
then the set of orthogonal vectors is - (B) Statement-I is True, Statement-II is True ;
[JEE 2005] Statement-II is NOT a correct explanation for
    Statement-I.
 
(A) (a , b1 , c3 ) (B) (a , b1 , c2 ) (C) Statement-I is True, Statement-II is False.
      (D) Statement-I is False, Statement-II is True.
(C) (a , b1 , c1 ) (D) (a , b2 , c2 )
91
19. The edges of a parallelopiped are of unit length Column–II
and are parallel to non–coplanar unit vectors â , 
(p)
1 6
bˆ , ĉ such that â . bˆ = bˆ . ĉ = ĉ . â = .
2 
Then, the volume of the parallelopiped is :- (q)
4
[JEE 2008] 
(r)
1 1 3 1 3
(A) (B) (C) (D)
2 2 2 2 3 
(s)
20. Let two non-collinear unit vectors â and b̂ form 2
an acute angle. A point P moves so that at any (t) 

time t the position vector OP (where O is the 23. Let P, Q, R and S be the points on the plane
origin) is given by â cost + bˆ sint. When P is with position vectors – 2 ˆi  ˆj, 4 ˆi,3 ˆi  3ˆj and

farthest from origin O, let M be the length of OP
 3 ˆi  2ˆj respectively. The quadrilateral PQRS
and û be the unit vector along OP . Then-
must be a [JEE 2010]
[JEE 2008]
(A) parallelogram, which is neither a rhombus
â  bˆ nor a rectangle
(A) û  and M = (1 + â . bˆ )1/2
â  bˆ (B) square
â  bˆ (C) rectangle, but not a square
(B) û  and M = (1 + â . bˆ )1/2
â  bˆ (D) rhombus, but not a square
 
â  bˆ 24. If a and b are vectors in space given by
(C) û  and M = (1 + 2 â . bˆ )1/2
â  bˆ  ˆi  2 ˆj  2 ˆi  ˆj  3kˆ
a and b  , then the value
5 14
â  bˆ      
(D) û  and M = (1 + 2 â . bˆ )1/2      
of 2a  b .  a  b  a  2b  is [JEE 2010]
â  bˆ  
    25. Two adjacent sides of a parallelogram ABCD
21. If a, b, c and d are unit vectors such that 
are giv en by AB  2 ˆi  10 ˆj  11kˆ and
     1
  
a  b . c  d  1 and a.c = , then :-
2

AD  ˆi  2 ˆj  2kˆ . The side AD is rotated by
[JEE 2009] an acute angle  in the plane of the parallelogram
   so that AD becomes AD'. If AD' makes a right
(A) a, b, c are non-coplanar
   angle with the side AB, then the cosine of the
(B) b, c, d are non-coplanar angle  is given by -
 
(C) b, d are non-parallel [JEE 2010]
   
(D) a, d are parallel and b, c are parallel 8 17 1 4 5
(A) (B) (C) (D)
9 9 9 9
22. Match the statements / expressions given in Col-  ˆ ˆ ˆ  ˆ ˆ ˆ  ˆ ˆ ˆ
26. (i) Let a  i  j  k, b  i  j  k and c  i  j  k be
umn I with the values given in Column II 
[JEE 2009] three vectors. A vector v in the plane of
Column–I    1
(A) Root(s) of the equation 2sin2 + sin22=2 a and b , whose projection on c is ,
3
(B) Points of discontinuity of the function f(x) is given by
6 x  3 x 
=   cos   where [y] denotes the (A) ˆi  3 ˆj  3kˆ (B) 3 ˆi  3 ˆj  kˆ

   
largest integer less than or equal to y (C) 3 ˆi  ˆj  3kˆ (D) ˆi  3 ˆj  3kˆ
(C) Volume of the parallelepiped with its edges (ii) The vector(s) which is/are coplanar with
represented by the vectors ˆi  ˆj , ˆi  2 ˆj and v ectors ˆi  ˆj  2kˆ and ˆi  2 ˆj  kˆ and
ˆi  ˆj  kˆ
perpendicular to the vector ˆi  ˆj  kˆ is/are
  
(D) Angle between vectors a and b where a, b
and c are unit v ectors satisf ying (A) ˆj  kˆ (B) ˆi  ˆj
    (C) ˆi  ˆj (D) ˆj  kˆ
a  b  3c  0
92
  
(iii) Let and c  ˆi  2 ˆj  3kˆ be
ˆ b  ˆi  ˆj
a  ˆi  k, List-II
 1. 100
three given vectors. If r is a vector such 2. 30
     3. 24
that r  b  c  b and r.a  0 , then the
 4. 60
value of r.b is [JEE 2011] Codes
   P Q R S
27. (i) If a, b and c are unit vectors satisfying (A) 4 2 3 1
      (B) 2 3 1 4
| a  b | 2  | b  c |2  | c  a |2  9 , then
  
(C) 3 4 1 2
| 2 a  5 b  5 c | is (D) 1 4 3 2
 
(ii) If a and b are v ectors such that
   31. Let L1 and L2 denote the lines
and a  (2 ˆi  3 ˆj  4 k)
| a  b |  29 ˆ  (2 ˆi  3 ˆj  4k)
ˆ b ,


then a possible value of (a  b).(7ˆi  2 ˆj  3k) ˆ is r  î   (  î  2ˆj  2k̂ ),   R and
[JEE 2012] 
r   ( 2î  ĵ  2k̂ ),   R
(A) 0 (B) 3 (C) 4 (D) 8

respectively. If L3 is a line which is perpendicular

28. Let PR  3 ˆi  ˆj  2kˆ and SQ  ˆi  3 ˆj  4kˆ determine to both L1 and L2 and cuts both of them, then
diagonals of a parallelogram PQRS and which of the following options describe(s) L3?
 [JEE 2019]
PT  ˆi  2 ˆj  3kˆ be another vector. Then the 
volume of the parallelepiped determined by the (A) r  t ( 2 î  2 ĵ  k̂ ), t  R
  
vectors PT, PQ and PS is [JEE-Ad. 2013]  2
(B) r  ( 2î  ĵ  2k̂ )  t ( 2î  2ˆj  k̂ ), t  R
(A) 5 (B) 20 (C) 10 (D) 30 9
29. Consider the set of eight v ect ors  1
(C) r  ( 2î  k̂ )  t ( 2î  2ˆj  k̂ ), t  R

V  a ˆi  bjˆ  ckˆ : a, b, c  {1,1}  . Three non-coplanar 3
vectors can be chosen from V in 2p ways. Then  2
(D) r  ( 4î  ĵ  k̂ )  t ( 2î  2ˆj  k̂ ), t  R
p is [JEE-Ad. 2013] 9
30. Match List-I with List-II and select the correct 32. Three lines are given by
answer using the code given below the lists.

[JEE-Ad. 2013] r   î,   R
List-I 
P. Volume of parallelepiped determined by r  (î  ĵ),   R and
   
vectors a, b and c is 2. Then the volume r   (î  ĵ  k̂ ),   R
of the parallelepiped determined by vectors
      Let the lines cut the plane x + y + z = 1 at the
   
2 a  b ,3 b  c and  c  a  is
points A, B and C respectively. If the area of the
Q. Volume of parallelepiped determined by
 
vectors a, b and is 5. Then the volume of triangle ABC is  then the value of 6 2
the parallelepiped determined by vectors equals ___ [JEE 2019]
     
  
3 a  b , b  c and 2  c  a  is
33. Three lines
R. Area of a triangle with adjacent sides
  
determined by vectors a and b is 20. Then L1 : r   î,  R ,
the area of the triangle with adjacent  

L 2 : r  k  ĵ,  R and

sides determined by vectors 2a  3 b and   
L3 : r  î  ĵ  vk̂ ,   R
 

a  b is  are given. For which point(s) Q on L2 can we
S. Area of a parallelogram with adjacent sides find a point P on L1 and a point R on L3 so that
  P, Q and R are collinear? [JEE 2019]
determined by vectors a and b is 30. Then
(A) k̂  ĵ (B) k̂
the area of the parallelogram with
adjacent sides determined by vectors 1 1
   (C) k̂  ĵ (D) k̂  ĵ
 
a  b and a is
2 2
93
   40. A variable plane at a distance of 1 unit from the
34. Let a  2î  ˆj  k̂ and b  î  2 ĵ  k̂ be two
origin cut the coordinate axis at A, B & C. If
vectors. Consider a vector centroid of triangle ABC is D(x, y, z) satisfy the
  
c   a   b,  ,   R . 1

1

1
 k , then value of k is -
relation
   x 2
y 2
z2

If the projection of c on the vector a  b is  [JEE 2005]
3 2, then the minimum v alue of (A) 3 (B) 1 (C) 1/3 (D 9

c  ( a  b ) · c equals [JEE 2019] 41. Find the equation of the plane containing the
line 2x – y + z – 3 = 0, 3x + y + z = 5 and at a
1
3- DIMENSIONAL GEOMETRY distance of from the point (2, 1, –1)
6
35. (A) Find the equation of the plane passing [JEE 2005]
through the points (2,1,0),(5,0,1) and 42. A plane passes through (1, –2, 1) and is per-
(4,1,1) pendicular to two planes 2x – 2y + z = 0 and
x – y + 2z = 4. The distance of the plane from
(B) If P is the point (2, 1, 6) then find the point
the point (1, 2, 3) is - [JEE 2006]
Q such that PQ is perpendicular to the
(A) 0 (B) 1 (C) 2 (D) 2 2
plane in (A) and the mid point of PQ lies
on it. [JEE 2003] 43. Match the following [JEE 2006]
x 1 y 1 z 1
 
Column-I
36. If the lines and
2 3 4 (A) Two rays in the first quadrant x + y = |a|
x3 yk z and ax – y = 1 intersects each other in the
  are intersecting each other
1 2 1 interval a  (a0, ), the value of a0 is
then ‘k’ is - [JEE 2004 ]
2 9 3 (B) Point () lies on the plane x + y + z
(A) (B) (C) 1 (D)  
9 2 2 ˆ kˆ  (kˆ  a)  0 ,then =
=2.Let a  ˆi  ˆj  k,

37. T is a parallelopiped in which A, B, C and D are 1 0

2
(C)  (1  y )dy   (y2  1)dy
vertices of one face. And the face just above it
0 1
has corresponding vertices A' , B', C', D'. T is
now compressed to S with face ABCD remain- (D) If sinAsinBsinC + cosAcosB = 1, then the
value of sinC =
ing same and A', B', C', D' shifted to A'', B'', C''
Column-II
and D'' in S. The volume of S is reduced to 90%
of T. Prove that locus of A'' is a plane. (p) 2
[JEE 2004] (q) 4/3
1 0
38. A plane is parallel to two lines whose direction (r)  1  x dx   1  x dx
ratios (1, 0, –1) & (–1, 1, 0) and it contains the 0 1

point (1, 1, 1). If it cuts the coordinate axes at (s) 1

A, B, C. then find the volume of tetrahedron 44. Match the following [JEE 2006]
OABC, where O is the origin. [JEE 2004]
Column-I

39. P1 and P2 are planes passing through origin. L1  1 
and L2 are two line on P1 and P2 respectively (A)  tan 1  2i2   t , then tant =
i 1
such that their intersection is origin. Show that
there exists points A, B, C, whose permutation (B) Sides a, b, c of a triangle ABC are in A.P.
A', B', C' can be chosen such that (i)
a b c
A is on L1, B on P1 but not on L1 and C not on and cos 1  , cos 2  , cos 3  ,
b c a c a b
P1 (ii) A' is on L2, B' on P2 but not on L2 and
C' not on P2. [JEE 2004] 2 1 
then tan  tan 2 3 =
2 2
94
 (C) A line is perpendicular to x + 2y + 2z = 0 47. Consider three planes [JEE 2008]
and passes through (0, 1, 0).The P1 : x – y + z = 1
perpendicular distance of this line from the P2 : x + y – z = –1
origin is P3 : x – 3y + 3z = 2
Column-II Let L1, L2, L3 be the lines of intersection of the
(p) 0 planes P2 and P3, P3 and P1, and P1 and P2,
respectively.
(q) 1 Statement-I : At least two of the lines L1, L2 and
L3 are non-parallel.
5 Statement-II : The three planes do not have a
(r)
3 common point.
(s) 2/3 (A) Statement-I is True, Statement-II is True ;
Statement-II is a correct explanation for
45. Consider the following linear equations
Statement-I.
ax + by + cz = 0; bx + cy + az = 0; cx + (B) Statement-I is True, Statement-II is True ;
ay + bz = 0 Statement-II is NOT a correct explanation for
[JEE 2007]
Statement-I.
Column-I (C) Statement-I is True, Statement-II is False.
2 2 2
(A) a + b + c  0 and a + b + c (D) Statement-I is False, Statement-II is True.
2 2 2
B) a + b + c  0 and a + b + c 
2 2 2 Paragraph for Question 48 to 50
(C) a + b + c  0 and a + b + c 
(D) a + b +
2 2 2
c = 0 and a + b + c = x 1 y  2 z 1
Consider the lines L1 :   , L2
Column-II 3 1 2
x2 y2 z3
(p) the equations represent planes meeting :  
1 2 3
only at = ab + bc + ca a single point 48. The unit vector perpendicular to both L1 and L2
(q) the equations represent the line x = y = is :- [JEE 2008]
zab + bc + ca
ˆi  7 ˆj  7kˆ ˆi  7 ˆj  5 kˆ
(r) the equations represent identical planes. (A) (B)
99 5 3
ab + bc + ca
ˆi  7 ˆj  5 kˆ 7 ˆi  7 ˆj  kˆ
(s) the equations represent the whole of the (C) (D)
5 3 99
three ab + bc + ca dimensional space
49. The shortest distance between L1 and L2 is :-
46. Consider the planes 3x – 6y – 2z = 15 and [JEE 2008]
2x + y – 2z = 5. [JEE 2007] 17 41 17
(A) 0 (B) (C) (D)
Statement-I : The parametric equations of the 3 5 3 5 3
line of intersection of the given planes are 50. The distance of the point (1, 1, 1) from the plane
x = 3 + 14t, y = 1 + 2t, z = 15t.
passing through the point (–1, –2, –1) and whose
Statement-II : The vector 14 ˆi  2 ˆj  15 kˆ is normal is perpendicular to both the lines L1 and
parallel to the line of intersection of given planes. L2 is :- [JEE 2008]
(A) Statement-I is True, Statement-II is True ; 2 7 13 23
Statement-II is a correct explanation for (A) (B) (C) (D)
75 75 75 75
Statement-I.
51. A line with positive direction cosines passes
(B) Statement-I is True, Statement-II is True ; through the point P (2, –1, 2) and makes equal
Statement-II is NOT a correct explanation for angles with the coordinate axes. The line meets
Statement-I.
the plane 2x + y + z = 9 at point Q. The length
(C) Statement-I is True, Statement-II is False.
of the line segment PQ equals [JEE 2009]
(D) Statement-I is False, Statement-II is True.
(A) 1 (B) 2 (C) 3 (D)2

95
52. Let P(3, 2, 6) be a point in space and Q be a 55. If the distance of the point P(1,–2,1) from the
point on the line plane x + 2y – 2z = , where  > 0, is 5, then
 ˆ ˆ ˆ  µ(3 ˆi  ˆj  5k)
ˆ . Then the value the foot of the perpendicular from P to the plane
r  (i  j  2k)
is- [JEE 2010]

of µ for which the vector PQ is parallel to the 8 4 7  4 4 1 
(A)  , ,   (B)  ,  , 
plane x – 4y + 3z = 1 is :- [JEE 2009] 3 3 3  3 3 3
 1 2 10  2 1 5 
1 1 1 1 (C)  , ,  (D)  ,  , 
(A) (B) – (C) (D) – 3 3 3  3 3 2 
4 4 8 8
56. If the distance between the plane Ax – 2y + z =
d and the plane containing the ines
53. Match the statements/ expressions given in Col-
umn I with the values given in Column II x 1 y  2 z  3 x 2 y 3 z4
  and 3  4  5 is 6 ,
[JEE 2009] 2 3 4

Column-I then |d| is [JEE 2010]

(A) The number of solutions of the equation 57. Match the statements in Column-I with the
values in Column-II. [JEE 2010]
 
xesinx – cosx = 0 in the interval  0, 
 2
Column-II
(B) Value(s) of k for which the planes kx + 4y
+ z = 0, 4x + ky + 2z = 0 and 2x + 2y + z = (A) A line from the origin meets the lines
0 intersect in a straight line
8
x  2 y 1 z  1 x
(C) Value(s) of k for which x  1  x  2 +   and 3  y  3  z  1 at
1 2 1 2 1 1

x  1 + x  2 = 4khas integer solution(s) P and respectively. If length PQ = d, then

d2 is
(D) If y' = y +1 and y(0) =1, then value(s) of y
(ln 2) (B) The values of x satisfying
Column-II 3 
tan 1 (x  3 )  tan 1 (x  3 )  sin 1   are
(p) 1 5 
 
(q) 2 (C) Non-zero vectors a, b and c satisfy
        
(r) 3 a.b  0 (b  a).(b  c)  0 and 2  b  c    b  a  .
  
(s) 4 If a  b  4c, then the possible values of
(t) 5  are
(D) Let f be the function on [–,] given by
54. Equation of the plane containing the straight 9x  x
x y z f(0)= 9 and f(x) = sin   sin   for
line   and perpendicular to the plane  2  2 
2 3 4
x  0.
x y z
containing the straight lines   and 2 
3 4 2 The value of  f(x)dx is
 
x y z Column-I
  is [JEE 2010]
4 2 3 (p) –4
(q) 0
(A) x + 2y – 2z = 0 (r) 4
(B) 3x + 2y – 2z = 0 (s) 5
(C) x – 2y + z = 0 (t) 6
(D) 5x + 2y – 4z = 0

96
58. (a) The point P is the intersection of the y z
61. Two lines L1 : x  5,  and
straight line joining the points Q(2,3,5) and 3   2
R(1,–1,4) with the plane 5x – 4y – z = 1. y z
L 2 : x  ,  are coplanar. Then  can
If S is the foot of the perpendicular drawn 1 2  
from the point T(2,1,4) to QR, then the take value(s) [JEE-Ad. 2013]
length of the line segment PS is - (A) 1 (B) 2 (C) 3 (D) 4
1 62. Consider the lines
(A) (B) 2 (C) 2 (D) 2 2
2
x 1 y z3 x4 y3 z3
L1 :   , L2 :   and the
(B) The equation of a plane passing through 2 1 1 1 1 2
the line of intersection of the planes x + 2y
planes P1:7x+y + 2z = 3, P2 : 3x + 5y – 6z
+ 3z = 2 and x – y + z = 3 and at a distance
= 4. Let ax + by + cz = d be the equation of
2
from the point (3, 1, –1) is the plane passing through the point of
3
intersection of lines L1 and L2 and perpendicular
(A) 5x – 11y + z = 17
to planes P1 and P2.
(B) 2x  y  3 2 1
Match List-I with List-II and select the correct
(C) x + y + z = 3 answer using the code given below the lists.
(D) x  2 y  1  2 [JEE-Ad. 2013]
x 1 y 1 z List-I List-II
(C) If the straight lines   and
2 k 2
P. a = 1.13
x 1 y 1 z
  are coplanar, then the plane(s) Q. b = 2.–3
5 2 k
R. c = 3.1
containing these two lines is(are)
[JEE 2012] S. d = 4.–2
(A) y + 2z = –1 (B) y + z = –1 Codes
(C) y – z = –1 (D) y – 2z = –1 P Q R S
59. Perpendiculars are drawn from points on the (A) 3 2 4 1
x  2 y 1 z (B) 1 3 4 2
line   to the plane x + y + z
2 1 3 (C) 3 2 1 4
= 3. The feet of perpendiculars lie on the line
[JEE-Ad. 2013] (D) 2 4 1 3
63. From a point P(, , ), perpendicular PQ and
x y 1 z  2 x y 1 z  2
(A)   (B)   PR are drawn respectively on the lines y = x,
5 8 13 2 3 5
z = 1 and y = –x, z = –1. If P is such that 
x y 1 z  2 x y 1 z  2
(C)   (D)   QPR is a right angle, then the possible value(s)
4 3 7 2 7 5
of  is (are) [JEE-Ad. 2014]
60. A line  passing through the origin is
(A) 2 (B) 1 (C) –1 (D)  2
perpendicular to the lines
 
ˆ   t   64. Let x, y and z be three vectors each of magni-
1 :  3  t  ˆi   1  2 t  ˆj   4  2 t  k,
tude 2 and the angle between each pair of
ˆ   s  
 2 :  3  2s  ˆi   3  2 s  ˆj   2  s  k,
 
Then , the coordinate(s) of the point(s) on 2 them is . If a is a nonzero vector perpendicu-
3
at a distance of 17 f rom the point of 
  
intersection of  and 1 is(are) -[JEE-Ad. 2013] lar to x and y  z and b is a non zero vector
  
7 7 5  perpendicular to y and z  x , then
(A)  , ,  (B) (–1,–1,0)
3 3 3  , [JEE-Ad. 2014]
       
7 7 8 
(D)  , , 
 
(A) b  b.z  z  x  (B) a   a.y   y  z 
(C) (1,1,1)
9 9 9        
 
(C) a.b    a.y  b.z (D) a   a.y   z  y 

97
65.
  
Let a, b and c be three non-coplanar unit vec-
   
67. Let PAR be a triangle. Let a = QR , b = PR ,
tors such that the angle between every pair of    
and c = PQ . If | a | = 12, | b | = 4 3 , c = 24,
       
them is . If a  b  b c  pa  qb  rc , where p, then which of the following is (are) true ?
3
q and r are scalars, then the v alue of [JEE-Ad. 2015]
2
p  2q  r 2 2 
is [JEE-Ad. 2014] | c |2 
(A)  | a | = 12
q2 2

66. Match the following : [JEE-Ad. 2014] | c |2 
(B)  | a | = 30
2 
List - I   
(C) | a  b  c  a | 48 3
(P) Let y(x) = cos (3cos –1x), x  [–1, 1],  
(D) a . b = –72
3
x   . Then
2 68. In R consider the planes P1 : y = 0 and P2 : x +
z = 1, Let P3 be a plane, different from P1 and
1  2 d2 y  x  dy  x  
 x  1 2
x  equal P2, which passes through the intersection of P1
y  x   dx dx 
and P2. If the distance of the point (0, 1, 0) from
(Q) Let A1, A2, ........, An (n > 2) be the vertices P3 is 1 and the distance of a point (, , ) from
of a regular P3 is 2, then which of the following relations is
(are) true ? [JEE-Ad. 2015]
polygon of n sides with its centre at the ori-
(A) 2 +  + 2 + 2 = 0 (B ) 2 – 
gin. Let ak be
+ 2 + 4 = 0 (C) 2 + 
the position vector of the point Ak, k = 1, 2,
– 2 – 10 = 0 (D) 2 – 
.....n. If
+ 2 – 8 = 07
 a    a .a  ,
n 1 n 1
k 1 k  a k 1  k 1 k k 1
69. In R3, let L be a straight line passing through the
then the minimum value of n is origin. Suppose that all the points on L are at a
(R) If the normal from the point P(h, 1) on the constant distance from the two planes P1 : x +
x 2 y2 2y – z + 1 = 0 and P2 : 2x – y + z – 1 = 0. Let M
ellipse   1 is perpendicular to the
6 3
be the locus of the feet of the perpendiculars
line x + y = 8, then the value of h is
(S) Number of positive solutions satisfying the drawn from the points on L to the plane P1. Which
equation of the following points lie(s) on M

 1  1  JEE-Ad. 2015]
1  1  2 
tan 1    tan    tan  2 
 2x  1   4x  1  x   5 2  1 1 1
(A)  0,  ,   (B)   ,  , 
is  6 3  6 3 6
List - II  5 1  1 2
(C)   , 0,  (D)   , 0, 
 6 6  3 3
(1) 1
(2) 2 70. Match the following [JEE-Ad. 2015]
(3) 8 Column - I
(4) 9 (A) In R2, if the magnitude of the projection
Codes : vector of the vector ˆi  ˆj on 3iˆ  ˆj is
P Q R S 3 and if a = 2 + 3 , then possible
(A) 4 3 2 1 value(s) of || is (are)
(B) 2 4 3 1 (B) Let a and b be real numbers such that the
(C) 4 3 1 2 function
(D) 2 4 1 3
2
3ax  2, x 1
f(x) =  2
bx  a , x 1

98
  
(C) Let   1 be a complex cube root of unity.. 72. Suppose that p, q and r are three non- copla-
If (3 – 3 + 22)4 + 3 (2 + 2 – 32)4 + 3 + (– nar vectors in R3. Let the components of a vec-
  
3 + 2 + 32)4 + 3 = 0, then possible tor s along p, q and r be 4, 3 and 5, respec-
value(s) of n is (are) 
tively. If the components of this vector s along
(D) Let the harmonic mean of two positive real  
( pq + r ), ( p  q + r ) and ( p  q + r ) are x, y
   
numbers a and b be 4.
and z respectively, then the value of 2x + y + z
If q is positive real number such that a, 5,
is [JEE-Ad. 2015]
q, b is an arighmetic number such that a,
5, q, b is an arithmetic progression, then 73. Consider a pyramid OPQRS located in the first
the value(s) of |q – a| is (are) octant (x  0, y  0, z  0) with O as origin,
Column - II and OP and OR along the x-axis and the y-
axis, respectively. The base OPQR of the pyra-
(p) 1
mid is a square with OP = 3. The point S is
(q) 2
directly above the midpoint T of diagonal OQ
(r) 3
such that TS = 3. Then [JEE-Ad. 2016]
(s) 4
(t) 5 
(A) the acute angle between OQ andOS is
3
71. Column I [JEE-Ad. 2015] (B) the equation of the plane containing the tri-
angle OQS is x – y = 0
(A) In a triangle XYZ, let a, b and c be the
(C) the length of the perpendicular from P to the
lengths of the sides opposite to the angles
X, Y and Z, respectively. If 2(a2 – b2) = c2
3
plane containing the triangle OQS is
2
sin(X  Y)
and  = ,then possible values (D) the perpendicular distance from O to the
sin Z
of n for which cos(n) = 0 is/are (B) 15
straight line containing RS is
In a triangle XYZ, let a, b and c the lengths 2
of the sides opposite to the angles X, Y 74. Let P be the image of the point (3, 1, 7) with
and Z, respectively. respect to the plane x – y + z = 3. Then the
If 1 + cos2X – 2cos2Y = 2sinX sinY, then equation of the plane passing through P and
a x y z
possible values of is/are containing the straight line   is
b 1 2 1
[JEE-Ad. 2016]
(C) In R2, let 3 iˆ  ˆj , ˆi  3 ˆj and ˆi  (1  )jˆ (A) x + y – 3z = 0 (B) 3x + z = 0
be the position vectors of X, Y and Z with (C) x – 4y + 7z = 0 (D) 2x – y = 0
respect to the origin O, respectively.
If the distance of Z from the bisector of the
75. Let û = u1 î + u2 ĵ + u3 k̂ be a unit vector in R3
  3
actue angle OX with OY is then pos
2 1 ˆ ˆ ˆ . Given that there exists
ˆ
and  (i  j  2k)
sible values of || is/are 6

(D) Suppose that F() denotes the area of the a vector v in R3 such that | uˆ   | = 1 and
region bounded by x = 0, x = 2, y2 = 4x and 
ˆ | uˆ   | = 1. W hi ch of the f ollowing
y = |x – 1| + |x – 2| + x, where  {0,
statement(s) is/are correct ? [JEE-Ad. 2016]
1}. Then the values of F() + 
(A) There is exactly one choice for such v
Column II
(B) There are infinitely many choices for such
(p) 1 
(q) 2 v
(r) 3 (C) If û lies in the xy-plane then |u1| = |u2|
(s) 5 (D) If û lies in the xz-plane then 2|u1| = |u3|
(t) 6

99
76. The equation of the plane passing through the  
point (1, 1, 1) and perpendicular to the planes 80. Let a and b be two unit vectors such that
2x + y – 2z = 5 and 3x – 6y – 2z = 7, is  
[JEE-Ad. 2017]
a · b = 0. For some x, y  R, let
(A) – 14x + 2y + 15z = 3      
c  xa  yb  (a  b) .If | c | = 2 and the vector
(B) 14x + 2y – 15z = 1 
(C) 14x + 2y + 15z = 31 c is inclined at the same angle  to both
(D) 14x – 2y + 15z = 27  
a and b , then the v alue of
77. Let O be the origin and let PQR be an arbitrary
8cos2 is _______. [JEE-Ad. 2018]
triangle. The point S is such that
3.00
OP · OQ  OR · OS = OR · OP  OQ · OS
81. Consider the cube in the first octant with sides
= OQ · OR  OP · OS . [JEE-Ad. 2017] OP, OQ and OR of length 1, along the x-axis,
y-axis and z-axis, respectively, where O(0, 0,
Then the triangle PQR has S as its
(A) circumcentre (B) orthocenter 1 1 1
(C) incentre (D) centroid 0) is the origin. Let S , ,  be the centre
2 2 2
Paragraph for question 74 & 75 of the cube and T be the vertex of the cube
opposite to the origin O such that S lies on
Let O be the origin, and OX , OY , OZ be   
three unit vectors in the directions of the sides the diagonal OT. If p  SP , q  SQ , r  SR
    
QR , RP , PQ , respectively, of a triangle PQR. and t  ST , then the value of ( p  q )  ( r  t )
[JEE-Ad. 2017] is______. [JEE-Ad. 2018]

78. OX OY =
(A) sin (P + R) (B) sin 2R
(C) sin (Q + R) (D) sin (P + Q)
79. If the triangle PQR varies, then the minimum
value of cos (P + Q) + cos (Q + R) + cos (R + P)
is
3 5 3 5
(A) B) (C) (D)
2 3 2 3

100
 ANSWER KEY

  
1. A. B B. A C. A 3. A.B B. C 5. v1  2ˆi , v2  ˆi  j , v3  3 ˆi  2 ˆj  4 kˆ

12. A. B B. C 9. D 11. A. C B. A 13. B

14.  = v –2 (â. v )â
w 15. A. A B. B,D 16. C 17. B 18. C

19. A 20. A 21. C 22. A q, s ; B  p, r, s, t ; C  T, D  R ;

23. A 24. 5 25. B 26. A. C B. A, D C. 9

27. A. 3 B. C 28. C 29. 5 30. C

31. BCD 32. 0.75 33. CD 34. 18.00

9
35. A. x + y – 2z = 3 B. 6, 5, –2 36. B 37. cubic unit 38. D
2

39. 2x –y + z – 3 = 0, 62x + 29y + 19z – 105 = 0 40. D

43. A  s ; B  p ; C  q, r ; D s ; 44. A q ; B s ; C  r ; 45. A  r ; B  q ; C  p ;D  s

46. D 47. D 48. B 49. D 50. C

51. C 52. A 53. A.  P ; B  Q, S ; C  Q, R, S, T ; D  R

54. C 55. A 56. 6 57. A  t ; B  p,r ; C  q,s ; D  r ;

58. A. A B. A C. B,C 59. D 60. B, D

61. A,D 62. A 63. C 64. ABC 65. 4

66. A 67. ACD 68. BD 69. AB

70. A  p, q ; B  p, q ; C  p, q, s, t ; D  q, t ; 71. A  p, r, s ; B  p ; C  p, q ; D  s, t ;

72. 9; 73. BCD 74. A 75. B, C

76. C 77. C 78. D 79. A 80. 3.00

81. 0.50

101