T E AC H I N G N OT E S

VECTORS
1st Lecture
1. General Definitions :
In the language of mathematics any directed line segment is called a vector. Physical
quantities having both magnitude and direction and which can be manupulated (i.e.
added or subtracted) by certain specific rules of vector addition and subtraction are
called vector Physical Quantities. Those which for their complete specification require
only magnitude along with an appropriate unit are called scalars.
ZERO 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 .
Note that zero vector has many properties similar to the number zero.
A boy throwing a ball up and catching it back in his hand, the displacement of the ball
is a null vector.

UNIT VECTOR : a vector of unit magnitude in the direction of vector a is called unit

 a
vector along a and is denoted by â symbolically â   . (dimensionless)
a
The concept of unit vector is merely to impart a direction to a physical quantity.
e.g. A rigid body rotates about an axis through the origin with an angular velocity

10 3 m/s. If  points in the direction ( î  ˆj  k̂ ), then
 (î  ĵ  k̂ )
  10 3 10( î  ˆj  k̂ ) .
3

Note that all rotational vectors are termed as Axial vectors.

Example:
If the sum of two unit vectors is a unit vector then find the
magnitude of their difference and the angle between â and b̂
 2 
[Ans:  3 ,  ]
 3 
EQUAL VECTORS : two vectors are said to be equal if they have the same magnitude,
same direction & represent the same physical quantity.
In fact vectors can be classified as (i) Free vectors and (ii) Localised vectors
(i) All such vectors are those which when transformed into space from one point to
another point without affecting their magnitude and direction, can be considered as
equal. i.e. the physical effects produced by them remains unaltered. e.g. displacement,
velocity
(ii) Localised vector : e.g. force , different physical effect if line of application is changed.
Note that number of distinct unit vectors in space perpendicular to a given plane is 2. (one
upward and one downward).
How many Perpendicular to given line in space? (Infinitely many, think of the line as
perpendicular to the xy plane. The unit vector might make any angle  with the x-axis.)
Are two vectors are equal if they have equal components in a given direction (not
necessarily)
In two given directions (non necessarily)
In three given directions (non necessarily)
In an arbitrary directions (yes)
Bansal Classes [1]
 COLLINEAR VECTORS : two vectors are said to be collinear if their directed line
segments are parallel disregards to their direction. Collinear vectors are also called
Parallel Vectors. If they have the same direction they are named as like vectors if
  
otherwise unlike vectors. (a , b , c are collinear )
   
Symbolically, two non zero vectors a and b are collinear if and only if, a  Kb ,
where K  R.
If K > 0 , like parallel vectors.
K < 0 , unlike parallel vectors
   
If a and b are non zero non collinear vectors then xa  yb  0  x = 0 and y = 0. Two
non zero non collinear vectors are also called as linearly independent vector or base
 
vectors. If a and b are collinear then they are known as linearly dependent vectors.
COPLANAR VECTORS : A given number of vectors are called coplanar if their line
segments are all parallel to the same plane. Note that “TWO VECTORS ARE ALWAYS IN THE
SAME PLANE”.

(Note that a plane is a surface such that a line segment joining any two points on the
    
surface lies wholly on it.) If a , b , c , d , .... are coplanar and n is a vector normal to
     
the plane then a . n  b . n  0  c . n .

POSITION VECTOR : To specify the position of an object in

3-D space use is made of P.V.'s. Let O be a fixed origin, then the

position vector of a point P is the vector OP . Hence p.v. of
general point P (x, y, z) is the vector extending from the origin to

P given by x î  y ĵ  z k̂ . If a & b are the position vectors of two
  
point A and B, then, AB = b  a = pv of B  pv of A.

Representation of a vector in space in terms of 3 orthonormal triad of unit vectors

With every point (x1, y1 ,z1) in space w.r.t. a fixed origin 'O' we can associate a directed
line segment whose initial point is the origin and terminal point is (x1,y1,z1). Thus
   
OP  OA  AN  NP = x1 î  y 1 j  z1 k̂
ˆ

Thus if A(x1 y1 z1) and B(x2 y2 z2) are two points in space

AB  ( x 2  x1 ) î  ( y  y1 )ˆj  (z 2  z1 ) k̂
2

; | a |2  a12  a 22  a 32
    
a  a1 î  a 2 ĵ  a 3 k̂

Thus any vector a can be expressed as a linear
combination of 3 orthonormal triad of unit vectors.

2. SECTION FORMULA :
 
If a & b are the position vectors of two points A & B then the p.v. of a point R which
divides AB in the ratio m : n is given by :
 
 m 2a  m1b  
r a b
. Note p.v. of mid point of AB = .
m1  m 2 2

Bansal Classes [2]

 Using section formula we can prove that   
   
(a) p.v. of the centroid of a triangle ABC =
3
(Concurrency of median)
  
a   b  c 
(b) Incentre of the  = and
abc
        
 a   b  c  a   b  c  a   b  c 
Excentres of the  are abc
; abc and a bc
  
(c) Circumcentre of the  =  sin 2A   sin 2B   sin 2C
(use loving triangles)  sin 2A
  
 tan A   tan B   tan C
(d) Orthocenter of the  =
 tan A
(Use the fact that distances of orthocentre from the vertices are 2R cosA , 2R cosB ,
2R cosC and from the sides are 2 R cosB cosC , 2R cosC cosA , 2R cosA cosB)
 
b cos C b  c cos B c
p.v. of D are
a
 
2R sin B cos C b  2R sin C cos B c
=
2R sin A
 
 (sin B cos C) b  (sin C cos B) c
d =
sin A
    
 qa  pd sin A ( 2 R cos B cos C) a  ( 2R cos A ) ·[ (sin B cos C) b  (sin C cos B) c ]
h = = sin A · 2R (cos A  cos B cos C)  sin A sin B sin C
pq
     
(cos B cos C cos A ) [ a tan A  b tan B  c tan C ] a tan A  b tan B  c tan C
= =
 sin A  tan A
Geometrical results with vectors:
(a) Straight line joining the mid points of two non parallel
sides of a trapezium is | | to the parallel sides and half
of their sum.
(b) Tetrahedron (a pyramid on a triangular base)
(b1) Lines joining the vertices of a tetrahedron to the centroids of
the opposite faces are concurrecnt and this point (P)of
   
 a  b c d
concurrency with p.v. g 
4
is called the centre of the tetrahedron (say G).
(b2) In a tetrahedron, straight lines joining the mid points of each pair of opposite edges are
also concurrent at the centre of the tetrahedron.
(c) Four diagonals of any parallelopiped (A prism
whose base is a | |gm) and the join of the mid point
of each pair of opposite edges are concurrent and
are bisected at the point of concurrence.
(See the adjacent figure) P is called the
centre of the parallopiped with p.v.
     
a  b c OA  OB  OC
i.e.
2 2

Bansal Classes [3]

Example:
Centre of the parallelopiped formed by
  
PA  î  2 ĵ  2 k̂ ; PB  4î  3ˆj  k̂ and PC  3î  5ˆj  k̂
is given by the p.v. (7, 6, 2). Then find the p.v. of the point P.
[Ans. 4î  2ˆj  k̂ ]
(d) The diagonals of the three faces of the parallopiped drawn from the same vertex are
prolonged half their lengths. Show that the three points thus obtained are coplanar
with the opposite vertex.

3. VECTOR ADDITION :
    
If two vectors a & b are represented by OA &OB , then their sum a  b is a vector

represented by OC , where OC is the diagonal of the parallelogram OACB.
         
a  b  b  a (commutative) (a  b)  c  a  ( b  c) (associativity)
         
a  0a 0  a a  ( a )  0  ( a ) a
4. 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 :
     
m ( a )  ( a )m  m a m ( n a )  n ( m a )  ( mn ) a
      
( m  n )a  m a  n a m (a  b )  m a  m b

Home work after 1st lecture : Shanti / Ex. 10.1 and 10.2 (NCERT)
2nd Lecture
6. VECTOR EQUATION OF A LINE :
It is possible to express the position vectors of points on given lines and planes in
terms of some fixed vectors and variable scalars called parameters, such that
(i) For arbitrary value of the parameter, the resulting position vector represent point on the
locus in question, and
(ii) Conversely, the position vectors of each point on the locus correspond to a definite
value(s) of the parameter.

Parametric vector equation of a line passing through

     
two point 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  t b
These two equations prove to be very useful in vector algebra.
Important Note :
(1) Two lines in a plane are either intersecting or parallel conversely two intersecting or
parallel lines must be in the same plane
(2) However in space we can have two neither parallel nor intersecting lines.
Such non coplanar lines are known as skew lines. If two lines are parallel and have a
common point then they are coincident.

Bansal Classes [4]

EXAMPLES :
(1) Find the p.v. of the point of intersection of the lines

(a) r  î  ˆj  10k̂   ( 2 î  3 ĵ  8k̂ )  intersecting lines.Coplanar (Ans:  = 2 and  =1)
and r  4 î  3 ĵ  k̂   ( î  4ˆj  7 k̂ ) 



(b) r   3î  6ˆj   (4î  3ˆj  2k̂ )  non intersecting lines
and r  2î  7 k̂   (4 î  ˆj  k̂ ) 


(c) 
r  t (3î  ˆj  k̂ ) 
and r  2î  s(6 î  2 ĵ  2k̂ )



r  2k̂  (3î  2ˆj  k̂ )  lines coincide. Lines are parallel and they
(d) and r  3î  2 ĵ  3k̂  (6î  4ˆj  2k̂ ) have a point in common. By comparing coefficient

[ Ans:  = 2 + 1 , where   R ]
(2) Through the middle point M of the side AD of | | gm
ABCD, a straight line BM is drawn intersecting AC at
R and CD produced at Q. use vectors to prove that
QR = 2RB.
(fig - 2)

(3) In AOB , E is the mid point of OB and F divides

OP 3
BA in the ratio 1 : 2 use vectors to prove that  .
PF 2

(fig - 3)
(4) In a triangle ABC, D divides BC in the ratio
3 : 2 and E divides CA in the ratio 1 : 3. The lines AD
and BE meet at H and CH meets AB in F.
Find the ratio in which F divides AB.
(Ans 2 : 1)
(fig. 4)
(5) In ABC , EF is drawn | | to BC with E on AB and F on AC. If BF and CE meet at L.
Prove that AL bisects BC.

Vector equation of the bisectors of the angles between the lines

   
 
r  a  b and r  a  c are
r  a  t ( b̂  ĉ ) and
   
r  a  s ( b̂  ĉ)
[Ans. One bisector is  
r  a  t (b̂  ĉ)
other bisector is r  a  s(b̂  ĉ) ]
Example:
Use vectors to prove that the internal (external) bisectors of a triangle divide the opposite
base internally (externally) in the ratio of the side containing the angle.

Note: We can make use of the vector equations to prove the concurrency of angle
bisectors by finding the point of intersection of two angle bisectors and then satisfy the
point of intersection in the third equation.
Home work after 2nd lecture : Shanti

Bansal Classes [5]

 3rd Lecture
7. TEST OF COLLINEARITY :
  
Three points A,B,C with position vectors a , b , c respectively are collinear,,
if and only if there exist scalars x , y , z not all zero simultaneously such that:
  
x a  y b  z c  0 , where x+ y+ z = 0.
[Proof : Explain

Example:
7(a) Find whether the following points are collinear or not
 
(i) 2î  5 ĵ  4k̂ ; î  4ˆj  3k̂ ; 4î  7ˆj  6k̂ (Collinear) [ 3 AB   BC ]
(ii) 3î  4ˆj  3k̂ ;  4î  5ˆj  6k̂ ; 4î  7ˆj  6k̂ (non- collinear)
Note: Collinearly can also be checked by first finding the equation of line through two
points and satisfying the third point.
7(b) ABC is a scalene triangle ; AD , AD' are the bisectors of the angle A meeting BC in D,
D' respectively ; A' is the mid point of DD' ; B' , C' are the points on CA and AB
similarly obtained. Show that the points A', B' , C' are collinear.
  
7(c) Vectors P , Q , act at 'O' (origin) have a resultant R . If any transversal cuts their line
P Q R
of action at A, B, C respectively, then show that   .
OA OB OC

Scalar Product (Useful Results)

     
(1) a . b  | a || b | cos  for two non zero vectors a & b if
   
(i) a .b  0   is acute and a   b
 
(ii) a .b  0   = /2
   
(iii) a .b  0   is obtuse and a   b
A point P moves in space such that P A & P B < 0 then
the locus of the point P is the interior of the
sphere with AB as diameter.
 
(2) Dot product is commutative i.e. a . b  b . a
Conventionally a . a  | a |2  a 2
   

î . î  ˆj. ˆj  k̂. k̂ 1 ; î . ˆj  ˆj. k̂  k̂. î  0

(3) Dot product is distributive

    
a . ( b  c )  a.b  a.c (refer text book for proof)
Simple identities to rembember are ,
       
(i) (a  b) . (a  b)  a 2  b2 (ii) (a  b) 2  a 2  2 a . b  b 2 = (a  b) . (a  b)
     
(iii) (a  b) 2  a 2  2 a . b  b 2 (iv) (a  b) 2  (a  b) 2  4 a . b
    1  
  
(v) (a  b  c 2 ) 2  a 2  b 2  c 2  2 ( a . b ) (vi) a . b  (a  b) 2  (a  b) 2 
    
4
  
Asking: (1) | a |  11 ; | b |  23 and | a  b |  30 , find | a  b | [ Ans: 20]
     
(2) If a  b  c  0 , | a |  3 ; | b | 1 and | c | 4 . Find  a . b   [Ans: –13]

Bansal Classes [6]

(4) General Expression for dot product
 
If a  a1î  a 2 ĵ  a 3k̂ & b  b1î  b 2 ĵ  b3k̂ then
 
  a .b a1b1  a 2 b 2  a 3b 3
a . b  a1b1  a 2b 2  a 3b3 & cos     =
| a || b | a12  a 22  a 32 . b12  b 22  b32
Compute d.c’s of the vector.
 
  a .b
(5) Projection of a on b =  . It can be +ve, – ve or zero.
|b|
  
  a .b 
Note that vector component of a along b =   2  b and vector component of

 b 
 
    a .b  
a perpendicular to b = a   b 2  b
 
 
Asking :
 
 Pr ojection of a on b
(1) If a  2î  3ˆj  6k̂ and b   2î  5ˆj  14k̂ . If  =
   . Find .
Pr ojection of b on a

|a| 7
[ Ans:  = | b |  15 ]
(2) Express the vector a  5î  2ˆj  5k̂ as the sum of two vectors such that one is parallel to
 
b  3î  k̂ and the other perpendicular to b . [Ans.  = 2]

General note :
   
(i) Maximum value of a . b  | a | | b | (ii) Minimum value of a . b   | a | | b |

(iii) Any vector a can be expressed as (a . î ) î  (a.ˆj) ˆj  (a.k̂ ) k̂
Home Work : Exercise 10.3 on dot product, NCERT for XII.
Illustrations on dot product
(1) Cosine formula for triangles
(2) Projection formula for triangles
(3) cos( + ) = cos cos  sin sin
(4) Use vectors to prove that a quadrilateral whose diagonals bisect each other at right
angle is a rhombous.
(5) Angle in a semicircle is a right angle.
(6) Acute angle between the diagonals of a cube. (Objective) [Ans: cos–1(1/3) ]
(7) Medians to the equal sides of an isosceles triangle are equal and converse.
(8) Concurrency of altitudes / right bisectors / medians / angle bisectors of a triangle
( By making use of vector equations or otherwise)
(9) If two circles intersect then the line joining their centres is
perpendicular to the common chord.
[Start: r 2  a 2 , ( r  c ) 2  b 2 ]
(10) O/G/C of a scalene triangle are collinear G divides the line joining O and C in the ratio
of 2 : 1. [ Take position vector w.r.t. circumcentre]

Bansal Classes [7]

(11) Vector equation of a ellipse / Hyperbola whose foci are the position vectors
 
c and – c and length of major / Transverse axis (T.A.) axis is 2a is.
[ Ans: ( r . c)  a 4  a 2 ( r 2  c 2 ) ]
  

 
12(a) Perpendicular distance of a point A (a ) from a line passing through the point b and

parallel to c and reflection of a point in a line.

 (a  b) . c  

[Ans. | AN |  b   2 c  a ]
c

 (a  b) . c 
Note : The p.v. of point N is b   2 c ]
c
12(b) Find the foot of the perpendicular from the origin on the line r   î  4k̂   (2î  ˆj  k̂ ) .
Also find the p.v. of its image in the line. [Ans. (1, –1, 3) ; (2, –2, 6)]
12(c) A line passes through a point with p.v. î  2ˆj  k̂ and is parallel to the vector î  2ˆj  2k̂ .
Find the distance of a point P (5, 0, – 4) from the line. [Ans. 5 ; wrong ans 5 , 26 ]

12(d) Equation of the altitude of a triangle A and

orthocentre of the triangle (a1, a2, a3)

     
(13) Two vectors e1 and e2 with | e1 | = 2 and | e2 | = 1 and angle between e1 and e2 is 60°.
   
The angle between 2t e1 + 7 e2 and e1 + t e2 belongs to the interval (90°, 180°). Find
the range of t.
 1 
[Ans.   7,  2  –  3.5  ]
 
(14) A triangle OAB is right angled
at O ; squares OALM and
OBPQ are constructed on the
sides OA and OB externally.
Show that the lines AP and BL
intersect on the altitude through
‘O’.
  
i.e. T.P.T. OR . (a  b)  0 .
(15) ABCD is a tetrahedron such that the perpendiculars AK , BL , CM and DN to the
opposite faces are concurrent. Prove that
(i) any two opposite edges of the tetrahedron are orthoganal.
(ii) k is the orthocentre of the triangle BCD.

(16) Obtain the equation to the locus of a point R ( r ) in space which is equidistant from
  
(i) two given points with p.v.’s a and b (ii) three given points with p.v.’ss a , b and c .
And interpret it geometrically.
  1 
   
[Ans.  r  2 (a  b) . (a  b )  0 ]
 

Bansal Classes [8]

(17) Radius of the sphere circumscribing and inradius of a regular tetrahedron whose edge
 3k k 
is k. (Objective) [ Ans:  ; ]
 24 24 

(18) Use vectors to prove that in a ABC cos2A + cos2B + cos2C > –3/2.Also prove that
1
the distance between the circumcentre and the centroid is R 2  (a 2  b 2  c 2 )
9
Linear combination
   
A vector r is said to be a linear combination of the vectors a , b , c ......
   
if  scalars x,y,z, ......... such that r  xa  yb  zc  ..........
Theorem in plane
 
If a and b are two non zero non collinear vetors then any vector r coplanar with them
  
can be expressed as a linear combination r  xa  yb . (Explain using a sketch)
EXAMPLES :
(1) Arc AC of the quadrant of a circle with centre as origin and radius
unity subtends a right angle at the origin. Point B divides the arc

AC in the ratio 1 : 2. Express the vector c in terms of
 
a and b .

[Ans. c  3 a  2 b ]

(2) Given that a  î  ˆj and b̂  î  2 ĵ are two vectors. Find a unit vector coplanar with a and
 
b and perpendicular to a . î  ˆj
[Ans. ± ]
 2
 
a b c
  
   a.a a.b a.c
(3) If a , b , c are coplanar vectors, prove that    =0
b.a b.b b.c

(4) Write the laws of reflection of a light beam from a mirror in

vector form using the directing unit vectors ê1 and ê2 of the
incident and reflected beams and the unit vector n̂ of
the outside normal to the mirror surface. [Ans : ê 2  ê1  2(ê1 . n̂ ) n̂ ]
Vector product ( Cross Product)
   
(1) a  b  | a | | b | sin  n̂ where n̂ is the unit vector perpendicular

to the plane containing the vectors a and b such that

a and b and n̂ forms a right handed screw system.


e.g. v  w  r (Tangent velocity of a particle moving

 
in a circle is cross product of its w and r vector..
Example :

Equation of a line which passes through the point with p.v. a and perpendicular to the
    
lines r  b   p and r  c   q . [ Ans : r  a  t ( p  q) ]

Bansal Classes [9]

(2) Lagranges identity:
 
Since a  b is very frequently needed for which Lagranges identify very useful.
 
  2 2 2  2 a.a a.b  
| a  b |  a b  (a.b) = a.b b.b
This is very useful when only | a  b | is to be

computed.

Example : For any vector a , | a  î |2  | a  ˆj | 2  | a  k̂ |2  2a 2

(3) 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 and (iii) a , b, c form a right handed system
(4) Properties of cross product
      
(i) a  b  0  a  b (a  0 ; b  0) i.e. a and b are Collinear / Linearly dependent.

   
however if a  b = 0 and a · b = 0  a  0 or b  0
   
(ii) ab  ba ( not commutative)
      
(iii) a  (b  c)  a  b  a  c ( distributive to be proved later using triple product )
     
(iv) (a  b)  c  a  ( b  c ) vector product is not associative.
(v) î  î  ˆj  ˆj  k̂  k̂  0 and î  ˆj  k̂ ; ˆj  k̂  î ; k̂  î  ˆj
  
(5) Expression for a  b Where a  a1 î  a 2 ĵ  a 3 k̂  and b  b1 î  b 2 ĵ  b3 k̂

î ˆj k̂
  = a1 a 2
ab a3
b1 b 2 b3

or Find the equation of the line through the point with p.v. 2î  3ˆj and perpendicular to the
 
vectors A  î  2ˆj  3k̂ and B  3î  4ˆj  5k̂ .
e.g. If ( 2î  6ˆj  27 k̂ )  (î  ˆj  k̂ ) = 0, find  and . [Ans. 3, 27/2]
   
(6) Geometrical interpretation of | a  b | = | a | | b | sin  denotes the area

of parallelogram whose two adjacent sides are the vectors a & b .
 
Example : Area of a parallelogram / quad. if diagonal vectors d1 & d 2 are known
1    
A= d1  d 2 If d1  2î  3ˆj  6k̂ and d 2  3î  4 ĵ  k̂ .
2
 
|ab|
Note: Area of the triangle =
2
 
ab  
(7) Unit vector perpendicular to the plane of is    & a vector of magnitude
a&b
| ab|
 
 r ab
r perpendicular to the plane of a & b is =   
 
| ab|
 
  ab
(8) If  is the angle between a & b then sin =  
| a || b |
Home Work : Exercise 10.4, NCERT for XII.

Bansal Classes [10]

Examples on Vector Product & S.D.
  
(i) For a non zero vector a , if a .b = a .c and a  b = a  c . Prove that b  c .
 

(ii) Find
(i) A vector of magnitude 6 perpendicular to the plane ABC
(ii) Area of triangle ABC
(iii) Length of the altitude from A (AB = AC = 11 )
(iv) Equation of the plane ABC
[Ans. (i)  ( 2î  ĵ  k̂ ) ; (ii) 2 6 ; (iv) r · (2î  ˆj  6k̂ ) = 3]
sin A sin B sin C
(iii) (a) Use vectors to prove sine law [ i.e. = = ]
a b c
 
[Hint: a  b  c  0 , now take cross with a and b .]
(b) sin(+) = sin cos + cos sin
(c) Heros Formula   s(s  a ) (s  b) (s  c)
1    
= | a  b | or 2  | a  b | or 42  | a  b |2
2
     
 4 2  a 2 b 2  (a . b ) 2 = ( | a | | b |  (a . b) ) ( | a | | b |  (a . b) )
      
Put a . b by using ( a  b ) 2  ( c ) 2 or a 2  b 2  2a . b  c 2
 
(iv) Let a  î  4ˆj  2k̂ ; b  3î  2ˆj  7 k̂ and c  2î  ˆj  4k̂ . Find the vector d which is
 
perpendicular to both a and b and satisfy c · d = 15.
(v) Perpendicular distance of a point P from a line using cross product.
  
(a  b)  c
d  .
|c|

If (a  b)  c = 0  P lies on the line
Four vertices O, A, B, C of a tetrahedron satisfy OA  OB  î  ĵ  k̂ ; OB  OC  î ;
OC  OA  î  ĵ , find CA  CB .

(vi) Find the unknown vector R satisfying
        
R  B  C  B and R . A  0 where A  2î  k̂ ; B  î  ĵ  k̂ and C  4î  3ˆj  7 k̂
(9) Interpretation of vector product as vector area
(a) Vector area of plane figures
With every closed bound surface which has been described in a certain specific manner
and whose boundaries do not cross, it is possible to associate a directed line segment

c such that


(i) | c | = no. of units of area enclosed by the plane figure

(ii) The support of c is perpendicular to the area and

(iii) The sense of description of the boundaries and the direction of c is in
accordance with the R.H. screw rule.

Bansal Classes [11]

(b) Vector area of a plane  (Triangle)
 1  
Vector area of OAB is   (a  b)
2

If a , b , c are the position vectors then the vetor area of ABC is
 1

2
(c  b )  (a  b )
 1  

2
(a  b)  (b  c)  (c  a )
Note :
  
(1) If 3 points with position vectors a , b and c are collinear then a  b  b  c  c  a = 0
  
e.g. Prove that 3 points with position vectors a  b , a  b and a   b are collinear for    R .

(2) Unit vector perpendicular to the plane of the ABC when a, b, c are the p.v. of its
     
a b  bc  ca 
angular point is n̂   , where a , b, c are the position vectors of the
2
angular points of the triangle ABC.
EXAMPLES:
  
(i) Prove the identity (a  b )  (a  b)  2(a  b ) and give its geometrical interpretation.

Twice the vector area of a | | gm whose two adjacent sides are the vectors a and b is


equal to the area of the parallelogram whose two adjacent sides are the diagonal vectors
of the original parallelogram.
 
(ii) Prove that Area of a plane quadrilateral whose diagonals vectors are d1 and d 2 is
1  
d1  d 2 .
2
      
(iii) Let OA  a , OB  10 a  2 b and OC  b where O, A & C are noncollinear points. Let 'p'
denote the area of the quadrilateral OABC, and let 'q' denote the area of the parallelogram
with OA and OC as adjacent sides. If p = kq. Find k. [Ans. 6]
(10) Shortest distance between 2 skew lines
Note that
– 2 lines in a plane if not | | must intersect and 2 lines in a plane if not intersecting must be
parallel. Convertely 2 intersecting or partallel must be coplanar.
– In space, however we come across situation when two lines neither intersect nor | | ,
Two such lines (like the flight paths of two planes) in space are known as skew lines
or non coplanar lines. S.D. between two such skew lines is the segment intercepted
betweeen the two lines and perpendicular to both.
Method 1: Two ways to determine the S.D.
  
L1 : r  a  p

L2 : r  b  q
  
n pq
  
AB  (b  a )
     
 AB . n
 (b  a ) . (p  q)
S.D. = | Projection of AB on n | = | n | =  
| pq |

If S.D. = 0  lines are intersecting and hence coplanar..

Bansal Classes [12]

 10
Illustration: r  î  ˆj   ( 2î  ĵ  k̂ ) ; r  2î  ˆj  k̂  (3î  5ˆj  2k̂ ) [Ans: ]
59
H.W. Vector equation of two lines are
(a) r  (1  t )î  ( t  2) ˆj  (3  2 t )k̂ ; r  (s  1) î  (2s  1) ˆj  (2s  1)k̂ . Find SD.
 

3 2 8
(b) r  î  2 ˆj  k̂   (î  ˆj  k̂ ) ; r  2î  ˆj  k̂  ( 2î  ˆj  2k̂ ) . Find SD.
  [Ans: , ]
2 29
  
Method 2 : p.v. of N1 = a  p ; p.v. of N2 = b  q
  
N 1 N 2 = ( b  a )  ( q   p )
 

   
now N1N 2 . p = 0 and N1 N 2 . q = 0 (two linear equations to get the unique values of  and .)
One p.v’s of N1 and N2 are known we can also determine the equation to the line of
shortest distance and the S.D.
Shortest Distance between two parallel lines
  
 (a  b )  c
d = | a  b | sin 
|c|
e.g. Find the distance between the lines L1 and L2 given by

r  ( î  2ˆj  4k̂ )   ( 2î  3ˆj  6k̂ )

293
and 
r  (3î  3ˆj  5k̂ )   (2î  3ˆj  6k̂ ) [Ans. ]
7

Examples: For the tetrahedron as shown in the figure compute the following
(i) p.v. corresponding to the point of concurrency of the join of the mid points of each
   
 a bcd 
pair of opposite edges. 
[Ans:  ]
4 
 
(ii) p.v. of the foot N from the vertex A and the perpendicular distance of A from the face
1
BCD. [Ans. (9î  7ˆj  11k̂ ); h  6 ]
4
(iii) Image of A in plane face BCD.
(iv) Altitude of tetrahedron from the vertex A.

(v) Volume of tetrahedron [Ans. V = 6]

1
[ Area of the base × h ]
3

(vi) Unit vectors normal to the plane face ABC and ADC.
(vii) Acute angle between the planes ABC and ADC.
(viii) S.D. between the skew lines AD and BC and the angle between them.
(ix) A unit vector | | to the plane EFM and perpendicular to the vector î  ˆj  k̂

b
[Ans.  ]
|b|
(x) Equation of the plane through ABC.
Home Work: Misc. on chapter 10 (NCERT)

Bansal Classes [13]

Product of 3 or more vectors :

(1) When 3 vectors are involved with a dot or a cross between them, then 6 different
symbols are
     
(a .b) c ; (a .b) . c ; (a .b)  c (1 , 2, 3)
        
(a  b ) c ; ( a  b) . c ; ( a  b)  c (4 , 5 , 6)
First is scalar multiple of a vector ; 2, 3, 4 are meaningless
5 and 6 are scalar and vector triple product respectively.
(2) SCALAR TRIPLE PRODUCT :
Definition :
       
(a  b) . c = | a | | b | sin  n̂ . c = | a | | b | | c | sin  cos 

  
where   a ^ b ;   n ^ c

but | a | | b | sin  = area of | |gm OACB and | c | cos  = h


Hence (a  b) . c geometrically describes the volume
of the parallelopiped whose 3.

cotorminous edges are the vectors a , b and c
Note :
     
(a) (a  b) . c is also known as box product and is written [a b c]

If the vector c also lies in the plane of a , b then  ( angle between n̂ and c ) is 900 and
 
(b)
       
[â b c] =0. Hence for three vectors if [a b c] =0  a , b , c are in the same plane &
conversely.
(c) If â , b̂, ĉ are unit vectors s.t. their box product is unity i.e. [â b̂ ĉ ] = 1
  
 sin cos = 1.This is possible only if  = 900 and  = 00 i.e. a , b , c are mutually
perpendicular to each other and converse e. g. [î ˆj k̂ ] = 1.
Example
   
Ex.1 If u  2î  ˆj  k̂ ; v  î  ˆj  k̂ and w is a unit vector then the maximum value of [u v w ] is
  
[Ans. [u v w ] max = 14 ]
 
(3) General expression for [a b c ] : When a , b , c are expressed in terms of î , ˆj , k̂
î ˆj k̂ a1 a2 a3
[a b c ] = (a  b) . c = a 1 a2 a3

. (c1î  c 2 ĵ  c3k̂ ) = b1 b2 b3
    
b1 b2 b3 c1 c2 c3

Note : If 3 directed line segments are in the same plane then the value of this determinant
vanishes.This can be used to determine the equation of a plane through 3 non collinear
points.

(4) Properties of STP

(a) Scalar triple product of three vectors when two of them are collinear / linearly dependent

or equal is zero. (two rows identical  determinant is zero) e.g. [a b c ] = 0.
(b) If the cyclic order of vector retains then the value of the STP does not change i.e.
   
[a b c ] = [ b c a ] = [ c a b ] = – [ a c b ]

Bansal Classes [14]

(c) The position of dot and cross can be interchanged provided the cyclic order of the

vectors a , b , c remains undisturbed.
  
we have (a  b) . c = ( b  c ) . a = ( c  a ) . b ....(1)
 
Also (a  b ) . c = c . (a  b )
     
(b  c) . a = a . (b  c ) As dot is commutative.
     
(c  a ) . b = b . (c  a )
    
(d) [a  b c d] = [a c d ]  [b c d ] Let 
a  a1 î  a 2 ˆj  a 3k̂ and so on
Proof :

(e) For right handed system


[a b c ]  0

and for left handed system where a , b , c are non- coplanar

[a b c ]  0
Home Work :
Illustrations
(1) Find the value of  for which three points with p.v.’s A(1,0,3) ; B(–1,3,4) ; C(1,2,1)
and D (, 2,5) are in the same plane.Also find the equations of the plane through ABC.
[ 3 N.C. points ]
2(a) Find the value of p for which the vectors (p  1)î  3 ĵ  pk̂ ; pî  (p  1)ˆj  3k̂ and
 3î  pˆj  ( p  1) k̂ are linearly dependent/coplanar . [ p = 1]
    
2(b) Show that a  b  [î a b ] î  [ ĵ a b ] ĵ  [k̂ a b ] k̂
  
(3)(a) [a  b , b  c , c  a ]  2 [a b c ] . This identity can be geometrically interpreted as:
(Volume of a cuboid whose three coterminous edges are the face diagonals of the
cuboid is twice the volume of the cuboid, whose three coterminous edges are the
   
vectors a , b , c ). This is also conclusive that if a , b , c are coplanar then a  b , b  c ,
 
c  a are also coplanar..
 
(b) [ a  b b  c c  a ] is always zero.
         
(c) Show that the lines R  R 0  tA and R  R i  sB intersect if ( R 0  R i )·(A  B) = 0 i.e.
     
[R 0 A B] = [R i A B] .
    
Note: If a , b, c are non coplanar then a  b , b  c and c  a will also be non coplanar..
  
l . a l . b l . c
              
(4)(a) Prove that [l m n ] [ a b c] = m . a m.b m.c , where l , m
   , n & a , b, c are non coplanar vectors.
n.a n.b n.c
  
p.a q.a a
    
(b) Prove that : [a b c ] (p  q ) = p.b q.b b
p.c q.c c
[Proof(a):
[Proof(b)
(5)(a) Let P be a point not on the plane that passes through Q, R and S. Show that the

[a b c] 
distance d from P to the plane is d = | a  b | where a  Q R ; b  Q S and c  Q P

Bansal Classes [15]

  1  1    
(5)(b) Volume of a tetrahedron OABC = V  [a b c]  OA , OB , OC  where O is the origin.
6 6  
[Proof:

Very Important Note : If S1, S2, S3 and S4 are the areas of the four triangular faces of
the tetrahedron with volume V. If r is the radius of the sphere touching the four faces
1
then V = (S1  S2  S3  S4 ) r
3
(RHS denotes the volume of four smallers tetrahedrons each with hight r and area of
the bases on S1, S2, S3 and S4 )
(6) To express scalar triple product of three vectors in terms of any three non coplanar

vectors l , m and n
    
Let a  a1 l  a 2 m

 a 3n ; b  b1 l  b 2 m  b3n ; c  c1 l  c 2 m  c3n
    

a1 a 2 a3
then  b b2 b3   
[a b c ] = 1 [l m n ]
c1 c 2 c3
 
(7) To prove the distributive property of vector product i.e. a  (b  c )  a  b  a  c
(3) Vector Triple Product
  
Definition: (a  b)  c is a vector which is coplanar with a and b and perpendicular to c .
   
Hence (a  b)  c = xa  yb ....(1) [ linear combination of a and b )
    
c . (a  b)  c = x (a.c)  y(b.c )


0 = x (a.c )  y(b.c ) ....(2)
x y
   
b.c a.c
 
 x =  ( b.c ) and y = –  (a.c)
  
Substituting the values of x and y in (a  b )  c   ( b . c ) a   (a . c ) b

This is an identity and must be true for all values of a , b, c

Put = a  î ; b  ˆj and c  î
( î  ˆj)  î   (ˆj. î ) î   ( î . î ) ˆj
ˆj    ĵ   = –1
        
hence (a  b)  c  (a . c) b  ( b . c ) a dot product of the extremes × (middle vector) – ( )
  
  ( a  b)  c
Note : Unit vector coplanar with a and b and perpendicular to c is ±   
| (a  b )  c |
  
  (a  b)  a
|||ly unit vector coplanar with a and b perpendicular to a is ±   

| (a  b )  a |
   
Examples a.a a.b a.c
     
       
(1) [a  b b  c c  a ] = [a b c]2 = b.a b.b b.c Note that if a , b , c are non coplanar
  
c.a c.b c.c
   
vectors then a  b , b  c and c  a will also be non coplanar vectors.

Bansal Classes [16]

    
V
1  a ( b c) 
(2) If V
 2  b  (c  a ) then prove that
V3  c  (a  b) 
     
(1) V1 , V2 , V3 are coplanar (2) V1 , V2 , V3 form the sides of a triangle.
     
(3) V1  V2  V3 is a null vectors (4) V1, V2 , V3 are linearly dependent
(3) If â  (b̂  ĉ)  1 b̂ where b̂ and ĉ are non collinear then find the angle between â and b̂ ;
2
between â and ĉ . [ Ans: /2 ; /3]
   
(4) î  (a  î )  ˆj (a  ˆj)  k̂  (a  k̂ )  2 a
(4) Scalar Product of Four Vector
   
      a . c a .d
(1) (a  b) . ( c  d ) = (a . c ) ( b . d )  ( a . d ) ( b . c ) =    
     
b.c b.d
[Proof: ]
 
(i) If we put c  a and d  b , Lagranges identity is established
(ii) This expression can be used to prove that the acute angle between the two planes faces
1
of a regular tetrahedron is cos–1
3
(iii) If a, b, c, d are four lines in space and (ad) represents any plane parallel to a and d and
so on, and if (ad)  (bc) and (bd)  (ca), prove that (cd) is  to (ab)
   
(2) Prove that : (a) {(a  b)  (a  c )}. d = (a . d) [a b c ] [Ans. zero ]
 
if a  î  2ˆj  3k̂ ; b  î  ˆj  4k̂ ; c  4î  3ˆj  6k̂ ; d  3î  6ˆj  5k̂
       
(b) d.[a  {b  (c  d )}] = ( b . d ) [a c d ] [Ans. – 1190 ]
(5) Vector Product of Four Vector
    
(1) V  (a  b)  ( c  d )
      
u  (c  d ) = [a b d] c  [a b c] d ....(1) (where u  a  b )
          
again V = (a  b)  ( c  d ) = (a . v ) b  (b.v ) a = [a c d ] b  [ b c d ] a
 ....(2)

V
       
from (1) and (2) [a b d ] c  [a b c ] d = [a c d ] b  [ b c d ] a ....(3)
   
Note that (a  b)  (c  d ) = 0  plane containing the values a & b and c & d are parallel.
   
|||ly ( a  b )  ( c  d ) = 0  the two planes are perpendicular..
 
(i) equation (3) is suggestive that if a , b, c, d are four vectors no 3 three of them are coplanar
then each one of them can be expressed as a linear combination of other.
 
(ii) If a , b, c, d are p.v.’s of four points then these four points are in the same plane if
   
[a b d ]  [a b c ]  [a c d ]  [ b c d ]
( refer to the article of condition of coplanarity of 4 points)
       
(2) Prove that : (a  b)  (c  d)  (a  c )  (d  b)  (a  d )  (b  c )  2 [b c d ] a [T/S]

Bansal Classes [17]

(4) Condition for coplanarity of four points
 
4 points with pv’s a, b, c, d are coplanar iff  scalars x, y, z and t not all simultaneously
 
zero and satisfying xa  yb  zc  td  0 where x + y + z + t = 0.
Case I : Let the four points A, B, C, D are in the same plane
 
 the vectors b  a , c  a and d  a are in the same plane.
 
hence d  a = l (b  a )  m( c  a )
   
(l  m  1) a l b m c 1 d  
or      –  –  +  = 0  xa  yb  zc  td  0 where, x + y + z + t = 0 and
x y z t
x, y, z, t not all simultaneous zero.
 
Case II : Let x a  y b  z c  t d  0 where x + y + z + t = 0 and not all simultaneously zero
 
Let t  0 (  y  z  t ) a  y b  z c  t d  0 [ putting x = – y – z – t ]
    
(d  a ) t  y( b  a )  z ( c  a )  0
   
d  a , b  a and c  a are coplanar points A, B, C, D are coplanar

 

(1) Linear combination/linear dependence and independence (Base vectors)

Linear combination
(1) Theorem in plane (already done)

(2) Theorem in space : If a , b, c are 3 non zero non

coplanar vectors then any vector r can be

expressed as a linear combination : r  x a  y b  z c
Examples
  
(1) Express the non coplanar vectors a , b, c in terms of b  c , c  a , a  b .
  
(2) Express b  c , c  a , a  b in terms of 3 non coplanar vectors a , b, c .
(3) Show that the p.v. of circumcentre of a tetrahedron OABC is (where ‘O’ is the origin)
        
a 2 ( b  c )  b 2 ( c  a )  c 2 (a  b )

2 [a b c ]
   
(4) Given the vector a and b orthogonal to each other find the vector V in terms of a and b

    
satisfying V . a  0 ; V . b  1 and [V a b ]  1
[Ans. V  12 1
   
b    2 (a  b) ]
b (a  b )
(2) Real definition of linearly independence
  
If V1 , V2 ,.........Vn are vectors and 1,  2 ,......... n are scalar and if the linear combination
  
1V1   2 V2  ........   n Vn  0 , necessarily implies 1 = 2 = .......n = 0, we say that
  
V1 , V2 ,.........Vn are said to constitutes a linearly independent set of vectors.
Note:
(1) 2 non zero , non collinear vectors are linearly independent.
[Proof: ]

(2) Three non zero , non coplanar vectors are linearly independent i.e. [a b c ]  0 .
[Proof: ]

Bansal Classes [18]

(3) Four or more vectors in 3D space are always linearly dependent.
[Proof: ]
(3) Reciprocal system of vectors
        
(a) If a , b , c and a ' , b ' , c ' are 2 sets of non coplanar vectors such that a .a '  b. b '  c. c ' = 1,
     
then a , b , c and a ' , b ' , c ' are said to be constitute a reciprocal system of vectors.
(b) Reciprocal system of vectors exists only in case of dot product.
 
(c) It is possible to define a ' , b ' , c ' in terms of a , b , c as.
     
 b c  ca  ab
a'    ;

[a b c ]
b '    ; c'    

[a b c ] [a b c ]
[ a b c ]  0 
        
      a  ( b  c )  b  ( c  a )  c  (a  b)
Note: (i) a  a ' b  b' c  c'  0 i. e. 
[ a b c]
      
(ii) (a  b  c).(a ' b' c' )  3 (as a · b  a · c  0 etc)
    1      
(iii) If [a b c ]  V then [a ' b' c' ]   [ a b c ] [a ' b ' c ' ] = 1
v
  
      abc 
(iv) a '  b '  b '  c '  c '  a ' =    , [a b c ]  0
[a b c ]
Home-Work
  
(a) If v1 , v 2 and v 3 are noncoplanar vectors, let
     
 V2  V3  V3  V1  V1  V2
k1 =    , k 2 =    and k 3 =   
v1 ·( v 2  v 3 ) v1 ·( v 2  v 3 ) v1 ·( v 2  v 3 )

(These vectors occurs in the study of crystallograpy. Vectors of the form

  
n1v1  n 2 v 2  n 3 v 3 , where each ni is an integer, form a lattice for a crystal. Vectors written
  
similarly in terms of k1 , k 2 and k 3 form the reciprocal lattice)
 
(i) Show that k i is perpendicular to v j if i  j.
 
(ii) Show that k i · vi = 1 for i = 1, 2, 3.
   1
(iii) Show that k1 · ( k 2  k 3 ) = v ·( v  v )
1 2 3

(b) Find the set of vector reciprocal to 2î

 3ˆj  k̂ , î  ĵ  2k̂ and  î  2ˆj  2k̂
    
   
 a b c
 
 bc 1 1 1
now a  
 [Ans:
(2î  k̂ ),  (8î  3ˆj  k̂ ), (7 î  3ˆj  5k̂ ) ]
[a b c ] 3 3 3
      
(c) Find a set of vectors reciprocal to the vectors a , b , and a  b ; Let c  a  b
 
      2  bc
[a b c] = (a  b ). c = a  b ; a '     etc.
[a b c ]
Isolating an known vectors
Satisfying a given relationship with some known vectors:
There is no general method for solving such equations, however dot or cross with

known or unknown vectors or dot with a  b , generally isolates the unknown vector..
Use of linear combination also proves to be advantageous.

Bansal Classes [19]

Examples:

(1) Solve for x
  
x .a  c ....(1) where c is a non-zero scalar ; a & b are non-zero vectors.
  
and ax b ....(2)
  
 ca  a  b
[Ans. x   2 ]
a
    
(2) Find the unknown vector R satisfying KR  A  R  B K 0 ....(1)
(3) Solve the following simultaneous equations
     
xy  a ....(1) ; x  y  b ....(2) ; x .a  1 ....(3)
  
    (a  b)  a
[Ans. y  a  x  a   2 ]
a
 
(4) Solve for x : x  a  (x.b) a  c
  
When a and c are non zero non collinear and a . b  0 .
     
(5) Let x , y , z be unit vectors such that x  y  z  a ; x  ( y  z )  b and ( x  y )  z  c
     

  3   7     
a . x  ; a . y  and | a |  2 . Find x , y , z in terms of a , b , c . [ REE - 96]
2 4
 1       4  
[Ans: x  (3a  4b  8c ) ; y   4c ; z  (c  b) ]
3 3
Application of vectors in geometry (For CBSE)

Bansal Classes [20]

 3 - D Coordinate Geometry
k x 2  x1 k y 2  y1 k z 2  z1
(1) Distance / Section Formula – x  ; y ; z
k 1 k 1 k 1
(2) DIRECTION COSINES :

Let a = a1i + a2j + a3k the angles which this vector makes with the +ve directions
OX,OY & OZ are called DIRECTION ANGLES & their cosines are called the
Hence if , ,  are the direction angles then the d.c's are
a a a
cos   1 , cos   2 , cos   3 .
a a a
Note that, cos²  + cos²  + cos²  = 1
Components of the unit vector denotes the dc’s of the vector.
cos , cos, cos are popularly denoted by l , m and n.
Note that sin2 + sin2 + sin2 = 2
2 2 1 2 2 1
(i) D.C's of the vector 2î  2ˆj  k̂ are ,– ,– or  , , 
3 3 3 3 3 3

(ii) Note that the locus of all points P for which OP represents a vector with
1
direction cosine cos = (O is origin) is "Cone concentric with the x-axis".
2
1 1 1 1 1 
(iii) Number of unit vectors with cos = and cos = (Two)  î  ˆj  k̂ 
2 2 2 2 2 
(iv) (Asking) There exists a vector with direction angles  = 300 and  = 300 (False)

(3) Direction ratios and its relationship with Direction Cosines

If a, b, c are 3 numbers which are proportional to direction cosines are called the
l m n
direction ratio’s of a line i.e. if     (say) l = a ; m = b ; n = c
a b c
now (a2 + b2 + c2) 2 = 1 (as l2 + m2 + n2 = 1)
a b c
l=  ;m=  ;n= 
2
a b c 2 2 2
a b c 2 2
a  b2  c2
2

Note
(1) Direction ratios of a line joining two points A and B are proportional to
x2 – x1 ; y2 – y1 ; z2 – z1

(2) Since AB = (x2 – x1) î + (y2 – y1) ˆj + (z2 – z1) k̂ . Hence the direction ratios of a

vector a  a1î  a 2 ĵ  a 3 k̂ are proportional to the numbers a1 , a2 and a3.
(3) If a line is having direction cosines l, m, n  it is travelling along the vector l î  m ˆj  n k̂ .
Hence angle between two lines with direction cosines l1 , m1 , n1 and l2, m2 , n2 is
a1a 2  b1b 2  c1c 2
cos = l1 l2 + m1 m2 + n1 n2 or in terms of direction ratios , cos =
a12  a 22  a 32  b12  b 22  b32
(a) If L1 is perpendicular to L2 then l1 l2 + m1 m2 + n1 n2 = 0 or a1a2 + b1b2 + c1c2 = 0
l1 m1 n1 a1 b1 c1
(b) If L1 is parallel to L2 then l  m  n or a  b  c
2 2 2 2 2 2

Bansal Classes [21]

Examples:
(1) A variable line has dc's l, m, n and l + l , m + m, n + n in two adjacent positions. If
 be the angle between the lines in these two positions then prove that
()2 = (l)2 + (m)2 + (n)2
(2) Find the direction cosines of a line perpendicular to two lines whose dr's are 1, 2, 3 and –2,
1, 4
1 2 1
[Ans. l  ;m n ]
6 6 6
(3) The direction cosines l , m, n of two lines are connected by the relations
l + m + n = 0 and 2lm + 2ln – mn = 0. Find them and the angle between them.
 1 1 2   1 2 1  2  
[ Ans  , ,  or  , ,    or ]
 6 6 6   6 6 6  3 3
(4) A line makes angle , , ,  with four diagonals of a cube. Prove that
4
cos2 + cos2 + cos2 + cos2 =
3
PLANES :
Definition : A plane is a surface such that a line segment joining any two points on the
surface lies wholly on it.
A linear equation in three variables of the type Ax + By + Cz + D = 0 denotes the
general equation of a plane.Where A, B, C are not simultaneously zero. Dividing by A
B C D
we get x  y  z   0 . Thus equation of the plane involves only 3 arbitrary
A A A
constants, hence in order to determine a unique plane 3 independent conditions are
needed.
Note: Equation of xy, yz and zx planes are z = 0 , x = 0 and y = 0.
Different forms of the equations of planes
n = a ^i + b ^j + c ^k


(1) A point and a vector normal to it

 
AR .n  0 
   R( r )
( r  r0 ).n  0 ....(1) 
r0
    A
r . n  r0 . n
 
Hence r . n  q is the general equation of a plane in vector form

If r  x î  y ˆj  z k̂ and r0  x1î  y1 ĵ  z1k̂
then (1) becomes a(x – x1) + b(y – y1) + c (z – z1) = 0
This is the equation of the plane containing the point (x1 y1 z1) when a î  b ĵ  c k̂ is a
vector normal to it. where a, b, c are the dr’s of a normal to the plane.
Examples:
(i) Find the equation of the plane through the point (2, –3, 1) and || to the plane
3x – 4y + 2z = 5.
(ii) Through the point (1, 0, –2) and perpendicular to the planes 2x + y – z = 2 and
x – y – z = 3.
(iii) The feet of normal from origin on a plane is ,  and . Find the equation of the plane.

Bansal Classes [22]

(iv) Find the equation of the plane passing through the points A(2, 2, 1) and B(1, –2, 3) and
perpendicular to the plane x – 2y + 3z + 4 = 0. A B , î  2ˆj  3k̂ and x – 2, y – 2, z – 1
are coplanar.
Asking: Two planes are given by equations x + 2y – 3z = 0 and 2x + y + z + 3 = 0. Find
(i) DC’s of their normals and the acute angle between them.
(ii) Length of the perpendiculars from the point (2, 2, 1)
(iii) DC’s of their line of intersection.
(iv) Equation of the plane perpendicular to both of them through the point (2, 2, 1)
1 2 3 2 1 1 1
[Ans. (i) , , ; , , and cos  =
14 14 14 6 6 6 2 21
3 10 5 7 3
(ii) p1 = ; p2 = ; (iii) ; ;
14 6 83 83 83
(iv) 5x – 7y – 3z + 7 = 0
for (ii) take a point (x 1 , y 1 , z 1 ) and find the projection of the vector
 
( x1  2)î  ( y1  2)ˆj  (z1  1) k̂ on n1 or n 2 ]

(2) Two intersecting lines determine a unique plane

Let the equations of the two lines are
  
r a  p 
  
r  a   q 
  
Now n  p  q is a vector perpendicular to the plane .
Hence the equation of the plane is,
          
( r  a ) . ( p  q )  0 or r  a p q  = 0 or r p q  [a p q] ....(1)
     
Note that vectors r  a , p and q are coplanar. Hence r  a = p  q  
....(2)
    
or r  a   p  q denotes a plane containing the point a and is parallel to two non collinear
 
vecors p and q . The equation (2) is known as parametric equation.
 
Note that (1) can be obtained from (2) by taking dot with p  q in (2)
         
r · p  q = a · ( p  q ) i.e. [ r p q ] = [a p q ]
e.g. Express the equation of a plane r  î  2ˆj   (2î  ˆj  3k̂ )  (3î  4ˆj  k̂ ) in
(i) scalar dot product forms (ii) cartesian form
Examples
Find whether the two lines

r  î  ˆj  k̂  (3î  ˆj) 

(i)(a) intersect or not
& r  4 î  k̂  (2 î  3 k̂ )


If they do, find the equation of the plane containing them. If they don’t find the S.D.
between them and also the intercept made by them on a line with dr’s (2, 1, 2).
5 12
[Ans: Lines are not intersecting ; ; ]
94 11
 ˆ
r  2 j  k̂  (î  ˆj  k̂ )
(i)(b) 
r  2î  3ˆj  6k̂  (2î  ˆj  5k̂ )
[ Ans: Intersecting at 2ˆj  k̂ ; equation of plane is 2x + y – z – 1 = 0]

Bansal Classes [23]

Asking:
(a) Condition for coplanarity of two lines.
   
r  a   b and r  c   d ( b and d are non collinear)
   

Note that lines must be intersecting

    
( r  a ) . b  d  0 is the equation of the plane and point c must satisfy is
         
 ( c  a ) . ( b  d )  0 is the required condition i.e. [a b d ]  [c b d ] .
   
Alternatively : [ a  c b d ] = 0  [ a b d ] = [ c b d ]
   
(b) Plane containing two parallel lines r  a   c and r  b   c

vector normal to plane n  (a  b)  c
   
 ( r  a ).{(a  b)  c}  0
       
( r  a ).{a  c  b  c}  0  [ r a c ]  [ r c b]  [a b c ]  0
    
[ Alternatively: r  a , a  b and c are coplanar
hence r  a a  b c = 0 will give the same result]
 
(3) Normal form of the plane
A unit vector n̂ normal to the plane from the origin is known and perpendicular distance
of the plane from the origin is d.

Projection of r on n̂ = d

r . n̂  d ....(1) (d > 0)
i.e. lx + my + nz = d (cartesian form)
equation (1) helps us to know the distance of the plane
from the origin and also the dc’s of the normal vector.
Example : r . (6î  3 ĵ  2k̂ )  1  0
6 3 2 1
(Dc’s are – , , and perpendicular distance of the plane from the origin is )
7 7 7 7
(4) Intercept form of the plane
x y z
  = 1 and
a b c
A (a, 0, 0) ; B (0, b, 0) and (0, 0, c)
1
Note: (1) Vector area =
2

ab ( î  ˆj)  bc ( ˆj k̂ )  ca (k̂  î ) 
1
=
2

bc î  ca ˆj  ab k̂ 
1 2 2
i.e. Area of the ABC = a b  b 2c2  c2a 2
2

(5)(a) Perpendicular distance of a point ‘P’ from a plane Ax + By + Cz + D = 0

 
d= Projection of RP on n P (x1, y1, z1)

 d

RP . n Ax1  By1  Cz1  D R
n =Ai + Bj + Ck = 0
d= 
|n | = r0 (x0, y0, z0)
A 2  B2  C 2
Note that the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z = d2 = 0 are
a1 b1 c1 d1
(i) Parallel if   
a 2 b2 c2 d 2

Bansal Classes [24]

 (ii) Perpendicular if a1a2 + b1b2 + c1c2 = 0
a1 b1 c1 d1
(iii) identical if   
a 2 b2 c2 d 2

d1  d 2
Distance between the parallel planes
a 2  b2  c2
(Take any point ‘P’ (x1 y1 z1) on one plane and from P draw perpendicular on the other plane)
Examples:
(i) Find the equation of the plane which is parallel to the plane x + 5y – 4z + 5 = 0 and the
sum of whose intercepts on the co-ordinate axes is 19 units. Also find the distance
25
between these planes. [ Ans: x + 5y – 4z = 20; ]
42
(ii) Find the equation of the plane parallel to 2x – 6y + 3z = 0 and at a distance of 2 from
the point (1, 2, –3).
(iii) A plane which always remain at a constant distance p from the origin cuts the
co-ordinate axes at A, B, C. Find the locus of
(a) Centroid of the plane face ABC
(b) Centre of the tetrahedron OABC
[ Ans : x–2 + y–2 + z–2 = 9p–2 ; x–2 + y–2 + z–2 = 16 p–2 ]
(6) Equation of the bisector planes between the planes
1 : a1x + b1y + c1z + d1 = 0 and ....(1)
2 : a2x + b2y + c2z + d2 = 0 is ....(2)
a1x  b1y  c1z  d1 a 2 x  b 2 y  c2 z  d 2
a12  b12  c12
=± a 22  b 22  c 22
Acute/Obtuse angle bisectors can be easily isolated by finding
a 1a 2  b 1 b 2  c1c 2
cos  = a 12  b 12  c12 • a 22  b 22  c 22
where  is the angle between any one of
the two given planes and any one of the two bisector planes.
1 1
if < | cos  | < 1   is acute ; if 0 < cos  <  is obtuse
2 2
Vectorially :
 
Let r . n1  q1 and r . n 2  q 2 be the given planes. Perpendicular distance of any point

r on either bisecting planes from the two given planes being equal, hence
     
| r .n1  q1 | | r .n 2  q 2 | r .n1  q1 r .n 2  q 2   n1 n 2   q1 q 
        or r .           2 
| n1 | | n2 | | n1 | | n2 |  | n 1 | | n 2 |   | n1 | | n 2 | 
where same sign is to be taken throughout.
(7) Family of planes
The equation P1 + P2 = 0 gives the family of planes containing the line of intersection
of P1 = 0 and P2 = 0 for all   R .
 
Vectorially: To find the equation of the plane coaxal with the planes r . n1  q1 and
 
r . n2  q2 ....(1)

and passing through the point with position vector a

 a .n1  q1 
[Ans. ( r .n1  q1 )  q  a . n ( r . n 2  q 2 )  0 or
       
( r . n1  q1 ) (q 2  a . n 2 )  ( r . n 2  q 2 ) (q1  a . n1 )  0 ]
2 2

Bansal Classes [25]

Examples :

(i) Find the equation of the plane containing the line of intersection of the planes r .n1  q1 ;
  
r .n 2  q 2 and is parallel to the line of intersection of the planes r .n 3  q 3 and r .n 4  q 4
(ii) The plane x – y – z = 4 is rotated through 900 about its line of intersection with the
plane x + y + 2z = 4. Find its equation in the new position. [Ans: 5x + y + 4z = 20]
(iii) Find the reflection of the plane P 1 : 2x – 3y + 6z + 1 = 0 in the plane
P2: 14x – 2y – 5z + 3 = 0.
 
Angle between two planes – The angle  , between the two planes r . n1  q1 and
   
r . n 2  q 2 , being equal to the
 angle
 between the vectors n1 and n 2 which are normal to the
1 n .n
planes, we have  = cos  1 2 .
| n1 | | n 2 |

Angle between a Line and a Plane – The angle  , between any line r  a  b t and any
  
plane r . n  q , being equal to the complement of the angle between the normal vector n , of
 
 1  n.b 
the plane and the direction vectors b of the line, we have  = sin    
 |n| |b | 

Straight Lines
Symmetrical form of straight line:
x  x1 y  y1 z  z1 x  x1 y  y1 z  z1   
(1) (a)     or     (derived from r  r0  v )
a b c l m n
x  x1 y  y1 z  z1
(b)  
x 2  x1 y 2  y1 z 2  z1
(Two point form)
where a, b, c are the dr’s of line or the vector along which the line is travelling.
Note :
x  2 y 1
(i)  & z=2 represent a line parallel to xy plane at a distance 2 units.
3 2
[ r  2î  ˆj  2k̂   (3î  2 ˆj) ]
x  1 2y  3 z5 3
(ii)  = passes the 1,  3 & – 5 with dr's 1, & 2.
1 3 2 2 2
x y z x y z x y z
(iii) Note that equation of x - axis   ; y-axis   ; z-axis  
1 0 0 0 1 0 0 0 1
‘0’ in denominator shows that the line is ar to that axis.
Example :
Convert the equation 3x + 1 = 6y – 2 = 1 – z in vector form and find its direction ratios.
 î ˆj
[Ans: r     k̂   ( 2î  ˆj  6k̂ ) ]
3 3
Unsymmetrical form of straight line:
The equations a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 together represents
a line in unsymmetrical form.
Procedure to convert Unsymmetrical Form of straight line to Symmetrical Form
Let the direction ratios of the line of intersection (AB) of two planes
a1x + b1y + c1z + d1 = 0 .....(1) and a2x + b2y + c2z + d2 = 0 .....(2) are a, b, c
Direction ratios of normal to plane (1) are a1, b1 , c1 and
Direction ratios of normal to plane (2) are a2 , b2 , c2

Bansal Classes [26]

Line AB lies in both the planes (1) and (2)
hence normals to (1) and (2) are perpendicular to AB.
Hence aa1 + bb1 + cc1 = 0 and aa2 + bb2 + cc2 = 0
these two will give the proportional values of a, b, c.
Let the line AB cuts the xy plane at (x1 , y1 , 0)
Hence a1x1 + b1y1 = – d1 and a2x1 + b2y1 = – d2 This will give a point on the line AB
x  x1 y  y1 z  0
 equation of AB is   .
a b c
Note : 3 planes arx + bry + crz = dr r = 1, 2, 3
(1) Can intersect at a point  system of equations in 3 variables having unique solution.
(2) Can intersect coaxially  system of equations in 3 variables having infinite solutions.
(3) May not have a common point  system of equations in 3 variables having no solution.
Example: Obtain the equations of the line of intersection of the planes x + y – 2z = 8 and
3x – y + 4z = 12 in the symmetrical form.
x 5 y 3 z
[Ans.   ]
1  5 1
Other Examples
(1) Find the points in which the line x = 1 + 2t ; y = – 1 – t and z = 3t meets the coordinate
planes i.e. xy , yz , zx plane P, Q, R. [(Ans: (1, – 1, 0) ; (0, –1/2, –3/2 ) ; (–1, 0 , –3)]
(2) Find the equation of the line through (1,4, –2) and parallel to the planes
6x + 2y + 2z + 3 = 0 &x + 2y – 6z + 4 = 0
(3) Show that the straight lines
L1 : 3x + 2y + z –5 = 0 = x + y – 2z – 3
and L2 : 8x – 4y – 4z = 0 = 7x + 10y – 8z
are at right angle
(4) Find the equation of the straight line which passes through the point (2, –1, –1) ; is
parallel to the plane 4x + y + z + 2 = 0 and is perpendicular to the line of intersection of
the planes 2x + y = 0 = x – y + z.
x  2 y 1 z 1
[Ans.   ]
1 13 9
(5)(a) Find the distance of the point A(1, 0, –3) from the plane P : x – y – z = 9 measured
x2 y2 z6
parallel to the line L :   . [Ans : 7 ]
2 3 6
(b) find also the distance of A from L parallel to the plane P.
(b) find the plane parallel to given plane P are containing the point A now compute the
intersection of the line L with the plane to get Q. Compute PQ. ]
Asking Problems
x  x1 y  y1 z  z1 x  x2 y  y2 z  z2
(a) If the lines l1

m1

n1 and l2

m2

n 2 are coplanar/ intersect if

x1  x 2 y1  y 2 z1  z 2
l1 m1 n1
= 0
l2 m2 n2

l1 m1 n1
x y z x y z x y z
(b) If the lines l  m  n ; l  m  n ; l  m  n are coplanar then l2 m2 n2 = 0
1 1 1 2 2 2 3 3 3
l3 m3 n3

Bansal Classes [27]

   
(c) Find the equation of the line through the point with p.v. c and | | to the plane r . n  1 and
 
perpendicular to the line r  a  t b [Ans. r  c   (n  b) ]

(d) Find the equation of the lines passing through the point with p.v. a and parallel to the
   
line of intersection of the planes r . n1 = 1 and r . n 2 = 1.
[Ans. r  a   (n 1  n 2 ) ]
(7) Show that the S.D. between the lines (Elegant approach of computing S.D.)
x 1 y  2 z  3
  .... (1)
2 3 4
x 2 y4 z 5 1
and   ....(2) is
3 4 5 6

(8) Find the parametric equation for the line which passes through the point (0, 1, 2) and is
perpendicular to the line x = 1 + t, y = 1 – t and z = 2t and also intersects this line.
[Ans. x = – 3t; y =1 + t; z = 2 + 2t]

   
(9) Find the equation of the line of intersection of the planes r . n1 = 1 and r . n 2 = 1 in
vector form.

  n 2  n  n    n 2  n  n  
[Ans. r   2 1 2 2  n1   1
   2  2
1 2
n  t n 1  n 2  where t is a parameter ]
 n1  n 2   n1  n 2 

(10) Find the equation of the two lines through the origin which intersect the line
x 3 y3 z  x y z x y z
  at an angle of . [Ans:   or   ]
2 1 1 3 1 2 1 1 1  2

(11) Prove that the three planes 2x + y – 4z – 17 = 0 ; 3x + 2y – 2z – 25 = 0 and

2x – 4y + 3z + 25 = 0 intersect at a point and find the coordinates of the points.
y z x z
(12) If 2d be the S.D. between the lines   1 ; x = 0 and   1 ; y = 0 then prove that
b c a c
1 1 1 1
2
 2
 2
 .
d a b c2
Line of Greatest slope in a plane
It is a line in the plane and perpendicular to the line of
intersection of the given plane with the horizontal plane.
PQ is the line of greatest slope . Its direction cosines
can be determined by the facts that
(i) It lies in G-plane
(ii) It is perpendicular to AB, the line of intersection of G and H plane.
Example
Assuming the plane 4x – 3y + 7z = 0 to be horozontal find the equation of the line of
greatest slope through the point (2, 1, 1) in the plane 2x + y – 5z = 0.
x  2 y 1 z 1
[Ans :   ]
3 1 1

Bansal Classes [28]

Asking:
Q. Determine whether each statement is true or false
(a) Two lines parallel to a third line are parallel. [T]
(b) Two lines perpendicular to a third line are parallel. [F]
(c) Two planes parallel to a third plane are parallel. [T]
(d) Two planes perpendicular to a third plane are parallel. [F]
(e) Two lines parallel to a plane are parallel. [F]
(f) Two lines perpendicular to a plane are parallel. [T]
(g) Two planes parallel to a line are parallel. [F]
(h) Two planes perpendicular to a line are parallel. [T]
(i) Two planes either intersect or are parallel. [T]
(j) Two lines either intersect or are parallel. [F]
(k) A plane and a line either intersect or are parallel. [T]

Bansal Classes [29]