DOT PRODUCT OF VECTORS

JEE MAINS - VOL - I DOT PRODUCT OF VECTORS

DOT OR SCALAR PRODUCT OF VECTORS
 If a and b are parallel and are in the same direction
SYNOPSIS
( a , b are like vectors ) then a.b  a b

 Let a and b be two nonzero vectors and  be  If a and b are parallel and are in the opposite
the angle between them, then the scalar product or direction ( a , b are unlike vectors ) then
dot product of a and b denoted by a . b is defined a.b   a b
as a . b  a b cos   Let a , b are two unit vectors and a.b  cos 

a.b  If a , b are any two vectors, then a . b  0 either
 If ( a , b )=  , then cos  
a b a  0 (or) b  0 (or) a  b . However if a , b are
 Let i , j , k be unit orthogonal vectors i.e two non-zero perpendicular vectors, then a .b  0

  Geometrical interpretation of scalar
i  j  k  1 and  i , j   j , k  k , i   , product :
2
Let a and b be two vectors represented by
then i . i = j . j = k . k = 1
OA and OB respectively . Let  be the angle
i . j = j .k = k. i = 0
between OA and OB . Draw
j . i = k. j = i .k = 0
 If a  a1 i  a2 j  a3 k, b  b1 i  b2 j  b3 k Then BL  OA and AM  OB . From triangle OBL
and OAM, We have OL  OB cos and
(i) a.b  a1b1  a 2 b 2  a 3b3
OM  OA cos  . Here OL and OM are known
2
(ii) a.a  a12  a 2 2  a 32 (iii) a.a  a
as projections of b on a and a on b .
 Let a = a1 i +a2 j +a3 k , b = b1 i +b2 j +b3 k
a .b  a b cos   a  OB cos    a  OL 
and let ( a , b )=  then
= (magnitude of a ) (projection of b on a )
a1b1  a 2 b2  a 3 b3
cos  
2 2
a1  a 2  a 3
2 2
b1  b2  b3
2 2
again a .b  a b cos   b  a cos  
 Let ( a , b )=  , then  b  OA cos    b  OM 
(i) θ  90  a.b  0 (ii)   90  a.b  0 0
= (magnitude of b ) (projection of a on b )
(iii)   90 0  a .b  0 ( a is perpendicular to b )
W.E-1: If a  i   j  2k and

b  8 i  6 j  k are perpendicular then  is
Sol: Let a  i   j  2k and
b
b  8 i  6 j  k Since, a , b are at right
angles, a .b  0 
 ( i   j  2k ).(8 i  6 j  k )  0
a
 8  6  2  0    1

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,DOT PRODUCT OF VECTORS JEE MAINS
JEE MAINS
- C.W -- VOL
VOL--I I

 Let a and b be two nonzero vectors. Then 2
 (3i  2 j  5k )  (2 i  j  2k )
Component vector of b on a (or) orthogonal 3

(b.a)a 13i  4 j  11k

projection of b on a is 2 3
a
 The length of the orthogonal projection of b on a
WE-2: Orthogonal projection of
(a.b)
b = 2i  3 j  6k on a  i  2 j  2k is is a
Sol: Orthogonal projection
[( 2 i  3 j  6 k ). ( i  2 j  2 k )]( i  2 j  2 k )  The length of the orthogonal projection of a on b
 2
i  2 j  2k
(a.b)
is b
(2  6  12) 8
 ( i  2 j  2k )  ( i  2 j  2k )
9 9 WE-5: The length of orthogonal projection of
a  2 i  3 j  k on b  4 i  4 j  7k is
 Component vector of a on b (or) orthogonal
Sol: The length of the orthogonal projection of a
(a.b)b
projection of a on b is b 2 (a.b) 8  12  7 27
on b is b   3
16  16  49 9
WE-3: If a  2 i  j  2k , b  5 i  3 j  k  The scalar product is commutative i.e., a . b = b . a
then orthogonal projection of a on b is  The scalar product is distributive over vector
addition i.e., a .( b + c ) = a . b + a . c , ( b + c ). a
(a.b)b
 = b.a +c .a
Sol: Orthogonal projection of a on b is b
2


  l a  .b  a .  l b   l  a .b  where l is a scalar
(10  3  2) (5i  3 j  k ) 9(5i  3 j  k )
   a .a  0; a .b  a b ; a  b  a  b
25  9  1 35
 The orthogonal projection of b in the direction
a b  a  b
(b.a)a  Cauchy schwartz in equality : Let
perpendicular to that of a is b  2
a a1 , a2 , a3 and b1 , b2 , b3 be real numbers. Then
WE-4: The orthogonal projection of
 a1b1  a2b2  a3b3 2 
b  3i  2 j  5k on a vector perpendicular to
a  2 i  j  2k is a 2
1  
 a22  a32 b12  b22  b32 and equality

Sol: Orthogonal projection of b on a a1 a2 a3
holds If b  b  b
1 2 3
 3i  2 j  5k 
2 2 2
(3i  2 j  5k ).(2 i  j  2k )  a  b  a  b  2a.b
(2 i  j  2k )
4 1 4 2 2 2
 a  b  a  b  2a.b
6  2  10
 3 i  2 j  5k  (2 i  j  2k ) 2 2 2 2
9  a  b  c  a  b  c  2(a.b  b.c  c.a)

36

, JEE MAINS - VOL - I DOT PRODUCT OF VECTORS

 Let l1 , m1 , n1 be the direction cosines of a and let equal to the work done by the resultant force.

l 2 , m2 , n2 be the direction cosines of b and let  A line makes angles  ,  ,  ,  , with the four

(a, b)   then Cos  l1l2  m1m2  n1n2 diagonals of a cube then cos 2   cos 2  
 The vector equation to the plane which is at a cos 2   cos 2  
distance of p units from the origin and n̂ is a unit  If r is any vector then
vector perpendicular to the plane is r . nˆ  p r  (r . i ) i  (r . j ) j  (r .k )k .
 If the origin lies on the plane then its equation is
 If a , b are two vectors then
r .n  0
 The vector equation of a plane passing through the i) a .a  0 ii) a . b  | a || b |
point A  a  and perpendicular to the vector n is
iii) a  b  a  b iv) a  b  a  b
 r  a  .n  0  The cartesian equation of the plane passing through
W.E-6 : The vector equation of the plane passing the point A( x1 , y1 , z1 ) and perpendicular to the
through the point  3, 2,1 and
vector m  ai  bj  ck is
perpendicular to the vector  4,7, 4  is a( x  x1 )  b( y  y1 )  c( z  z1 )  0 .

 
Sol. r   3i  2 j  k  .  4 i  7 j  4k   0  The equation of the plane passing through the point
A( x1 , y1 , z1 ) and whose normal has d.r.s a,b,c is
 r .  4 i  7 j  4k   a( x  x1 )  b( y  y1 )  c( z  z1 )  0 .
 Angle between any two diagonals of a cube is
 3i  2 j  k  .  4 i  7 j  4k   12  14  4
cos 1 (1/ 3) .
 r .  4 i  7 j  4 k   6  Angle between a diagonal of a cube and a diagonal
 In a parallelogram, if its diagonals are equal then it of a face of the cube which are passes through the
is a rectangle. same corner is cos 1 .
 In a parallelogram, the sum of the squares of the  Angle between a diagonal of a cube and edge of a
diagonals is equal to the sum of the squares of the
sides.  1 
cube is cos1  
 3
 If F be the force and s be the displacement
inclined at an angle  with the direction of the force, Angle between a line and a plane :
then work done F .S  i) The angle between a line and a plane is the
complement of the angle between the line and
 If a constant force F acting on a particle displaces normal to the plane. If  is the anlge between a
it from A to B, then work done, W  F . AB line r  a  tb and a plane r .m  d then
 If F is the resultant of the forces F1 , F2 ......Fn b .m
cos  900     sin  
then work done in displacing the particle from A to b m .

B is W  F1  F2  ....  Fn . AB  WE-7: The angle between the line
 If a number of forces are acting on a particle, the r  ( i  3 j  3k )  t (2i  3 j  6 k )
sum of the work done by the seperate forces is and the plane r .( i  j  k )  5 is
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