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DOT PRODUCT OF VECTORS
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DOT PRODUCT OF VECTORS
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JEE MAINS - VOL - I DOT PRODUCT OF VECTORSDOT 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
35
,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 cos1
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
37
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