vector

416 Vectors
r =
i j k
x y z
r r r
=
1 2 3
i j k
a ( r)
=
a1 a2 a3
r3 r2 r3
r1 r2 r1
y z x z x y
r
r r
r r r
i j k
y z
x z
x y
3 2 3 1 2 1
r2 r1 r3 r1 r2 r1 r3
r2
a2 a2 a3 a3 i a1 a1 a3 a3
j
=
x y
x z
x y y z
r3 r1 r3
r2
a1 a1 a2 a2
k
x z y z
r
1 r
2 r3 r
1 r
2 r3
= a1i a2 i a3 i a1 j a2 j a3
j
x x x y y y
r
1 r
2 r3
r
1 r
2 r3
ak 1 a2k a3k
a1 i a1 j ak 1
z z z
x x x
r
1 r
2 r3 r
1 r
2 r3
a2 i a2 j a2k a3 i a3 j a3k
y y y z z z
= i j k ( a11 r a2r2 a33 r ) a1 a2 a3 ( r1 i r2 j
r3
k)
x y z x y z
= ( a. r) ( a. )
r
Proved.
Example 61. If r is the distance of a point (x, y, z) from the origin, prove that
1 1
Curl k grad
grad kgrad
.
= 0, where k is the unit vector in the direction OZ.
r r (U.P., I Semester, Winter 2000)
Solution. r 2 = (x – 0) 2 + (y – 0) 2 + (z – 0) 2 = x 2 + y 2 + z 2
1
r
= (x 2 + y 2 + z 2 ) – 1/2
grad 1 1 2 2 2 1/2
= = i j k ( x y z )
r
r x y z
=
1 ( 2 2 2 3/2
x y z (2 xi 2 yj 2 zk
)
2
k × grad 1 r
2 2 2 3/2
= – ( x y z ) ( xi yj
zk)
=
2 2 2 3/2
k [ ( x y z ) ( xi yj
zk)]
curl
1
k
grad
r
=
=
2 2 2 3/2
( x y z ) ( xj
yi)
1
k
grad
r