vector

400 Vectors
Solution. div ( ur ) =
2 2 2
i j k .[( x y z )( xi yj
zk)]
x y z
=
2 2 2 2 2 2 2 2 2
i j k .[( x y z ) xi( x y z ) yj( x y z ) zk]
x y z
=
3 2 2 2 3 2 2 2 3
( x xy xz ) ( x y y yz ) ( x z y z z )
x y z
= (3x 2 + y 2 + z 2 ) + (x 2 + 3y 2 + z 2 ) + (x 2 + y 2 + 3z 2 ) = 5 (x 2 + y 2 + z 2 ) = 5 u Ans.
Example 36. Find the value of n for which the vector
r xi y j zk .
Solution. Divergence F =
=
n
r r is solenoidal, where
n 2 2 2 n/2
. F . r r .( x y z ) ( xi yj
zk)
2 2 2 n/2 2 2 2 n/2 2 2 2 n/2
i j k .[( x y z ) xi( x y z ) yj( x y z ) zk]
x y z
= 2
n (x 2 + y 2 + z 2 ) n/2 – 1 (2x 2 ) + (x 2 + y 2 + z 2 ) n/2 + 2
n (x 2 + y 2 + z 2 ) n/2 – 1 (2y 2 )
+ (x 2 + y 2 + z 2 ) n/2 + 2
n (x 2 + y 2 + z 2 ) n/2 – 1 (2z 2 ) + (x 2 + y 2 + z 2 ) n/2
= n(x 2 + y 2 + z 2 ) n/2 – 1 (x 2 + y 2 + z 2 ) + 3 (x 2 + y 2 + z 2 ) n/2
= n(x 2 + y 2 + z 2 ) n/2 + 3(x 2 + y 2 + z 2 ) n/2 = (n + 3) (x 2 + y 2 + z 2 ) n/2
n
If r r is solenoidal, then (n + 3) (x 2 + y 2 + z 2 ) n/2 = 0 or n + 3 = 0 or n = –3. Ans.
Example 37. Show that
( a. r) a n( a. r)
r
. (M.U. 2005)
n n n 2
r
r r
Solution. We have,
Let =
x
a.
r
r
=
n
=
But r 2 = x 2 + y 2 + z 2 r
2r
x
( a i a j a k).( xi yj
zk)
1 2 3
a.
r ax 1 ay 2 az
3
n
n
r r
n
n1
r . a ( ax a y az) nr ( r/ x)
1 1 2 3
2n
n
r
r
= 2x
r x
x
r
n
=
axa y az
1 2 3
n
ar 1 ( ax 1 a2y az 3 ). nr . x a1
nax ( 1 a2y a3zx
)
=
2n
=
x
r
n
n 2
r r
=
i j k
x y z
=
1 n
( a1 i a
n
2 j a3k )
n 2 [( ax 1 a2y az 3 )( xi yj
zk
)]
r
r
=
a n
( ar . ) r
n n 2
r r
n
2
r