vector

Vectors 419
g g g
Solution. We have, g = i j k
x y z
g g g
f g = f i f j f k
x y z
g g g
div (f g) = f
f
f
x x y y z z
2 2 2
g g g f g f g f g
= f
2 2 2
x y z
x x y y z z
2 2 2
f f f g g g
= f
g i j k . i j k
2 2 2
x y z
x y z x y z
= f 2 g + f.g Proved.
Example 64. For a solenoidol vector F , show that curl curl curl curl F = 4
F .
(M.D.U., Dec. 2009)
Solution. Since vector F is solenoidal, so div F = 0 ... (1)
We know that curl curl F = grad div ( F – 2
F ) ... (2)
Using (1) in (2), grad div F = grad (0) = 0 ... (3)
On putting the value of grad div F in (2), we get
curl curl F = – 2
F ... (4)
Now, curl curl curl curl F = curl curl (– 2
F ) [Using (4)]
= – curl curl ( 2
F ) = – [grad div ( 2
F ) – 2 ( 2
F ) ] [Using (2)]
= – grad ( . 2
F ) + 2
( 2
F ) = – grad ( 2
. F
) + 4
F [ . F = 0]
= 0 + 4 F = 4 F [Using (1)] Proved.
EXERCISE 5.9
1. Find the divergence and curl of the vector field V = (x 2 – y 2 ) i + 2xy j + (y 2 – xy) k .
Ans. Divergence = 4x, Curl = (2y – x) i + y j + 4y k
2. If a is constant vector and r is the radius vector, prove that
(i) ( a. r)
a (ii) div ( r a) 0 (iii) curl ( r
a) 2
a
where r
= xi yj zk and a a1i a2 j a3k
.
3. Prove that:
(i) .(A) = .A + (.A)
(ii) (A.B) = (A.)B + (B.)A + A × ( × B) + B × ( × A) (R.G.P.V. Bhopal, June 2004)
(iii) × (A × B) = (B.)A – B(.A) – (A.)B + A(.B)
4. If F = (x + y + 1) i + j – (x + y) k , show that F.curl F = 0.
(R.G.P.V. Bhopal, Feb. 2006, June 2004)
Prove that
5. ( F) ( ) F ( F )
6. .( F G) G.( F) F.( G)
7. Evaluate div ( A r)
if curl A
= 0. 8. Prove that curl ( a r)
= 2a