vector

396 Vectors
x 1 y 3 z
=
i.e., parallel to the vector
2 2 1
2i 2 j k .
On comparing the coefficients of (1) and (2)
...(2)
2a
c
= 2 b a 2 c b
2 2 1
2a + c = – 2b – a 3a + 2b + c = 0 ...(3)
and 2b + a = – 2(2c + b)
2b + a = – 4c – 2b a + 4b + 4c = 0 ...(4)
Rewriting (3) and (4), we have
3a 2b c
0
b
c
a 4b 4c
0
4 11 10
= k (say)
a = 4k, b = –11k and c = 10k.
Now, we have
(2a + c) 2 + (2b + a) 2 + (2c + b) 2 = (15) 2
(8k + 10k) 2 + (–22k + 4k) 2 + (20k – 11k) 2 = (15) 2
k =
a =
5
9
20
, b =
9
Example 32. If r xi y j zk,
show that :
r
1
(i) grad r =
r
r
Solution. (i) r = xi yj zk r =
r
2r
x
r
Similarly,
y
= 2x
= y r
grad r = r =
and
r
(ii) grad .
3
r
r
x
x
r
r
z
z
r
55
and c =
9
50
9
Ans.
(Nagpur University, Summer 2002)
2 2 2
x y z r 2 = x 2 + y 2 + z 2
r r r
i j k r i j k
x y z
x y z
x y z xi yj
zk r
= i j k Proved.
r r r r r
1
(ii) grad
r
= 1 1
1 1 1
i j k = i
j k
r x y zr
xr yr zr
= 1 r 1 r 1 r
i j k
2 2
2
r x r y
r z
=
1 x 1 y 1 z
i j k
2 2
2
r r r r r r = xi yj
zk r
Proved.
3 3
r r
2 2
Example 33. Prove that f () r f ´´ () r f´( r)
. (K. University, Dec. 2008)
r
Solution.