vector

424 Vectors
=
3 7 10
t t t
9 – 28 60
3 7 10
1
0
= 3 – 4 + 6 = 5 Ans.
Example 70. Evaluate
S
2
A. nˆ
ds where A ( x y ) iˆ
– 2xj ˆ 2yzkˆ
and S is the surface of
the plane 2x + y + 2z = 6 in the first octant. (Nagpur University, Summer 2000)
Solution. A vector normal to the surface “S” is given by
(2x y 2) z = ˆ
ˆ
ˆ
i j k (2x y 2) z 2iˆ ˆj 2kˆ
x y z
And ˆn = a unit vector normal to surface S
Z
2iˆ
ˆj 2kˆ
2 1 2 N
=
i ˆ ˆ j k ˆ
41
4 3 3 3
K
kˆ
2 1 2 2
n
. n ˆ = ˆ
k.
iˆ
ˆj kˆ
–
3 3 3 3
O
M Y
3
A . nds
dx dy
R
ˆ = A.
nˆ
S
R
k ˆ L
. n
Where R is the projection of S.
X
Now, A . nˆ
= [( x y 2 ) iˆ – 2xj ˆ 2 yzkˆ].
2
iˆ 1 ˆj 2 kˆ
3 3 3
2 2 2 4 2 2 4
= ( x y ) – x yz y yz
...(1)
3 3 3 3 3
Putting the value of z in (1), we get
on the plane 2x y 2z
6,
A.
nˆ
2 2 4 6
2 x y
= y y
(6 2 x y)
3 3 2 z
2
A.
nˆ
= 2 y ( y 6 –2 x – y) 4 y (3 – x)
...(2)
3 3
M
Hence,
A
. nds
dx dy
ˆ = A.
n
S
R
| k ˆ
...(3)
. n |
Putting the value of A.
nˆ
from (2) in (3), we get
A
. nds 4 3
3 62x
ˆ = (3 – ). 2 (3 )
S y x dx dy y x dydx
R 3 2
0 0
=
3
0
2
y
2(3– x)
2
6–2x
0
dx
=
3 (3 – )(6 –2 ) 2 4 3
(3 – )
3
0 0
4
3
x x dx x dx
(3 – x)
= 4. –(0–81) 81
4(–1)
0
iy ˆ ˆjx
Example 71. Compute F. dr,
where F
c
2 2
and c is the circle x 2 + y 2 = 1 traversed
x y
counter clockwise.
O
2x + 3y = 6
L
Ans.
X