vector

Vectors 391
Normal to the surface (1), λ x yz(λ2)
x
ˆ
ˆ
ˆ 2
i j k x yz ( 2) x
x y z
iˆ(2x2) ˆj ( z) kˆ
( y)
Normal at (1, –1, 2) = î (2 – – 2) – ĵ (–2) + ˆk ...(3)
Normal at the surface (2)
= î ( – 2) + ĵ z (2) + ˆk
ˆ
ˆ
ˆ 2 3
i j k (4x yz
4)
x y z
= î (8 × y) + ĵ (4x2 ) + ˆk (3z 2 )
Normal at the point (1, –1, 2) = – 8 î + 4 ĵ + 12 ˆk ...(4)
Since (3) and (4) are orthogonal so
iˆ( 2) ˆj(2 ) kˆ. 8iˆ4ˆj12kˆ
0
8( 2) 4(2 ) 12 0 8 168120
8 20160
4( 2 54) 0
2 540
25 4
...(5)
Point (1, – 1, 2) will satisfy (1)
2
(1) ( 1)(2) ( 2)(1) + 2 = + 2 = 1
Putting = 1 in (5), we get
9
25 4
2
9
Hence and =1
Ans.
2
Example 23. Find the angle between the surfaces x 2 + y 2 + z 2 = 9 and z = x 2 + y 2 – 3 at
the point (2, –1, 2). (Nagpur University, Summer 2002)
Solution. Normal on the surface (x 2 + y 2 + z 2 – 9 = 0)
=
2 2 2
i j k ( x y z 9) (2xi 2yj
2 zk)
x y z
Normal at the point (2, –1, 2) = 4i 2 j 4 k
...(1)
Normal on the surface (z = x 2 + y 2 2 2
– 3) = i j k ( x y z 3)
x y z
= 2xi 2yj
k
Normal at the point (2, –1, 2) = 4i 2
j k
...(2)
Let be the angle between normals (1) and (2).
(4 i 2 j 4 k).(4 i 2 j k)
= 16 416 16 41cos
16 + 4 – 4 = 6 21cos 16 = 6 21cos