Question
Find the equation of the plane passing through (1, 1, – 1) and perpendicular to planes x + 2 y + 3 z – 7 = 0, 2 x –3 y + 4 z = 0.
Solution
The equation of the plane through (1, 1, – 1) is
a (x – 1) + b (y – 1) + c (z + 1) = 0 ...(1)
∴ it is perp. to the planes
x + 2 y + 3 z = 7 and 2 x – 3 y + 4 z = 0
∴ a + 2 b + 3 c = 0 ...(2)
and 2 a – 3 b + 4 c = 0 ...(3)
solving (2) and (3), we get
∴ a = 17 k, b = 2 k, c = –7 k
Putting these values of a, b, c in (1), we get,
17 k (x – 1) + 2 k (y – 1) –7 k (z + 1) = 0
or 17 (x – 1) + 2(y – 1) –7(z + 1) = 0
or 17x – 17 + 2y – 2 – 7 z – 7 = 0
or 17 x + 2 y – 7 z – 26 = 0, which is required equation of plane.
