Question
Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to the planes 2x + 3 y – 2 z = 5 and x + 2 y – 3 z = 8.
Solution
The equation of any plane through (1, – 1, 2) is
a(x – 1) + b (y + 1) + c (z – 2) = 0 ...(1)
∴ it is perpendicular to the planes
2x + 3 y – 2 z = 5 and x + 2 y – 3 z = 8
∴ 2 a + 3 b – 2 c = 0 ...(2)
and a + 2 b – 3 c = 0 ...(3)
Solving (2) and (3), we get,
Putting values of a, b, c in (1), we get,
5 k (x – 1) – 4 k (y + 1) – k (z – 2) = 0
or 5 (x – 1) – 4 (y + 1) – 1 (z – 2) = 0
or 5 x – 5 – 4 y – 4 – z + 2 = 0
or 5 x – 4 y – z – 7, which is required equation of plane.
