Question
Find the equation of the plane through the points (1, –1, 2) (2, –2, 2) and perpendicular to the plane 6 x – 2 y + 2 z = 9.
Solution
The equation of any plane through (1, – 1, 2) is
a (x – 1) + b (y + 1) + c (z – 2) = 0 ....(1)
∴ it passes through (2, – 2, 2)
∴ a (2 – 1) + b (– 2 + 1) + c (2 – 2) = 0
∴ a – b + 0 c = 0 ....(2)
Also plane (1) is perpendicular to the plane 6 x – 2 y + 2 c = 9
∴ 6 a – 2 b + 2 c = 0 ⇒ 3 a – b + c = 0 ....(3)
From (2) and (3), we get,
∴ a = k, b = k, c – 2 k
Putting these values of a, b, c in (1), we get,
k(x – 1) + k (y + 1) – 2 k (z – 2) = 0
or x – 1 + y + 1 – 2 z + 4 = 0
or x + y – 2 z + 4 = 0,
which is required equation of the plane.
