Question
Find the equation of the plane through the three points (1, 1, 1), (1, – 1, 1) and (–7, – 3,–5).
Solution
The equation of any plane through (1, 1, 1) is
a (x – 1) + b (y – 1) + c (z – 1) = 0 ...(1)
∴ it passes through (1, –1, 1)
∴ a(1 – 1) + b (– 1 – 1) + c(1 – 1) = 0
∴ 0 a – 26 + 0 c = 0 ...(2)
Also (1) passes through (–7, –3, –5)
∴ a (–7–1) + b (–3–1) + c (–5 – 1) = 0
∴ – 8 a – 4 b – 6 c = 0
∴ 4 a + 2 b + 3 c = 0 ...(3)
From (2) and (3), we get,
∴ a = 3k, b = 0, c = – 4k
Putting values of a, b, c in (1), we get,
3k (x – 1) + 0 (y – 1) – 4k (z – 1) = 0
or 3 (x – 1) – 4(z – 1) = 0 or 3 x – 3 – 4z + 4 = 0
or 3x – 4 z + 1 = 0
which is required equation of plane.
