Question

Find the equation of the plane passing through the points (2, 2, –1), (3, 4, 2) and (7, 0, 6).

Answer 1

The general equation of the plane passing through (2, 2, –1) is

a(x – 2) + b(y – 2) + c(z + 1) = 0   ...(i)

It will pass through (3, 4, 2) and (7, 0, 6), we get

a(3 – 2) + b(4 – 2) + c(2 + 1) = 0

a + 2b + 3c = 0   ...(ii)

And a(7 – 2) + b(0 – 2) + c(6 + 1) = 0

5a – 2b + 7c = 0   ...(iii)

Solving equations (ii) and (iii) by cross multiplication method, we get

`a/(14 - (-6)) = b/(15 - 7) = c/(-2 - 10)`

⇒ `a/20 = b/8 = c/(-12)`

⇒  `a/5 = b/2 = c/(-3) = λ`   ...(Let)

a = 5λ, b = 2λ, c = –3λ

Substituting these values in equation (i), we get

5λ(x – 2) + 2λ(y – 2) – 3λ(z + 1) = 0

⇒ 5x – 10 + 2y – 4 – 3z – 3 = 0

⇒ 5x + 2y – 3z – 17 = 0

The required equation of plane is 5x + 2y – 3z = 17.