Question

Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.

Answer 1

Equation of the plane passing through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) with its normal’s d’ratios is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0  ......(i)

If the plane is passing through the given points (2, 1, – 1) and (– 1, 3, 4)

Then a(x₂ – x₁) + b(y₂ – y₁) + c(z₂ – z₁) = 0

⇒ a(– 1 – 2) + b(3 – 1) + c(4 + 1) =

⇒ – 3a + 2b + 5c = 0  ......(ii)

Since the required plane is perpendicular to the given plane x – 2y + 4z = 10

Then 1.a – 2.b + 4.c = 10  ......(iii)

Solving (ii) and (iii) we get,

`"a"/(8 + 10) = (-"b")/(-12 - 5) = "c"/(6 - 2) = lambda`

 a = 18λ, b = 17λ, c = 4λ

Hence, the required plane is

18λ(x – 2) + 17λ(y – 1) + 4λ(z + 1) = 0

⇒ 18x – 36 + 17y – 17 + 4z + 4 = 0

⇒ 18x + 17y + 4z – 49 = 0