Question

Find the equation of a plane passing through the point (– 1, – 1, 2) and perpendicular to the planes 3x + 2y – 3z = 1 and 5x – 4y + z = 5.

Answer 1

We know that solution of a plane passing through (x₁, y₁, z₁) is given as - 

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

The required plane passes through (– 1, – 1, 2), so the equation of plane is 

a(x + 1) + b(y + 1) + c(z – 2) = 0

⇒ ax + by + cz = 2c – a – b …… (1)

Now, the required plane is also perpendicular to the planes,

3x + 2y – 3z = 1 and 5x – 4y + z = 5

We know that planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are at right angles

 if, a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (a) Using (a) we have,

3a + 2b – 3c = 0 …… (b)

5a – 4b + c = 0 …… (c)

Solving (b) and (c) we get

∴a = – 10λ, b = – 18λ, c = – 22λ

Putting values of a, b, c in equation (1) we get,

(– 10λ)x + (– 18λ)y + (– 22λ)z = 2(– 22)λ – (– 10λ) – (– 18λ)

⇒ – 10λx – 18λy – 22λz = – 44λ + 10λ + 18λ

⇒ – 10λx – 18λy – 22λz = – 16λ

Dividing both sides by (– 2λ) we get

5x + 9y + 11z = 8

So, the equation of the required planes is 5x + 9y + 11z = 8