Question

Find the equation of the plane which contains the line of intersection of the planes x + 2y + 4z = 4 and 2x – 3y – z = 9 and which is perpendicular to the plane 4x – 3y + 5z = 10.

Answer 1

Let the equation of required plane be

(x + 2y + 4z – 4) + λ(2x – 3y – z – 9) = 0

where λ is a scalar.

i.e. (1 + 2λ)x + (2 – 3λ)y + (4 – λ)z – 4 – 9λ = 0

The d.r’s of the normal are

1 + 2λ, 2 – 3λ, 4 – λ

This plane is perpendicular to the plane

4x – 3y + 5z = 10, the d.r’s of whose normal are 4, –3, 5

∴ 4(1 + 2λ) – 3(2 – 3λ) + 5(4 – λ) = 0

∴ 4 + 8λ – 6 + 9λ + 20 – 5λ = 0

∴ 12λ = – 18

∴ λ = `-3/2`

Required equation is

`(x + 2y + 4z - 4) - 3/2 (2x - 3y - z - 9)` = 0

i.e. –4x + 13y + 11z + 19 = 0

i.e. 4x – 13y – 11z – 19 = 0