Question

Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

Answer 1

The given planes are

P₁ : 5x + 3y + 6z + 8 = 0

P₂ : x + 2y + 3z – 4 = 0

P₃ : 2x + y – z + 5 = 0

Equation of the plane passing through the line of intersection of P₂ and P₃ is

(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0

⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0   ......(i)

Plane (i) is perpendicular to P₁, then

5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0

⇒ 5 + 10λ + 6 + 3λ + 18 – 6λ = 0

⇒ 7λ + 29 = 0

∴ `lambda = (-29)/7`

Putting the value of λ in equation (i), we get

`[1 + 2((-29)/7)]x + [2 - 29/7]y + [3 + 29/7]z - 4 + 5((-29)/7)` = 0

⇒ `(-15)/7x - 15/7y + 50/7z - 4 - 145/7` = 0

⇒ – 15x – 15y + 50z – 28 – 145 = 0

⇒ – 15x – 15y + 50z – 173 = 0

⇒ 51x + 15y – 50z + 173 = 0