Question

Find the equation of the plane passing through the intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Find also the perpendicular distance of (1, 1, 1) from this plane.

Answer 1

We know that equation of plane passing through the intersection of planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So, equation of plane passing through the intersection of planes

x – 2y + z – 1 = 0 and 2x + y + z – 8 = 0 is

(x – 2y + z – 1) + k(2x + y + z – 8) = 0 ……(1)

⇒ x(1 + 2k) + y(– 2 + k) + z(1 + k) + (– 1 – 8k) = 0

We know that line

is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0

Given the plane is parallel to line with direction ratios 1, 2,1

1 × (1 + 2k) + 2 × (– 2 + k) + 1 × (1 + k) = 0

⇒ 1 + 2k – 4 + 2k + 1 + k = 0

⇒ k = \(\cfrac25\)

Putting the value of k in equation (1)

⇒ 9x – 8y + 7z – 21 = 0

We know that the distance (D) of point (x₁, y₁, z₁) from plane ax + by + cz – d = 0 is given by

So, distance of point (1,1,1) from plane (1) is

Taking the mod value we have

D = \(\cfrac{13}{\sqrt{194}}\)