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 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from the plane.

Answer 1

Equation of plane through the line of intersection of planes in Cartesian form is

A₁x + B₁y +C₁z + D₁ + λ(A₂x + B₂y + C₂z + D₂) = 0(1)

For the standard equation of planes,

A₁x + B₁y + C₁Z + D₁ and A₂x + B₂y + C₂z + D₂

So, putting in equation (1), we have 

x-2y + z-1 + λ(2x + y + z-8) = 0 

(1 + 2λ)x + (-2 + λ)y + (1 + λ)z-1-8λ = 0 (2) 

For plane the normal is perpendicular to line given parallel to this i.e.

A₁A₂ + B₁B₂ + C₁C₂ = 0

Where A₁, B₁, C₁ are direction ratios of plane and A₂, B₂, C₂ are of line. 

(1 + 2λ).1 + (-2 + λ).2 + (1 + λ).1 = 0 

1 + 2λ-4 + 2λ + 1 + λ = 0  

-2 + 5λ = 0

λ = 2/5

Putting the value of λ in equation (2)

9x - 8y + 7z-21 = 0 

9x - 8y + 7z = 21 

For the equation of plane Ax + By + Cz = D and point (x1,y1,z1), a distance of a point from a plane can be calculated as

So, the required equation of the plane is 9x - 8y + 7z = 21, and distance of the plane from (1,1,1) is

d = \(\frac{13}{\sqrt{194}}\)