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.
Equation of plane through the line of intersection of planes in Cartesian form is
(1)
For the standard equation of planes,
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.
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
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
