Question

Find the equation of the plane through the line of intersection of the planes
x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z= 0

Answer 1

The equation of the plane through the intersection of the planes,
x+y+z=1 and 2x+3y+4z=5, is
(x+y+z-1)+λ(2x+3y+4z-5)=0
\(\Rightarrow\) (2\(\lambda\)+1)x+(3\(\lambda\)+1)y+(4\(\lambda\)+1)z-(5\(\lambda\)+1)=0...(1)

The direction ratios, a₁, b₁, c₁ of this plane are (2\(\lambda\)+1), (3\(\lambda\)+1), and (4\(\lambda\)+1).
The plane in equation(1)is perpendicular to x-y+z=0
Its direction ratios, a₂, b₂, c₂, are 1, -1, and 1.

Since the planes are perpendicular, a₁a₂+b₁b₂+c₁c₂=0
\(\Rightarrow\) (2\(\lambda\)+1)-(3\(\lambda\)+1)+(4\(\lambda\)+1)=0
\(\Rightarrow \) 3\(\lambda\)+1=0
\(\Rightarrow\) \(\lambda\)=-\(\frac{1}{3}\)

Substituting \(\lambda\) =-\(\frac{1}{3}\) in equation (1), we obtain

\(\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3}=0\)

\(\Rightarrow\) x-z+2=0

This is the required equation of the plane.