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.