Question

Find the equation of the plane passing 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 plane through the line of intersection of planes $x+y+z=1$ and $2x+3y+4z=5$ is

$(x+y+z-1)+\lambda (2x+3y+4z-5)=0$

$(2\lambda +1)x+(3\lambda +1)y+(4\lambda +1)z-(5\lambda +1)=0$ .....(1)

The direction ratios of this plane are $(2\lambda +1),(3\lambda +1),(4\lambda +1)$.

The plane in equation 1 is perpendicular to $x-y+z=0$.

Therefore,

$(2\lambda +1)-(3\lambda +1)+(4\lambda +1)=0$

$3\lambda +1=0$

$\lambda =\frac{-1}{3}$

Putting this value in equation 1, we get,

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

$x-z+2=0$

This is the required equation of the plane.