Question

Find the equation of the plane 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 plane $x+y+z=1$ and $2x+3y+4z=5$ is

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

$\Rightarrow$ $(2\lambda +1)x+(3\lambda +1)y+(4\lambda +1)z-(5\lambda +1)=\mathrm{0......}\left(1\right)$

The direction ratios $a_1,b_1,c_1$ 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 $a_2,b_2,c_2$ are $1,-1$ and $1$

Since the planes are perpendicular,

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

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

$\Rightarrow$ $3\lambda +1=0$

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

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.