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 $x-y+z=3$

Answer 1

Equation of plane passing through intersections of the planes $P_1:x+y+z-1=0$

and $P_2:2x+3y+4z-5=0$ is given by,

$P_2+\lambda P_1=0$

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

Now for above plane to be perpendicular to the given plane $x-y+z=0$

Dot product of their normal vector should be zero

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

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

Hence the required plane is,

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

$\Rightarrow -x+z-2=0$

$\Rightarrow x-z+2=0$