Question

Find the image point of plane $x+y+z=2$ in the plane (w.r.t. plane) $2x-y+2z-1=0$.

Answer 1

Plane1:$x+y+z=2$

Plane2:$2x-y+2z-1=0$

The image of plane1 also passes through line of intersection of plane1 & plane2.

$\therefore$ image equation,

$(x+y+z-2)+\lambda (2x-y+2z-1)=0\Rightarrow (2\lambda +1)x+(1-\lambda )y+(1+2\lambda )-2-\lambda =0$

The angle between plane1 & plane2 be $\theta$,

$\cos \theta =\frac{2\times 1+1\times -1+1\times 2}{\sqrt{1^2+1^2+1^2}\sqrt{2^2+1^2+2^2}}=\frac{1}{\sqrt{3}}$

$\therefore$ the angle between image and plane2 will also be $\theta$

$\therefore \cos \theta =\frac{(2\lambda +1)2-(1-\lambda )+2(1+2\lambda )}{\sqrt{{(2\lambda +1)}^2+{(1-\lambda )}^2+{(1+2\lambda )}^2}\sqrt{2^2+2^2+1^2}}=\frac{1}{\sqrt{3}}\Rightarrow \frac{9\lambda +3}{3\sqrt{{(2\lambda +1)}^2+{(1-\lambda )}^2+{(1+2\lambda )}^2}}=\frac{1}{\sqrt{3}}\Rightarrow \lambda =0,\frac{-2}{3}$

the equation of image of plane1 is w.r.t plane2 is,

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