Question

Find the equation of the image of the plane $x-2y+2z-3=0$ in the plane $x+y+z-1=0$

Answer 1

$P_1=x-2y+23-3=0$

$P_2=x+y+z-1=0$

Let P', be a plane s.f. $P_2$ bisects the

plane $P_1$, then the distance between

$P_2$ from $P_1$ and P', are equal i.e

$\frac{ax+by+cz+d}{\sqrt{a^2+b^2+c^2}}=\frac{x+y+z-1}{\sqrt{1+1+1}}=\frac{x+y+z-1}{\sqrt{3}}$

Let $\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{3}}=k$ ,thus

$ax+by+cz+d=k(x+y+z-1)...(\star )$

$a-k=1,b-k=-2,c-k=2,d-k=-3$

$\Rightarrow a=1+k,b=k-2,c=k+2,d=k-3$

putting this in $(\star )$ , we get

$\frac{\sqrt{(1+k{)}^2+(k-2{)}^2}+(k+2{)}^2}{\sqrt{3}}=k\Rightarrow (1+k{)}^2+(k-2{)}^2+(k+2{)}^2=3k^2$

$\Rightarrow 3k^2+2k+9=3k^2\Rightarrow k=-9/2.$

$\Rightarrow a=\frac{-7}{2},b=\frac{-13}{2},c=\frac{-15}{2},d=\frac{-15}{2}$

$\therefore e{q}^n$ of plane = $\frac{-7}{2}x-\frac{13}{2}y-\frac{15}{2}z-\frac{15}{2}=0$

$\Rightarrow 7x+13y+5z+15=0$