Question

A ray $M$ is sent along the line $\frac{x-0}{2}=\frac{y-2}{2}=\frac{z-1}{0}$ and is reflected by the plane $x=0$ at point $A$. The reflected ray is again reflected by the plane $x+2y=0$ at point $B$. The initial ray and final reflected ray meets at point $J$. Then absolute value of sum of the co-ordinates of point $J$ is

Answer 1

Line $\frac{x-0}{2}=\frac{y-2}{2}=\frac{z-1}{0}=r$
Point $A(2r,2r+2,1)$ lies on plane $x=0$
$\Rightarrow r=0$
Point $A$ is $(0,2,1)$


Let point on plane $x+2y=0$ be $B(-2\alpha ,\alpha ,\beta )$
Image of $B(-2\alpha ,\alpha ,\beta )$ w.r.t. plane $x=0$ is $B^{\prime }(2\alpha ,\alpha ,\beta )$
Direction ratios of $A{B}^{\prime }$ i.e., $2\alpha ,\alpha -2,\beta -1$ are proportional to $2,2,0$
$(\because A{B}^{\prime }$ is a part of ray $M)$
$\Rightarrow \frac{2\alpha }{2}=\frac{\alpha -2}{2}=\frac{\beta -1}{0}$
$\Rightarrow \alpha =-2,\beta =1$
$\therefore$ Points are $B(4,-2,1)$ and $B^{\prime }(-4,-2,1)$

Image of $A(0,2,1)$ with respect to plane $x+2y=0$
$\frac{x-0}{1}=\frac{y-2}{2}=\frac{z-1}{0}=-2\times \left(\frac{0+4}{1^2+2^2}\right)=-\frac{8}{5}$
$\Rightarrow A^{\prime }(-\frac{8}{5},-\frac{6}{5},1)$
Equation of $A^{\prime }B$ is
$\frac{x-4}{-28/5}=\frac{y+2}{4/5}=\frac{z-1}{0}=k$
Any point on above line is $(\frac{-28k}{5}+4,\frac{4k}{5}-2,1).$
This lies on the line $\frac{x-0}{2}=\frac{y-2}{2}=\frac{z-1}{0}$
$\therefore \frac{-\frac{28}{5}k+4}{2}=\frac{\frac{4k}{5}-2-2}{2}$
$\Rightarrow 8=\frac{4k}{5}+\frac{28k}{5}$
$\Rightarrow k=\frac{5}{4}$
$\therefore$ Point $J$ is $(-3,-1,1)$
$|-3-1+1|=3$