Question

A ray of light comes along the line $L=0$ and strikes the plane mirror kept along the plane $P=0$ at $B$. $A(2,1,6)$ is a point on the line $L=0$ whose image about $P=0$ is $A^{\prime }$. It is given that $L=0$ is $\frac{x-2}{3}=\frac{y-1}{4}=\frac{z-6}{5}$ and $P=0$ is $x+y-2z=3$.

If $L_1=0$ is the reflected ray, then its equation is

• A. $\frac{x+10}{4}=\frac{y-5}{4}=\frac{z+2}{3}$
• B. $\frac{x+10}{3}=\frac{y+15}{5}=\frac{z+14}{5}$
• C. $\frac{x+10}{4}=\frac{y+15}{5}=\frac{z+14}{3}$
• D. None of these

Answer 1

The correct option is C $\frac{x+10}{4}=\frac{y+15}{5}=\frac{z+14}{3}$

The equation of the reflected ray $L_1=0$ is the line joining $A^{\prime }(x_2,y_2,z_2)$ and $B(-10,-15,-14)$.
$\frac{x+10}{16}=\frac{y+15}{20}=\frac{z+14}{12}$
$\Rightarrow \frac{x+10}{4}=\frac{y+15}{5}=\frac{z+14}{3}$