Question

A mirror and a source of light are situated at a origin $O$ and a point on $OX$ , respectively. A ray of light from the source strikes the mirror and is reflected . If the direction ratios of the normal to the plane are $1,-1,1$, then find the $DCs$ of the reflected ray.

Answer 1

Let the source of light be situated at $A(a,0,0)$

Let $OA$ be the incident ray and $OB$ be the reflected ray

Let $ON$ be the normal to the mirror at $O$

So,

$\angle AON=\angle NOB=\frac{\theta }{2}$ (say) ...$\left(1\right)$

$D{R}^{\prime }s$ of $OA$ are $a,0,0$ and so its $D{C}^{\prime }s$ are $1,0,0$

$D{R}^{\prime }s$ of $ON$ are $\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

So from $\left(1\right)$ we can say that

$cos\left(\frac{\theta }{2}\right)=\frac{1}{\sqrt{3}}$

Let $l,m,n$ be the $D{C}^{\prime }s$ of the reflected ray $OB$

$\Rightarrow \frac{l+1}{2cos\left(\frac{\theta }{2}\right)}=\frac{1}{\sqrt{3}}$ ........$\left(2\right)$

$\Rightarrow \frac{m+0}{2cos\left(\frac{\theta }{2}\right)}=\frac{-1}{\sqrt{3}}$ ........$\left(3\right)$

$\Rightarrow \frac{n+0}{2cos\left(\frac{\theta }{2}\right)}=\frac{1}{\sqrt{3}}$ ........$\left(4\right)$

On substituting the value of $cos\left(\frac{\theta }{2}\right)=\frac{1}{\sqrt{3}}$ in $\left(2\right),\left(3\right),\left(4\right)$ , we get

$\Rightarrow l=\frac{2}{3}-1$

$\Rightarrow l=\frac{-1}{3}$

$\Rightarrow m=\frac{-2}{3}$

$\Rightarrow n=\frac{2}{3}$

Therefore, $D{C}^{\prime }s$ of the reflected ray are $\frac{-1}{3},\frac{-2}{3},\frac{2}{3}$