Question

A ray of light is incident on a plane mirror, along the direction given $\overrightarrow{A}=2\hat{i}-3\hat{j}+4\hat{k}$. Find the unit vector along the reflected ray. Take normal to mirror along the direction of $\overrightarrow{B}=3\hat{i}-6\hat{j}+2\hat{k}$

• A. $\frac{-94\hat{i}+237\hat{j}+68\hat{k}}{49\sqrt{29}}$
• B. $\frac{-94\hat{i}+68\hat{j}-237\hat{k}}{49\sqrt{29}}$
• C. $\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$
• D. None of these

Answer 1

Answer: (D)

None of these

Consider the problem

First, we find the angle between :

$\begin{array}{c}\overrightarrow{A}=2i-3j+4k\\ \overrightarrow{B}=3i-6j+3k\\ \overrightarrow{A}.\overrightarrow{B}=AB\cos \theta \\ =\left(2i-3j+4k\right).\left(3i-6j+3k\right)\cos \theta \\ =\sqrt{4+9+16}\sqrt{9+36+4}\cos \theta \\ \cos \theta =\frac{32}{7\sqrt{29}}\end{array}$

Now, the angle between $B$ and reflected ray is same

And on checking the angles of a unit vector with $B$ in given points

None option is matched

So,

the answer is none of these.

Hence, option $D$ is the correct answer.