Question

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

• A. $\frac{-94\hat{i}+237\hat{j}+68\hat{k}}{49\sqrt{29}}$
• B. $\frac{-94\hat{i}+68\hat{j}-273\hat{k}}{49\sqrt{29}}$
• C. $\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$
• D. None of these

Answer 1

Answer: (A)

Given,
Incident ray is represented by vector $A$
$A=2\hat{i}-3\hat{j}+4\hat{k}$
Similarly, normal is represented by
$B=3\hat{i}-6\hat{j}+2\hat{k}$
Let reflected ray be represented by vector $R$.
$R=?$
Here, normal can be represented by any vector pointing up from the mirror or downwards.
We can verify, $A\cdot B=6+18+8=32(+ve)$
Since, $A\cdot B>0$, angle between incident
Unit vector along $A,$
$\hat{A}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}...(i)$
Unit vector along $B$,
$\hat{B}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{49}}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}...(ii)$
Let $ON$ and $OM$ represent the unit vectors, hence
$ON=OM-MN$ (from triangle law ) ...(iii)
$|MN|=2$ (projection length of vector $ON$ along $PN$ )
$=2(\hat{A}\cdot \hat{B})=2\frac{32}{7\sqrt{29}}$
From Eq. (iii), $\hat{R}=\hat{A}-2(\hat{A}\cdot \hat{B})\hat{B}$
Substituting the values of Eqs. (i) and (ii), we get
$R=\left(\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}\right)-2\left(\frac{32}{7\sqrt{29}}\right)\left(\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}\right)$
$=\left(\frac{49(2\hat{i}-3\hat{j}+4\hat{k})-(192\hat{i}-384\hat{j}+128\hat{k})}{49\sqrt{29}}\right)$
$=\left(\frac{98\hat{i}-147\hat{j}+196\hat{k}-192\hat{i}+384\hat{j}-128\hat{k}}{49\sqrt{29}}\right)$
$=\left(\frac{-94\hat{i}+237\hat{j}+68\hat{k}}{49\sqrt{29}}\right)$