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 $O N$ and $O M$ represent the unit vectors, hence
$O N =O M-M N$ (from triangle law ) ...(iii)
$|M N|=2$ (projection length of vector $O N$ along $P N$ )
$=2(\hat{ A } \cdot \hat{ B })=2 \frac{32}{7 \sqrt{29}}$
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
$R =\left(\frac{2 \hat{ i }-3 \hat{ j }+4 \hat{ k }}{\sqrt{29}}\right)-2\left(\frac{32}{7 \sqrt{29}}\right)\left(\frac{3 \hat{ i }-6 \hat{ j }+2 \hat{ k }}{7}\right)$
$=\left(\frac{49(2 \hat{ i }-3 \hat{ j }+4 \hat{ k })-(192 \hat{ i }-384 \hat{ j }+128 \hat{ k })}{49 \sqrt{29}}\right)$
$=\left(\frac{98 \hat{ i }-147 \hat{ j }+196 \hat{ k }-192 \hat{ i }+384 \hat{ j }-128 \hat{ k }}{49 \sqrt{29}}\right)$
$=\left(\frac{-94 \hat{ i }+237 \hat{ j }+68 \hat{ k }}{49 \sqrt{29}}\right)$