Explanation

(d) Let $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ be as shown in figure. Let $\theta$ be the angle of incidence, which is also equal to the angle of reflection. Resolving these vectors in rectangular components, we have
[image:4_image149.emf]
$\overrightarrow{A} =$sin$\theta\widehat{i} - cos\theta\widehat{j}$
$\overrightarrow{B} =$ sin$\theta\widehat{i} + cos\theta\widehat{j}$
$\overrightarrow{B} - \overrightarrow{A} = 2cos\theta\ \widehat{j}$
or $\overrightarrow{B} = \overrightarrow{A} + 2cos\theta\ \widehat{j}$
Now $\overrightarrow{A}.\overrightarrow{C} =$ $2cos\theta\widehat{j}$ or $\overrightarrow{B} = \overrightarrow{A}cos\theta\ \widehat{j}$
$\therefore\ \ \ \overrightarrow{B} = \overrightarrow{A} - 2(\overrightarrow{A}.\overrightarrow{C})\widehat{j}$ or $\overrightarrow{B} = \overrightarrow{A} - 2(\overrightarrow{A}.\overrightarrow{C})\overrightarrow{C}$ (as $\widehat{j} = \overrightarrow{C}$)