\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

The unit vector ${\bf \hat{n}}$ is along the bisector of the angle
between the two unit vectors ${\bf \hat{d}}$ and ${\bf \hat{d}}'$.
Show that $$2{\bf (\hat{n} \cdot \hat{d})\hat{n} = \hat{d} +
\hat{d}'}.$$  Hence prove that a ray of light emerges parallel to
itself after successive reflections in each of three mutually
perpendicular mirrors.

\vspace{.25in}

{\bf Answer}

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\begin{picture}(4,2)
\put(0,1){\vector(1,0){2}}

\put(0,1){\line(2,1){1}}

\put(0,1){\line(2,-1){1}}

\multiput(1,0.5)(0,.2){6}{\circle*{0.05}}

\put(1,1){\circle*{0.075}}

\put(1.3,0.8){\makebox(0,0){$P$}}

\put(2.2,1){\makebox(0,0){$\hat{n}$}}

\put(1,1.7){\makebox(0,0){$A$}}

\put(1,0.3){\makebox(0,0){$B$}}

\put(.5,1.5){\makebox(0,0){$\hat{d}$}}

\put(.5,0.5){\makebox(0,0){$\hat{d}'$}}

\end{picture}


$P$ is the mid point of $AB$  So $\ds (n \cdot d)\hat{n} =
\frac{\bf d + d'}{2} \Rightarrow 2(n \cdot d) \hat{n} = d + d'$


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\begin{picture}(4,5)
\put(2,2){\line(1,0){2}}

\put(2,2){\line(0,1){2.5}}

\put(2,2){\line(-1,-1){1}}

\put(3,1.5){\line(-2,-1){.5}}

\put(3,1.5){\vector(-2,-1){.25}}

\put(2.8,1.2){\makebox(0,0){$d_1$}}

\put(2.5,1.25){\vector(0,1){.5}}

\put(2.55,1.8){\makebox(0,0){$n_1$}}

\put(2.5,1.25){\line(-1,2){1}}

\put(2.5,1.25){\vector(-1,2){.75}}

\put(1.5,2.7){\makebox(0,0){$d_2$}}

\put(1.5,3.25){\vector(1,0){.4}}

\put(1.85,3.05){\makebox(0,0){$n_2$}}

\put(1.5,3.25){\line(3,1){1.5}}

\put(1.5,3.25){\vector(3,1){.75}}

\put(2.3,3.3){\makebox(0,0){$d_3$}}

\put(3,3.75){\vector(-1,0){.5}}

\put(2.3,3.75){\makebox(0,0){$n_3$}}

\put(3,3.75){\line(1,1){.5}}

\put(3,3.75){\vector(1,1){.25}}

\put(3.25,3.75){\makebox(0,0){$d_4$}}
\end{picture}

\begin{eqnarray} d_2 + (-d_1) & = & 2(n_1 \cdot d_2)n_1 \\ d_3 +
(-d_2) & = & 2(n_1 \cdot d_3)n_2 \\ d_4 +(-d_3) & = & 2(n_3 \cdot
d_4)n_3 \end{eqnarray} Adding (1), (2), (3) \begin{eqnarray}d_4 -
d_1 & = & 2(n_1 \cdot d_2)n_1 +2(n_2 \cdot d_3)n_2 + 2(n_3 \cdot
d_4)n_3 \end{eqnarray}

From (3) $n_2 \cdot d_4 - n_2 \cdot d_3 = 0 $ since $n_2 \cdot n_3
= 0$

From (2) $n_1 \cdot d_3 - n_1 \cdot d_2 = 0$

From(3) $n_1 \cdot d_4 - n_1 \cdot d_3 = 0$ So $n_1 \cdot d_4 =
n_1 \cdot d_2$

So (4) becomes $$d_4 - d_1 = 2(n_1 \cdot d_4 )N-1 + 2(n_2 \cdot
d_4)n_2 + 2(n_3 \cdot d_4)n_3 = 2d_4$$ So $$d_4 = -d_1$$




\end{document}