<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

{\bf Question}

From the relations \begin{eqnarray*} {\bf e}_r &amp; = &amp; {\bf i} \sin
\theta \cos \phi + {\bf j} \sin \theta \sin \phi + {\bf k} \cos
\theta \\ {\bf e}_\phi &amp; = &amp; - {\bf i} \sin \phi + {\bf j} \cos
\phi \\ {\bf e}_\theta &amp; = &amp; {\bf i} \cos \theta \cos \phi + {\bf
j} \cos \theta \sin \phi - {\bf k} \sin \theta \end{eqnarray*}

derive \begin{eqnarray*} {\bf i} &amp; = &amp; {\bf e}_r \sin \theta \cos
\phi - {\bf e}_\phi \sin \phi + {\bf e}_\theta \cos \theta \cos
\phi \\ {\bf j} &amp; = &amp; {\bf e}_r \sin \theta \sin \phi + {\bf
e}_\phi \cos \phi + {\bf e}_\theta \cos \theta \sin \phi \\ {\bf
k} &amp; = &amp; {\bf e}_r \cos \theta - {\bf e}_\theta \sin \theta
\end{eqnarray*}

\vspace{.25in}

{\bf Answer}

\begin{eqnarray*} {\bf i} &amp; = &amp; ({\bf i} \cdot {\bf e}_r){\bf e}_r
+ ({\bf i} \cdot {\bf e}_\phi) {\bf e}_\phi + ({\bf i} \cdot {\bf
e}_\theta) {\bf e}_\theta \\ &amp; = &amp; \sin \theta \cos \phi\, {\bf
e}_r - \sin \phi\, {\bf e}_\phi + \cos \theta \cos \phi\, {\bf
e}_\theta
\end{eqnarray*}

From the given expressions for ${\bf e}_r$, ${\bf e}_\theta$,
${\bf e}_\phi$ in terms of ${\bf i,j,k},$

{\bf j} and {\bf k} can be found by an analogous procedure.


\end{document}
</pre></body></html>