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{\bf Question}

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

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

\vspace{.25in}

{\bf Answer}

\begin{eqnarray*} {\bf i} & = & ({\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 \\ & = & \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.


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