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\begin{document}
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\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl Identities}}
\end{center}

\textbf{Question}

It is given that $\un{r} = x\un{i} + y\un{j} +z\un{k}$ and that
$\un{c}$ is a constant vector.

Show that
\begin{eqnarray*}
\nabla\bullet(\un{c} \times \un{r}) & = & 0\\
\nabla\times(\un{c}\times\un{r}) & = & 2\un{c}\\
\nabla(\un{c}\bullet\un{r}) & = & \un{c}
\end{eqnarray*}


\textbf{Answer}

\begin{eqnarray*}
\nabla\bullet\un{r} & = & 3\\
\nabla\times\un{r} & = & \un{0}\\
\nabla r = \frac{\un{r}}{r}
\end{eqnarray*}
$\un{c}$ is a constant vector, hence its div and curl both equal zero.
\begin{eqnarray*}
\Rightarrow
\nabla\bullet(\un{c}\times\un{r}) & = & (\nabla \times \un{c})\bullet
\un{r} - \un{c}\bullet(\nabla \times \un{r}) = \un{0}\\
\nabla \times (\un{c}\times\un{r}) & = & (\nabla \bullet \un{r})\un{c}
+ (\un{r}\bullet \nabla)\un{c} - (\nabla\bullet\un{c})\un{r} -
(\un{c}\bullet\nabla)\un{c}\\
& = & 3\un{c} + \un{0} - \un{0} - \un{c} = 2\un{c}\\
& & \\
\nabla(\un{c}\bullet\un{r}) & = & \un{c}\times(\bullet \times \un{r})
+ \un{r} \times(\nabla \times \un{c}) + (\un{c}\bullet\nabla)\un{r} +
(\un{r}\bullet\nabla) \un{c}\\
& = & \un{0} + \un{0} + \un{c} + \un{0} = \un{c}
\end{eqnarray*}

\end{document}