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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
$\un{F}$ is smooth.

Show that
$$\nabla\times (\un{F}\times\un{r}) = \un{F} -
(\nabla\bullet\un{F})\un{r} + \nabla (\un{F}\bullet\un{r}) -
\un{r}\times(\nabla\times\un{F}).$$

In particular, if $\nabla\bullet\un{F}=0$ and
$\nabla\times\un{F}=\un{0}$, then
$$\nabla\times(\un{F}\times\un{r}) = \un{F} +
\nabla(\un{F}\bullet\un{r}).$$


\textbf{Answer}

\begin{eqnarray*}
\nabla\times(\un{F}\times\un{r}) & = & (\nabla\bullet\un{r})\un{F} +
(\un{r}\bullet\nabla) \un{F} - (\nabla\bullet\un{F})\un{r} -
(\un{F}\bullet\nabla) \un{r}\\
\nabla(\un{F}\bullet\un{r}) & = & \un{F} \times (\nabla \times \un{r})
+ \un{r}\times(\nabla\times\un{F}) + (\un{F}\bullet\nabla)\un{r} +
(\un{r}\bullet \nabla) \un{F}.
\end{eqnarray*}

It can be seen that from the given $\un{r}$, $\nabla \bullet \un{r}=3$
and $\nabla \times \un{r} = \un{0}$, and that
$$(\un{F}\bullet \nabla)\un{r} = F_1 \frac{\pa \un{r}}{\pa x} + F_2
\frac{\pa \un{r}}{\pa y} + F_3\frac{\pa\un{r}}{\pa z} = \un{F}.$$

Using all of this leads us to
\begin{eqnarray*}
\nabla\times(\un{F}\times\un{r}) - \nabla (\un{F}\bullet \un{r} & = &
3\un{F} - 2(\un{F} \bullet \nabla) \un{r} -
(\nabla\bullet\un{F})\un{r}\\
& & - \un{r}\times (\nabla \times \un{F})\\
& = & \un{F} - (\nabla \bullet \un{F})\un{r} - \un{r}\times (\nabla
\times \un{F}).
\end{eqnarray*}
If $\nabla\bullet\un{F} = 0$ and $\nabla\bullet\un{F} = \un{0}$ then
$$\nabla\times(\un{F}\times\un{r}) - \nabla(\un{F}\bullet \un{r}) =
\un{F}.$$

\end{document}