= Solution
{c}
Write $\partial_i=\partial/\partial x_i$ and use $\times$ for the paper's wedge notation, which denotes the <cross product>. The given expression for the <curl> and the <product rule> give
$$
\nabla\times(\mathbf v\times\mathbf w)
=\sum_i\mathbf e_i\times\bigl((\partial_i\mathbf v)\times\mathbf w
+\mathbf v\times(\partial_i\mathbf w)\bigr).
$$
Apply the <vector triple product identity> $\mathbf a\times(\mathbf b\times\mathbf c)=\mathbf b(\mathbf a\cdot\mathbf c)-\mathbf c(\mathbf a\cdot\mathbf b)$ separately to the two terms:
$$
\begin{aligned}
\mathbf e_i\times((\partial_i\mathbf v)\times\mathbf w)
&=w_i\partial_i\mathbf v-\mathbf w\,\partial_i v_i,\\
\mathbf e_i\times(\mathbf v\times\partial_i\mathbf w)
&=\mathbf v\,\partial_i w_i-v_i\partial_i\mathbf w.
\end{aligned}
$$
Summing over the Cartesian <basis vectors> identifies the two <directional derivatives> and the two <divergences>:
$$
\boxed{\nabla\times(\mathbf v\times\mathbf w)
=(\mathbf w\cdot\nabla)\mathbf v+\mathbf v(\nabla\cdot\mathbf w)
-(\mathbf v\cdot\nabla)\mathbf w-\mathbf w(\nabla\cdot\mathbf v)}.
$$
Here, for example, $((\mathbf w\cdot\nabla)\mathbf v)_j=\sum_iw_i\partial_i v_j$. This proves the <curl of a cross product> directly from the formula supplied.
Back to article page
