## Proof of Identity for Operator Consisting of Cross Product of a Vector With the Gradient

Theorem<
If
$\mathbf{v}$
and
$\mathbf{w}$
are vectors then
$(\mathbf{v} \times \mathbf{\nabla}) \cdot \mathbf{w} = \mathbf{v} \cdot (\mathbf{\nabla} \times \mathbf{w})$

Proof
Let
$\mathbf{v}=v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}$
and
$\mathbf{w}=w_1 \mathbf{i} + w_2 \mathbf{j} + w_3 \mathbf{k}$

\begin{aligned} (\mathbf{v} \times \mathbf{\nabla}) \cdot \mathbf{w} &= ((v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}) \times (\frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k} +))\cdot (w_1 \mathbf{i} + w_2 \mathbf{j} + w_3 \mathbf{k}) \\ &= ((v_2 \frac{\partial}{\partial z} - v_3 \frac{\partial}{\partial y})\mathbf{i} + (v_3 \frac{\partial}{\partial x} - v_1 \frac{\partial}{\partial z})\mathbf{j} + (v_1 \frac{\partial}{\partial y} - v_2 \frac{\partial}{\partial x})\mathbf{k}) \cdot (w_1 \mathbf{i} + w_2 \mathbf{j} + w_3 \mathbf{k}) \\ &= (v_2 \frac{\partial}{\partial z} - v_3 \frac{\partial}{\partial y})w_1 + (v_3 \frac{\partial}{\partial x} - v_1 \frac{\partial}{\partial z})w_2 + (v_1 \frac{\partial}{\partial y} - v_2 \frac{\partial}{\partial x})w_3 \\ &= v_2 \frac{\partial w_1}{\partial z} - v_3 \frac{\partial w_1}{\partial y} + v_3 \frac{\partial w_2}{\partial x} - v_1 \frac{\partial w_2}{\partial z} + v_1 \frac{\partial w_3}{\partial y} - v_2 \frac{\partial w_3}{\partial x} \\ &= v_1 (\frac{\partial w_3}{\partial y} - \frac{\partial w_2}{\partial z}) + v_2 (\frac{\partial w_1}{\partial z} - \frac{\partial w_3}{\partial x}) + v_3 (\frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y}) \\ &= \mathbf{v} \cdot (\mathbf{\nabla} \times \mathbf{w} ) \end{aligned}