Is the following true: $[\nabla \times(\vec{A} \times \vec{B})] \cdot \vec{C}=(\vec{C} \times \vec{\nabla}) \cdot(\vec{A} \times \vec{B})$?

Question

Is the following equation correct? $$[\nabla \times(\vec{A} \times \vec{B})] \cdot \vec{C}=(\vec{C} \times \vec{\nabla}) \cdot(\vec{A} \times \vec{B})$$

If so, how can this be shown?

Answer 1

Having $\vec A\times \vec B$ just muddies the waters. Just try to see instead why $$(\nabla\times\vec A)\cdot\vec C = (\vec C\times\nabla)\cdot\vec A$$ for any vector fields $\vec A$ and $\vec C$.

Let's write out the right-hand side in terms of components. Writing $\vec A = (A_1,A_2,A_3)$ and $\vec C = (C_1,C_2,C_3)$, this is \begin{align*} \big(C_2\frac{\partial}{\partial z}& - C_3\frac{\partial}{\partial y}\big)A_1 + \big(C_3\frac{\partial}{\partial x} - C_1\frac{\partial}{\partial z}\big)A_2 + \big(C_1\frac{\partial}{\partial y} - C_2\frac{\partial}{\partial x}\big)A_3 \\ &=C_1\big(\frac{\partial A_3}{\partial y} - \frac{\partial A_2}{\partial z}\big) + C_2\big(\frac{\partial A_1}{\partial z} - \frac{\partial A_3}{\partial x}\big) + C_3\big(\frac{\partial A_2}{\partial x}- \frac{\partial A_1}{\partial y}\big) \\ &= (\nabla\times\vec A)\cdot \vec C, \end{align*} as desired.