Answer to Question #278280 in Calculus for Pial

Question

If A and B are vector fields, prove the following:

∇.(A×B)=B⋅(∇×A)−A⋅(∇×B).

Answer 1

$∇\cdot(A×B)=B⋅(∇×A)−A⋅(∇×B)$

$A×B=\begin{vmatrix} i & j &k\\ A_x & A_y&A_z\\ B_x & B_y&B_z \end{vmatrix}=(A_yB_z-A_zB_y)i+(A_zB_x-A_xB_z)j+(A_xB_y-A_yB_x)k$

$∇\cdot(A×B)=\frac{\partial}{\partial x}(A_yB_z-A_zB_y)+\frac{\partial}{\partial y}(A_zB_x-A_xB_z)+\frac{\partial}{\partial z}(A_xB_y-A_yB_x)=$

$=(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})B_z+(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})B_y+(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})B_x-$

$-(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y})A_z-(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x})A_y-(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z})A_x=$

$=B⋅(∇×A)−A⋅(∇×B)$