Question #278280

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

∇.(A×B)=B⋅(∇×A)−A⋅(∇×B).



1
Expert's answer
2021-12-13T16:52:58-0500

(A×B)=B(×A)A(×B)∇\cdot(A×B)=B⋅(∇×A)−A⋅(∇×B)


A×B=ijkAxAyAzBxByBz=(AyBzAzBy)i+(AzBxAxBz)j+(AxByAyBx)kA×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


(A×B)=x(AyBzAzBy)+y(AzBxAxBz)+z(AxByAyBx)=∇\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)=


=(AyxAxy)Bz+(AxzAzx)By+(AzyAyz)Bx=(\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-


(ByxBxy)Az(BxzBzx)Ay(BzyByz)Ax=-(\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)=B⋅(∇×A)−A⋅(∇×B)


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