Hi, I'm having trouble proving that:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \nabla \cdot (\textbf{A} \times \textbf{B}) = \textbf{B}\cdot

(\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times

\textbf{B}) [/tex]

This is how I proceeded:

[tex]\textbf{A} \times \textbf{B} = \overrightarrow{i}(A_y B_z - A_z

B_y) + \overrightarrow{j} (A_z B_x - A_x B_z) + \overrightarrow{k}

(A_x B_y - A_y B_x)[/tex]

[tex]\nabla \cdot (\textbf{A} \times \textbf{B}) = \frac{\partial A_y B_z - A_z

B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +

\frac {\partial A_x B_y - A_y B_x}{\partial z}[/tex]

[tex]\nabla \times \textbf{A} = \overrightarrow{i}(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})

+ \overrightarrow{j}(\frac {\partial A_x}{\partial z} - \frac

{\partial A_z}{\partial x}) + \overrightarrow{k}(\frac {\partial

A_y}{\partial x} - \frac {\partial A_x}{\partial y}) [/tex]

[tex] \textbf{B} \cdot (\nabla \times \textbf{A}) = (\frac{\partial A_z B_x}{\partial y} - \frac{\partial A_y B_x}{\partial z})

+ (\frac {\partial A_x B_y}{\partial z} - \frac {\partial A_z

B_y}{\partial x}) + (\frac {\partial A_y B_z}{\partial x} - \frac

{\partial A_x B_z}{\partial y}) = [/tex]

[tex] = \frac{\partial A_y B_z - A_z

B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +

\frac {\partial A_x B_y - A_y B_x}{\partial z} = \nabla \cdot

(\textbf{A} \times \textbf{B}) [/tex]

[tex] \textbf{A} \cdot (\nabla \times \textbf{B}) = (\frac{\partial B_z A_x}{\partial y} - \frac{\partial B_y A_x}{\partial z})

+ (\frac {\partial B_x A_y}{\partial z} - \frac {\partial B_z

A_y}{\partial x}) + (\frac {\partial B_y A_z}{\partial x} - \frac

{\partial B_x A_z}{\partial y}) = -\nabla \cdot (\textbf{A} \times

\textbf{B}) [/tex]

Which finally yields:

[tex] \textbf{B}\cdot (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times \textbf{B}) = 2\nabla \cdot (\textbf{A} \times \textbf{B}) [/tex]

Where did I make a mistake?

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# Homework Help: Divergence of (A x B)

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