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Proving vector identities

1. The problem statement, all variables and given/known data

Prove the following vector identity:

[tex]\nabla[/tex]x(AxB) = (B.[tex]\nabla[/tex])A - (A.[tex]\nabla[/tex])B + A([tex]\nabla[/tex].B) - B([tex]\nabla[/tex].A)

Where A and B are vector fields.

2. Relevant equations

Curl, divergence, gradient

3. The attempt at a solution

I think I know how to do this: I have to expand out the LHS and the RHS and show that they equal one another. To do this I need to use the product rule when taking the gradient of components with more than one term multiplied together.

What I don't understand is what's going on on the RHS: doesn't (B.[tex]\nabla[/tex])A = B([tex]\nabla[/tex].A) ? (Obviously this can't be the case since then all the components would cancel to zero.) So how does this really work?
 
Last edited:

hunt_mat

Homework Helper
1,669
11
No, since if
[tex]
\begin{displaymath}
\begin{array}{rcl}
\mathbf{A} & = & A_{x}\mathbf{i}+A_{x}\mathbf{j}+A_{z}\mathbf{k} \\
\mathbf{B} & = & B_{x}\mathbf{i}+B_{x}\mathbf{j}+B_{z}\mathbf{k}
\end{array}
\end{displaymath}
[/tex]
Then:
[tex]
\begin{displaymath}
\begin{array}{rcl}
\mathbf{B}\cdot\nabla & = & B_{x}\partial_{x}+B_{y}\partial_{y}+B_{z}\partial_{z} \\
\nabla\cdot\mathbf{A} & = & \partial_{x}A_{x}+\partial_{y}A_{y}+\partial_{z}A_{z}
\end{array}
\end{displaymath}
[/tex]
Using the above compare one component of the two vector expressions and see if they're the same.
 

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