Is (B.\nabla)A the same as B(\nabla.A)?

[featheredteap]

Homework Statement

Prove the following vector identity:

$$\nabla$$x(AxB) = (B.$$\nabla$$)A - (A.$$\nabla$$)B + A($$\nabla$$.B) - B($$\nabla$$.A)

Where A and B are vector fields.

Homework Equations

Curl, divergence, gradient

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.$$\nabla$$)A = B($$\nabla$$.A) ? (Obviously this can't be the case since then all the components would cancel to zero.) So how does this really work?

 

[hunt_mat]

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