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Homework Help: Proving vector identities

  1. Aug 19, 2010 #1
    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: Aug 19, 2010
  2. jcsd
  3. Aug 19, 2010 #2

    hunt_mat

    User Avatar
    Homework Helper

    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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