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Orthogonal transformation

  1. Jul 7, 2006 #1
    I've been banging my head against this problem for some time now, and I just can't solve it. The problem seems fairly simple, but for some reason I don't get it.

    Given the coordinate transformation matrix

    [tex]A=\left( \begin{array}{ccc}\cos{\alpha}&0&-\sin{\alpha}\\0&1&0\\\sin{\alpha}&0&\cos{\alpha}\end{array}\right)[/tex]

    show how

    [tex]\mathbf{B}=\mathbf{r} \times \hat{z}[/tex]

    transforms. Now how do I do this? For example, I've tried writing out the cross product, which becomes
    [tex]\mathbf{B} = -y\hat{x} + x\hat{y}[/tex]
    and then simply transforming this vector using the above matrix A, but it doesn't seem to work.

    Any hints on how to think about this?
    Last edited: Jul 7, 2006
  2. jcsd
  3. Jul 7, 2006 #2


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    Why do you say "it doesn't seem to work"?
    [tex]\left( \begin{array}{ccc}\cos{\alpha}&0&-\sin{\alpha}\\0&1&0\\\sin{\alpha}&0&\cos{\alpha}\end{array}\right)\left(\begin{array}{c}-y\\x\\0 \end{array}\right)= \left(\begin{array}{c}y cos(\alpha)\\ x\\y sin(\alpha)\end{array}\right)[/tex]
    Is that what you got?

    The transformation is, of course, a rotation around the y-axis, through the angle [itex]\alpha[/itex] so that the y component, here x, stays the same.
    Last edited: Jul 7, 2006
  4. Jul 7, 2006 #3
    well yes that is what I got, but in the answer sheet it says the answer is supposed to be

    [tex]\mathbf{B}' = y' \cos{\alpha} \hat{x}' - (x' \cos{\alpha} + z'\sin{\alpha}) \hat{y}' + y' \sin{\alpha}\hat{z}'[/tex]

    which I simply don't understand, both how to get the y-component, or why the x,y,z components are all primed? Is it not supposed to be the same vector in a different coordinate system?

    This problem is in the chapter about tensors and such, if that makes any difference...
    Last edited: Jul 7, 2006
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