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Math Help - help with vectors

  1. #1
    Newbie omeganeu's Avatar
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    help with vectors

    Hello every one I have this vector relation
    E=A+(HXA) where E,A and H are vectors I want to proof that
    A=(E-(HXA))/
    any one remeber how we can do this using vector identities or simple geometry or both of them (may be)?
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  2. #2
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    Quote Originally Posted by omeganeu View Post
    Hello every one I have this vector relation
    E=A+(HXA) where E,A and H are vectors I want to proof that
    A=(E-(HXA))/
    any one remeber how we can do this using vector identities or simple geometry or both of them (may be)?
    As stated this is not true.
    It is true that A=E- \alpha(H \times A).
    Because H is a vector H^2 has no meaning.
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  3. #3
    Newbie omeganeu's Avatar
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    H is the vector and H is its magnitude

    actually I found this relation in one book but I do not remember how I can get it
    Last edited by mr fantastic; April 20th 2009 at 04:44 AM. Reason: Merged posts
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  4. #4
    Newbie omeganeu's Avatar
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    the correct equation is A=(E-(HXE))/(1+H) I am really sorry for that mistake
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  5. #5
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    Quote Originally Posted by omeganeu View Post
    the correct equation is A=(E-(HXE))/(1+H) I am really sorry for that mistake
    Even with the correction it still not true unless we know that A~\&~H are perpendicular.
    Did you leave that out?
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  6. #6
    Newbie omeganeu's Avatar
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    actually E and H should be perpendicular to each other. E has two component in the x-and the y- axis and H is parallel to the z-axis
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  7. #7
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    Quote Originally Posted by omeganeu View Post
    actually E and H should be perpendicular to each other.
    If that is the case then clearly A~\& ~H are also perpendicular. Form which we get using a well known fact
    H \times \left( {H \times A} \right) = \left( {H \cdot A} \right)H - \left( {H \cdot H} \right)A =  - \left\| H \right\|^2 A.
    Using that fact, we get:
    \begin{array}{rcl}<br />
   {H \times E} &  =  & {\left( {H \times A} \right) - \alpha \left\| H \right\|^2 A}  \\<br />
   {} &  =  & {\frac{1}<br />
{\alpha }\left( {E - A} \right) - \alpha \left\| H \right\|^2 A}  \\<br />
   {A + \alpha ^2 \left\| H \right\|^2 A} &  =  & {E - \alpha \left( {H \times E} \right)}  \\ \end{array}

    The result follows at once.
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  8. #8
    Newbie omeganeu's Avatar
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    Smile

    yes finally I get it! thank you very much for the great help and sorry for the mistakes.
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