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Math Help - vector identity

  1. #1
    Super Member Random Variable's Avatar
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    vector identity

    EDIT: I'm trying to show that  \frac{| v \times w |}{|v|}= \Big |w - \frac{v \cdot w}{|v|^{2}} v \Big|

    My initial thought was to get the right side to look like  |w| |\sqrt{(1-\cos^{2} \theta}| , but I can't seem to do it.


    By the way, what happened to all my thanked posts? We're they lost when the forum was updated a few months ago? I haven't been been here in a while.
    Last edited by Random Variable; September 27th 2010 at 04:45 PM. Reason: stated the problem incorrectly
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  2. #2
    A Plied Mathematician
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    \displaystyle{\left|w - \frac{v \cdot w}{|v|^{2}} v \right|=\sqrt{\left(w - \frac{v \cdot w}{|v|^{2}} v\right)\cdot\left(w - \frac{v \cdot w}{|v|^{2}} v\right)}=\sqrt{w\cdot w-2\frac{(v\cdot w)^{2}}{|v|^{2}}+(v\cdot w)^{2}}}

    \displaystyle{=|w|\sqrt{1-2\cos^{2}(\theta)+|v|^{2}\cos^{2}(\theta)}}.

    This will equal the desired RHS if and only if |v|=1. Is that stated in the problem somewhere?

    [EDIT]: TheEmptySet's computations are more correct than mine. The v squared terms do cancel.
    Last edited by Ackbeet; September 27th 2010 at 07:35 PM. Reason: Incorrect.
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  3. #3
    Behold, the power of SARDINES!
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    Remember that for any vectors \vec{a},\vec{b} we have

    |\vec{a}|^2=\vec{a}\cdot \vec{a} and
    (\vec{a}-\vec{b})\cdot ((\vec{a}-\vec{b}))=|\vec{a}|^2-2(\vec{a}\cdot \vec{b})+|\vec{b}|

    Using this we have

    \displaystyle \bigg| \vec{w}-\frac{\vec{v}\cdot \vec{w}}{|\vec{v}|^2}\vec{v}\bigg|^2=\left( \vec{w}-\frac{\vec{v}\cdot \vec{w}}{|\vec{v}|^2}\vec{v}\right)\cdot \left( \vec{w}-\frac{\vec{v}\cdot \vec{w}}{|\vec{v}|^2}\vec{v}\right)=
    \displaystyle |\vec{w}|^2-\frac{2(\vec{v}\cdot \vec{w})}{|\vec{v}|^2}+\left( \frac{\vec{v}\cdot \vec{w}}{|v|^2}\right)^2(\vec{v}\cdot \vec{v})
    \displaystyle |\vec{w}|^2-|\vec{w}|^2\cos^2(\theta)=|\vec{w}|^2(1-\cos^2(\theta))=|w|^2\sin^2(\theta)
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