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Thread: A system of cross and dot products

  1. #1
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    A system of cross and dot products

    This was found in a textbook on classical mechanics so I thought it best to post it here rather then in the calc forum.

    \bold{b}\cdot\bold{v}=\lambda
    \bold{b}\times\bold{v}=\bold{c}

    \bold{b}, \bold{c},and \lambda are known. Express \bold{v} in their terms.

    I can't think of which rules would give me an answer. Any help would be appreciated.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Sorry!!! Fool me!

    Btw: two wonderful quotes.
    Last edited by Also sprach Zarathustra; Oct 27th 2010 at 11:35 AM.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Are they maybe vectors? Then it'll be different to what Also sprach Zarathustra posted.
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  4. #4
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    Actually, this does belong to the calculus forum.
    Recall that u\times (v\times w)=(u\cdot w)v-(u\cdot v)w so
    b\times (b\times v)=(b\cdot v)b-(b\cdot b)v=b\times c
    Now solve for v.
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  5. #5
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    Thank you Thus Spake Zarathustra

    And thank you Plato I was never aware of that formula. I'll be sure to prove it before I move on.

    So the identity you've given when combined with the equations gives

    \lambda \bold{b}-b^2\bold{v}=\bold{b}\times \bold{c}
    which rearranged gives

    \bold{v}=\dfrac{\lambda \bold{b}-\bold{b}\times \bold{c}}{b^2}

    Which matches the answer provided by the text.
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  6. #6
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    One small detail. It should be ||b||^2=b\cdot b.
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  7. #7
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    Oh yes of course. I had figured that notation was standard but I meant ||\bold{b}||=b. I'll make it more clear in the future. Thank you.
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