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Triple product vector problem

  1. Aug 24, 2011 #1
    Find [itex](\vec{a}\times \vec{b})\cdot \vec{c}[/itex] if [itex]\vec{a}=3\vec{m}+5\vec{n}[/itex], [itex]\vec{b}=\vec{m}-2\vec{n}[/itex], [itex]\vec{c}=2\vec{m}+7\vec{n}[/itex], [itex]|\vec{m}|=\frac{1}{2}[/itex], [itex]|\vec{n}|=3[/itex], [itex]\angle(\vec{m},\vec{n})=\frac{3\pi}{4}[/itex]

    This is my approach:
    [itex](\vec{a}\times\vec{b})\cdot\vec{c}=[(3\vec{m}+5\vec{n})\times(\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})=[3\vec{m}\times\vec{m}-6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}-10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n})=\bf(-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})[/itex]

    I stuck here. I don't know the coordinates of [itex]\vec{m}[/itex] and [itex]\vec{n}[/itex].
    Maybe the whole approach is wrong. I don't have any other idea on solving this problem so I need your help.
     
  2. jcsd
  3. Aug 24, 2011 #2

    Ray Vickson

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    What are mxm and nxn? Go back and _look at the definition_ of a vector product.

    RGV
     
  4. Aug 24, 2011 #3
    mxm and nxn by the definition gives us 0. That doesn't change anything.
     
  5. Aug 24, 2011 #4

    Ray Vickson

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    OK, I see now that you got rid of those terms; the results were hidden behind a popup that appeared on my screen before (but not now). Anyway, you now want to evaluate 11 mxn . (2m + 7n). Now go back and look at the definition of mxn; in particular, pay attention to the directions in which various vectors are pointing.

    RGV
     
  6. Aug 25, 2011 #5
    Hello Ray,

    I did these operations: [itex](-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})=(-11\vec{m}\times\vec{n})\cdot2\vec{m}+(-11\vec{m}\times\vec{n})\cdot7\vec{n}[/itex][itex]=[/itex]
    [itex]=[/itex][itex]-22(\vec{m}\times\vec{n})\cdot\vec{m}-77(\vec{m}\times\vec{n})\cdot\vec{n}[/itex]

    So, according to definition, the vector [itex]\vec{m}\times\vec{n}[/itex] is normal with the plane spanned by vectors [itex]\vec{m}[/itex] and [itex]\vec{n}[/itex], as a result I got 0 at the and. Is this correct? Are these operations I did allowed?
     
  7. Aug 25, 2011 #6

    tiny-tim

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    hello chmate! :smile:

    erm :redface:

    before embarking on long calculations, always check the obvious

    hint: what is b + c ? :wink:
    see above! :rolleyes:
     
  8. Aug 25, 2011 #7
    Hi tinytim,

    I see now that the vectors are lineary dependent so 0 is the right answer.
    I just want to have this question answered, are these operations I did legal?

    Thank you
     
  9. Aug 25, 2011 #8

    tiny-tim

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    let's see …
    yes, that's fine :smile:

    (and (A x B).B is always zero)
     
  10. Aug 25, 2011 #9

    Ray Vickson

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    Yes, everything is legal. You just used things like A*(B+C) = A*B + A*C and A*(rB) = r(A*B) for scalar r; these are true if * is either the dot product or the cross-product. You also used AxB = -BxA which is specific to the cross product. Of course, you could have written the answer right away, without any calculations, because you had (UxV).W, where U, V and W are all linear combinations of m and n, so all lie in the same plane containing m and n---hence UxV is perpendicular to W.

    RGV
     
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