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Math Help - Please verify.

  1. #1
    Member courteous's Avatar
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    Red face Please verify.

    Vectors a, b, c ( \vec{i}=i) make up tetrahedron with volume V=3. What's the volume of a tetrahedron made up of vectors a\times b, b\times c, a\times c?
    Solutions aren't given, so I kindly ask for a recheck.

    Setup: V=\frac{1}{6}(a,b,c)=3\Rightarrow a\cdot(b\times c)=18\Rightarrow b\times c=\frac{18}{a}

    Attempt:
    (a\times b, b\times c, a\times c)=((a\times b)\times(a\times c))\cdot b\times c=(-\langle b, a\times c\rangle\cdot a)\cdot b\times c=(-\langle b, a\times c\rangle\cdot a)\frac{18}{a}=

    =-18\langle b, a\times c \rangle=-18(b\cdot (a\times c))=-18(b,a,c)=18(a,b,c)=54


    Isn't the volume of the second tetrahedron too large?
    Last edited by courteous; August 27th 2009 at 06:10 AM. Reason: typo
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    What does it mean to divide by a vector? What does the expression \frac{18}{a} mean?
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  3. #3
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    Let [\mathbf a,\mathbf b,\mathbf c] refer to the scalar triple product \mathbf a\times\mathbf b\cdot\mathbf c so that the volume of the tetrahedron is V=|[\mathbf a,\mathbf b,\mathbf c]|/6.

    Volume of new tetrahedron is V'=|[\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b]|/6.

    As is well known, [\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b]=[\mathbf a,\mathbf b,\mathbf c]^2.

    Hence V'=6V^2=54 as claimed. Although it came close, the method shown by courteous was ultimately incorrect -- that answer is right for the wrong reasons.
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  4. #4
    Member courteous's Avatar
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    Red face

    Quote Originally Posted by halbard View Post
    As is well known, [\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b]=[\mathbf a,\mathbf b,\mathbf c]^2.
    Can you expand the "well known" (google is useless)?

    BTW, did you intentionally write c\times a instead of a\times c (or it doesn't matter for given problem)?
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