Results 1 to 4 of 4

Thread: Please verify.

  1. #1
    Member
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Red face Please verify.

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

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

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

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


    Isn't the volume of the second tetrahedron too large?
    Last edited by courteous; Aug 27th 2009 at 05:10 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    What does it mean to divide by a vector? What does the expression $\displaystyle \frac{18}{a}$ mean?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2009
    Posts
    91
    Let $\displaystyle [\mathbf a,\mathbf b,\mathbf c]$ refer to the scalar triple product $\displaystyle \mathbf a\times\mathbf b\cdot\mathbf c$ so that the volume of the tetrahedron is $\displaystyle V=|[\mathbf a,\mathbf b,\mathbf c]|/6$.

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

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

    Hence $\displaystyle V'=6V^2=54$ as claimed. Although it came close, the method shown by courteous was ultimately incorrect -- that answer is right for the wrong reasons.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Red face

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

    BTW, did you intentionally write $\displaystyle c\times a$ instead of $\displaystyle a\times c$ (or it doesn't matter for given problem)?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Verify that x = 1 => fg(x) = gf(x)
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Jun 27th 2010, 06:05 PM
  2. verify
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Jan 25th 2010, 04:33 PM
  3. Please verify
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jan 14th 2010, 10:46 PM
  4. Verify
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 7th 2008, 06:08 PM
  5. verify this one
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Oct 6th 2005, 05:26 AM

Search Tags


/mathhelpforum @mathhelpforum