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Cross product invariant under SO(3)-matrices?

  1. Apr 30, 2010 #1
    Hi there!

    I'm trying to prove the following obvious statement, but am somehow stuck :(

    Let [tex]\vec a,\ \vec b\in\mathbb{R^3}[/tex] , let M be in SO(3) and x be the cross product

    prove: [tex]M(\vec a\times\vec b)=M\vec a\times M\vec b[/tex]

    I tried using the epsilon tensor, as in physics, but it doesn't really produce an opportunity, as you can convince yourself:

    [tex](M\vec a\times M\vec b)_i=\varepsilon_{ijk}(M\vec a)_j(M\vec b)_k=\varepsilon_{ijk}M_{jl} a_l M_{km}b_m...[/tex]

    where as usual summation is over repeated indices

    Now I want to use the fact that M is orthogonal, i.e.
    and preserves the orientation of the basis but I don't know where exactly this has to come into the proof...

    What I want to end up with is

    [tex]...=M_{ij}\varepsilon_{jkl}a_kb_l=(M(\vec a\times\vec b))_i[/tex]

    The statement seems to me obvious and can be envisioned very quickly by the right-hand-rule; I don't know why establishing it makes real problems

    If you have any idea, I'd be glad to see it :)

    With regards,
  2. jcsd
  3. Apr 30, 2010 #2
    Hi Marlin,

    I'll give you a hint (mainly 'cause I'm too lazy to latex up the answer)
    The 'preserving the orientation of the basis' bit is equivalent to saying det(M)=1
    Rewrite the determinant using the epsilon tensor then a little algebraic rearrangement should get you the result.
    (The formula I'm thinking of can be found http://en.wikipedia.org/wiki/Cross_product#Algebraic_properties"...)
    Last edited by a moderator: Apr 25, 2017
  4. May 1, 2010 #3
    Last edited by a moderator: Apr 25, 2017
  5. May 1, 2010 #4
    Thanks a lot :)

    this was very helpful :)
  6. May 5, 2010 #5
    More Generally
    Where A is any invertable 3x3 matrix and M(A) is the matrix of the minors of A

    [tex]A\vec a\times A\vec b=M(A)(\vec a\times\vec b)[/tex]

    I got this result by playing with the algebra. Hope I didn't make a mistake.:rolleyes:
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