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Homework Help: Prove whether Cross-Product is associative

  1. Dec 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Using the definition [itex]\bold{A}\times \bold{B}=AB\sin\theta[/itex] show whether or not [itex](\bold
    A\times\bold B)\times\bold C=\bold A\times(\bold B\times\bold C)[/itex]

    I know this is probably easy, but I am missing the obvious here. So I started like this:

    Assuming [itex]\theta[/itex] lies between A and B and [itex]\phi[/itex] lies between [itex]\bold{A}\times \bold{B}[/itex] and C then [itex](\bold
    A\times\bold B)\times\bold C=(AB\sin\theta)C\sin\phi[/itex] ... wait is that it?

    Since that mess is associative then so is the cross-product?

    Granted, I should probably include an angle between B and C to be thorough...
  2. jcsd
  3. Dec 27, 2008 #2


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    Homework Helper

    yes you should include an angle between B and C, put it as [itex]\alpha[/itex] and the angle between A and (B x C) as [itex]\beta[/itex]. You'd get:

    [tex]|\bold A\times(\bold B\times\bold C)|= Asin\beta(BCsin\alpha)[/tex]

    Now in only a certain case would this be true,right?
  4. Dec 27, 2008 #3
    Isn't it just enough to come up with a counterexample to show that it's not true or were you specifically told to use the definition?
  5. Dec 27, 2008 #4
    When alpha=beta ?
  6. Dec 27, 2008 #5


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    No, that's not a complete proof. You are correct that the length of (AxB)xC is [itex]|A||B||C| sin(\theta)sin(\phi)[/itex] where [itex]\theta[/itex] is the angle between A and B and [itex]\phi[/itex] is the angle between (AxB) and C. But you haven't even looked at Ax(BxC). That would have length [itex]|A||B||C|sin(\alpha)sin(\beta)[/itex] where [itex]\alpha[/itex] is the angle between B and C and [itex]\beta[/itex] is the angle between A and BxC. Is see no reason to assume those angles are the same as [itex]\theta[/itex] and [itex]\phi[/itex] and no reason to assume those lengths are the same. In fact, even if you could prove those lengths are the same, that would not prove the vectors are the same.

    That is so complicated I might tend to assume it is false and look for a counter example, as NoMoreExams said. If fact, it might be something as simple as ((i + j)xk) x k and (i+ j) x (k x k). What are those?
  7. Dec 27, 2008 #6
    Hmm. Well I think the latter is 0 since k x k=0. I know that that the angle between (i+j) and k is non-zero since the angles between i and k and j and k are non-zero, hence the angle between (i+j) x k and k is non-zero, hence the cross product is non-zero.

    Is that what you are thinking Halls?
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