Prove whether Cross-Product is associative

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SUMMARY

The discussion centers on proving whether the cross-product is associative, specifically evaluating the expression \((\bold{A} \times \bold{B}) \times \bold{C} = \bold{A} \times (\bold{B} \times \bold{C})\). Participants analyze the lengths of the resulting vectors based on angles \(\theta\), \(\phi\), \(\alpha\), and \(\beta\) between the vectors involved. It is concluded that the assumption of equal angles does not hold, and thus, the cross-product is not associative. A counterexample is suggested involving the vectors \((i + j)\) and \(k\).

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  • Understanding of vector operations, specifically cross-products.
  • Familiarity with trigonometric functions and their application in vector analysis.
  • Knowledge of vector angles and their geometric implications.
  • Basic proficiency in linear algebra concepts.
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  • Study the properties of vector cross-products in detail.
  • Learn about vector identities and their proofs, focusing on non-associative operations.
  • Explore counterexamples in vector mathematics to solidify understanding of associative properties.
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Saladsamurai
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Homework Statement


Using the definition \bold{A}\times \bold{B}=AB\sin\theta show whether or not (\bold<br /> A\times\bold B)\times\bold C=\bold A\times(\bold B\times\bold C)



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

Assuming \theta lies between A and B and \phi lies between \bold{A}\times \bold{B} and C then (\bold<br /> A\times\bold B)\times\bold C=(AB\sin\theta)C\sin\phi ... 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...
 
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yes you should include an angle between B and C, put it as \alpha and the angle between A and (B x C) as \beta. You'd get:

|\bold A\times(\bold B\times\bold C)|= Asin\beta(BCsin\alpha)

Now in only a certain case would this be true,right?
 
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?
 
rock.freak667 said:
yes you should include an angle between B and C, put it as \alpha and the angle between A and (B x C) as \beta. You'd get:

|\bold A\times(\bold B\times\bold C)|= Asin\beta(BCsin\alpha)

Now in only a certain case would this be true,right?

When alpha=beta ?
 
Saladsamurai said:

Homework Statement


Using the definition \bold{A}\times \bold{B}=AB\sin\theta show whether or not (\bold<br /> A\times\bold B)\times\bold C=\bold A\times(\bold B\times\bold C)




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

Assuming \theta lies between A and B and \phi lies between \bold{A}\times \bold{B} and C then (\bold<br /> A\times\bold B)\times\bold C=(AB\sin\theta)C\sin\phi ... 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...
No, that's not a complete proof. You are correct that the length of (AxB)xC is |A||B||C| sin(\theta)sin(\phi) where \theta is the angle between A and B and \phi is the angle between (AxB) and C. But you haven't even looked at Ax(BxC). That would have length |A||B||C|sin(\alpha)sin(\beta) where \alpha is the angle between B and C and \beta is the angle between A and BxC. Is see no reason to assume those angles are the same as \theta and \phi 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?
 
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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