Prove whether Cross-Product is associative

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Homework Help Overview

The discussion revolves around proving whether the cross product is associative, specifically examining the equality of \((\bold{A} \times \bold{B}) \times \bold{C}\) and \(\bold{A} \times (\bold{B} \times \bold{C})\) using the definition of the cross product involving angles between vectors.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the implications of including angles between vectors and question whether a counterexample could suffice to demonstrate non-associativity. Some participants suggest that certain conditions must hold for the equality to be true.

Discussion Status

The discussion is ongoing, with participants offering insights into the complexity of the proof and suggesting alternative approaches, including the possibility of finding counterexamples. There is no explicit consensus on the method to be used or the outcome of the proof.

Contextual Notes

Participants note the need to consider angles between the vectors involved, which may affect the validity of the associative property in this context. There is also mention of homework constraints regarding the use of definitions and the requirement for thoroughness in the proof.

Saladsamurai
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Homework Statement


Using the definition [itex]\bold{A}\times \bold{B}=AB\sin\theta[/itex] show whether or not [itex](\bold<br /> 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<br /> 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...
 
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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?
 
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 [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?

When alpha=beta ?
 
Saladsamurai said:

Homework Statement


Using the definition [itex]\bold{A}\times \bold{B}=AB\sin\theta[/itex] show whether or not [itex](\bold<br /> 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<br /> 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...
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?
 
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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