# Thread: Proof that Cross Product is not Associative using the definition of Cross Product

1. ## Proof that Cross Product is not Associative using the definition of Cross Product

Does this work as a direct proof?

If the cross product is associative, then (A x B) x C = A x (B x C).

Consider when A = B.

One the left side we have:

(A x A) x C = 0 x C = 0·C·sinθ·n(AxA)xC= 0, where θ is the angle between vector A and C and n(AxA)xC is the unit vector perpendicular to their plane.

Then we have on the right side:

A x (B x C) = A x BC·sinφ·n(BxC) = ABC·sinφ·sinω·n(Ax(BxC)

, where φ is the angle between B and C in their plane, n(BxC) is the unit vector perpendicular to their plane, ω is the angle between A and the unit vector n(BxC) in their plane, and n(Ax(BxC) is the unit vector perpendicular to A and the unit vector n(BxC) in their plane.

This can only be zero when at least one of φ, ω, A, B, or C = 0.

Since there are cases where none of φ, ω, A, B, or C is equal to 0, the statement (A x B) x C = A x (B x C) is not true for all vectors A, B and C and therefore we can conclude that the cross product is not associative.

Is this a valid proof? Would it be considered an attempt at a direct proof? I used a particular example, but I didn't use a contradiction, so I'm thinking yes, but I wouldn't know.

2. ## Re: Proof that Cross Product is not Associative using the definition of Cross Product

Originally Posted by FaradaysGlassBlock
Does this work as a direct proof?

If the cross product is associative, then (A x B) x C = A x (B x C).

Consider when A = B.

One the left side we have:

(A x A) x C = 0 x C = 0·C·sinθ·n(AxA)xC= 0, where θ is the angle between vector A and C and n(AxA)xC is the unit vector perpendicular to their plane.

Then we have on the right side:

A x (B x C) = A x BC·sinφ·n(BxC) = ABC·sinφ·sinω·n(Ax(BxC)
Where have you used the fact that A= B?

, where φ is the angle between B and C in their plane, n(BxC) is the unit vector perpendicular to their plane, ω is the angle between A and the unit vector n(BxC) in their plane, and n(Ax(BxC) is the unit vector perpendicular to A and the unit vector n(BxC) in their plane.

This can only be zero when at least one of φ, ω, A, B, or C = 0.

Since there are cases where none of φ, ω, A, B, or C is equal to 0, the statement (A x B) x C = A x (B x C) is not true for all vectors A, B and C and therefore we can conclude that the cross product is not associative.

Is this a valid proof? Would it be considered an attempt at a direct proof? I used a particular example, but I didn't use a contradiction, so I'm thinking yes, but I wouldn't know.

3. ## Re: Proof that Cross Product is not Associative using the definition of Cross Product

Originally Posted by FaradaysGlassBlock
Does this work as a direct proof?

If the cross product is associative, then (A x B) x C = A x (B x C).
The cross product is NOT associative.

Consider this fact: $P\times(Q\times S)= (P\cdot S)Q-(P\cdot Q)S~\&~P\times Q=-Q\times P$

4. ## Re: Proof that Cross Product is not Associative using the definition of Cross Product

Originally Posted by HallsofIvy
Where have you used the fact that A= B?
I'm sorry that was a type-o. I was really pressed for time and didn't get a chance to proof read due to getting kicked off the computer by my brother right when I was finishing. I wish I could edit the first post.

In the first part, where I do the left side, A = B. In the second I forgot to include that change.

Here is what my post should have looked like: (I also cleaned up the subscripts which were a bit ridiculous)

Originally Posted by FaradaysGlassBlock

If the cross product is associative, then (A x B) x C = A x (B x C).

Consider when A = B.

One the left side we have:

(A x A) x C = 0 x C = 0·C·sinθ·n1= 0, where θ is the angle between vector A and C and n1is the unit vector perpendicular to their plane.

Then we have on the right side:

A x (A x C) = A x AC·sinφ·n2 = A2C·sinφ·sinω·n3

, where φ is the angle between B and C in their plane, n2 is the unit vector perpendicular to their plane, ω is the angle between A and the unit vector n2 in their plane, and n3 is the unit vector perpendicular to A and the unit vector n2 in their plane.

This can only be zero when at least one of φ, ω, A, or C = 0.

Since there are cases where none of φ, ω, A, or C is equal to 0, the statement (A x A) x C = A x (A x C) is not true, and therefore the statement (A x B) x C = A x (B x C) is not true for all vectors A, B and C and therefore we can conclude that the cross product is not associative.

Originally Posted by Plato
The cross product is NOT associative.

Consider this fact: $P\times(Q\times S)= (P\cdot S)Q-(P\cdot Q)S~\&~P\times Q=-Q\times P$
Yeah that's what I was trying to show. I guess that WAS an attempt at a proof by contradiction. First I assumed that the cross product was associative, then I tried to show that it led to a contradiction by using the definition directly and substituting A for B.

5. ## Re: Proof that Cross Product is not Associative using the definition of Cross Product

Originally Posted by FaradaysGlassBlock
Yeah that's what I was trying to show. I guess that WAS an attempt at a proof by contradiction. First I assumed that the cross product was associative, then I tried to show that it led to a contradiction by using the definition directly and substituting A for B.
Do you really understand that, $P\times(Q\times S)= (P\cdot S)Q-(P\cdot Q)S~?$
That says $P\times(Q\times S)$ is a linear combination of $Q~\&~S$. Do you get that?
So $(P\times Q)\times S$ is a linear combination of $P~\&~Q$. Do you get that?

So what does that mean?

6. ## Re: Proof that Cross Product is not Associative using the definition of Cross Product

Yeah that's what I was trying to show. I guess that WAS an attempt at a proof by contradiction
You have shown a counterexample, and one exception disproves the rule.

In what context are you studying vector triple products, have you studied any linear algebra to understand what Plato is saying?

Note that the vector triple product a x (b x c) lies in a plane parallel to b and c

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