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While trying to find cases that showed the cross product is not associative, I found some that were. I'm trying to show that

$(\mathbf{A}\times \mathbf{B}) \times \mathbf{C} \ne \mathbf{A}\times (\mathbf{B} \times \mathbf{C})$

And if

$\mathbf{A} = \hat{x}$

$\mathbf{B} = \hat{y}$

$\mathbf{C} = \hat{x}$

I find the the inequality is actually equal, both sides equal to $\hat{y}$. For one example where the inequality is true, I used

$\mathbf{A} = \hat{x}+\hat{y}+\hat{z}$.

To double check my work, I also tested the cases in python:

import numpy as np
# TRUE:
a = np.array((1,0,0))  
# FALSE:
#a = np.array((1,1,1))  
b = np.array((0,1,0))
c = np.array((1,0,0))

ab_c = np.cross(np.cross(a,b),c)
a_bc = np.cross(a,np.cross(b,c))
print "(a x b) x c =? a x (b x c)"
print ab_c,"=?",a_bc
print np.all(ab_c == a_bc)
  1. Are my calculations correct?
  2. Is there any significance to this?
  3. How does this relate to a Lie algebra? Does it? An answer to another question, What's the opposite of a cross product?, referred to a Lie Bracket.

EDIT After thinking about Mariano's answer, I realized I've tricked myself and there is no significance in the above result. The inequality is false when $\mathbf{A} = \mathbf{C}$, which in reality is my first case.

$(\mathbf{A} \times \mathbf{B}) \times \mathbf{A} = [-(\mathbf{B} \times \mathbf{A})] \times \mathbf{A} = -\mathbf{A} \times [-(\mathbf{B} \times \mathbf{A})] = \mathbf{A} \times (\mathbf{B} \times \mathbf{A})$

So my example is testing a case that is not valid for the inequality $(\mathbf{A} \times \mathbf{B}) \times \mathbf{C} \ne \mathbf{A} \times (\mathbf{B} \times \mathbf{C})$, which assumes the three vectors are unique.

In regards to Mariano's valid answer, the fact that $(c\times a) \times b = 0 $ is true in my case because $ c = a $.

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2 Answers 2

up vote 9 down vote accepted

A longuish comment

The cross product satisfies the Jacobi identity $$(a\times b)\times c+(b\times c)\times a+(c\times a)\times b=0.$$ Using this and the fact that it is antisymmetric, you can easily see that $$(a\times b)\times c=a\times(b\times c)\iff(c\times a)\times b=0.$$

This immediately explains your example where the equality holds and the one where it doesn't.

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After thinking about Mariano's answer, I realized I've tricked myself and there is no significance in the above result. The inequality is false when $\mathbf{A} = \mathbf{C}$, which in reality is my first case.

$(\mathbf{A} \times \mathbf{B}) \times \mathbf{A} = [-(\mathbf{B} \times \mathbf{A})] \times \mathbf{A} = -\mathbf{A} \times [-(\mathbf{B} \times \mathbf{A})] = \mathbf{A} \times (\mathbf{B} \times \mathbf{A})$

So my example is testing a case that is not valid for the inequality $(\mathbf{A} \times \mathbf{B}) \times \mathbf{C} \ne \mathbf{A} \times (\mathbf{B} \times \mathbf{C})$, which assumes the three vectors are unique.

In regards to Mariano's valid answer, the fact that $(c\times a) \times b = 0 $ is true in my case because $ c = a $.

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1  
You should probably edit this into the question, as it is not an answer. –  Mariano Suárez-Alvarez Oct 26 '11 at 18:15
    
Why is it not an answer? Putting this in the question would basically say there is no question to ask. –  Yann Oct 26 '11 at 18:32
1  
There was no question to ask :) You just needed to see that there wasn't! –  Mariano Suárez-Alvarez Oct 26 '11 at 18:34
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