I need to show that the cross product is not associative without using components. I understand how to do it with components, which leads to an immediate counterexample, but other than that I am not sure how to do it.
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Consider two non-zero perpendicular vectors $\def\v#1{{\bf#1}}\v a$ and $\v b$. We have $$(\v a \times\v a)\times\v b=\v0\times\v b=\v0\ .$$ However $\v a\times\v b$ is perpendicular to $\v a$, and is not the zero vector, so $$\v a\times(\v a\times \v b)\ne\v 0\ .$$ Therefore $$(\v a \times\v a)\times\v b\ne\v a\times(\v a\times \v b)\ .$$
David
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1I was going to go with an explicit example of this, but this works too! I was going to illustrate this with $a = e_1$ and $b= e_2$. – Cameron Williams Jan 19 '16 at 03:49
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The cross product does satisfy the Jacobi identity (vector algebra relations)
$$ a \times (b \times c) + b \times (c \times a) + c \times (a \times b) = 0.$$
Move the 3rd term to the right and use anti-commutativity ($\,u \times v = -v \times u\,$) to get
$$ a \times (b \times c) + b \times (c \times a) = (a \times b) \times c. $$
However the term $\, b \times (c \times a) \,$ is not always zero which prevents associativity.
Somos
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