Vector Cross Product is not Associative

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Theorem

The vector cross product is not associative.

That is, in general:

$\mathbf a \times \paren {\mathbf b \times \mathbf c} \ne \paren {\mathbf a \times \mathbf b} \times \mathbf c$

for $\mathbf a, \mathbf b, \mathbf c \in \R^3$.


Proof

Proof by Counterexample:

Let $\mathbf a = \begin {bmatrix} 1 \\ 0 \\ 0 \end {bmatrix}$, $\mathbf b = \begin {bmatrix} 1 \\ 1 \\ 0 \end {bmatrix}$, $\mathbf c = \begin {bmatrix} 1 \\ 1 \\ 1 \end {bmatrix}$

be vectors in $\R^3$.

\(\ds \mathbf a \times \paren {\mathbf b \times \mathbf c}\) \(=\) \(\ds \mathbf a \times \paren {\begin {bmatrix} 1 \\ 1 \\ 0 \end {bmatrix} \times \begin {bmatrix} 1 \\ 1 \\ 1 \end {bmatrix} }\)
\(\ds \) \(=\) \(\ds \mathbf a \times \begin {bmatrix} 1 \\ -1 \\ 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 1 \\ 0 \\ 0 \end {bmatrix} \times \begin {bmatrix} 1 \\ -1 \\ 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 \\ 0 \\ -1 \end {bmatrix}\)
\(\ds \paren {\mathbf a \times \mathbf b} \times \mathbf c\) \(=\) \(\ds \paren {\begin {bmatrix} 1 \\ 0 \\ 0 \end {bmatrix} \times \begin {bmatrix} 1 \\ 1 \\ 0 \end {bmatrix} } \times \mathbf c\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 \\ 0 \\ 1 \end {bmatrix} \times \mathbf c\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 \\ 0 \\ 1 \end {bmatrix} \times \begin {bmatrix} 1 \\ 1 \\ 1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} -1 \\ 1 \\ 0 \end {bmatrix}\)

$\blacksquare$


Sources

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