TITLE: When are $3$ vectors associative in triple cross products?
QUESTION [8 upvotes]: The question I am trying to show under what conditions
$$\vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\times\vec{B})\times\vec{C}.$$
I have found that right-hand side of the above equation is equal to
\begin{align}
(\vec{A}\times\vec{B})\times\vec{C}
&=-\vec{C}\times(\vec{A}\times\vec{B})\\
&= \vec{C}\times(\vec{B}\times\vec{A}).
\end{align}
This is similar to the left-hand side of the original equation.
The conclusion I arrived at was, in order for the equality to be true, either $\vec{A}, \vec{B}$ or $\vec{C}$ have to be zero, or $\vec{A}$ must be equal to $\vec{C}$.
Is this correct? If not, or if I am missing anything, please let me know.

REPLY [7 votes]: Your reasoning is correct. The reason is that the cross product is anti-commutative. Therefore, the only situations wherein you will find the desired properties are when one of the vectors is zero, or when $A \propto C$
To prove this, note that $A\times (B\times C) = (A\cdot C)B-(A\cdot B)C$ and likewise $(A\times B)\times C = -(C\cdot B)A+(C\cdot A)B$
Assume the two are equal, and note that the dot product commutes, and we find
$$ (A\cdot B) C = (C\cdot B) A $$
Then, $$\frac{A}{A\cdot B} = \frac{C}{C \cdot B} \implies C = kA.$$
So $C$ is some scalar multiple of $A$.