Can the signs × and · be interchanged in the scalar triple…

Can the signs $\times$ and $\cdot$ be interchanged in the scalar triple product, i.e. is $(\vec a\times\vec b)\cdot\vec c=\vec a\cdot(\vec b\times\vec c)$?

Step 1: Both sides as a determinant

\[ (\vec a\times\vec b)\cdot\vec c=c_1(a_2b_3-a_3b_2)+c_2(a_3b_1-a_1b_3)+c_3(a_1b_2-a_2b_1), \]

\[ \vec a\cdot(\vec b\times\vec c)=a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1). \]

Both expand to the same six terms, the expansion of

\[ \begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}. \]

(Equivalently: $\vec a\cdot(\vec b\times\vec c)=(\vec b\times\vec c)\cdot\vec a=[\vec b,\vec c,\vec a]=[\vec a,\vec b,\vec c]$ by a cyclic shift.)

\[(\vec a\times\vec b)\cdot\vec c=\vec a\cdot(\vec b\times\vec c)\quad\text{— yes, they can be interchanged.}\]

Step 2: Remarks

• This is why the product is written simply as $[\vec a,\vec b,\vec c]$ or $\vec a\vec b\vec c$.
• The order of the vectors must stay the same; only the signs $\times,\cdot$ move.
• The bracketing is forced: $\vec a\times(\vec b\cdot\vec c)$ is meaningless (vector $\times$ scalar), and $(\vec a\cdot\vec b)\times\vec c$ is meaningless too.