Scalar triple product - why equivalent to determinant?

Here's a slightly different perspective.

We use the mnemonic device...

$$ \langle a_1,a_2,a_3 \rangle \times \langle b_1,b_2,b_3 \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$

...to remember the formula for the cross product.

If you think about the dot product in the following (somewhat odd way), the triple scalar product identity becomes a triviality: Dot product-ing replaces standard unit vectors with corresponding vector components:

$$ (a_1{\bf i} + a_2{\bf j}+a_3{\bf k}) \bullet (b_1{\bf i} + b_2{\bf j}+b_3{\bf k}) = b_1\fbox{$a_1$}+b_2\fbox{$a_2$}+b_3\fbox{$a_3$}$$

So ${\bf i,j,k}$ have been replaced by $a_1,a_2,a_3$ respectively. Thus

$$ \langle c_1,c_2,c_3 \rangle \bullet (\langle a_1,a_2,a_3 \rangle \times \langle b_1,b_2,b_3 \rangle) = \begin{vmatrix} \fbox{$c_1$} & \fbox{$c_2$} & \fbox{$c_3$} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$

I am pretty sure you don't want this answer, but I couldn't resist.

The term on the left is an alternating $3$-linear form in $(a,b,c)$.

So it is equal to a constant times the determinant on the right..

Compute the constant with the canonical basis. It's $1$.

So the lhs is equal to the rhs.

Note: see here if you want to read about this characterization of the determinant and see here if you want a proof of the fact that the space of $n$-linear alternating forms is one-dimensional.

The easiest demonstration is just to cmpute both sides of the equation:

$$b\times c=(b_2c_3-b_3c_2,b_3c_1-b_1c_3,b_1c_2-b_2c_1)$$ $$a\cdot(b\times c)=a_1b_2c_3-a_1b_3c_2+a_2b_3c_1-a_2b_1c_3+a_3b_1c_2-a_3b_2c_1$$

That's the determinant. Note that each term of the sum is one possible permutation multiplied by its parity, which happens to be the definition of determinant.