vector identity for triple scalar product

Question

How do you prove the identity: a$\cdot($b$\times$c$) =$ c$\cdot($a$\times$b$) = -$b$\cdot$(a$\times$c)
I'm thinking that this is because of row swaps in the determinant of a matrix with a,b,c as its row vectors. But I'm not sure how to write out the proof, and why its related to the determinant of that matrix

Answer 1

To proved the identities, you only need following properties from dot and cross products:

1. $u \times v$ is orthogonal to $u$, i.e. $u \cdot ( u \times v ) = 0$.
2. dot product is linear in its arguments.
3. cross product is linear in its arguments.

Let's take the identity $a \cdot (b \times c) = -b\cdot( a \times c)$ as an example, $$\begin{align} a\cdot(b \times c) &\stackrel{*1}{=} a \cdot (b \times c) - b \cdot (b\times c)\\ &\stackrel{*2}{=} (a-b)\cdot(b \times c)\\ &\stackrel{*1}{=} (a-b)\cdot(b \times c) + (a-b)\cdot((a-b)\times c)\\ &\stackrel{*2}{=} (a-b)\cdot(b \times c + (a-b)\times c))\\ &\stackrel{*3}{=} (a-b)\cdot((b + (a - b))\times c)\\ &= (a-b)\cdot(a \times c)\\ &\stackrel{*1}{=} (a-b)\cdot(a \times c ) - a \cdot( a\times c)\\ &\stackrel{*2}{=} ((a-b)-a)\cdot(a \times c)\\ &= (-b)\cdot( a \times c)\\ &\stackrel{*2}{=} - b \cdot( a \times c) \end{align} $$

We can view the determinant as a scalar function over $n \times n$ matrices.
It is characterized by three of its properties:

• $\det(I_n) = 1$
• viewing an $n \times n$ matrix as begin composed on $n$ columns, the determinant is an $n$-linear function. i.e. linear with respect to each columns.
• the $n$-linear function is an alternating form. The determinant vanishes whenever any two columns are identical.

Any function that satisfies these three properties will coincide with the determinant.

The properties for dot and cross products mentioned at beginning of this answer implies the triple product has the $2^{nd}$ and $3^{rd}$ properties. Together with the fact

$$(1,0,0) \cdot ((0,1,0) \times (1,0,0)) = 1$$

We find triple product has all three properties required to be the determinant. This means when $A$ is an $n\times n$ matrices with column vectors $v_1, v_2, v_3$, we can express it as a triple product: $$\det(A) = v_1 \cdot (v_2 \times v_3)$$