Proof that the cross product is perpendicular to its factors

Proof

We compute both dot products from the coordinate formula.

1. Let $a = (a_1, a_2, a_3)$ and $b = (b_1, b_2, b_3)$. By the definition of the cross product, $a \times b = (a_2 b_3 - a_3 b_2, \; a_3 b_1 - a_1 b_3, \; a_1 b_2 - a_2 b_1)$.
2. By the definition of the dot product, $$a \cdot (a \times b) = a_1 (a_2 b_3 - a_3 b_2) + a_2 (a_3 b_1 - a_1 b_3) + a_3 (a_1 b_2 - a_2 b_1).$$
3. We expand the right side. It is $a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_1 a_2 b_3 + a_1 a_3 b_2 - a_2 a_3 b_1$.
4. The six terms cancel in pairs: $a_1 a_2 b_3$ with $-a_1 a_2 b_3$, $a_1 a_3 b_2$ with $-a_1 a_3 b_2$, and $a_2 a_3 b_1$ with $-a_2 a_3 b_1$. So $a \cdot (a \times b) = 0$.
5. In the same way, $$b \cdot (a \times b) = b_1 (a_2 b_3 - a_3 b_2) + b_2 (a_3 b_1 - a_1 b_3) + b_3 (a_1 b_2 - a_2 b_1).$$
6. Expanded, this is $a_2 b_1 b_3 - a_3 b_1 b_2 + a_3 b_1 b_2 - a_1 b_2 b_3 + a_1 b_2 b_3 - a_2 b_1 b_3$.
7. Again the six terms cancel in pairs. So $b \cdot (a \times b) = 0$.