Problem on proving that the cross product gives a perpendicular vector

Show that $\bfa \times \bfb$ is perpendicular to $\bfa$ by computing a dot product.
• Solution

We recall the relationship of dot products and perpendicular vectors:
Hence, we need to show that $\bfa \cdot (\bfa \times \bfb) = 0$ for all vectors $\bfa$ and $\bfb$.
In order to directly compute the dot product, we need to find the components of all vectors involved.
Let $\bfa = \langle a_1, a_2, a_3\rangle$ and $\bfb = \langle b_1, b_2, b_3 \rangle.$
Recall that
To compute the dot product with $\bfa$, recall that
Hence, \begin{align} \bfa \cdot (\bfa \times \bfb) = &a_1(a_2 b_3 - a_3 b_2) - a_2(a_1 b_3 - a_3 b_1) \\&+ a_3(a_1 b_2 - a_2 b_1).\end{align}
Expanding, we get that \begin{align}\bfa\cdot(\bfa \times \bfb) = &a_1 a_2 b_3 - a_1 a_3 b_2 - a_1 a_2 b_3 + a_2 a_3 b_1 \\&+ a_1 a_3 b_2 - a_2 a_3 b_1.\end{align}
Which simplifies to $$\bfa \cdot (\bfa \times \bfb) = 0.$$