Section 12.4: Cross Product

We constructed the cross product \(\mathbf{a} \times \mathbf{b}\) so that it would be perpendicular to both a and b. This is one of the most important properties of a cross product, so let’s emphasize and verify it in the following theorem and give a formal proof.

Theorem 8 The vector \(\mathbf{a} \times \mathbf{b}\) is orthogonal to both a and b.

PROOF In order to show that \(\mathbf{a} \times \mathbf{b}\) is orthogonal to a, we compute their dot product as follows: \[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} a_1 - \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} a_2 + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} a_3 \] \[ = a_1(a_2b_3 - a_3b_2) - a_2(a_1b_3 - a_3b_1) + a_3(a_1b_2 - a_2b_1) \] \[ = a_1a_2b_3 - a_1a_3b_2 - a_1a_2b_3 + a_2a_3b_1 + a_1a_3b_2 - a_2a_3b_1 = 0 \] A similar computation shows that \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0\). Therefore \(\mathbf{a} \times \mathbf{b}\) is orthogonal to both a and b.