= Solution
For nonzero vectors with angle $\theta$, $|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta$ and $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$. Hence $\sin^2\theta=1-\cos^2\theta$ gives <Lagrange's identity>
$$
\boxed{|\mathbf a\times\mathbf b|^2=|\mathbf a|^2|\mathbf b|^2-(\mathbf a\cdot\mathbf b)^2.}
$$
If either vector is zero both sides vanish, so this covers every case. Alternatively, expansion of the three <cross product> components gives the same identity without introducing an angle.
Back to article page
