Prove that for any two vectors $\mathbf a$ and $\mathbf b$, $\lvert \mathbf a \times \mathbf b \rvert^2 + (\mathbf a \cdot \mathbf b)^2 = \lvert \mathbf a \rvert^2 \, \lvert \mathbf b \rvert^2$.
Can someone offer me advice on how to prove this in an easier way? So far, I'm solving it in a really complicated way, by labelling $\mathbf a$ as $(x,y,z)$ and $\mathbf b$ as $(a,b,c)$, then multiplying them out.
So, for the $\lvert \mathbf a \times \mathbf b \lvert^2$ term, I found $(yc-bz)^2 + (za-xc)^2 + (xb-ya)^2$, and then $(\mathbf a \cdot \mathbf b)^2=(ax+yb+zc)^2$.
Is there an easier way?