TITLE: Proof for vectors involving cross and dot product
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: I prefer this purely vector method (using the Einstein summation convention) to the trig based method in the answer by Eli Rose.
$$
\begin{align*}
&\left| a\times b\right|^2+\left(a\cdot b\right)^2\\
=\;&\varepsilon_{ijk}a_jb_k\varepsilon_{ilm}a_lb_m+a_jb_ja_kb_k\\
=\;&a_jb_ka_lb_m(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})+a_jb_ja_kb_k\\
=\;&a_ja_jb_kb_k-a_jb_ka_kb_j+a_jb_ja_kb_k\\
=\;&a_ja_jb_kb_k=\left|a\right|^2\left|b\right|^2
\end{align*}
$$
(Where $\varepsilon$ is the Levi-Civita symbol, $\delta$ is the Kronecker delta, and I've used the relation $\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$)