How can I prove this eqn. in 3D vectors?

Question

If $\vec A, \vec B \in \mathbb{R}^3$ Prove that: $\|\vec A×\vec B\|^2 + (\vec A \cdot \vec B)^2 = \|\vec A\|^2 \|\vec B\|^2$

There was no reference to such questions in my academic book, please explain your answer in details.

Answer 1

Let $\theta$ be the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$, recall that one has: $$\begin{align}\left\|\overrightarrow{A}\times \overrightarrow{B}\right\|&=\left\|\overrightarrow{A}\right\|\left\|\overrightarrow{B}\right\||\sin(\theta)|\\\overrightarrow{A}\cdot \overrightarrow{B}&=\left\|\overrightarrow{A}\right\|\left\|\overrightarrow{B}\right\|\cos(\theta).\end{align}$$ Whence the result.