Answer
We have proven the property
Work Step by Step
For two vectors $\vec{v}\ \text {and} \ \vec{u}$ in 3d-space, we get that:
\[
\begin{aligned}
\vec{u} \cdot \vec{v} &=\|\vec{v}\|\|\vec{u}\| \cos \theta \Rightarrow \cos \theta=\frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|\|\vec{u}\|} \\
\|\vec{u} \times \vec{v}\| &=\|\vec{u}\|\|\vec{v}\| \sin \theta \Rightarrow \sin \theta=\frac{\|\vec{u} \times \vec{v}\|}{\|\vec{u}\|\|\vec{v}\|}
\end{aligned}
\]
where $\theta$ is the angle between $\vec{v}$ , $\vec{u}$. From the second equation in (1) we get that:
\[
\begin{aligned}
\sin \theta=\frac{\|\vec{u} \times \vec{v}\|}{\|\vec{v}\|\|\vec{u}\|} \Rightarrow\|\vec{v} \times \vec{u}\|^{2} &=\sin ^{2} \theta \|\vec{v}\|^{2}\|\vec{u}\|^{2} \\
&=\left(-\cos ^{2}+1)\theta\|\vec{u}\|^{2}\|\vec{v}\|^{2}\right \\
&=\left\|\vec{u}^{2}\right\| \vec{v}\left\|^{2}-\right\| \vec{u}\left\|^{2}\right\| \vec{v} \|^{2} \cos ^{2} \theta \\
&=-\|\vec{u}\|^{2}\|\vec{v}\|^{2}\left(\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|\|\vec{v}\|}\right)^{2}+\|\vec{v}\|^{2}\|\vec{u}\|^{2}\\
&=-(\vec{u} \cdot \vec{v})^{2}+\|\vec{v}\|^{2}\|\vec{u}\|^{2}
\end{aligned}
\]
We have proven that
\[
\|\vec{v} \times \vec{u}\|^{2}=-(\vec{u} \cdot \vec{v})^{2}+\|\vec{u}\|^{2}\|\vec{v}\|^{2}
\]
