If \(\mathrm{u}=\mathrm{a}-\mathrm{b}\) and \(\mathrm{v}=\mathrm{a}+\mathrm{b}\) and \(|\mathrm{a}|=|\mathrm{b}|=2\), then
\(|\mathbf{u} \times \mathbf{v}|\) is equal to

\(|\mathbf{u} \times \mathbf{v}|=|(\mathbf{a}-\mathbf{b}) \times(\mathbf{a}+\mathbf{b})|\)
\(
=2|\mathbf{a} \times \mathbf{b}| \quad(\because \mathbf{a} \times \mathbf{a}=\mathbf{b} \times \mathbf{b}=0)
\)
and \(|\mathbf{a} \times \mathbf{b}|^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=(a b \sin \theta)^{2}+(a b \cos \theta)^{2}\)
\(
=a^{2} b^{2}
\)
\(
\begin{array}{l}
\Rightarrow |\mathbf{a} \times \mathbf{b}|=\sqrt{a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
\text { So, } \quad|\mathbf{u} \times \mathbf{v}|=2|\mathbf{a} \times \mathbf{b}|
\end{array}
\)
\(
\begin{array}{l}
=2 \sqrt{a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
=2 \sqrt{2^{2} 2^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
\left.=2 \sqrt{16-(\mathbf{a} \cdot \mathbf{b})^{2}} \because|\mathbf{a}|=|\mathbf{b}|=2\right)
\end{array}
\)