Q62 / 150 · MHT CET 2011 · Mathematics
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
- Correct answerA$2 \sqrt{16-(a \cdot b)^{2}}$
- B$\sqrt{16-(\mathrm{a} \cdot \mathbf{b})^{2}}$
- C$2 \sqrt{4-(\mathbf{a} \cdot \mathbf{b})^{2}}$
- D$2 \sqrt{4+(\mathbf{a} \cdot \mathbf{b})^{2}}$
Solution
$|\mathbf{u} \times \mathbf{v}|=|(\mathbf{a}-\mathbf{b}) \times(\mathbf{a}+\mathbf{b})|$
$$
=2|\mathbf{a} \times \mathbf{b}| \quad(\because \mathbf{a} \times \mathbf{a}=\mathbf{b} \times \mathbf{b}=0)
$$
and $|\mathbf{a} \times \mathbf{b}|^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=(a b \sin \theta)^{2}+(a b \cos \theta)^{2}$
$$
=a^{2} b^{2}
$$
$$
\begin{array}{l}
\Rightarrow \quad|\mathbf{a} \times \mathbf{b}|=\sqrt{a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
\text { So, } \quad|\mathbf{u} \times \mathbf{v}|=2|\mathbf{a} \times \mathbf{b}|
\end{array}
$$
$$
\begin{array}{l}
=2 \sqrt{a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
=2 \sqrt{2^{2} 2^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
\left.=2 \sqrt{16-(\mathbf{a} \cdot \mathbf{b})^{2}} \because|\mathbf{a}|=|\mathbf{b}|=2\right)
\end{array}
$$
Hence, the correct answer is option: 'A'.