Q61 / 180 · COMEDK 2018 · Mathematics
If $\mathbf{u}=\mathbf{a}-\mathbf{b}, \mathbf{v}=\mathbf{a}+\mathbf{b}$, and $|\mathbf{a}|=|\mathbf{b}|=2$, then $|\mathbf{u} \times \mathbf{v}|$ is
- Correct answerA$2 \sqrt{16-(\mathbf{a} \cdot \mathbf{b})^{2}}$
- B$2 \sqrt{4-(\mathbf{a} \cdot \mathbf{b})^{2}}$
- C$\sqrt{16-(\mathbf{a} \cdot \mathbf{b})^{2}}$
- D$\sqrt{4-(\mathbf{a} \cdot \mathbf{b})^{2}}$
Solution
Given that, $\mathbf{u}=\mathbf{a}-\mathbf{b}, \mathbf{v}=\mathbf{a}+\mathbf{b}$ and $|\mathbf{a}|=|\mathbf{b}|=2$
Now, $|\mathbf{u} \times \mathbf{v}|=|\mathbf{a}-\mathbf{b} \times \mathbf{a}+\mathbf{b}|$ $=2(\mathbf{a} \times \mathbf{b})=2|\mathbf{a}||\mathbf{b}| \sin \theta$ $=2 \times 2 \times 2 \sin \theta=8 \sin \theta=8 \sqrt{1-\cos ^{2} \theta}$ $=8 \sqrt{1-\left(\frac{a \cdot b}{|a||b|}\right)^{2}}$
$$
=8 \sqrt{1-\left(\frac{\mathbf{a} \cdot \mathbf{b}}{4}\right)^{2}}=8 \sqrt{1-\left(\frac{\mathbf{a} \cdot \mathbf{b}}{16}\right)^{2}}
$$
$$
=\frac{8}{4} \sqrt{16-(a \cdot b)^{2}}=2 \sqrt{16-(a \cdot b)^{2}}
$$
Hence, the correct answer is option: 'A'.