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If $\overrightarrow{\mathbf{u}}=\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}},|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|=2$ then $|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|$ is equal to
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Verified Answer
The correct answer is:
$2 \sqrt{16-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$
We have, $\overrightarrow{\mathbf{u}}=\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$
$\Rightarrow \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
$=0-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-0$
$=2 \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$
$\Rightarrow \quad|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
$=2 \sqrt{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2}$
$=2 \sqrt{|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2} \sin ^2 \theta|\hat{\mathbf{n}}|^2$
$\{\because \hat{\mathbf{n}}=$ unit vector $|\hat{\mathbf{n}}|=1\}$
$=2 \sqrt{4 \cdot 4 \cdot \sin ^2 \theta \cdot 1}$
$=2 \sqrt{16\left(1-\cos ^2 \theta\right)}$
$=2 \sqrt{16-16 \cos ^2 \theta}$
$=2 \sqrt{16-16\left(\frac{\vec{a} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}\right)^2}$
$\because\{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta\}$
$=2 \sqrt{16-16 \frac{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}{|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2}}$
$=2 \sqrt{16-16 \frac{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}{4 \cdot 4}}$
$\Rightarrow \quad 2 \sqrt{16-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$
$\Rightarrow \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
$=0-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-0$
$=2 \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$
$\Rightarrow \quad|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
$=2 \sqrt{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2}$
$=2 \sqrt{|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2} \sin ^2 \theta|\hat{\mathbf{n}}|^2$
$\{\because \hat{\mathbf{n}}=$ unit vector $|\hat{\mathbf{n}}|=1\}$
$=2 \sqrt{4 \cdot 4 \cdot \sin ^2 \theta \cdot 1}$
$=2 \sqrt{16\left(1-\cos ^2 \theta\right)}$
$=2 \sqrt{16-16 \cos ^2 \theta}$
$=2 \sqrt{16-16\left(\frac{\vec{a} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}\right)^2}$
$\because\{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta\}$
$=2 \sqrt{16-16 \frac{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}{|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2}}$
$=2 \sqrt{16-16 \frac{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}{4 \cdot 4}}$
$\Rightarrow \quad 2 \sqrt{16-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$
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