Q44 / 160 · TS EAMCET 2010 · Mathematics

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

  • Correct answer
    A
    $2 \sqrt{16-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$
  • B
    $\sqrt{16-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$
  • C
    $2 \sqrt{4-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$
  • D
    $\sqrt{4-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$

Solution

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 $|\hat{\mathbf{n}}|=1\}$ $=2 \sqrt{4 \cdot 4 \cdot \sin ^2 \theta \cdot 1}$ $=2 \sqrt{16\left(1-\cos ^2 \theta\right)}$ $=2 \sqrt{16-16 \cos ^2 \theta}$ $=2 \sqrt{16-16\left(\frac{\vec{a} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}\right)^2}$ $\because\{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta\}$ $=2 \sqrt{16-16 \frac{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}{|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2}}$ $=2 \sqrt{16-16 \frac{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}{4 \cdot 4}}$ $\Rightarrow \quad 2 \sqrt{16-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2}$

Hence, the correct answer is option: 'A'.

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