Q93 / 200 · Manipal (MET) 2010 · Mathematics
Let $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$, if $\overrightarrow{\mathbf{c}}$ is a vector such that $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}=|\overrightarrow{\mathbf{c}}|,|\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|=2 \sqrt{2}$ and the angle between $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ is $30^{\circ}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to
- A$\frac{2}{3}$
- Correct answerB$\frac{3}{2}$
- C2
- D3
Solution
We have, $\vec{a} \cdot \vec{c}=|\vec{c}|$ and $|\vec{c}-\vec{a}|=2 \sqrt{2}$
$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}| \text { and }|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})=8$
$\Rightarrow |\vec{c}|^2+9-2|\vec{c}|=8$
$\Rightarrow (|\overrightarrow{\mathbf{c}}|-1)^2=0$
$\Rightarrow |\vec{c}|=1$
$\Rightarrow |(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}|=|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \sin 30^{\circ}$
$=\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{3}{2}$
$[\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}]$
Hence, the correct answer is option: 'B'.