Q114 / 170 · Manipal (MET) 2015 · Mathematics
Let $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$. If $\mathbf{c}$ is a vector such that $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $30^{\circ}$. Then, $\mid(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$ is equal to
- A$\frac{2}{3}$
- Correct answerB$\frac{3}{2}$
- C$2$
- D$3$
Solution
We have, $\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
and $\quad|\times \mathbf{b}|=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=3$
$\therefore \quad|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \sin 30^{\circ}$
$=\frac{3}{2}|c|$
Given,
$|c-a|=2 \sqrt{2}$
$\begin{array}{rr}\Rightarrow & |c-a|^2=8 \\ \Rightarrow & |c|^2+|a|^2-2(\mathbf{a} \cdot \mathbf{c})=8 \\ \Rightarrow & |c|^2+9-2|\mathbf{c}|=8 \\ \Rightarrow & |c|^2-2|\mathbf{c}|+1=0 \\ \Rightarrow & |c|=1\end{array}$
Hence, $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=\frac{3}{2}$
Hence, the correct answer is option: 'B'.