Q75 / 150 · MHT CET 2023 (11 May Shift 1) · Mathematics
Let $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. If $\overline{\mathrm{c}}$ is a vector such that $\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|,|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2}$ and the angle between $(\overline{\mathrm{a}} \times \overline{\mathrm{b}})$ and $\overline{\mathrm{c}}$ is $\frac{\pi}{6}$, then $|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|$ is
- Correct answerA$\frac{3}{2}$
- B$\frac{2}{3}$
- C$1$
- D$\frac{3}{4}$
Solution
$$
\begin{aligned}
& |\overline{\mathrm{c}}-\overline{\mathrm{a}}|^2=2 \sqrt{2} \\
& \Rightarrow|\overline{\mathrm{c}}|^2+|\overline{\mathrm{a}}|^2-2(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=8 \\
& \Rightarrow|\overline{\mathrm{c}}|^2+9-2|\overline{\mathrm{c}}|=8 \\
& \Rightarrow\left(|\overline{\mathrm{c}}|^{-1}\right)^2=0 \\
& \Rightarrow|\overline{\mathrm{c}}|^2=1
\end{aligned}
$$
Now,
$$
\begin{aligned}
|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}| & =|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})||\overline{\mathrm{c}}| \sin \frac{\pi}{6} \\
& =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|(1)\left(\frac{1}{2}\right) \quad \ldots[\text { From (i) }] \\
& =\frac{3}{2} \quad \ldots[\because \overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}]
\end{aligned}
$$
Hence, the correct answer is option: 'A'.