Let $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and…

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
  1. $\frac{3}{2}$
  2. $\frac{2}{3}$
  3. $1$
  4. $\frac{3}{4}$

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, $\begin{aligned} |(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}| & =|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})||\overline{\mathrm{c}}| \sin \frac{\pi}{6} \\ & =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|(1)\left(\frac{1}{2}\right) \quad \ldots[\text { From (i) }] \\ & =\frac{3}{2} \quad \ldots[\because \overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}] \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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