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 $\frac{2 \pi}{3}$, then $|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=$
  1. $\frac{\sqrt{3}}{2}$
  2. $\frac{3\sqrt{3}}{2}$
  3. ${3\sqrt{3}}$
  4. ${4\sqrt{3}}$

Solution

$\begin{aligned} & |\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2} \\ & \Rightarrow|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=8 \\ & \left.\Rightarrow|\overline{\mathrm{c}}|^2+9-2|\overrightarrow{\mathrm{c}}|=8 \quad \ldots[\because \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}| \text { (given })\right] \\ & \Rightarrow(|\overrightarrow{\mathrm{c}}|-1)^2=0 \\ & \Rightarrow \mid \overrightarrow{\mathrm{c}}=1\end{aligned}$ Now, $\begin{aligned}|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}| & =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \sin \frac{2 \pi}{3} \\ & =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|(1)\left(\frac{\sqrt{3}}{2}\right) \\ & =\frac{3 \sqrt{3}}{2} \\ & \ldots[\because \overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}]\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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