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 $60^{\circ}$, then the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|$ is
  1. $\frac{\sqrt{3}}{2}$
  2. $\frac{3 \sqrt{3}}{2}$
  3. $\frac{5 \sqrt{3}}{2}$
  4. $\frac{\sqrt{3}}{4}$

Solution

$\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ $\begin{aligned} & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|:=\sqrt{4+4+1}=3 ...(i)\\ & |\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2} \\ & \Rightarrow\left|(\overline{\mathrm{c}}-\overline{\mathrm{a}})^2\right|=8 \end{aligned}$ $\begin{aligned} & \Rightarrow|\vec{c}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=8 \\ & \Rightarrow|\bar{c}|^2+9-2|\bar{c}|=8 \\ & \Rightarrow|\vec{c}|^2-2|\vec{c}|+1=0 \\ & \Rightarrow(|\vec{c}|-1)^2=0 \\ & \Rightarrow|\vec{c}|=1 ...(ii)\end{aligned}$ $\ldots[\because \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|]$
Now, $|(\bar{a} \times \bar{b}) \times \bar{c}|$ $=|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \cdot|\overline{\mathrm{c}}| \sin 60^{\circ}$ $\begin{aligned} & =(3)(1)\left(\frac{\sqrt{3}}{2}\right) \\ & =\frac{3 \sqrt{3}}{2} \end{aligned}$ ...[From (i) and (ii)]

Asked in: MHT CET 2024 (09 May Shift 1)

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