Let $\overline{\mathrm{a}}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ is a…

Let $\overline{\mathrm{a}}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ 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 $30^{\circ}$, then $|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|$ is equal to
  1. $\frac{3}{2}$
  2. $\frac{2}{3}$
  3. $-\frac{3}{2}$
  4. $-\frac{2}{3}$

Solution

$\begin{aligned} & \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right| \\ & =2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=\sqrt{4+4+1}=3 ...(i)\\ & |\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2} \\ & \Rightarrow(\overline{\mathrm{c}}-\overline{\mathrm{a}})^2=8 \end{aligned}$ $\begin{aligned} & \Rightarrow|\overline{\mathrm{c}}|^2+|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=8 \\ & \Rightarrow|\overline{\mathrm{c}}|^2+9-2|\overline{\mathrm{c}}|=8 \\ & \Rightarrow|\overline{\mathrm{c}}|^2-2|\overline{\mathrm{c}}|^2+1=0 \\ & \Rightarrow\left(|\overline{\mathrm{c}}|-1\right)^2=0 \\ & \Rightarrow \mid \overline{\mathrm{c}}=1 ...(i)\\ & \text {Now, }|(\overline{\mathrm{a}} \times \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \times \overline{\mathrm{c}}=|\overline{\mathrm{c}}|] \\ & =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \cdot|\overline{\mathrm{c}}| \sin 30^{\circ} \\ & =(3)(1)\left(\frac{1}{2}\right) \\ & =\frac{3}{2} \end{aligned}$ ...[From (i) and (ii)]

Asked in: MHT CET 2024 (15 May Shift 1)

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