Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}$ and $\bar{c}$ be a vector such that…

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}$ and $\bar{c}$ be a vector such that $|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=4, \quad|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=3$ and the angle between $\overline{\mathrm{c}}$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}$ is $\frac{\pi}{6}$, then $\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}$ is equal to
  1. $-3$
  2. $\frac{3}{2}$
  3. $3$
  4. $\frac{-3}{2}$

Solution

$\begin{array}{ll} & \overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ \therefore \quad & |\overline{\mathrm{a}}|=\sqrt{4+1+4}=3 \\ & \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=\sqrt{4+4+1}=3\end{array}$ $\text { Angle between } \overline{\mathrm{c}} \text { and } \overline{\mathrm{a}} \times \overline{\mathrm{b}} \text { is } \frac{\pi}{6} \quad \ldots \text { [Given] }$ $\begin{aligned} \therefore \quad & \sin \frac{\pi}{6}=\frac{|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}} \| \overline{\mathrm{c}}|} \\ & \frac{1}{2}=\frac{3}{3 \times|\overline{\mathrm{c}}|} \Rightarrow|\overline{\mathrm{c}}|=2\end{aligned}$ Now, $|\bar{c}-\bar{a}|=4$.. [Given] $\begin{aligned} & \Rightarrow|\bar{c}|^2+|\bar{a}|^2-2 \bar{a} \cdot \bar{c}=16 \\ & \Rightarrow 4+9-2 a \cdot c=16 \\ & \Rightarrow a \cdot c=\frac{-3}{2}\end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 2)

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