If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}),…

If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{5}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$, then the value of $(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\}$ is
  1. 5
  2. -3
  3. -5
  4. 3

Solution

$\therefore \quad \overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are perpendicular unit vectors. $\begin{aligned} & \text { Now, }(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\} \\ & =[2 \bar{a}-\bar{b} \bar{a} \times \bar{b} \bar{a}+2 \bar{b}] \\ & =-[\bar{a} \times \bar{b} 2 \bar{a}-\bar{b} \bar{a}+2 \bar{b}] \\ & =-(\bar{a} \times \bar{b}) \cdot\{(2 \bar{a}-\bar{b}) \times(\bar{a}+2 \bar{b})\} \end{aligned}$ $\begin{aligned} & =-(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot 5(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \\ & =-5|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=-5|\overline{\mathrm{a}}|^2|\overline{\mathrm{~b}}|^2 \\ & =-5\end{aligned}$ $\begin{aligned} & \cdots \cdot[\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}}] \\ & \cdots \cdot[\because|\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|=1]\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 2)

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