If $\hat{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\hat{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$…

If $\hat{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\hat{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$, then the value of $(2 \hat{a}-\hat{b}) \cdot[(\hat{a} \times \hat{b}) \times(\hat{a}+2 \hat{b})]$ is
  1. 5
  2. 3
  3. -5
  4. -3

Solution

Here, $\hat{a} \cdot \hat{b}=0$ $\therefore \quad \hat{a}$ and $\hat{b}$ are perpendicular unit vectors. Now, $(2 \hat{a}-\hat{b}) \cdot\{(\hat{a} \times \hat{b}) \times(\hat{a}+2 \hat{b})\}$ $=\left[\begin{array}{lll} 2 \hat{a}-\hat{b} & \hat{a} \times \hat{b} & \hat{a}+2 \hat{b} \end{array}\right]$ $\begin{aligned} & =-[\hat{a} \times \hat{b} \quad 2 \hat{a}-\hat{b} \quad \hat{a}+2 \hat{b}] \\ & =-(\hat{a} \times \hat{b}) \cdot\{(2 \hat{a}-\hat{b}) \times(\hat{a}+2 \hat{b})\} \\ & =-(\hat{a} \times \hat{b}) \cdot 5(\hat{a} \times \hat{b}) \\ & =-5|\hat{a} \times \hat{b}|=-5|\hat{a}|^2|\hat{b}|^2 \\ & =-5\end{aligned}$ $\begin{aligned} & \ldots[\because \hat{a} \perp \hat{b}] \\ & \ldots[\because|\hat{a}|=|\hat{b}|=1]\end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 2)

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