If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{\imath}+\hat{k}), \overline{\mathrm{b}}=\frac{1}{7}(2 \hat{\imath}+3…

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

Solution

Given vectors $\bar{a} \& \bar{b}$ are unit vectors. $\begin{array}{l} \text { Now }(2 \bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})] \\ \quad=-(2 \bar{a}-\bar{b}) \cdot[(\bar{a}+2 \bar{b}) \times(\bar{a} \times \bar{b})] \\ \quad=-(2 \bar{a}-\bar{b}) \cdot[\bar{a} \times(\bar{a} \times \bar{b})+2 \bar{b} \times(\bar{a} \times \bar{b})] \\ \quad=-(2 \bar{a}-\bar{b})[(\bar{a} \cdot \bar{b}) \cdot \bar{a}-(\bar{a} \cdot \bar{a}) \cdot \bar{b}+2(\bar{b} \cdot \bar{b}) \cdot \bar{a}-2(\bar{b} \cdot \bar{a}) \cdot \bar{b}] \\ \quad=-(2 \bar{a}-\bar{b})[(\bar{a} \cdot \bar{b}) \bar{a}-\bar{b}+2 \bar{a}-2(\bar{b} \cdot \bar{a}) \bar{b}] \\ \text { Here } \quad \bar{a} \cdot \bar{b}=\frac{3(2)}{7 \sqrt{10}}-\frac{1(6)}{7 \sqrt{10}}=0 \end{array}$ Hence given expression becomes $\begin{array}{l} =-(2 \bar{a}-\bar{b})[-\bar{b}+2 \bar{a}]=-(2 \bar{a}-\bar{b})^{2} \\ =-\left[4 \bar{a}^{2}+\bar{b}^{2}-4 \bar{a} \cdot \bar{b}\right]=-(4+1)=-5 \end{array}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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