Q59 / 150 · MHT CET 2020 (12 Oct Shift 1) · Mathematics
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
- A7
- Correct answerB-5
- C5
- D-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
$\begin{array}{l}
=-(2 \bar{a}-\bar{b})[-\bar{b}+2 \bar{a}]=-(2 \bar{a}-\bar{b})^{2} \\
=-\left[4 \bar{a}^{2}+\bar{b}^{2}-4 \bar{a} \cdot \bar{b}\right]=-(4+1)=-5
\end{array}$
Hence, the correct answer is option: 'B'.