Q45 / 90 · JEE Main 2011 · Mathematics
If $\overrightarrow{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})$, then the value of $(2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot[(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})]$ is
- A$-3$
- B$5$
- C$3$
- Correct answerD$-5$
Solution
$(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot\{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})\}=(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot\{[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})] \overline{\mathrm{b}}-[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \overline{\mathrm{a}}]\}$
$=-5(\overline{\mathrm{a}})^2(\overline{\mathrm{b}})^2+5(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2=-5$
Hence, the correct answer is option: 'D'.