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45. If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})$, then the value of $(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})]$ is
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The correct answer is:
$-5$
Since $\bar{a} \cdot \bar{b}=0$
$\therefore \quad \overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are perpendicular unit vectors.
Now, $(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})]$
$\begin{aligned} & =[2 \overline{\mathrm{a}}-\overline{\mathrm{b}} \overline{\mathrm{a}} \times \overline{\mathrm{b}} \overline{\mathrm{a}}+2 \overline{\mathrm{b}}] \\ & =-[\overline{\mathrm{a}} \times \overline{\mathrm{b}} 2 \overline{\mathrm{a}}-\overline{\mathrm{b}} \overline{\mathrm{a}}+2 \overline{\mathrm{b}}] \\ & =-(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot\{(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})\} \\ & =-5(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})\end{aligned}$
$\begin{array}{ll}=-5|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2=-5|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2 & \ldots . \cdot[\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}}] \\ =-5 & \ldots . \cdot[\because|\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|=1]\end{array}$
$\therefore \quad \overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are perpendicular unit vectors.
Now, $(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})]$
$\begin{aligned} & =[2 \overline{\mathrm{a}}-\overline{\mathrm{b}} \overline{\mathrm{a}} \times \overline{\mathrm{b}} \overline{\mathrm{a}}+2 \overline{\mathrm{b}}] \\ & =-[\overline{\mathrm{a}} \times \overline{\mathrm{b}} 2 \overline{\mathrm{a}}-\overline{\mathrm{b}} \overline{\mathrm{a}}+2 \overline{\mathrm{b}}] \\ & =-(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot\{(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})\} \\ & =-5(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})\end{aligned}$
$\begin{array}{ll}=-5|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2=-5|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2 & \ldots . \cdot[\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}}] \\ =-5 & \ldots . \cdot[\because|\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|=1]\end{array}$
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