If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are vectors in space given by…
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are vectors in space given by $\overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}}$ and $\overrightarrow{\mathbf{b}}=\frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}}$, then the value of $(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]$ is
Solution
From the given information, it is clear that
$
\begin{gathered}
\overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}} \\
\Rightarrow \quad|\overrightarrow{\mathbf{a}}|=1,|\overrightarrow{\mathbf{b}}|=1, \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 \\
\text { Now, }(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]
\end{gathered}
$
$
\begin{aligned}
& =(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot\left[a^2 \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}+2 b^2 \cdot \overrightarrow{\mathbf{a}}\right. \\
& =[2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}] \cdot[\overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{a}}]=4 \overrightarrow{\mathbf{a}}^2+\overrightarrow{\mathbf{b}}^2 \\
& =4 \cdot 1+1=5 \quad \text { [as } a \cdot b=0]
\end{aligned}
$