If \(\bar{a}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}\), then the angle between the vectors \((2 \bar{a}+\bar{b})\) and \((\bar{a}+2 \bar{b})\) is

\(\begin{aligned}
& \text { Given, } 2 \overline{\mathrm{a}}+\overline{\mathrm{b}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\
& \overline{\mathrm{a}}+2 \overline{\mathrm{~b}}=5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\
& |2 \overline{\mathrm{a}}+\overline{\mathrm{b}}|=\sqrt{(4)^2+(-1)^2+(5)^2}=\sqrt{42} \\
& |\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}|=\sqrt{5^2+4^2+1^2}=\sqrt{42}
\end{aligned}\)
\(\therefore \quad\) Angle between \((2 \bar{a}+\bar{b})\) and \((\bar{a}+2 \bar{b})\) is given by
\(\begin{aligned}
\cos \theta & =\frac{(2 \bar{a}+\bar{b})(\bar{a}+2 \bar{b})}{|2 \bar{a}+\bar{b}||\bar{a}+2 \bar{b}|} \\
& =\frac{(4 \hat{i}-\hat{j}+5 \hat{k})(5 \hat{i}+4 \hat{j}+\hat{k})}{\sqrt{42} \cdot \sqrt{42}}=\frac{21}{42} \\
\cos \theta & =\frac{1}{2} \\
\Rightarrow \theta & =\frac{\pi}{3}
\end{aligned}\)