If \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) are two unit vectors such that \(5 \bar{a}+4 \bar{b}\) and \(\bar{a}-2 \bar{b}\) are perpendicular to each other, then the between \(\bar{a}\) and \(\bar{b}\) is

Let \(\theta\) be the angle between \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\).
Since \(\quad \overline{\mathrm{c}}=\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}\) and \(\overline{\mathrm{d}}=5 \overline{\mathrm{a}}+4 \overline{\mathrm{~b}}\). are perpendicular to each other.
\(\begin{aligned}
& \therefore \quad \overline{\mathrm{c}} \cdot \overline{\mathrm{~d}}=0 \\
& \Rightarrow(\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}) \cdot(5 \overline{\mathrm{a}}+4 \overline{\mathrm{~b}})=0 \\
& \Rightarrow 5(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}})-6(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}})-8(\overline{\mathrm{~b}} \cdot \overline{\mathrm{~b}})=0 \\
& \quad \Rightarrow 5|\overline{\mathrm{a}}|^2-6|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta-8|\overline{\mathrm{~b}}|^2 \\
& \quad \Rightarrow 5-6 \cos \theta-8=0 \\
& \Rightarrow \cos \theta=-\frac{1}{2} \\
& \quad \Rightarrow \theta=\frac{2 \pi}{3}
\end{aligned}\)