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 angle between \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) is

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