If \( \vec{a} \) and \( \vec{b} \) and are two unit vectors such that \( \vec{a}+2 \vec{b} \) and \( 5 \vec{a}-4 \vec{b} \) are perpendicular to each other, then the angle between \( \vec{a} \) and \( \vec{b} \) is:
Step-by-step Solution:
\[ \text{We have} \] \[ \overrightarrow{a} + 2\overrightarrow{b} \text{ is perpendicular to } 5\overrightarrow{a} - 4\overrightarrow{b} \] \[ \Rightarrow \text{ their dot product has to be } 0. \] \[ \Rightarrow 5a^2 - 4 \overrightarrow{a} \cdot \overrightarrow{b} + 10 \overrightarrow{a} \cdot \overrightarrow{b} - 8b^2 = 0 \] \[ \Rightarrow 5 + 6 \cos \theta - 8 = 0 \] \[ \Rightarrow \cos \theta = \frac{1}{2} \] \[ \text{So, } \theta = 60^\circ \]