If $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are two unit vectors such that…
If $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are two unit vectors such that $\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}$ and $5 \bar{a}-4 \bar{b}$ are perpendicular to each other, then the angle between $\bar{a}$ and $\bar{b}$ is
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\cos ^{-1}\left(\frac{1}{3}\right)$
$\quad \cos ^{-1}\left(\frac{3}{7}\right)$
Solution
Let $\theta$ be the angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$.
Since $\overline{\mathrm{c}}=\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}$ and $\overline{\mathrm{d}}=5 \overline{\mathrm{a}}-4 \overline{\mathrm{~b}}$ are perpendicular to each other.
$\begin{aligned}
\therefore \quad & \overline{\mathrm{c}} \cdot \overline{\mathrm{~d}}=0 \\
& \Rightarrow(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}) \cdot(5 \overline{\mathrm{a}}-4 \overline{\mathrm{~b}})=0 \\
& \Rightarrow 5(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}})+6(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}})-8(\overline{\mathrm{~b}} \cdot \overline{\mathrm{~b}})=0 \\
& \Rightarrow 5|\overline{\mathrm{a}}|^2+6|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta-8|\overline{\mathrm{~b}}|^2=0 \\
& \Rightarrow 5+6 \cos \theta-8=0 \\
& \Rightarrow \cos \theta=\frac{1}{2} \\
& \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}$