If $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are two unit vectors such that $5 \bar{a}+4 \bar{b}$…
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
$\frac{2 \pi}{3}$
$\cos ^{-1}\left(\frac{2}{3}\right)$
$\frac{\pi}{3}$
$\cos ^{-1}\left(\frac{1}{3}\right)$
Solution
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.
$\begin{aligned}
& \therefore \quad \overline{\mathrm{c}} \cdot \overline{\mathrm{~d}}=0 \\
& \Rightarrow(\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}) \cdot(5 \overline{\mathrm{a}}+4 \overline{\mathrm{~b}})=0 \\
& \Rightarrow 5(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}})-6(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}})-8(\overline{\mathrm{~b}} \cdot \overline{\mathrm{~b}})=0 \\
& \quad \Rightarrow 5|\overline{\mathrm{a}}|^2-6|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta-8|\overline{\mathrm{~b}}|^2 \\
& \quad \Rightarrow 5-6 \cos \theta-8=0 \\
& \Rightarrow \cos \theta=-\frac{1}{2} \\
& \quad \Rightarrow \theta=\frac{2 \pi}{3}
\end{aligned}$