Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\bar{c}=\hat{a}+2 \hat{b}$ and $\bar{d}=5…
Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\bar{c}=\hat{a}+2 \hat{b}$ and $\bar{d}=5 \hat{a}+4 \hat{b}$ are perpendicular to each other, then the angle between $\hat{a}$ and $\hat{b}$ is
$\frac{\pi}{6}$
$\cos ^{-1}\left(\frac{13}{14}\right)$
$\frac{\pi}{3}$
$\quad \cos ^{-1}\left(\frac{-13}{14}\right)$
Solution
Let $\theta$ be the angle between $\hat{a}$ and $\hat{b}$.
Since $\bar{c}=\hat{a}+2 \hat{b}$ and $\bar{d}=5 \hat{a}+4 \hat{b}$ are perpendicular to each other.
$\begin{aligned}
& \therefore \quad \overline{\mathrm{c}} \cdot \hat{\mathrm{~d}}=0 \\
& \Rightarrow(\hat{\mathrm{a}}+2 \hat{\mathrm{~b}}) \cdot(5 \hat{\mathrm{a}}+4 \hat{\mathrm{~b}})=0 \\
& \Rightarrow 5(\hat{\mathrm{a}} \cdot \hat{\mathrm{a}})+14(\hat{\mathrm{a}} \cdot \hat{\mathrm{~b}})+8(\hat{\mathrm{~b}} \cdot \hat{\mathrm{~b}})=0 \\
& \Rightarrow 5|\hat{a}|^2+14|\hat{a}||\hat{\mathrm{b}}| \cos \theta+8|\hat{\mathrm{~b}}|^2=0 \\
& \Rightarrow 5+14 \cos \theta+8=0 \\
& \Rightarrow \cos \theta=-\frac{13}{14} \\
& \Rightarrow \theta=\cos ^{-1}\left(-\frac{13}{14}\right)
\end{aligned}$