Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, then find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$, given that $\left(\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right)$ is a unit vector. [CBSE 2014]

Answer 1

Let the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ be $\theta$.

It is given that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|=1$.

$$\begin{gathered} \left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|=1 \\ \Rightarrow {\left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|}^2=1 \\ \Rightarrow {\left|\sqrt{3}\overrightarrow{a}\right|}^2-2\sqrt{3}\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2=1 \\ \Rightarrow 3{\left|\overrightarrow{a}\right|}^2-2\sqrt{3}\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta +{\left|\overrightarrow{b}\right|}^2=1 \end{gathered}$$
$$\begin{gathered} \Rightarrow 3\times 1-2\sqrt{3}\times 1\times 1\times \cos \theta +1=1 \\ \Rightarrow 2\sqrt{3}\cos \theta =3 \\ \Rightarrow \cos \theta =\frac{\sqrt{3}}{2}=\cos \frac{\mathrm{\pi }}{6} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{6} \end{gathered}$$

Thus, the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\mathrm{\pi }}{6}$.