Question

If $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are unit vectors, then what is the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{for}\sqrt{3}\overrightarrow{a}-\overrightarrow{b}$ to be a unit vector?
​$$\begin{gathered} \left(\mathrm{a}\right)\frac{\mathrm{\pi }}{6} \\ \left(\mathrm{b}\right)\frac{\mathrm{\pi }}{4} \\ \left(\mathrm{c}\right)\frac{\mathrm{\pi }}{3} \\ \left(\mathrm{d}\right)\frac{\mathrm{\pi }}{2} \end{gathered}$$

Answer 1

Given:
$\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are unit vectors
$\sqrt{3}\overrightarrow{a}-\overrightarrow{b}$ is a unit vector


$$\begin{gathered} \mathrm{if}\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector} \\ \Rightarrow \left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|=1 \\ \Rightarrow {\left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|}^2=1 \\ \Rightarrow \left(\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right)·\left(\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right)=1 \\ \Rightarrow \sqrt{3}\overrightarrow{a}·\sqrt{3}\overrightarrow{a}+\overrightarrow{b}·\overrightarrow{b}-2\sqrt{3}\overrightarrow{a}·\overrightarrow{b}=1 \\ \Rightarrow 3\overrightarrow{a}·\overrightarrow{a}+\overrightarrow{b}·\overrightarrow{b}-2\sqrt{3}\overrightarrow{a}·\overrightarrow{b}=1 \\ \Rightarrow 3{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\sqrt{3}\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \right)=1 \\ \Rightarrow 3\left(1\right)+\left(1\right)-2\sqrt{3}\left(\left(1\right)\left(1\right)\cos \theta \right)=1\left(\because \left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=1\right) \\ \Rightarrow 4-2\sqrt{3}\left(\cos \theta \right)=1 \\ \Rightarrow 2\sqrt{3}\left(\cos \theta \right)=4-1 \\ \Rightarrow 2\sqrt{3}\left(\cos \theta \right)=3 \\ \Rightarrow \cos \theta =\frac{3}{2\sqrt{3}} \\ \Rightarrow \cos \theta =\frac{\sqrt{3}}{2} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, the correct option is (a).