If unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are inclined at an angle $2\theta$ such that $|\overrightarrow{a}-\overrightarrow{b}|<1$ and $0\le \theta \le \pi ,$ then $\theta$ lies in the interval

$\Rightarrow |\overrightarrow{a}-\overrightarrow{b}{|}^2=2-2\cos 2\theta (\because \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1)$

$\Rightarrow |\overrightarrow{a}-\overrightarrow{b}{|}^2=4{\sin }^2\theta \Rightarrow |\overrightarrow{a}-\overrightarrow{b}|=2|\sin \theta |$

Now, $|\overrightarrow{a}-\overrightarrow{b}|<1\Rightarrow 2|\sin \theta |<1$

$\Rightarrow |\sin \theta |<\frac{1}{2}\Rightarrow \theta \in \left[0,\frac{\pi }{6}\right]\text{ or }\theta \in \left(\frac{5\pi }{6},\pi \right)$