Question

If $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$are two unit vectors inclined at an angle θ, such that $\left|\overrightarrow{a}+\overrightarrow{b}\right|<1,$ then
(a) $\mathrm{\theta }<\frac{\mathrm{\pi }}{3}$

(b) $\mathrm{\theta }>\frac{2\mathrm{\pi }}{3}$

(c) $\frac{\mathrm{\pi }}{3}<\mathrm{\theta }<\frac{2\mathrm{\pi }}{3}$

(d) $\frac{2\mathrm{\pi }}{3}<\mathrm{\theta }<\mathrm{\pi }$

Answer 1

(d) $\frac{2\mathrm{\pi }}{3}<\mathrm{\theta }<\mathrm{\pi }$

$$\begin{gathered} \mathrm{We}\mathrm{have} \\ \left|\overrightarrow{a}+\overrightarrow{b}\right|<1 \\ \Rightarrow \sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \theta }<1 \\ \Rightarrow \sqrt{1^2+1^2+2\times 1\times 1\times \cos \theta }<1 \\ \Rightarrow \sqrt{2+2\cos \theta }<1 \\ \Rightarrow \sqrt{2\left(1+\cos \theta \right)}<1 \\ \Rightarrow \sqrt{2\times 2{\cos }^2\frac{\theta }{2}}<1 \\ \Rightarrow 2\left|\cos \frac{\theta }{2}\right|<1 \\ \Rightarrow \left|\cos \frac{\theta }{2}\right|<\frac{1}{2} \\ \Rightarrow \frac{\mathrm{\pi }}{3}<\frac{\theta }{2}<\frac{2\mathrm{\pi }}{3} \\ \Rightarrow \frac{2\mathrm{\pi }}{3}<\mathrm{\theta }<\frac{4\mathrm{\pi }}{3} \\ \mathrm{But}\mathrm{here}\theta \mathrm{cannot}\mathrm{be}\mathrm{more}\mathrm{than}\mathrm{\pi }. \\ \therefore \frac{2\mathrm{\pi }}{3}<\mathrm{\theta }<\mathrm{\pi } \end{gathered}$$