Question

If $\overline{a}$ and $\overline{b}$ are two unit vector inclined at angle $\theta$ to each other, then $\left|\overline{a}+\overline{b}\right|<1$ if:

• A. $\theta =\frac{\pi }{6}$
• B. $\theta =\frac{\pi }{2}$
• C. $\theta =\frac{\pi }{3}$
• D. $\frac{2\pi }{3}<\theta \le \pi$

Answer 1

The correct option is D $\frac{2\pi }{3}<\theta \le \pi$

$|\overline{a}+\overline{b}|=\sqrt{{\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+2\left|\overline{a}\right|.\left|\overline{b}\right|.\cos \theta }$

$⟹|\overline{a}+\overline{b}|=\sqrt{1+1+2\cos \theta }$

$\because \overline{a}\&\overline{b}$ are unit vectors.

$⟹(\cos \theta =2\cos {}^2\frac{\theta }{2}-1)$

$⟹1+\cos \theta =2\cos {}^2\frac{\theta }{2}$

$⟹|\overline{a}+\overline{b}|=\sqrt{2(1+\cos \theta )}$

$=\sqrt{2\left(2\cos {}^2\frac{\theta }{2}\right)}$

$=2\sqrt{\cos {}^2\frac{\theta }{2}}$

$=2\cos \frac{\theta }{2}$

According to question:

$⟹2\cos \frac{\theta }{2}<1$

$⟹\cos \frac{\theta }{2}<\frac{1}{2}$

$⟹\frac{\theta }{2}ϵ(\frac{\pi }{3},\frac{\pi }{2}]$ -for first quadrant.

$⟹\frac{\pi }{3}<\frac{\theta }{2}\le \frac{\pi }{2}$

$⟹\frac{2\pi }{3}<\theta \le \pi$