Q. Let $\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}$ be unit vectors inclined at an angle $2\alpha(0 \leq \alpha \leq \pi)$ each other, then $\left| \ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right| < 1$, if
A) $\alpha = \frac{\pi}{2}$
B) $\alpha < \frac{\pi}{3}$
C) $\alpha > \frac{2\pi}{3}$
D) $\frac{\pi}{3} < \alpha < \frac{2\pi}{3}$
Explanation
(d)
Given, $\left| \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right| < 1$
$\Rightarrow \ \sqrt{1 + 1 + 2\cos{2\alpha}} < 1$
$\Rightarrow \ \sqrt{2(1 + \cos{2\alpha)}} < 1$
$\Rightarrow \ \sqrt{4\cos^{2}\alpha} < 1$
$\Rightarrow {|cos}{\alpha| < \frac{1}{2}}$
$\Rightarrow \frac{\pi}{3} < \alpha < \frac{2\pi}{3}\ \ (\because 0 \leq \alpha \leq \pi)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.