Question

If the unit vectors $a$ and $b$ are inclined at $2\theta$ and $|a-b|<1$, then if $0\le \theta \le \pi ,\theta$ lies in the interval

• A. $[0,\frac{\pi }{6})$
• B. $(\frac{5\pi }{6},\pi ]$
• C. $[\frac{\pi }{6},\frac{\pi }{2}]$
• D. $(\frac{\pi }{2},\frac{5\pi }{6}]$

Answer 1

Answer: (A), (B)

The correct options are
A $(\frac{5\pi }{6},\pi ]$
B $[0,\frac{\pi }{6})$

Since $a$ and $b$ are unit vectors, we have
$|a-b|=\sqrt{{(a-b)}^2}=\sqrt{(a-b).(a-b)}=\sqrt{a^2+b^2-2a.b}=\sqrt{1+1-2\cos 2\theta }=\sqrt{2(1-\cos 2\theta )}=\sqrt{2\left(2{\sin }^2\theta \right)}=2\left|\sin \theta \right|$
Therefore $|a-b|<1$ implies
$\left|\sin \theta \right|<\frac{1}{2}\Rightarrow \theta \in [0,\frac{\pi }{6})$ or $(\frac{5\pi }{6},\pi ]$