Q38 / 110 · VITEEE 2024 · Mathematics

If the unit vectors $\mathbf{a}$ and $\mathbf{b}$ are inclined at $2 \theta$ and $|\mathbf{a}-\mathbf{b}| \lt 1$, then if $0 \lt \theta \lt \pi, \theta$ lies in the interval.

  • Correct answer
    A
    $\left(\frac{5 \pi}{6}, \pi\right]$
  • B
    $\left[0, \frac{\pi}{6}\right]$
  • C
    $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$
  • D
    $\left(\frac{\pi}{2}, \frac{5 \pi}{6}\right]$

Solution

$\because \mathrm{a}$ and b are unit vectors. We have, $\begin{aligned} |a-b| & =\sqrt{(a-b)^2} \\ & =\sqrt{(a-b) \cdot(a-b)} \\ & =\sqrt{|a|^2+|b|^2-2 a \cdot b} \end{aligned}$ $\begin{aligned} &=\sqrt{1+1-2 \cos 2 \theta} \\ &=\sqrt{2(1-\cos 2 \theta)}=2|\sin \theta| \\ & \because|a-b| \lt 1 \Rightarrow \mid \sin \theta \left\lvert\, \lt \frac{1}{2}\right. \\ & \text { Hence, } \theta \in\left[0, \frac{\pi}{6}\right) \text { or }\left(\frac{5 \pi}{6}, \pi\right]\end{aligned}$

Hence, the correct answer is option: 'A'.

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