Vector Algebra
Unit Vectors and Angles
Grade 12
Question:
<p>If the unit vectors <span class="math">\mathbf{e}_1</span> and <span class="math">\mathbf{e}_2</span> are inclined at an angle <span class="math">2\theta</span> and <span class="math">|\mathbf{e}_1 - \mathbf{e}_2| < 1</span>, then for <span class="math">\theta \in [0, \pi]</span>, <span class="math">\theta</span> may lie in the interval</p>
<p>(a) <span class="math">\left[0, \frac{\pi}{6}\right]</span></p>
<p>(b) <span class="math">\left[\frac{\pi}{6}, \frac{\pi}{2}\right]</span></p>
<p>(c) <span class="math">\left[\frac{5\pi}{6}, \pi\right]</span></p>
<p>(d) <span class="math">\left[\frac{\pi}{2}, \frac{5\pi}{6}\right]</span></p>
Step-by-Step Solution
Key Concept: Use the formula for magnitude of difference of vectors and the relation with angle between them to set up an inequality.
Solution: Given that \mathbf{e}_1 and \mathbf{e}_2 are unit vectors inclined at angle 2\theta : |\mathbf{e}_1 - \mathbf{e}_2|^2 = |\mathbf{e}_1|^2 + |\mathbf{e}_2|^2 - 2\mathbf{e}_1 \cdot \mathbf{e}_2 = 1 + 1 - 2\cos(2\theta) = 2(1 - \cos(2\theta)) = 4\sin^2(\theta) Therefore: |\mathbf{e}_1 - \mathbf{e}_2| = 2|\sin(\theta)| Given |\mathbf{e}_1 - \mathbf{e}_2| < 1 : 2|\sin(\theta)| < 1 |\sin(\theta)| < \frac{1}{2} For \theta \in [0, \pi] , this gives \theta \in \left[0, \frac{\pi}{6}\right] \cup \left[\frac{5\pi}{6}, \pi\right] The answer is (a).
Correct Answer: A