Let \(\vec a\) and \(\vec b\) are unit vectors. If \(\sqrt{3}\vec a-\vec b\) is a unit vector, then the angle between \(\vec a\) and \(\vec b\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → π/6
Let \(\theta\) be the angle between \(\vec a\) and \(\vec b\). Since \(|\vec a| = |\vec b| = 1\), \(\vec a\cdot\vec b = \cos\theta\). Squaring the magnitude of the unit vector \(\sqrt{3}\vec a - \vec b\):
\(|\sqrt{3}\vec a - \vec b|^2 = 1\)
\(3|\vec a|^2 - 2\sqrt{3}\,\vec a\cdot\vec b + |\vec b|^2 = 1\)
\(3 - 2\sqrt{3}\cos\theta + 1 = 1\)
\(2\sqrt{3}\cos\theta = 3\)
\( \displaystyle \cos\theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \)
\( \displaystyle \theta = \frac{\pi}{6} \)
NCERT Exemplar:
Vector Algebra
