If \(\vec{a}, \vec{b}\) and \(\sqrt{3}\vec{a}-\vec{b}\) are three unit vectors, then the angle between \(\vec{a}\) and \(\vec{b}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pi/6\)
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). Since \(|\vec{a}| = |\vec{b}| = 1\) and \(|\sqrt{3}\vec{a} - \vec{b}| = 1\), squaring the magnitude with the dot product gives:
\(|\sqrt{3}\vec{a} - \vec{b}|^2 = 3|\vec{a}|^2 - 2\sqrt{3}\,\vec{a}\cdot\vec{b} + |\vec{b}|^2\)
\(1 = 3 - 2\sqrt{3}\cos\theta + 1\)
\(2\sqrt{3}\cos\theta = 3\)
\(\displaystyle \cos\theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}\)
\(\displaystyle \theta = \frac{\pi}{6}\)