If \(\vec{a}\) and \(\vec{b}\) are unit vectors such that \(|\vec{a}-\vec{b}| = 1\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{3}\)
Let \(\theta\) be the angle between the vectors. Since \(|\vec{a}| = |\vec{b}| = 1\), squaring the given magnitude gives
\(|\vec{a}-\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b}\)
\(1 = 1+1-2\cos\theta\)
\( \displaystyle \cos\theta = \frac{1}{2} \)
\( \displaystyle \theta = \frac{\pi}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{6}\) — Incorrect. It gives \(|\vec{a}-\vec{b}|^2 = 2-\sqrt{3}\), not 1.
Option (3) \(\displaystyle\frac{\pi}{2}\) — Incorrect. It gives \(|\vec{a}-\vec{b}| = \sqrt{2}\).
Option (4) \(\displaystyle\frac{2\pi}{3}\) — Incorrect. It gives \(|\vec{a}-\vec{b}| = \sqrt{3}\); the angle \(\frac{2\pi}{3}\) is the one for \(|\vec{a}+\vec{b}| = 1\).