#### Question

If \[\vec{a}\] and \[\vec{b}\] are two unit vectors such that \[\vec{a} + \vec{b}\] is also a unit vector, then find the angle between \[\vec{a}\] and \[\vec{b}\]

#### Solution

Let the angle between \[\vec{a}\] and \[\vec{b}\] be \[\theta\].

It is given that \[\left| \vec{a} \right| = \left| \vec{b} \right| = \left| \vec{a} + \vec{b} \right| = 1\]

\[\left| \vec{a} + \vec{b} \right| = 1\]

\[ \Rightarrow \left| \vec{a} + \vec{b} \right|^2 = 1\]

\[ \Rightarrow \left| \vec{a} \right|^2 + 2\left| \vec{a} \right|\left| \vec{b} \right|\cos\theta + \left| \vec{b} \right|^2 = 1\]

\[ \Rightarrow 1 + 2 \times 1 \times 1 \times \cos\theta + 1 = 1\]

\[ \Rightarrow 2\cos\theta = - 1\]

\[ \Rightarrow \cos\theta = - \frac{1}{2} = \cos\frac{2\pi}{3}\]

\[ \Rightarrow \theta = \frac{2\pi}{3}\]

Thus, the angle between \[\vec{a}\] and \[\vec{b}\] is \[\frac{2\pi}{3}\]