Step-by-Step Solution

1. Given that \(\vec{p}\) and \(\vec{q}\) are unit vectors, we have \(|\vec{p}| = 1\) and \(|\vec{q}| = 1\).

2. We are given that \((\vec{p} + \vec{q})\) is a unit vector. Therefore, \(|\vec{p} + \vec{q}| = 1\).

3. We know that \(|\vec{p} + \vec{q}|^2 = (\vec{p} + \vec{q}) \cdot (\vec{p} + \vec{q}) = |\vec{p}|^2 + 2(\vec{p} \cdot \vec{q}) + |\vec{q}|^2\).

4. Since \(|\vec{p} + \vec{q}| = 1\), we have \(|\vec{p} + \vec{q}|^2 = 1^2 = 1\).

5. Also, \(\vec{p} \cdot \vec{q} = |\vec{p}| |\vec{q}| \cos{\alpha} = (1)(1) \cos{\alpha} = \cos{\alpha}\).

6. Substituting the values, we get \(1 = 1^2 + 2\cos{\alpha} + 1^2\), which simplifies to \(1 = 1 + 2\cos{\alpha} + 1\).

7. This gives us \(2\cos{\alpha} = -1\), so \(\cos{\alpha} = -\frac{1}{2}\).

8. The angle \(\alpha\) for which \(\cos{\alpha} = -\frac{1}{2}\) is \(\alpha = \frac{2\pi}{3}\).