If $\hat a,\hat b$ and $\hat c$ are three units vectors and $\hat a+\hat b+\hat c=\vec 0$, then the angle between $\hat a$ and ($-\hat b$) is

$\frac{\pi}{3}$

The correct answer is Option (3) → $\frac{\pi}{3}$

Given: $\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{a} + \vec{b} = -\vec{c}$

Take magnitude on both sides:

$|\vec{a} + \vec{b}| = |\vec{c}| = 1$ (since $\vec{c}$ is a unit vector)

Now, use the identity: $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$

Since $\vec{a}$ and $\vec{b}$ are unit vectors:

$1^2 = 1 + 1 + 2\vec{a} \cdot \vec{b} \Rightarrow 1 = 2 + 2\vec{a} \cdot \vec{b}$

$\Rightarrow \vec{a} \cdot \vec{b} = -\frac{1}{2}$

Now, angle between $\vec{a}$ and $-\vec{b}$:

$\vec{a} \cdot (-\vec{b}) = -\vec{a} \cdot \vec{b} = -(-\frac{1}{2}) = \frac{1}{2}$

Since $\vec{a} \cdot (-\vec{b}) = |\vec{a}||\vec{b}|\cos\theta = \cos\theta$

$\cos\theta = \frac{1}{2} \Rightarrow \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$