If $\vec a+\vec b+\vec c=0$ and $|\vec a| = 5,|\vec b|=3, |\vec c| = 7$, then the acute angle between $\vec a$ and $\vec b$ is

$\frac{\pi}{3}$

The correct answer is Option (4) → $\frac{\pi}{3}$

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

⟹ $\vec{a} + \vec{b} = -\vec{c}$

Now take modulus square on both sides:

$|\vec{a} + \vec{b}|^2 = |\vec{c}|^2$

$|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{c}|^2$

Substitute values: $25 + 9 + 2\vec{a} \cdot \vec{b} = 49$

⟹ $34 + 2\vec{a} \cdot \vec{b} = 49$

⟹ $2\vec{a} \cdot \vec{b} = 15$

⟹ $\vec{a} \cdot \vec{b} = \frac{15}{2}$

Now, use the dot product formula: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$

$\frac{15}{2} = 5 \cdot 3 \cdot \cos\theta = 15\cos\theta$

⟹ $\cos\theta = \frac{1}{2}$

⟹ $\theta = \frac{\pi}{3} = 60^\circ$

Answer: $60^\circ$