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

$\frac{\pi}{3}$

The correct answer is Option (2) → $\frac{\pi}{3}$

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

$\vec{a}+\vec{b}=-\vec{c}$ squaring both sides

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

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

$9+25+2|\vec{a}||\vec{b}|\cos θ=49$ θ → angle between a and b

$2×3×5\cos θ=15$

$\cos θ=\frac{1}{2}⇒θ=\frac{\pi}{3}$