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

$\frac{\pi}{3}$

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

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

Taking dot product with itself:

$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c})$

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

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

$2\vec{a}\cdot\vec{b} = 49 - 34 = 15 ⇒\vec{a}\cdot\vec{b} = \frac{15}{2}$

$\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{15/2}{3 \times 5} = \frac{15}{30} = \frac{1}{2}$

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