Given vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ such that $\vec{a} + \vec{b} + \vec{c} = 0$, $|\vec{a}| = 3$, $|\vec{b}| = 5$, and $|\vec{c}| = 7$. Show that the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$.

1. Problem Description

Given vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ such that $\vec{a} + \vec{b} + \vec{c} = 0$, $|\vec{a}| = 3$, $|\vec{b}| = 5$, and $|\vec{c}| = 7$. Show that the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$.

2. Solution Steps

Since $\vec{a} + \vec{b} + \vec{c} = 0$, we have $\vec{c} = -(\vec{a} + \vec{b})$.

Taking the magnitude squared of both sides, we get $|\vec{c}|^2 = |-(\vec{a} + \vec{b})|^2 = |\vec{a} + \vec{b}|^2$.

Expanding the magnitude squared, we have:

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

We also know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

So, $|\vec{c}|^2 = |\vec{a}|^2 + 2|\vec{a}| |\vec{b}| \cos{\theta} + |\vec{b}|^2$.

Substituting the given values:

$7^2 = 3^2 + 2(3)(5) \cos{\theta} + 5^2$.

$49 = 9 + 30 \cos{\theta} + 25$.

$49 = 34 + 30 \cos{\theta}$.

$15 = 30 \cos{\theta}$.

$\cos{\theta} = \frac{15}{30} = \frac{1}{2}$.

$\theta = \arccos(\frac{1}{2}) = 60^\circ$.

3. Final Answer

The angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$.