Step 1: Conceptual Basis:
The problem involves three vectors and their magnitudes, with a given vector relationship. The objective is to determine the angle between two of these vectors, denoted as $\vec{a}$ and $\vec{b}$. The dot product formula, which connects the dot product, vector magnitudes, and the cosine of the angle between vectors, is the appropriate tool.
Step 2: Fundamental Equation:
The angle $\theta$ between vectors $\vec{a}$ and $\vec{b}$ is defined by the equation $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. The value of $\vec{a} \cdot \vec{b}$ can be derived from the provided vector equation.
Step 3: Derivation and Calculation:
Given the equation $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
To isolate the term involving $\vec{a}$ and $\vec{b}$, we rearrange the equation as $\vec{a} + \vec{b} = -\vec{c}$.
Taking the dot product of both sides with themselves yields the squared magnitudes:
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c}) \]\[ |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 = |\vec{c}|^2 \]Expanding the left side:
\[ \vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b} = |\vec{c}|^2 \]\[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{c}|^2 \]Substituting the given magnitudes: $|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7$.
\[ 3^2 + 2(\vec{a} \cdot \vec{b}) + 5^2 = 7^2 \]\[ 9 + 2(\vec{a} \cdot \vec{b}) + 25 = 49 \]\[ 34 + 2(\vec{a} \cdot \vec{b}) = 49 \]Solving for the dot product:
\[ 2(\vec{a} \cdot \vec{b}) = 49 - 34 = 15 \]\[ \vec{a} \cdot \vec{b} = \frac{15}{2} \]Now, calculating the angle $\theta$ between $\vec{a}$ and $\vec{b}$ using the dot product formula:
\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{15/2}{3 \times 5} = \frac{15/2}{15} = \frac{1}{2} \]For $\theta$ in the range $[0, \pi]$, the angle where $\cos\theta = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.
Step 4: Conclusion:
The angle between vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$.