To address this problem, we must ascertain the resultant magnitude of the vector sum or difference of two vectors, \(\vec{A}\) and \(\vec{B}\). These vectors possess equal magnitudes and are oriented at an angle \(\theta\).
We commence with the formula for vector addition of two vectors \(\vec{A}\) and \(\vec{B}\), assuming they have equal magnitudes, denoted by \(R\):
\(|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}\)
Given that \(|\vec{A}| = |\vec{B}| = R\), we substitute these values into the aforementioned formula:
\(|\vec{A} + \vec{B}| = \sqrt{R^2 + R^2 + 2 \cdot R \cdot R \cdot \cos \theta}\)
Upon simplification, the expression becomes:
\(|\vec{A} + \vec{B}| = \sqrt{2R^2(1 + \cos \theta)}\)
Employing the trigonometric identity \(1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right)\), we perform the substitution to obtain:
\(|\vec{A} + \vec{B}| = \sqrt{2R^2 \cdot 2 \cos^2 \left(\frac{\theta}{2}\right)}\)
This simplifies to:
\(|\vec{A} + \vec{B}| = 2R \cos \left(\frac{\theta}{2}\right)\)
Consequently, the correct result is:
\(|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\)
This result aligns with the provided correct option: \(|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\).
An examination of the alternative options further substantiates this conclusion:
Therefore, Option 3 is the accurate and valid choice.