To solve this problem, we need to analyze the condition given:$\\$The magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors.$\\$Let's denote the two vectors as $\mathbf{A}$ and $\mathbf{B}. \\$The magnitude of the sum of the vectors is given by:$\\ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \\$The magnitude of the difference of the vectors is given by:$\\ |\mathbf{A} - \mathbf{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta} \\$According to the problem, these magnitudes are equal:$\\ \sqrt{A^2 + B^2 + 2AB \cos \theta} = \sqrt{A^2 + B^2 - 2AB \cos \theta} \\$Squaring both sides to eliminate the square roots, we get:$\\ A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta \\$Simplifying the equation, we have:$\\ 2AB \cos \theta = -2AB \cos \theta \\$Adding $2AB \cos \theta$ to both sides, we obtain:$\\ 4AB \cos \theta = 0 \\$Since $A$ and $B$ are magnitudes of vectors and cannot be zero, we have:$\\ \cos \theta = 0 \\$The angle $\theta$ for which $\cos \theta = 0$ is $90^\circ. \\$Therefore, the angle between the vectors is $90^\circ. \\$This corresponds to Option 4.