Concept:
Relating vector sum and difference magnitudes to the angle between the vectors.

Formula:
$$|\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos\theta$$
$$|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB\cos\theta$$

Solution:
Set the squared magnitudes equal to each other:
$$A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta$$
Subtract \(A^2 + B^2\) from both sides:
$$2AB\cos\theta = -2AB\cos\theta$$
$$4AB\cos\theta = 0$$
Since the vectors are non-zero (\(A \neq 0, B \neq 0\)):
$$\cos\theta = 0 \implies \theta = 90^\circ$$

Why other options are incorrect:

• \(0^\circ\): The sum is at a maximum magnitude (\(A+B\)), and the difference is at a minimum magnitude (\(A-B\)).
• \(180^\circ\): The sum is at a minimum magnitude, and the difference is at a maximum magnitude.
• \(60^\circ\): The cosine of 60° is 0.5, which does not satisfy the equality.