Step-by-step Solution:

(A) \( |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \) We use the magnitude formula: \[ |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \] \[ |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta} \] For both magnitudes to be equal: \[ A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta \] \[ 4AB \cos \theta = 0 \] \[ \cos \theta = 0 \] \[ \theta = 90^\circ \] So, the correct match for (A) is (III) 90°. --- (B) \( |\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B} \) We use the definitions: \[ |\vec{A} \times \vec{B}| = AB \sin \theta \] \[ \vec{A} \cdot \vec{B} = AB \cos \theta \] For these to be equal: \[ AB \sin \theta = AB \cos \theta \] Dividing by \( AB \) (assuming \( A, B \neq 0 \)): \[ \sin \theta = \cos \theta \] \[ \tan \theta = 1 \] \[ \theta = 45^\circ \] So, the correct match for (B) is (I) 45°. (C) \( \vec{A} \cdot \vec{B} = \frac{AB}{2} \) Using the dot product formula: \[ \vec{A} \cdot \vec{B} = AB \cos \theta \] Given: \[ AB \cos \theta = \frac{AB}{2} \] Dividing by \( AB \): \[ \cos \theta = \frac{1}{2} \] \[ \theta = 60^\circ \] So, the correct match for (C) is (IV) 60°. (D) \( |\vec{A} \times \vec{B}| = \frac{AB}{2} \) Using the cross product formula: \[ |\vec{A} \times \vec{B}| = AB \sin \theta \] Given: \[ AB \sin \theta = \frac{AB}{2} \] Dividing by \( AB \): \[ \sin \theta = \frac{1}{2} \] \[ \theta = 30^\circ \] So, the correct match for (D) is (II) 30°. --- Final Matching: \[ (A) \to (III) 90^\circ, \quad (B) \to (I) 45^\circ, \quad (C) \to (IV) 60^\circ, \quad (D) \to (II) 30^\circ \]