Source: Extra Practice

If ∣A⃗+B⃗∣=∣A⃗−B⃗∣|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|, then the angle between A⃗\vec{A} and B⃗\vec{B} is:

Options

Option A is correct

90∘90^\circ

Option B

60∘60^\circ

Option C

0∘0^\circ

Option D

180∘180^\circ

Explanation

Squaring both sides of the given equation ∣A⃗+B⃗∣=∣A⃗−B⃗∣|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|, we get: A2+B2+2ABcos⁡θ=A2+B2−2ABcos⁡θ  ⟹  4ABcos⁡θ=0A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta \implies 4AB\cos\theta = 0. Since A⃗\vec{A} and B⃗\vec{B} are non-zero, cos⁡θ=0\cos\theta = 0, which yields θ=90∘\theta = 90^\circ. The other options are incorrect because any angle other than 90∘90^\circ would make the magnitudes of the sum and difference unequal.