Source: Extra Practice

Two vectors A⃗\vec{A} and B⃗\vec{B} lie in a plane. Another vector C⃗\vec{C} is added to them such that the resultant is zero. If ∣A⃗∣=3|\vec{A}| = 3, ∣B⃗∣=4|\vec{B}| = 4, and ∣C⃗∣=5|\vec{C}| = 5, what is the angle between A⃗\vec{A} and B⃗\vec{B}?

Options

Option A is correct

90∘90^\circ

Option B

180∘180^\circ

Option C

0∘0^\circ

Option D

60∘60^\circ

Explanation

Given that the resultant of the three vectors is zero, we have A⃗+B⃗+C⃗=0\vec{A} + \vec{B} + \vec{C} = 0, which implies A⃗+B⃗=−C⃗\vec{A} + \vec{B} = -\vec{C}. Taking the magnitude on both sides: ∣A⃗+B⃗∣2=∣−C⃗∣2  ⟹  A2+B2+2ABcos⁡θ=C2|\vec{A} + \vec{B}|^2 = |-\vec{C}|^2 \implies A^2 + B^2 + 2AB\cos\theta = C^2, where θ\theta is the angle between A⃗\vec{A} and B⃗\vec{B}. Substituting the given values: 32+42+2(3)(4)cos⁡θ=52  ⟹  9+16+24cos⁡θ=25  ⟹  25+24cos⁡θ=25  ⟹  24cos⁡θ=0  ⟹  cos⁡θ=03^2 + 4^2 + 2(3)(4)\cos\theta = 5^2 \implies 9 + 16 + 24\cos\theta = 25 \implies 25 + 24\cos\theta = 25 \implies 24\cos\theta = 0 \implies \cos\theta = 0. Therefore, θ=90∘\theta = 90^\circ. The options 180∘180^\circ, 0∘0^\circ, and 60∘60^\circ are incorrect because they would yield resultant magnitudes different from 55 units.