Source: Extra Practice

If ∣A⃗×B⃗∣=3(A⃗⋅B⃗)|\vec{A} \times \vec{B}| = \sqrt{3}(\vec{A} \cdot \vec{B}), then the value of ∣A⃗+B⃗∣|\vec{A} + \vec{B}| is:

Options

Option A is correct

(A2+B2+AB)1/2(A^2 + B^2 + AB)^{1/2}

Option B

(A2+B2+3AB)1/2(A^2 + B^2 + \sqrt{3}AB)^{1/2}

Option C

(A2+B2+2AB)1/2(A^2 + B^2 + 2AB)^{1/2}

Option D

(A2+B2)1/2(A^2 + B^2)^{1/2}

Explanation

We are given ∣A⃗×B⃗∣=3(A⃗⋅B⃗)  ⟹  ABsin⁡θ=3ABcos⁡θ|\vec{A} \times \vec{B}| = \sqrt{3}(\vec{A} \cdot \vec{B}) \implies AB\sin\theta = \sqrt{3}AB\cos\theta. Dividing both sides by ABcos⁡θAB\cos\theta, we get tan⁡θ=3  ⟹  θ=60∘\tan\theta = \sqrt{3} \implies \theta = 60^\circ. Now, the magnitude of the vector sum is given by ∣A⃗+B⃗∣=A2+B2+2ABcos⁡θ=A2+B2+2ABcos⁡(60∘)=A2+B2+2AB(0.5)=(A2+B2+AB)1/2|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \sqrt{A^2 + B^2 + 2AB\cos(60^\circ)} = \sqrt{A^2 + B^2 + 2AB(0.5)} = (A^2 + B^2 + AB)^{1/2}. Therefore, option 1 is correct.