Source: Extra Practice

If A⃗\vec{A} and B⃗\vec{B} are two non-zero vectors such that A⃗×B⃗=B⃗×A⃗\vec{A} \times \vec{B} = \vec{B} \times \vec{A}, then the angle between them is:

Options

Option A is correct

00 or π\pi rad

Option B

π/2\pi/2 rad

Option C

π/4\pi/4 rad

Option D

π/3\pi/3 rad

Explanation

By vector cross product definition, we know that B⃗×A⃗=−(A⃗×B⃗)\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B}). If the given condition holds, then A⃗×B⃗=−(A⃗×B⃗)  ⟹  2(A⃗×B⃗)=0  ⟹  A⃗×B⃗=0\vec{A} \times \vec{B} = -(\vec{A} \times \vec{B}) \implies 2(\vec{A} \times \vec{B}) = 0 \implies \vec{A} \times \vec{B} = 0. This means the magnitude ABsin⁡θ=0AB\sin\theta = 0. Since A⃗\vec{A} and B⃗\vec{B} are non-zero vectors, sin⁡θ=0  ⟹  θ=0\sin\theta = 0 \implies \theta = 0 or π\pi radians. The other options are incorrect because they would yield a non-zero cross product where A⃗×B⃗=−B⃗×A⃗≠B⃗×A⃗\vec{A} \times \vec{B} = -\vec{B} \times \vec{A} \neq \vec{B} \times \vec{A}.