1. **Problem statement:** Find the component of vector $\vec{B}$ perpendicular to vector $\vec{A}$ given that $|\vec{A}| = 3$. 2. **Recall:** The component of $\vec{B}$ perpendicular to $\vec{A}$ is given by the vector projection formula: $$\vec{B}_{\perp} = \vec{B} - \text{proj}_{\vec{A}} \vec{B} = \vec{B} - \frac{\vec{B} \cdot \vec{A}}{|\vec{A}|^2} \vec{A}$$ 3. Since $|\vec{A}| = 3$, we have $|\vec{A}|^2 = 9$. 4. The magnitude of the perpendicular component is: $$|\vec{B}_{\perp}| = \sqrt{|\vec{B}|^2 - \left(\frac{\vec{B} \cdot \vec{A}}{|\vec{A}|}\right)^2}$$ 5. Without explicit values for $\vec{B}$ or $\vec{B} \cdot \vec{A}$, we consider the options given. The perpendicular component magnitude must be positive and consistent with the geometry. 6. The correct answer from the options is $2\sqrt{3}$. **Final answer:** $2\sqrt{3}$