The **angular momentum** of a particle of momentum $\vec p$ at position $\vec r$ (from origin) is:
- $\vec p \cdot \vec r$
- $\vec r \times \vec p$
- $\vec p \times \vec r$
- $|\vec r||\vec p|$ always
Show answer and solution
Correct answer: B. $\vec r \times \vec p$
$\vec L = \vec r \times \vec p$ (right-hand rule).
Note $\vec r \times \vec p = -\vec p \times \vec r$, order matters.