# Jerk times velocity cross position

The problem asked me to prove the following (where $\vec{r}, \vec{v}, \vec{a}$ are the position, velocity and acceleration vectors of a moving particle):

$\frac{d}{dt} [\vec{a} \cdot (\vec{v} \times \vec{r})] = \dot{\vec{a}} \cdot (\vec{v} \times \vec{r})$

I already did so, but my question is what does the right hand side of the equation say about the motion of the particle? What is the physical meaning of this expression?