Physical Interpretation of a Vector Quantity

[Bashyboy]

Homework Statement

If $\mathbf{r}, \mathbf{v},\mathbf{a}$ denote the position, velocity, and acceleration of a particle, prove that

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

Homework Equations

The Attempt at a Solution

I have already proven the result, but am now wondering the what the physical significance of this vector quantity is. I believe I have some idea:

The vectors $\mathbf{v}$ and $\mathbf{r}$ create their own two dimensional subspace (a plane) in $\mathbb{R}^3$, with $\mathbf{v} \times \mathbf{r}$ being normal to this plane. The dot product of this with the acceleration vector $\mathbf{a}$ gives the projection of the acceleration vector onto the normal vector, it is a measure of how parallel they are. If the dot product is zero, then this implies $\mathbf{a}$ is orthogonal to the vector $\mathbf{v} \times \mathbf{r}$, and that it lies in the plane $\mathbf{v}$ and $\mathbf{r}$ create. If it is not zero, then the this implies that the acceleration vector has a component lying in the plane, and a component parallel to the normal of the plane. This causes the particle to spiral away from the plane.

Does this seem correct?

 

[AlephZero]

Can you define your notation better?

The question says $\mathbf{a}$ denotes the position. In your solution you say it is the acceleration vector.

What do $\mathbf{b}$ and $\mathbf{c}$ have to do with the question?

And what are $\mathbf{v}$ and $\mathbf{r}$?

 

[Bashyboy]

Whoops, I am terribly sorry. I just edited my original post.

 

[AlephZero]

Thanks. In your first version of the question, I wasn't sure if v and r were supposed to be a rotating coordinate system or something similar.

For the geometrical interpretation, See http://en.wikipedia.org/wiki/Frenet–Serret_formulas