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