Vector multiplication question

[Nabeshin]

I'm not sure where to put this question, since it deals with both physics and math, so I figured here would be a good starting point.

In the book of astrodynamics I'm currently reading, I came across this expansion:

$$(\vec{r}\times\vec{v})\times\vec{r}=[\vec{v}(\vec{r}\cdot\vec{r})-\vec{r}(\vec{r}\cdot\vec{v})]$$

Can anyone explain how this result is arrived at? If any physical significance is needed, r is a position vector and v its derivative with respect to time, the velocity vector.

 

[element4]

Well, the standard way is just to use the vector triple product (Lagranges formula, see http://en.wikipedia.org/wiki/Triple_product#Vector_triple_product"):

$$\vec{a}\times (\vec{b}\times \vec{c})= \vec{b}(\vec{a}\cdot\vec{c}) - \vec c (\vec a\cdot \vec b)$$
(also know as the BAC CAB rule)

This relation can easily be proved, by calculating both sides (just insert $$\vec {a} = (a_1,a_2,a_3)$$ and so on).

In your case we have, $$\vec a = \vec c \equiv \vec r$$ and $$\vec b \equiv \dot{\vec r} = \vec v$$.

 

[Nabeshin]

Ah, thank you very much :) Simply a formula I was not familiar with.

 

[HallsofIvy]

? The "triple product" is just the same as the product asked about so I don't see how that answers your question, except to say "yes, it is a well known calculation". Of course, the way to see that it is true is to use some general <a, b, c>, <x, y, z> for $\vec{r}$ and $\vec{v}$.

For a little more "intuitive" insight, we know that $\vec{r}\times\vec{v}$ is perpendicular to both $\vec{r}$ and $\vec{v}$ so the result of any vector cross $\vec{r}\times\vec{v}$ must be perpendicular to that vector and so in the plane spanned by $\vec{r}$ and $\vec{v}$ and can be written as a linear combination of them: $a\vec{r}+ b\vec{v}$. But $(\vec{r}\times\vec{v})\times \vec{r}$ must also be perpendicular to $\vec{r}$ so we must subtract off any projection onto $\vec{r}$. That is the "$\vec{r}(\vec{r}\cdot\vec{v}$" part.