Difference between Two Vectors, Spherical Coordinates

[Higgy]

Homework Statement

I'm doing a problem that involves expressing, for two arbitrary vectors \vec{x} and \vec{x'},

|\vec{x}-\vec{x'}|

in spherical coordinates (\rho,\theta,\phi).

Homework Equations

Law of Cosines:

c^{2}=a^{2}+b^{2}-2ab\cos\gamma

where \gamma is the angle between a and b.

The Attempt at a Solution

Using the law of cosines, we can write

|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'\cos\gamma)^{\frac{1}{2}}

but I can't figure out what the angle between the vectors \gamma would be. I imagine a plane formed by the two vectors, and the angle being between the two vectors in that plane. How I describe this mathematically, though, is confusing. Can anyone push me in the right direction?

EDIT:
Alright, I took a new approach. I decided to express the difference in Cartesian coordinates, and then convert to spherical coordinates. In doing so, I get:

|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'[\sin\theta\sin\theta'(\cos\phi\cos\phi'+\sin\phi\sin\phi')+\cos\theta\cos\theta'])^{\frac{1}{2}}

which gives me the \gamma I was looking for. It would be nice if there were an easy way to simply that, though...

 

[gabbagabbahey]

Usually, you are free to orient your coordinate system as you choose, and so a good choice is often to orient it such that \vec{x'} points in the z-direction and hence \theta'=0 and you get a much simpler expression:

|\vec{x}-\vec{x'}|=\rho^{2}+(\rho ')^{2}-2\rho\rho ' cos \theta

 

[Higgy]

Usually, you are free to orient your coordinate system as you choose, and so a good choice is often to orient it such that \vec{x'} points in the z-direction and hence \theta'=0 and you get a much simpler expression:

|\vec{x}-\vec{x'}|=\rho^{2}+(\rho ')^{2}-2\rho\rho ' cos \theta

That's useful a point.

In the larger problem that this is for, a rigid body (described by f(x')) is rotating around a particular axis with angular velocity \omega - so I am restricted to making the z-axis parallel to \vec{\omega}. Hence, the most general expression is required.

 

[gabbagabbahey]

In that case, I think the most you can do to simplify is to use the trig Identity \cos (\phi -\phi ')=\cos \phi \cos \phi '+\sin \phi \sin\phi '

 

[Higgy]

Right! Working it out right now. Thanks.