Transfomation of one coordinate system to a another

Question

I have a molecule with one coordinate system ( denote as x,y,z ) where the origin is center of mass of the molecule. I have to define another coordinate system (p,q,r) for a local motion. (shown in the figure.) I know the unit vectors of p,q,r axises respective to x,y,z coordinates and origin of the p,q,r coordinate system respect to x,y,z. Hence know the r vector. I know the point C respect to x,y,z coordinate system. Now I want to find the coordinate of point C respect to p,q,r coordinate system. Appreciate very much if someone can help me to solve this problem. Thank you.

Answer 1

$$ \vec{r} - \vec{R}_{c\,m}\quad\mbox{( No relativistic )} $$

Answer 2

It looks like you used the symbol $\mathbf{r}$ twice. So I'm going to use $\mathbf{v}$ to denote the origin of the $\mathbf{p}$-$\mathbf{q}$-$\mathbf{r}$ coordinate system.

The coordinates of $\mathbf{c}$ in the $\mathbf{p}$-$\mathbf{q}$-$\mathbf{r}$ coordinate system are $(\mathbf{c} -\mathbf{v})\cdot \mathbf{p}$, $(\mathbf{c} -\mathbf{v})\cdot \mathbf{q}$, $(\mathbf{c} -\mathbf{v})\cdot \mathbf{r}$.

This easy to prove: suppose $\alpha$, $\beta$, $\gamma$ are the desired coordinates. Then we know that $$ \mathbf{c} = \mathbf{v} + \alpha\mathbf{p} + \beta\mathbf{q} + \gamma\mathbf{r} $$ and so $$ \alpha\mathbf{p} + \beta\mathbf{q} + \gamma\mathbf{r} = \mathbf{c} - \mathbf{v} $$ taking dot products with $\mathbf{p}$, $\mathbf{q}$, $\mathbf{r}$ in turn gives $$ \alpha = (\mathbf{c} -\mathbf{v})\cdot \mathbf{p} \quad ; \quad \beta = (\mathbf{c} -\mathbf{v})\cdot \mathbf{q} \quad ; \quad \gamma = (\mathbf{c} -\mathbf{v})\cdot \mathbf{r}. $$

So, the short answer is that you should just calculate these three dot products. When you do it, make sure that you have expressed all the vectors ($\mathbf{c}$, $\mathbf{v}$, $\mathbf{p}$, $\mathbf{q}$, $\mathbf{r}$) in the original $x$-$y$-$z$ coordinate system.

You can write this using a transformation matrix, if you want to, but I don't think it makes anything any clearer.