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$\vec{r}_a$ is a positional vector from reference frame $a$. What is the position of same point from reference frame $b$ ? If required, assume position of origin of frame $a$ is $\vec{m}$ and unit point (i.e. $\langle 1,1,1\rangle_a$ ) is $\vec{n}$ from reference frame $b$.

I am studying Kleppner and Kolenkow and this is the first thing I asked myself. Unexpectedly its taking a while to figure it out. So help needed.

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Is that a Newtonian text? I'm basically asking if you are asking about change of frames in special relativity or Galilean? Either way, the answer is that it depends on the frames... What is the transformation between the two frames? This will tell you how coordinates of a point change. –  Will Jul 16 '13 at 16:45
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@Will K&K is Newtonian, so no special relativity. (I think it includes a bit of SR toward the end, but this would be a question from the beginning.) –  David Z Jul 16 '13 at 16:46
    
@Will The question is more coordinate geomtery than physics. for example i dont even consider time. I just want to interchange position. Anyway I did a successful thought experiment, just cant write it down in math. –  AirTycoon Jul 16 '13 at 17:10

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Assuming a non-relativistic context, if $\mathbf r_a$ and $\mathbf r_b$ are the vector positions of certain point, with respect to frame systems $A$ and $B$, and $\mathbf s$ is the vector position of the frame $B$ with respect to $A$. Then, if the axis of $A$ and $B$ are parallel and both right-handed (or both left-handed) $\mathbf r_b=\mathbf r_a-\mathbf s$ holds.

A more general transformation rule is a rigid transformation (translation plus rotation): when the axis of $A$ and $B$ are not parallel, a rotation matrix $R$ should be inlcuded. $\mathbf r_b= R(\mathbf r_a-\mathbf s)$.

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So you aren't considering rotations? –  Will Jul 16 '13 at 16:52
    
I dont how can rotations be avoided ? I am totally hypothesizing here, but i think at least two points are needed to calculate $\vec{r}_b$. –  AirTycoon Jul 16 '13 at 17:22
    
@Will sorry, I was thinking of the simplest case :), so you are right. I reedited. –  c.p. Jul 17 '13 at 7:05
    
@ato Well, if your axes are parallel, you need just a point: namely, where is $B$ as seen from $A$. But if your axis are not parallel, you need more information: an element of $O(3)$ -- which turns out be parametrized by three real coordinates (three angles, for instance). –  c.p. Jul 17 '13 at 7:10

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