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Given two points $\mathcal{A}$ and $\mathcal{B}$ in $\mathbb{R^3}$ whose position and velocity vectors are, respectively: $$\mathbf{r_A}=\begin{pmatrix}r_{A_{x}}\\r_{A_{y}}\\r_{A_{z}}\end{pmatrix}$$ $$\mathbf{r_B}=\begin{pmatrix}r_{B_{x}}\\r_{B_{y}}\\r_{B_{z}}\end{pmatrix}$$ $$\mathbf{v_A}=\begin{pmatrix}v_{A_{x}}\\v_{A_{y}}\\v_{A_{z}}\end{pmatrix}$$ $$\mathbf{v_B}=\begin{pmatrix}v_{B_{x}}\\v_{B_{y}}\\v_{B_{z}}\end{pmatrix}$$
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How would you calculate the relative velocity of $\mathcal{A}$ with respect to $\mathcal{B}$ and the opposite, given random position and velocity?

For example: if the two points are moving in the same direction, you would calculate the velocity of $\mathcal{A}$ relative to $\mathcal{B}$ with the following formula: $$\mathbf{v_{AB}}=\begin{pmatrix}v_{A_{x}}\\v_{A_{y}}\\v_{A_{z}}\end{pmatrix}-\begin{pmatrix}v_{B_{x}}\\v_{B_{y}}\\v_{B_{z}}\end{pmatrix}=\begin{pmatrix}v_{A_{x}}-v_{B_{x}}\\v_{A_{y}}-v_{B_{y}}\\v_{A_{z}}-v_{B_{z}}\end{pmatrix}$$

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  • $\begingroup$ Your question is strange because in the example at the bottom you give the correct answer. I do not see what your question is I'm afraid. (+1 though for setting your question out so neatly with a good diagram.) $\endgroup$
    – tom
    Commented May 19, 2016 at 21:49
  • $\begingroup$ Does the last formula also work when the two velocities are not collinear? $\endgroup$
    – Oliver
    Commented May 19, 2016 at 21:51
  • $\begingroup$ Ok I think I might see your question now... $\endgroup$
    – tom
    Commented May 19, 2016 at 21:52
  • $\begingroup$ Thank you and sorry for not making it clear in the first place. $\endgroup$
    – Oliver
    Commented May 19, 2016 at 21:54
  • $\begingroup$ No problem - I hope the answer is clear - if not comment... $\endgroup$
    – tom
    Commented May 19, 2016 at 22:04

1 Answer 1

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Your equation for the $\mathbf{v_{AB}}$, the velocity of $\mathcal{A}$ relative to $\mathcal{B}$ with the following formula is correct and general for all 'real' values of $v_{X_{i}}$ where $X=A, B$ and $i=x, y, z$.

$$\mathbf{v_{AB}}=\begin{pmatrix}v_{A_{x}}\\v_{A_{y}}\\v_{A_{z}}\end{pmatrix}-\begin{pmatrix}v_{B_{x}}\\v_{B_{y}}\\v_{B_{z}}\end{pmatrix}=\begin{pmatrix}v_{A_{x}}-v_{B_{x}}\\v_{A_{y}}-v_{B_{y}}\\v_{A_{z}}-v_{B_{z}}\end{pmatrix}$$

Note that the velocities will only be collinear if $\mathbf{v_{A} \times v_{B}} = \mathbf{0}$, which cannot be assumed for random values.

Final note is that this is not correct for particles moving at speeds close to the speed of light, where special relativity will give a better picture.

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  • $\begingroup$ So if for example vA = (0, 12, 0) (going all the way to the positive y semiaxis) and vB = (-5, -5, 0) (following the bisector of the negative x and y semiplane), is it correct to calculate vAB this way? vAB = (0, 12, 0) - (-5, -5, 0) = (0 - [-5], 12 - [-5], 0 - 0) = (5, 17, 0). $\endgroup$
    – Oliver
    Commented May 19, 2016 at 22:23
  • $\begingroup$ yes that is correct :-) $\endgroup$
    – tom
    Commented May 19, 2016 at 22:26

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