Given two points $\mathcal{A}$ and $\mathcal{B}$ in $\mathbb{R^3}$ whose position and velocity vectors are, respectively:
$$\mathbf{r_A}=\begin{pmatrix}r_{A_{x}}\\r_{A_{y}}\\r_{A_{z}}\end{pmatrix}$$
$$\mathbf{r_B}=\begin{pmatrix}r_{B_{x}}\\r_{B_{y}}\\r_{B_{z}}\end{pmatrix}$$
$$\mathbf{v_A}=\begin{pmatrix}v_{A_{x}}\\v_{A_{y}}\\v_{A_{z}}\end{pmatrix}$$
$$\mathbf{v_B}=\begin{pmatrix}v_{B_{x}}\\v_{B_{y}}\\v_{B_{z}}\end{pmatrix}$$
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:
$$\mathbf{v_{AB}}=\begin{pmatrix}v_{A_{x}}\\v_{A_{y}}\\v_{A_{z}}\end{pmatrix}-\begin{pmatrix}v_{B_{x}}\\v_{B_{y}}\\v_{B_{z}}\end{pmatrix}=\begin{pmatrix}v_{A_{x}}-v_{B_{x}}\\v_{A_{y}}-v_{B_{y}}\\v_{A_{z}}-v_{B_{z}}\end{pmatrix}$$
Best Answer
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.