Let $p$ be a vector-valued function
$$p: {\bf N} \longrightarrow {\bf R}^3$$
$$p: t \longmapsto (p_x,p_y,p_z)$$
whose values are the 3D position of something moving through 3D space (at discrete time steps).
We can approximate the velocity vector function
$$v: {\bf N} \longrightarrow {\bf R}^3$$
$$v: t \longmapsto (v_x,v_y,v_z)$$
as
$${\Delta p \over \Delta t}.$$
Now we're given a sequence of vectors $(p_x,p_y,p_z)$ and vectors $(v_x,v_y,v_z)$ and we want to approximate the scalar $\Delta t$. How could we do it?
If $p$ is a scalar function, then
$${\Delta p \over \Delta t} = v$$
so you just compute scalar division
$${\Delta p \over v} = \Delta t$$
but if $p$ and $v$ are vectors, what can we do? Vector division? For which vector product?
Another option is to compute the "$\Delta t$'s" in each dimension:
$$\Delta t_x = {\Delta p_x \over v_x}$$
$$\Delta t_y = {\Delta p_y \over v_y}$$
$$\Delta t_z = {\Delta p_z \over v_z}$$
But this gives 3 different values of $\Delta t$: $\Delta t_x$, $\Delta t_y$, and $\Delta t_z$. You could compute the norm of the vector $(\Delta t_x, \Delta t_y, \Delta t_z)$, but I don't know what that would mean.
Yet another option is to compute the scalar division of the norm $p_r $of $p$ and the norm $v_r$ of $v$:
$${\Delta p_r \over v_r} = \Delta t$$
but this gives yet another quantity.
Best Answer
When you write $\frac{\Delta p}{\Delta t} = v \in\mathbb{R}^3$, you are saying that you are doing this division coordinate-wise: $\frac{\Delta p}{\Delta t} = \frac{1}{\Delta t}\begin{pmatrix}\Delta p_1\\ \Delta p_2\\ \Delta p_3 \end{pmatrix} = \begin{pmatrix}\frac{\Delta p_1}{\Delta t}\\ \frac{\Delta p_2}{\Delta t}\\ \frac{\Delta p_3}{\Delta t} \end{pmatrix} = \begin{pmatrix}v_1\\ v_2\\ v_3\end{pmatrix} \implies \begin{cases}\Delta p_1/\Delta t=v_1\\ \Delta p_2/\Delta t=v_2\\ \Delta p_3/\Delta t=v_3 \end{cases}\implies \begin{cases}\Delta p_1=v_1\Delta t\\ \Delta p_2=v_2\Delta t\\ \Delta p_3=v_3\Delta t \end{cases}\implies \begin{cases}\Delta p_1/v1=\Delta t\\ \Delta p_2/v_2=\Delta t\\ \Delta p_3/v_3=\Delta t \end{cases}$
There are three equations all involving $\Delta t$. So, to invert this, you should divide each coordinate of $\Delta p$ by the corresponding coordinate of $v$. You say this gives you three different values, but this is not possible if $\frac{\Delta p}{\Delta t} = v$ holds like you described. Either your model $\frac{\Delta p}{\Delta t} = v$ is wrong or your data is wrong.