Okay, so I am modelling a problem where I have two objects $a$ and $b$, each with a certain initial position, ${x}_a (0)$ and ${x}_b (0)$, respectively, and each with a certain velocity ${\dot{x}_a}$ and ${\dot{x}_b}$. This problem reduces to a two dimensional problem such that ${x_a}, {x_b}, {\dot x_a}, {\dot x_b} \in \mathbb R^2$.
I'm trying to determine what direction for ${\dot x_a}$ is optimal given fixed $||{\dot x_a}||$ and fixing all other variables.
In the process of examining this problem, I did the following algebra:
The position after time $t$ of each of the objects is as follows:
$${x_a} (t) = {x_a} (0) + {\dot x_a}t$$ $${x_b} (t) = {x_b} (0) + {\dot x_b}t$$
Since we want to know when these two meet:
$$ {x_a} (0) + {\dot x_a}t - {x_b} (0) - \dot x_b t =0$$
Blindly doing algebra to this, I got the following function $f$ that maps $\dot x_a$ to $t$:
$$ t = f(\dot x_a) = \frac{x_b - x_a}{\dot x_a - \dot x_b}$$
Given this, it seems like it would be helpful to take $\nabla f$, but I'm not sure this is actually intelligible.
If I do unit analysis, it makes sense:
$$\frac{x_b - x_a}{\dot x_a - \dot x_b} = \frac{\text {m} - \text {m}}{\frac{\text {m}}{\text {s}} - \frac{\text {m}}{\text {s}}} = \frac{\text {m} }{\frac{\text {m}}{\text {s}} } = \text{m} \cdot \frac{\text{s}}{\text{m}} = \text{s} $$
But the problem is that I have a vector in $\mathbb R^2$ divided by another vector in $\mathbb R^2$ that should, in principle, give me scalar (or maybe time should be thought of as a vector in $\mathbb R^1$?). I'm not sure this actually works or if I've just abused the algebra. Is vector division actually a defined operation that I can use?
Further Developments:
So I tried a few things just to see what would happen, so I set up this simplified problem (dropping units for readability:
Object $q$ at time $t = 0 $ is located at $x(0) = \begin{bmatrix} 0 \\0 \end{bmatrix}$ and traveling at a velocity $\dot x = \begin{bmatrix} 10 \\12 \end{bmatrix}$. At what time does $x(t) = \begin{bmatrix} 5 \\6 \end{bmatrix}$?
Our function is:
$$ x(t) = x(0) + \dot x(t) $$
Plugging in the knowns:
$$\begin{bmatrix} 0 \\0 \end{bmatrix} + \begin{bmatrix} 10 \\12 \end{bmatrix}t = \begin{bmatrix} 5 \\6 \end{bmatrix}$$
We can get rid of the zero vector:
$$\begin{bmatrix} 10 \\12 \end{bmatrix}t = \begin{bmatrix} 5 \\6 \end{bmatrix}$$
Now to isolate the $t$, we do the following operation:
$$\begin{bmatrix} \frac{1}{10} && \frac{1}{12} \end{bmatrix}\begin{bmatrix} 10 \\12 \end{bmatrix}t = \begin{bmatrix} \frac{1}{10} && \frac{1}{12} \end{bmatrix}\begin{bmatrix} 5 \\6 \end{bmatrix} $$
$$2t = 1$$
$$t = \frac{1}{2}$$
All well and good, but then if we make $\dot x = \begin{bmatrix} 10 \\ 13 \end{bmatrix}$, meaning we miss $x(t) = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$, we get:
$$\begin{bmatrix} 10 \\13 \end{bmatrix}t = \begin{bmatrix} 5 \\6 \end{bmatrix}$$ $$\begin{bmatrix} \frac{1}{10} && \frac{1}{13} \end{bmatrix}\begin{bmatrix} 10 \\13 \end{bmatrix}t = \begin{bmatrix} \frac{1}{10} && \frac{1}{13} \end{bmatrix}\begin{bmatrix} 5 \\6 \end{bmatrix} $$
$$2t = \frac{1}{2} + \frac{6}{13} = \frac{25}{26}$$
$$t = \frac{25}{52}$$
But:
$$\frac{25}{52} \begin{bmatrix} 10 \\13 \end{bmatrix} \ne \begin{bmatrix} 5 \\6 \end{bmatrix}$$
What just happened? My understanding from a bit of the math I've learned in quantum mechanics was that:
$$ | \psi \rangle = | \phi \rangle \implies \langle \xi | \psi \rangle = \langle \xi | \phi \rangle $$
This would seem to make some kind of contradiction. What happing?
You are looking for $t$ satisfying $\vec{u} = t\vec{v}$ where $\vec{u}=x_b(0)-x_b(a)$ and $\vec{v} = \dot{x}_a-\dot{x}_b$.
If you divide component-wise and get the same quotient (i.e. $u_1/v_1 = u_2/v_2$), then $t$ is this common quotient. (e.g. if $\vec{u}=(2,3)$ and $\vec{v} = (4, 6)$, then $t=0.5$).
Otherwise $\vec{u}$ and $\vec{v}$ are not parallel, so there is no such $t$. In your original problem, this corresponds to the case where the two objects never meet.