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so I have this example;

The position of ship 1 is $(3t-10, t+4)$. The position of ship 2 is $(2t-3, -t+13)$. What is the closest distance the ships approach and what time does it occur?

Then I have;

$$d^2 = (3t-10-(2t-3))^2+(t+4-(-t+13))^2=(t-7)^2+(2t-9)^2=5t^2-50t+130$$ $$d^2=5|(t-5)^2-25|+130=5(t-5)^2+5$$ So $t=5$, and $d=\sqrt{5}$.

So I think I understand about this problem conceptually and that the closest approach is always perpendicular from the first point to the relative velocity line, and that distance is the one I want. I've seen geometric interpretations with relative velocity and my main problem is I have a disconnect between what I think is happening and what the above maths is telling me.

The first thing that confuses me is that the distance formula looks like it finds the hypotenuse of a triangle and I can't figure out why that would be the perpendicular line. I don't understand how the maths applies geometrically.

The second thing is that I don't understand why this tells me anything about the closest approach.

Thanks.

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    You put $-t+13$ in your distance formula when I think you meant $-t-13$. – 2017-02-17
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    Sorry yes, I actually got the position of the ship wrong, which should've been -t+13, I'll fix that now. – 2017-02-17
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    as for the "second thing", it tells you everything instead: just put $t=5$ in the equations and you get the position of either ship! – 2017-02-17

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What you are calculating is the distance $\Delta_t$ between ship 1 and ship 2 at time $t$. And you were correct to look at this distance and find the $t$ for which $\Delta_t$ is the smallest.

I think a source of the confusion is that the lines $\ell, \ell'$ that are the respective routes of ships 1 and 2 are not parallel ($\ell$ has slope $\frac{1}{3}$ and $\ell'$ has slope $-\frac{1}{2}$) so the distance between these two lines is zero at the unique point $\mathbf{x}$ in the plane where they intersect.

However, ship 1 arrives at $\mathbf{x}$ at a different time then ship 2. So the distance between the two lines is not necessarily the closest the two ships will ever get.

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    I do get the visualisation in the sense that they cross paths but not at the same time, I imagine it as if there was a string connecting the two ships. It's that I don't understand why the distance formula gives me the smallest $\delta_t$. From my understanding it seems that we would want the sin or cosine of the distance? Or is it that the distance between them has no sin or cosine so it's a collapsed triangle in a sense? – 2017-02-17
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    Given two point $(a,b),(c,d)$ in the plane, the formula $\sqrt{(a-c)^2 + (b-d)^2}$ is precisely the length of the straight line segment between said points, and this distance formula is motivated by the Pythagorean Theorem. – 2017-02-17
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    Yeah it was because I knew that, that it was confusing me because I associate that formula with the longest side of a triangle. However it makes sense now I understand it, so thank you for the help. – 2017-02-17
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Besides working in a absolute reference, as you did and as Joeb's answer already well explained, you could work in a relative reference frame, for instance positioned on ship 2.
As for the orientation of the axes, you can keep them aligned with the absolute (geographic) ones, or to tilt e.g. the $y$ axis towards the starting point, or towards the minimum distance between trajectories, etc.

Let' consider the case of having same orientation as the absolute. Then the parametric equations are: $$ \left\{ \begin{gathered} x = x_2 - x_1 = 3t - 10 - \left( {2t - 3} \right) = t -7 \hfill \\ y = y_2 - y_1 = t + 4 - \left( { - t + 13} \right) = 2t - 9 \hfill \\ \end{gathered} \right. $$ and the trajectory of ship 2 vs. 1 is $$ y = 2\left( {x +7} \right) - 9 = 2x +5 $$ The line through the origin and orthogonal to the trajectory is $$ y = - 1/2\;x $$ which gives a crossing point (minimum distance) of $$ H=(-2, 1) \quad d=\sqrt{5\,} $$ and inserting either the $x$ or the $y$ coordinate of $H$ into the parametric eq. you get $$ t=5 $$