Distance an arbitrary point is found along a given vector Say I have a vector in 2D space defined by two points $(x_1, y_1)$ and $(x_2, y_2)$: $$\vec{v}=(x_2 - x_1, y_2 - y_1)$$ I would like to find how far along that vector an arbitrary point $(x_3, y_3)$ is. This very woolly language$^*$, so I've attempted to create a diagram showing the sitation.

In this diagram, the quantity I'm interested in $a$, which I can calculate using Pythagoras' theorem if I know $b$ and $c$. I know $c$, which is the length of vector $(x_3 - x_1,y_3 - y_1)$, given by, $$c = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}$$ So, now I need to calculate $b$: the length of a vector – that I'll call $\vec{u}$ – that is perpendicular to $\vec{v}$ and passes through point $(x_3, y_3)$. For $\vec{u}$ and $\vec{v}$ to be perpendicular the dot product must be zero. That is,
$$\vec{v}\cdot \vec{u}=0$$
$$(x_4-x_3)(x_2-x_1)+(y_4-y_3)(y_2-y_1)=0$$
This is where I begin to falter: one equation with two unknowns, $y_4$ & $x_4$. I expect there is some obvious constraint on $\vec{u}$ that I should be using to eliminate an unknown, but my sleep-deprived mind is offering no help. Can someone point out what I've missed?
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$^*$I really want to use the word project to describe how my arbitrary point $(x_3, y_3)$ is placed along that vector. Is this the correct terminology?

You are right in calling this a projection. If $(x_1,y_1)$ is the origin, then you can project ${\bf u} = (x_3,y_3)$ onto $\bf v$ thus:
$${\rm proj}_{\bf v}{\bf u} = \frac{\bf u \cdot v}{\bf v \cdot v}{\bf v}.$$
If $(x_1,y_1)$ is not the origin, then just shift the frame of reference to make it so.
