Trying to understand homogeneous coordinates I am trying to understand how homogeneous coordinates work, and think I have an explanation but need to check it is correct.
For a homogeneous coordinate $[x,y,1]$, or $[x,y,3]$ does the last number essentially say which $z$-plane the point is lying on?
 A: For 2D homogeneous coordinates $[x,y,1]$ and $[x,y,3]$ there is no $z$-plane. So, no! The last number cannot and does not say that. 
I use simple 2D plane for explaining.
Remember, there is redundancy in homogeneous coordinates and you can always multiply or divide all the components with some number and it's still the same point. So $[x,y,1]$ and $[3x,3y,3]$ are the same point. 
But what is $[x,y,3]$ in relation to $[x,y,1]$? One way to put it is, that points $[x,y,1]$ and $[x,y,3]$ are in the same direction from the origin and the latter is 3 times closer to origin than the former. How? The first two components give you a direction (note that $(x,y)$ and $(3x,3y)$ give the same direction). The third component $w$ shows somehow "how far" in that direction is your point. It's $1/w$ "far" in units of the length of $(x,y)$.
It's common to explain homogeneous coordinates that if you make $w=1$ you get usual cartesian coordinates in the first two components. Yes, that's true. If $ \left[ x, y, w \right] = \left( \frac{x}{w}, \frac{y}{w} \right)$ then $ \left[ x, y, 1 \right] = \left( x, y \right)$.  But I like to think from another viewpoint. It's that if I scale $(x,y)$ so that its length becomes $1$, I get unit vector and reciprocal of distance.  $$\left[ k_1, k_2, \frac{1}{d} \right] = \left[ \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, \frac{w}{\sqrt{x^2+y^2}}\right]$$
If you expand unit vector, you must contract the distance the same amount to stay at the same place, and vice versa. That's why reciprocal.
