Understanding perspective transform matrix elements interpretation

I am representing 3D points (vectors) in the following way:

(* conversion from 3D point, represented by normal list of \
coordinates, to matrxi column, suitable for transforms *)
ToColumn[point_] := Transpose[{Append[point, 1]}];


(* conversion from matrix column, representing 3D point, to a list, \
representing the same point *)
ToPoint[column_] := Take[Transpose[column/column[[4, 1]]][[1]], 3];


I.e. forth element serves as the scale factor.

(Is this conventional representation and what is the name of it?)

I am representing perspective transform with the following matrix:

PerspectiveXYZ[{x_, y_, z_}] := {
{1, 0, 0, 0},
{0, 1, 0, 0},
{0, 0, 1, 0},
{x, y, z, 0}
};


so that

My question is: what is the sense of transform elements x, y and z?

I drew a cube of 8 points and transformed it with various values of these variables:

And found, that x and y controls projection plane orientation, while z controls both the scale and distance from origin point, while z=1 means projecting into some small region (1?), and that the smaller this value, the bigger is the scale, becoming infinite when z=0.

Is there any clear geometric interpretation of these values, especially z? May be they should be substituted with 1/z or something for better interpretation?

May be my vector model should be changed?

-
It's a fact that if you use "homogeneous coordinates" then a perspective (or projective) transformation from a (projective) plane to a (projective) plane can be represented by a $3 \times 3$ "homogeneous" matrix. This is one reason homogeneous coordinates are useful, though they seem unintuitive at first. –  littleO Jun 14 at 10:28
I am using 4x4 matrices. –  Suzan Cioc Jun 14 at 10:36
But warpPerspective accepts a $3 \times 3$ matrix, so why are you using $4 \times 4$ matrices? Also, I don't understand exactly what your question is. –  littleO Jun 14 at 11:51
See my update. 3x3 matrix gives the same result (I think) as 4x4 matrix with 3rd row/column excluded / turned to zero. My question is more general. –  Suzan Cioc Jun 14 at 11:55
Thanks. I'm confused now about why the $4 \times 4$ matrix isn't invertible, I had thought that it should be. Btw, a decent reference for this stuff is the book Multiple View Geometry in Computer Vision by Hartley and Zisserman. –  littleO Jun 14 at 19:43
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$Ax + By + Cz + D=0$