How to project a figure onto a plane?

Question

Programming

Hello!
I can't figure out how to project a figure onto a plane.
Let's say I define a quadrilateral in the XYZ plane with points:

A = (x1,y1,z1)
B = (x2,y2,z2)
C = (x3,y3,z3)
D = (x4,y4,z4)

I need to get something like this:

I want to set a figure in 3d space, and then display its projection on a 2d plane.
I want to take the projection of each point, get new 4 points with coordinates (x,y) and build already them in the Cartesian coordinate system. But I can't figure out how, having a point with coordinates (x,y,z) to translate into (x,y)
Please tell me, I will be grateful for any help

Answer 1

But I can't figure out how, having a point with coordinates (x, y, z), translate into (x, y)

Actually, this is an elementary task from the course of descriptive geometry (1st course).
And it largely depends on how you set the plane.

Answer 2

To find a point "in the plane", you need to define a coordinate system in it. That is, choose the origin point O(x0,y0,z0) and two basis vectors X=(x1,y1,z1) and Y=(x2,y2,z2). From what you say about "Cartesian coordinate system", the vectors should be of unit length and perpendicular to each other.
You add a Z vector to the system, in parallel to which the design is going on. From the condition it is not clear whether you are considering only an orthogonal projection, or the general case of a parallel projection.
In the case of orthogonal, everything is simple - you don’t even need to mess with matrices:
the vector Z is calculated as the vector product of X and Y, but we don’t need it at all: if the projected point P has coordinates (x, y, z), then its projection Q will have coordinates (in plane coordinate system)
x'=(PO,X)=(x-x0)*x1+(y-y0)*y1+(z-z0)*z1
y'=(PO,Y)=(x-x0)*x2+(y-y0 )*y2+(z-z0)*z2.
In the case of an oblique projection, the calculations are more complicated - you need to multiply the vector (PO) by the matrix inverse to the matrix composed of X,Y,Z. And there the main thing is not to get confused, where are the rows and where are the columns.

Answer 3

I didn’t quite understand the question, I want to clarify: is the plane on which you want to make a projection - is it an arbitrary plane or does it mean a plane formed by the coordinate axes (xy yz xz)?
because if the second - it turns out everything is trite. You just take the coordinates of the same points, place them on the plane. Visibility is determined by the Z coordinate. In this case, this is redundant information, because you are simply projecting a flat figure onto a plane.
Write in the comment if you did not quite understand me or I misunderstood you