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Back in 2001, I wrote my own 3D graphics engine, down to the individual pixel rendering, shading, camera tracking, Z buffer, hell even error diffusion dithering for 256 color palettes.
And I still don't know half the terms you just used.
I do know points, polygons, vectors, normals, roll, pitch, yaw, Lambert's Law shading, error diffusion feedback..
And my Calculus 2 teacher admired my works and told me I had the understanding of a Calculus 4 student.
impressive, I'd like to ask abou stuff like how long it took you and stuff. But in this discussion I'd like to mention that I didn't use any complicated terms, only orthogonal projection (middle school) and perpendicularity (elementary school).
I started from the ground up in December 1998 with a bare wireframe engine, largely inspired from a demo wireframe engine from another developer. I was 17 years old then so it was basically my after school project, not a school assignment, but my teachers were impressed.
I didn't quite just copy/paste his code though, I carefully read over his code and comments to the point that I understood how it all worked, and rewrote a much cleaner wireframe engine of my own that supported colored lines and even loading from files, which the original demo didn't support.
Later on I came across another demo, from the same developer I think, that demonstrated rendering solid triangle shaded 3D models. Again, I read over everything and rewrote everything from the ground up, largely looking to optimize the rendering technique for the highest number of polygons per second, and of course to be able to load different models from file.
Then I just started having a bit of fun with the polygon rendering, starting with an optimized integer based greyscale gouraud shading algorithm, which ran way faster than any similar demos I could find at the time. Note that this was all CPU driven, no fancy GPU at the time, the 3Dfx Voodoo was still a pretty new thing I couldn't afford..
Then I got the idea of trying to bring color to the project via error diffusion, since I was basically limited to 320x200x256 color display mode, unless I wanted to run a high end video mode at a snail's pace LOL! Error diffusion is slow though, so how did I speed that up?
Well, I did away with the gouraud shading and went back to treating each polygon as a single solid RGB color, shaded using the Lambert's Law technique. To speed up the error diffusion process, I'd only process 8 pixels into the diffusion algorithm, then as the polygon rendered, it would just pick randomly from that 8 pixel buffer.
Since I was programming in QuickBasic, arrays were limited to 64KB each, meaning that memory was very tight, and I actually had to allocate two arrays for the Z Buffer, one for the top half of the screen and another for the bottom half.
The inspiration for the camera tracking came from a rather unexpected source, a simple mouse string toy demo of all things LOL! I realized that if I used just one segment of that string algorithm, I could link the viewing angle to follow a point in the model, or with some creative adjustments, basically follow any arbitrary point.
I also made a side project crude CAD scripting thing of sorts, mainly meant to render a torus or sections of a torus with whatever dimensions I wanted. With the right inputs, that also allowed me to easily generate spheres, cylinders, cones and tubes.
I think I finished the original wireframe engine within just a couple or few days, but the other versions that had filled in polygons probably took me a week to start with, and the more advanced techniques probably took me around 2 months each, all in my spare time of course.
I didn't really have any final product in mind, I was just experimenting and learning ya know. When 3D GPUs started becoming a big and common thing, I didn't see much future for my little project, but I sure did learn a lot!
ADHD driven hard work could never disappoint huh? 
But what was the advantage of QuickBasic? Weren't C++ and Javascript around at the time? I only hear about them in this context
Things were different back then. QBasic was free yo, I couldn't afford $200 or whatever for paid development software. Besides, I was just starting to learn anyways.
Later on I did end up finding a pirated copy of the full QuickBasic 4.5 at least, which allowed more RAM usage for my programs.
Edit: In a parallel universe, if I could have afforded it, I might have otherwise started with Borland Delphi.
retro computing was so chad
so you asked an LLM a question and then asked if we should adjust our schooling based on that?
you're the one who might need schooling again, bruh
yeah, I'm starting all over again with university, so hopefully this will be eventually fixed. About the rest of the population though ...
Good luck.
Keep in mind there's no way to rush hours. Whenever you're learning something, you can ask yourself roughly how many hours you've spent on it. That should reflect your skill and there's no way around it.
You should read Flatland it's an awesome book and blew my mind as a young textile geek.
You just reminded me of the planiverse which was inspired by flatland.
Is a simple concept, thoroughly thought through, with coool drawings in it.
3 years ago, a university teacher proposed it to me on facebook and added it to "the list", but still didn't go back to 
We percieve the 3 dimensions we exist in, through a 2d mapping, i.e. our retinas. So I think we are limited in how much of 3d we can really grasp at a time.
Came here to say this. Even "3D movies" are actually just stereoscopic 2D (meaning two ever-so-slightly-different 2D images, one for each eye). True 3D vision would be, for lack of better term, x-ray vision.
I can't quite tell what the question is by your image. I have done a lot of descriptive geometry prior to CAD tools coming on the scene, and now work a lot with 3D geometry/topology problems, but what you describe is not going to be taught in schools because 95% of people will never need to know it. Honestly half, to three-quarters, of the people that run 3D CAD don't really understand the geometry; its just a result they get
Spoiler alert: the image has no relation to the question, it's just something OP picked to elicit that "3D geometry" feeling. There is a point A', a point B, point C, and point D, but no point P or H or n.
my lazyass had it hard to put correct labels. But judging by how many people ignored the proble an are just scolding me for using AI, fair is fair.
@TauZero@mander.xyz It is a Geogebra drawing I did to reason with the problem, I took a screenshot of the drawing to attached it.
In the drawing, the labels are different from the problem, but I just made a sphere whose diameter is [AP] (here point P has label A, while A has label A'),
then constructed the plane using A and two other points of the sphere (C and D in the picture),
I thought like "if that property from 2D geometry holds in 3D then any point in the intersection of plane and the sphere will satisfy the perpendicularity, and thus two of them will do for a counterexample".
And It is exactly what happened: Using Geogebra's tool of measuring angles it shows that the two points, C and D, that I picked up both satisfy the orthogonality condition (in the picture angle(A,C,A')=90Β°=angle(A,D,A'), but they can't be both the projection of P, right ? Counterexample! (the hypothesis was that a point on a plane that satisfy that condition is immediately THE projection of the point that isn't on the plane)
Yeah It is not the best thing but I wanted to attach something, and the drawing that I used was the best thing at hand.
Ah, I can see OP's line of thought now:
I like treating posts as puzzles, figuring out thread by thread WTF they are talking about. But dear OP, let me let you know, your picture and explanation of it are completely incomprehensible to everyone else xD. The picture is not an illustration to the question but a sketch of your search for a counterexample, with all points renamed of course, but also a sphere appearing out of nowhere (for you to invoke the inscribed-triangle-rule, also mentioned nowhere). Your headline question is a non-sequitur, jumping from talking about 4D (never to be mentioned again) into a ChatGPT experiment, into demanding more education in schools. You complain about geometry being hard but also simple. The math problem itself was not even your question, yet it distracted everyone else from whatever it is you were trying to ask. If you ever want to get useful answers from people other than crazed puzzleseekers like me, you'll need to use better communication!
~~
fyi: the orthogonal projection of a point P into a plane is a point H of that plane such that for any other point A of the plane: (PH) is orthogonal to (HA). One might think that finding that "(PH) is orthogonal to (HA)" for one such point A of the plane is enough, turns out it is not.
luckily an easier criterion exists: H is the orthogonal projection of P if (PH) is parallel to n the normal to the plane.
I think adding a bit of curvature to the six surfaces of a regular cube can throw off many. Then there's scale. Astronomical scales and milli or micro meter scales adds its own complexity by the simple fact that we lack regular language tools to capture the ideas and express them completely.
Where do we see curved surfaces? Everywhere from flight routes to space flight to deep sea diving.
Though I am not all tbat clear where we apply 3d geometry at micro scale or smaller, just a hunch that we may need them.
Language plays catch up. Is.
No for an orthogonal projection, because literally every point in the plane centered at H and normal to (AH) (so dihedrally perpendicular to the plane given in the problem) could potentially be P. In other words, it could project to H, or a point off of P perpendicularly to (AH)
You don't really need math for that one, it's just spacial reasoning, which you can't really directly teach. I suppose just the concept of solid angle vs. dihedral angle vs. face angle would be good for everyone to know. To formally prove it, it seems like you'd need linear algebra, which they don't usually teach in high school anyway.
Now, if you can use oblique projections as well, it's pretty trivial to find one that's "tilted" such that any point not already in the plane maps to a given H - the projection can proceed along any set of parallel lines through the space, and there's always a line between any point X and H. Mathematically, you use the fact that X-H must be in the kernel space of the projection, and the standard formula for constructing a projection operator from a basis complementary to the kernel space and one in the plane it projects to.
I couldn't make sense of the first paragraph, are you sure it is right ?
Pretty sure, yes. I'm probably just explaining badly.
There's a full 360 degrees of rays perpendicular to (AH) starting at H. That would be true of line to a point in 3D. In 2D there would be exactly 2 possibilities (left and right), while in 4D they would correspond to an ordinary sphere, and hyperspheres in higher dimensions yet.
Together, they take up a plane. Only points on a certain (infinite) line going through this new plane and H will actually orthogonally map to H, and it's the same one that's normal to to original plane. Let's call the line L.
If point P wasn't in this plane, (PH) couldn't be perpendicular to (AH). It is in the new plane, but we still don't know for sure it's on line L, so it's not true that that implies it projects to H.
~~I tried again, I don't find mistakes in your statements, I just don't see how they make up for "instant in-mind proofs" for the problem~~ I think I see it now, nevermind. Your got a very good visualization for 3D CanadPlus. It seems so intuitive that "the set of points that map to H with orthogonal projection is a straight line", but do you happen to have a pocket proof for that ?
Uhh, that the preimage of a point like H is a line? Off the top of my head, I'd use the fact it's a shifted copy of the kernel. Well, assuming without loss of generality that we're in a vector space and not just an affine space.
Using basic rules and notions from linear algebra and the theory of functions:
For a projection O in space V, your preimage L is defined as {lβV | O(l) = H}. Using the linearity of O you can turn that into {lβV | O(l-H) = 0}, which is equivalent to {yβV | O(y) = 0} by setting l=y+H. Definitionally, an affine subspace is constructed from the members of a subspace added to a constant like that. The kernel, {yβV | O(y) = 0}, is a subspace because any linear combination of vectors within it will, once you apply and distribute the operator using linearity again, turn into a sum of 0s, meaning the result must always be another member of the kernel.
All that's left is to prove it's a 1D affine subspace, AKA an infinite line. Every point w in the domain V is in some preimage, by the definition of a function, and so using the same math you can construct it as O(w)+k for some k in the kernel. O(r)=r for all r in the range by the definition of a projection, which you can use to both show it's a subspace and can't contain any basis of the kernel (expanding that out I'll leave as an exercise). So, the dimension of the range and the kernel have to add to that of the whole domain. This actually holds for all other linear operators as well.
Our space is 3D and the provided plane is 2D. 3-2=1, QED.
Probably there's a proof from the axioms of Euclidean geometry that doesn't need linear algebra, but I was never good at that sort of thing. It's also worth noting that any set defined purely by linear operators and affine linear subspaces will again be (describable as) affine linear. It's like a closure property.
How first reading felt:
How the second reading felt at the beginning:
How it ended up:
What is {yβV | O(y) = 0} ? If the plane doesn't pass through $0_V$ then how would that 0 be the image of some point ? Most likely you're using something from linear algebra that I didn't learn in my course (I didn't learn projection I think, only examples when learning matrices).
If the plane doesnβt pass through $0_V$ then how would that 0 be the image of some point ?
Answer, at risk of making it worse:
I was assuming this is a linear projection in a (non-affine) vector space, from the beginning. All linear operators have to to map the origin (which I've just called 0; the identity of vector addition) to itself, at least, because it's the only vector that's constant under scalar multiplication. Otherwise, O(0)*s=O(0*s) would somehow have a different value from O(0). That means it's guaranteed to be in the (plane-shaped) range.
I can make this assumption, because geometry stays the same regardless of where you place the origin. We can simply choose a new one so this is a linear projection if we were working in an affine space.
Can I ask why you wanted a proof, exactly? It sounds like you're just beginning you journey in higher maths, and perfect rigour might not actually be what you need to understand. I can try and give an intuitive explanation instead.
Does "all dimensions that aren't in the range must be mapped to a point/nullified" help? That doesn't prove anything, and it's not even precise, but that's how I'd routinely think about this. And then, yeah, 3-2=1.
(I didnβt learn projection I think, only examples when learning matrices).
Hmm. Where did the question in OP come from?
They're abstractly defined by idempotence: Once applied, applying them again will result in no change.
There's other ways of squishing everything to a smaller space. Composing your projection to a plane with an increase in scale to get a new operator gives one example - applied again, scale increases again, so it's not a projection.
It sounds like youβre just beginning you journey in higher maths
I'm actually old and lurked in university stuff for a long time and dropped out of engineering in university and started with math all anew, yet at the same time I'm still a beginner.
Hmm. Where did the question in OP come from?
I don't exactly remember How I started thinking about the "distance between plane and a point formula", I think I stumbled upon it while organizing my old bookmarks. Tried to make a proof, and in the process that question came, and when I couldn't solve it on the fly I though like "it's so over for me". Then ChatGPT also got it wrong and was like "It's so over for mankind". And I ended up making this post to share my despair. Actually many answers were eye opening.
Haha, I thought it was a homework question. It would be a pretty good one; it's not hard to answer, but the a proof touches on a lot of things. I probably would have gone about this differently if I hadn't thought I was addressing someone who's actively studying these things. Hopefully you still knew most of the terms I was using.
And the missing part, because including an exercise is low-key a dick move if you were just curious:
Any basis vector k can't be 0 (that would be dumb), so if O(k)=0 it fails idempotence and can't be in the range. Therefore, all kernel bases are not in the range.
For the range being a subspace, O(a+b)=O(a)+O(b)=a+b, and you can extend that to any linear combination of range vectors.
I guess you'd need to include the proof that vector (sub)spaces must have a basis to make it airtight, so we know the kernel has any dimensional at all. But, then it's just the pigeonhole principle, since you can choose a basis for the whole space made up from bases of the two subspaces.
Best of luck.
hhhhhhh homework in the summer ?
Although I know in Japan they give them such horrors
Hey, summer semesters do exist as well.
Wait is that not true? Why wouldn't H form a right angle with P and A?
AH would be perpendicular to n, and PH would be parallel to n, making them perpendicular to each other? Or am I misunderstanding the definition of a plane projection?
AH and PH do form a right angle, that's postulated in the problem. But P is only the projection of H onto the plane if PH is indeed parallel to n. Which is not necessary.
Imagine a nail patrols hammered into a piece of wood at an angle. The wood surface is the plane, the entry point is H and the head of the nail is P. A is anywhere on the line perpendicular to the nail on the board.
If you shine a light from above, you can see P', the projection of P as the end of the shadow cast by thaw nail. Unless the nail is straight, P' != H.
PH would be parallel to n
The question doesn't posit that.
OH! I see now. Perpendicular-ness is not commutative in 3d. Gotcha, thank you!
transitive you mean ?
Yes.
if (PH) is perpendicular to (AH) and n is perpendicular to (AH) ==> it doesn't really follow that (PH) is parallel to n, unlike in 2D geometry. ChatGPT also got the wrong implication at first.
Props to you for being one the few comments who actually understood the problem from my horrible statement/language though.
An argument could be said that 3D space is infinitely more complex than 2D space.
As with most things, skill comes with practice. It's not really worth forcing people to learn something that has a low likelihood of being relevant to their lives, but simple exposure to something like Blender or Godot would be great for making more people familiar with 3D software.
While 3D geometry is more difficult for me than 2D, I could almost immediately tell that the answer is no, there are infinitely many points H that satisfy this. The reason it's unintuitive is that our intuition about what "perpendicular" means comes from 2D and poorly translates to 3D.
The most intuitive explanation I can muster is this: imagine all possible planes that pass through both A and P. It should be obvious that there are infinitely many of them (I visualize it as a plane "rotating" around the AP axis). Each of these planes intersects the given plane since it passes through A. Think of the intersection line. It never passes through P (unless P is on the plane), so it is always possible to draw a perpendicular line from P to that intersection line. With one exception (when the perpendicular line falls on the A point), the point where the perpendicular falls satisfies the conditions for H. (I think all such points actually form a circle with AP' as the diameter, where P' is the parallel projection of P to the given plane, but I'm not 100% sure)
DUH! If this was math.stackexchange I'd choose this as answer
