I think 3D geometry has a lot of quirks and has so many results that un_intuitively don’t hold up. In the link I share a discussion with ChatGPT where I asked the following:
assume a plane defined by a point A=(x_0,y_0,z_0), and normal vector n=(a,b,c) which doesn’t matter here, suppose a point P=(x,y,z) also sitting on the space R^3. Question is:
If H is a point on the plane such that (AH) is perpendicular to (PH), does it follow immediately that H is the projection of P on the plane ?
I suspected the answer is no before asking, but GPT gives the wrong answer “yes”, then corrects it afterwards.
So Don’t we need more education about the 3D space in highschools really? It shouldn’t be that hard to recall such simple properties on the fly, even for the best knowledge retrieving tool at the moment.
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
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.
I assume it ran on CPU, including the rasterization? 🤯
Yes indeed! You gotta pull quite a few programming speed tricks to get any sort of acceptable rendering performance when running everything on the CPU, especially on an old 486…
I believe this has multiple reasons in education. Most 3D stuff is assumed to be self-taught later by the interested learner in private somehow.
- the cross product for 3D is introduced within 1 day. No tasks or assignment where given to me that improve my intuition and proofs on its properties.
a × b. Tbh the Cartesian product and determinants of 3x3 matrices are way more important than the time dedicated to them in average education systems. - rotation in 2D space is very intuitive “clockwise” and “counterclockwise” everyone learns that at age 11 minimum. While in 3D you suddenly need gimbals and
f****quarternions, sphere coordinates, which are also never properly explained to the average person. I needed to teach myself why sphere coordinates work the way they do, my teachers think it was obvious I think. Coding something up in a 3D game engine helps with this immensely imo. - right hand rule and its connection to the cross product. Its connection to physics and magnetic fields, forces that work on every charged particle in our universe. 🌌
Personally I am interested in a concept I call prime spaces. This to me means an intersection of geometry and number theory. Every entry in Matrix or Vector has to be a prime number. Geometry is connected to every other field of math.
Happy to see someone agreeing with me for change, the skill issue cult is exhausting (there are many other cool answers)
- the cross product for 3D is introduced within 1 day. No tasks or assignment where given to me that improve my intuition and proofs on its properties.
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 problemI 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 {l∈V | O(l) = H}. Using the linearity of O you can turn that into {l∈V | O(l-H) = 0}, which is equivalent to {y∈V | 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, {y∈V | 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{y∈V | 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 doesn’t 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 didn’t 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 you’re 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.
You should read Flatland it’s an awesome book and blew my mind as a young textile geek.
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
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
@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.
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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’),
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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’)=90°=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:
- you have a point A’ on a plane and a random point A
- you find a midpoint B and draw a sphere around it. A and A’ are now a diameter of the sphere
- pick two random points D and C at the intersection of the plane and the sphere
- by the “triangle inscribed in a circle/sphere where one side is a diameter” rule, such a triangle must be a right triangle
- therefore both angles ACA’ and ADA’ are right angles
- thus C and D both satisfy the conditions of the initial question (with all points renamed: A=P, (C or D)=H, A’=A)
- OP never defined what a projection is, it being “4th grade math”, but one of the requirements is being unique
- C and D cannot both be the projection, therefore the initial question must be answered “false”: just because AH is perpendicular to PH doesn’t make H a projection.
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!
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