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.
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.
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
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.
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 
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 {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.
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.
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.
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