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Everyone talks about the 4th dimension, BUT why is 3D geometry so hard for the average person compared to 2D ? (lemmy.ml)

submitted 1 day ago* (last edited 1 day ago) by zaknenou@lemmy.dbzer0.com to c/asklemmy@lemmy.ml

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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.

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[–] CanadaPlus@lemmy.sdf.org 1 points 12 hours ago* (last edited 12 hours ago) (1 children)

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.

[–] zaknenou@lemmy.dbzer0.com 1 points 9 hours ago (1 children)

I couldn't make sense of the first paragraph, are you sure it is right ?

[–] CanadaPlus@lemmy.sdf.org 1 points 9 hours ago* (last edited 8 hours ago) (1 children)

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.

[–] zaknenou@lemmy.dbzer0.com 1 points 8 hours ago* (last edited 8 hours ago)

~~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 ?