Asklemmy

48788 readers

723 users here now

A loosely moderated place to ask open-ended questions

Search asklemmy 🔍

If your post meets the following criteria, it's welcome here!

Open-ended question
Not offensive: at this point, we do not have the bandwidth to moderate overtly political discussions. Assume best intent and be excellent to each other.
Not regarding using or support for Lemmy: context, see the list of support communities and tools for finding communities below
Not ad nauseam inducing: please make sure it is a question that would be new to most members
An actual topic of discussion

Looking for support?

Looking for a community?

Lemmyverse: community search
sub.rehab: maps old subreddits to fediverse options, marks official as such
!lemmy411@lemmy.ca: a community for finding communities

~Icon~ ~by~ ~@Double_A@discuss.tchncs.de~

founded 6 years ago

MODERATORS

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

46 comments fedilink hide all child comments

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.

you are viewing a single comment's thread
view the rest of the comments

[–] CanadaPlus@lemmy.sdf.org 1 points 8 hours ago* (last edited 7 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 7 hours ago* (last edited 7 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 ?