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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

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.

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[–] BCsven@lemmy.ca 7 points 1 day ago (2 children)

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

[–] zaknenou@lemmy.dbzer0.com 2 points 15 hours ago* (last edited 15 hours ago) (1 children)

@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')=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.

[–] TauZero@mander.xyz 3 points 8 hours ago (1 children)

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!

[–] zaknenou@lemmy.dbzer0.com 2 points 8 hours ago* (last edited 8 hours ago)

~~
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.

[–] TauZero@mander.xyz 6 points 22 hours ago (1 children)

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.

[–] zaknenou@lemmy.dbzer0.com 2 points 15 hours ago* (last edited 15 hours ago)

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.