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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 2 days ago* (last edited 2 days 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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[–] balsoft@lemmy.ml 1 points 15 hours ago* (last edited 15 hours ago)

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)