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

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).

[–] CanadaPlus@lemmy.sdf.org 1 points 5 hours ago* (last edited 4 hours ago)

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.