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.
I couldn’t make sense of the first paragraph, are you sure it is right ?
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.
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 problemI 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 ?Uhh, that the preimage of a point like H is a line? Off the top of my head, I’d use the fact it’s a shifted copy of the kernel. Well, assuming without loss of generality that we’re in a vector space and not just an affine space.
Using basic rules and notions from linear algebra and the theory of functions:
For a projection O in space V, your preimage L is defined as {l∈V | O(l) = H}. Using the linearity of O you can turn that into {l∈V | O(l-H) = 0}, which is equivalent to {y∈V | O(y) = 0} by setting l=y+H. Definitionally, an affine subspace is constructed from the members of a subspace added to a constant like that. The kernel, {y∈V | O(y) = 0}, is a subspace because any linear combination of vectors within it will, once you apply and distribute the operator using linearity again, turn into a sum of 0s, meaning the result must always be another member of the kernel.
All that’s left is to prove it’s a 1D affine subspace, AKA an infinite line. Every point w in the domain V is in some preimage, by the definition of a function, and so using the same math you can construct it as O(w)+k for some k in the kernel. O(r)=r for all r in the range by the definition of a projection, which you can use to both show it’s a subspace and can’t contain any basis of the kernel (expanding that out I’ll leave as an exercise). So, the dimension of the range and the kernel have to add to that of the whole domain. This actually holds for all other linear operators as well.
Our space is 3D and the provided plane is 2D. 3-2=1, QED.
Probably there’s a proof from the axioms of Euclidean geometry that doesn’t need linear algebra, but I was never good at that sort of thing. It’s also worth noting that any set defined purely by linear operators and affine linear subspaces will again be (describable as) affine linear. It’s like a closure property.