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.

  • CanadaPlus@lemmy.sdf.org
    1·
    1 day ago

    Haha, I thought it was a homework question. It would be a pretty good one; it’s not hard to answer, but the a proof touches on a lot of things. I probably would have gone about this differently if I hadn’t thought I was addressing someone who’s actively studying these things. Hopefully you still knew most of the terms I was using.

    And the missing part, because including an exercise is low-key a dick move if you were just curious:

    Any basis vector k can’t be 0 (that would be dumb), so if O(k)=0 it fails idempotence and can’t be in the range. Therefore, all kernel bases are not in the range.

    For the range being a subspace, O(a+b)=O(a)+O(b)=a+b, and you can extend that to any linear combination of range vectors.

    I guess you’d need to include the proof that vector (sub)spaces must have a basis to make it airtight, so we know the kernel has any dimensional at all. But, then it’s just the pigeonhole principle, since you can choose a basis for the whole space made up from bases of the two subspaces.

    Best of luck.