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.

  • TauZero@mander.xyz
    3·
    13 hours ago

    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.comOPEnglish
      2·
      12 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.