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.
Wait is that not true? Why wouldn’t H form a right angle with P and A?
AH would be perpendicular to n, and PH would be parallel to n, making them perpendicular to each other? Or am I misunderstanding the definition of a plane projection?
if (PH) is perpendicular to (AH) and n is perpendicular to (AH) ==> it doesn’t really follow that (PH) is parallel to n, unlike in 2D geometry. ChatGPT also got the wrong implication at first.
Props to you for being one the few comments who actually understood the problem from my horrible statement/language though.
AH and PH do form a right angle, that’s postulated in the problem. But P is only the projection of H onto the plane if PH is indeed parallel to n. Which is not necessary.
Imagine a nail patrols hammered into a piece of wood at an angle. The wood surface is the plane, the entry point is H and the head of the nail is P. A is anywhere on the line perpendicular to the nail on the board.
If you shine a light from above, you can see P’, the projection of P as the end of the shadow cast by thaw nail. Unless the nail is straight, P’ != H.
The question doesn’t posit that.
OH! I see now. Perpendicular-ness is not commutative in 3d. Gotcha, thank you!
transitive you mean ?
Yes.