Proving that the orthogonal projection on the xy-plane of the corner of a cube looking upwards forms a triangle.

Question

By the corner of a cube looking upwards I mean one vertex V and the three vertices adyacent to V such that V has the minimum z-coordinate of the cube. I suppose that the projection on the xy plane of V should be contained inside the projection of the other vertices, because that is what my visual intuition suggests, but I'm trouble proving that.

My objective is to try to prove this analytically, preferably without using angles, for the unit cube. I thought about using vectors in the xy plane with V as the origin and expressing one of the vectors as the cross product of the other two so that I had less variables, but I'm having trouble since it can go both ways. I guess I should try to find a convex combination of the projection of the projection of those three vectors that gives the projection of V, but I'm not sure how to proceed.

Answer 1

Define the cube to originally have been a unit cube with a vertex on the origin and $3$ neighboring vertices $(0,0,1),(0,1,0)$, and $(1,0,0)$. Rotate the cube first by $0<\alpha<\frac{\pi}{2}$ about the y-axis and then by $0<\beta<\frac{\pi}{2}$ about the x-axis. Looking "down" from positive z onto the xy-plane, the cube is first rotating about the y-axis up toward you and tilting left, and then that rotated cube is rotating down on the origin to tilt left and down. The vertex at the origin remains at the origin and is the lowest point. This is a specific setup, but the outcome is entirely general, you just need a plane not parallel to any face for the projection, and we just generated that situation for the xy plane.

You want to know if the origin's neighboring vertices form a triangle around it when projected to the xy plane.

$(0,0,1)$: The first rotation moves it left, the second down, so its projection is in the 3rd quadrant.

$(0,1,0)$: The first rotation does nothing, the second pulls it down the y-axis some but not to the origin, so its projection is on the positive y-axis.

$(1,0,0)$: The first rotation pulls the projection along the x-axis but not to the origin, the second moves it down, so it is in the 4th quadrant.

Thus the origin is enclosed by a triangle formed by points in the 3rd and 4th quadrants, and by some positive point on the y-axis.

You can formalize this in several ways, but if you're having trouble, I would suggest starting with a 2D situation and proving that a unit square, rotated, has its inferior projection on x within the interval defined by the projections of its neighboring vertices. Hope this helps!

Answer 2

This is pretty clear geometrically. Why do you need an algebraic proof?

The projections of those three vertices of a cube will always form a triangle unless two of them are collinear in the $xy$-plane. Your "looking upward" hypothesis prevents that since if two of the projections were collinear the third vertex would coincide with its projection, so not be "above" the corner.

In fact, the corner must be an interior point of the triangle. If it were not then one of the lines containing an edge of the triangle joining the projections of two of the vertices would separate the corner from the projection of the third vertex, which happens only if the third vertex is "below" the $xy$-plane.