Conclusion that can be drawn from 4 points in space whose angles are $90^{\circ}$

Question

Four points A, B, C and D are in space such that angles $A\hat BC, B\hat CD, C\hat DA$ and $D\hat AB$ are all right angles, then

1. A, B, C, D cannot be coplanar
2. A, B, C, D are necessarily coplanar.
3. A, B, C, D may or may not be coplanar.
4. No such points exist.

I got an answer by pure imagination. Please provide a proof/reasoning so as to why you feel your answer is correct.

Answer 1

From $\angle{ABC}=90^\circ$, we may suppose that $$A(0,a,0),\quad B(0,0,0),\quad C(c,0,0)$$ It follows from $\angle{BCD}=90^\circ$ that the $x$-coordinate of $D$ is $c$.

It follows from $\angle{DAB}=90^\circ$ that the $y$-coordinate of $D$ is $a$.

Let $D(c,a,d)$. By the law of cosines, $$AC^2=DA^2+DC^2-2\cdot DA\cdot DC\cos\angle{CDA},$$ $$\Rightarrow \quad c^2+a^2=c^2+d^2+a^2+d^2\quad\Rightarrow\quad d=0.$$

The four points are on the plane $z=0$.

Answer 2

As ABC is right angle, line segment 1) line segment AB & BC must be on same plane. 2) line segment BC & CD must be on same plane 3) line segment CD & DA must be on same plane 4) line segment DA & AB must be on same plane. (any two lines are always on same plane) From points 1 and 2, AB,BC and DA has to share a common plane. From either point 2 or 3, CD must also be in same plane.

So all the line segments must be on same plane and so all points are coplanar.

Answer 3

Only the first three right angles are adequate to establish that the 4 points belong to a common (developable) plane. However as an aside, the four points need not strictly be confined to a plane in general.

We can apply Gauß - Bonnet theorem when going around a loop along geodesic arcs in 3 space. If the sum of angles turned at each intersection is $ \Sigma \varphi $ , integral curvature is $ \int K d A $, we have

$$ \Sigma \varphi + \int K d A = 2 \pi + \int K d A = 2 \pi , K =0, $$

we should have not only a plane but cones/cylinders and three types of developables associated with curves on surfaces.