End result: sum of changes in each dimension each squared equals distance squared.
In 3d we have every plane that has a right triangle may have Pythagorean Theorem apply to that right triangle. In 2 dimensions, we have x^2+y^2=distance^2. Since z dimension is perpendicular to all xy planes, we can do (previous distance)^2+z^2=distance^2. This also applies with change from any point in form for each dimension, end-start. Previous distance squared was x^2+y^2 and therefore this formula holds thus far.
Link to proof sourceFrom there we have that for any 3 points as specified with a 3d and no change in other dimensions vector to in a fourth dimension a distance away from that end of 3d vector from start is a right angle. It may be one of many but it is one. angle=arccos((vector1 dot product vector 2)/((magnitude of vector1) times (magnitude of vector2))). As long as distance is all in 3d and one is 4d only from it we have arccos(0/(magnitude stuff)). arccos(0)=90 degrees. We therefore have a right triangle and this applies as long as all in one set of dimensions for a vector and only in one that is not included in that for another.
What about coplanar? Any set of 3 points in 4d+ should have one plane that contains them all. It is an assumption I make but it seems correct. Also, what about if a magnitude is 0 for arccos? Is that undefined? Maybe but that indicates that one or both is irrelevant in our equation and can be not used. 0^2=0 and therefore any no change dimensions will not affect result. It is still square root whole for result of distance total. All can be 0 and this equation still works.