Given bounds(x,y,z-max/min)of a cuboid, how to get the remaining 2 corner coordinates

How can I get all corner/vertex coorinates of a cuboid (all angles between edges = 90degrees, but edges with different lenght) just having the bounds (xmin, xmax, ymin, ymax, zmin, zmax)?

If the cuboid is parallel to the a cartesian coorinate system it seems easy, but if the cuboid is not arranged parallel to the coordinate system it is a bit confusing...

Maybe someone has an Idea...

Re: Given bounds(x,y,z-max/min)of a cuboid, how to get the remaining 2 corner coordin

Hey kennyfujimoto.

What information do you have about the orientation of the cuboid (you will need some information if it is not axis-aligned)?

1 Attachment(s)

Re: Given bounds(x,y,z-max/min)of a cuboid, how to get the remaining 2 corner coordin

Attachment 28078

For this case the known vertices are:

Point : -0.364580 , -0.208230 , 0.604200

Point : -0.139230 , -0.663350 , 0.195500

Point : -0.150820 , -0.686420 , 0.217620

Point : 0.036428 , -0.203130 , 0.819640

Point : 0.048021 , -0.180060 , 0.797520

Point : 0.261790 , -0.658250 , 0.410940

And the "unknown" are

Point : 0.250193 , -0.681320 , 0.433066

Point : -0.352992 , -0.185162 , 0.582077

But my question was, how to get in a general case the 2 unkown vertices when you have 6 points.

To clarify my question: If you only have 3 vertices of a (2D) rectangle you can calculate with a bit of math the 4th vertex. Can you do the same with a cuboid just having 6 vertices?

Re: Given bounds(x,y,z-max/min)of a cuboid, how to get the remaining 2 corner coordin

You would have to exploit the symmetry.

Basically if you have three points, then for a parallelogram you get vectors X = B-A, and Y = C-A and point four will be A + X + Y. This assumes that the shape is parallelogram and that all points lie in the same plane.

You can apply the same kind of thinking to the other planes of the cuboid.

Re: Given bounds(x,y,z-max/min)of a cuboid, how to get the remaining 2 corner coordin

I will think about that. It might be helpfull. I hoped, that someone could do the thinking for me and post a magic-formula :-) But thanks anyway.