For a rectangular prism that is 3D comprised of points G H B and D. Are these points coplanar? Why?
A plane is a flat geometric surface. The prism is not flat, but is rather a 3-D pyramid. Therefore not all the points will lie in the same plane (are coplanar).
Three points that are not all in the same line uniquely define a single plane. Therefore any three of the points listed, GHBD, are coplanar. But the four together are not - one always lies outside the plane of the other three.
A straight line is detrmined by 2 distinct points .
The expert has already said well about your question. I only try to say in terms of the points that are away perpendicularly from the plane GHB.
A plane is detrmined any by 3 distinct points.
A plane is determined by any 2 distinct intersecting straight lines.
Any plane has a particular inclination to the axis of the rectangular prism.
Therefore, for any three distinct poins G, H, B it is a fixed plane, say GHB in 3 dimensional space.
If the point D is on this plane it is said to be coplanar. The set of all points on the plane are said to be coplanar.However, you can consider any point outside the plane GHB, in this 3 dimensional rectangular prism. The criteria, that any point is out side the plane is that it has a perpendicular distance from the plane GHB. If the point is on the plane itself,the there is no perdincular distance from the plane which is well know by common sense itself. This proves that you can find a lot of coplanar points and still more noncoplanar points in the 3 dimensional rectangular prism (or 3 D space).
Hope this helps.