In the previous chapter we have discovered that it is possible to pinpoint any location on a plane with two numbers. The space we live in is not a flat plane though. It would be nice to have a coordinates system adequate for this situation too.
To figure it out, we will simply add a third dimension to the two-dimensional world described in the previous chapter and study the consequences. Hopefully we stumble upon something that will allow us to introduce coordinates in 3D.
As usual, we will make some observations and perform some simple experiments to come up with a handful of claims from which everything else follows. Many things covered here may feel obvious and spending so much time proving them redundant. Please bear with me and treat this as an exercise in reasoning: it’s a good idea to refine our tools before we venture into areas where they will be desperately needed. Let’s start our quest by studying the relations between basic objects in 3D.
Planes and segments in 3d space
If we grab a plane and some random point in space, this point must be on the plane or in the space on either side of the plane. How can we tell whether two points are on the same or opposite sides of the plane? A similar question was answered in 2D space so instead of reinventing the wheel, we will simply introduce a 3D version of postulate 3:
Postulate 3-3D: If the shortest route linking point P with point Q intersects a plane, then P and Q are on the opposite sides of this plane i.e. every route in space that links P and Q must also intersect the plane. If on the other hand the straight segment PQ (the shortest route) doesn’t intersect the plane, both points are on the same side of this plane.
Can a straight line intersect a plane more than once? According to our definition of a plane if two different points are on the same plane then all points on the straight line linking them also belong to this plane. From this it follows that if a straight line has a common point with a plane then either it is the only common point or this line is on the plane.
Two planes in 3D space
Let’s construct a plane in space. We will call it plane p. Next, we choose some point A in space that is not on plane p, yet it is on some other plane p’ (points A, B, A’ B’ are all on plane p’ not indicated in the picture to avoid clutter).
Now let’s assume that planes p and p’ have one point O in common. We don’t claim that it is the sole point. The aim is to establish the consequences of planes having at least one point in common, and doing this by reasoning alone.
If we pass a straight line k between A and O, according to postulate 3-3D, line k must intersect plane p (if it didn’t, all points on line k would have to be on the same side of plane p and line k wouldn’t touch plane p at all).
Let’s choose on plane p’ some other point B that happens to be on the same side of plane p as point A, yet not on line k. If on line k we choose some point A’ that is on the other side of plane p with respect to point A, the straight segment A’B must pierce through plane p in some point O’ that is different than O. Why? Because A’OB is not the shortest route between A’ and B. Points O and O’ are members of plane p and plane p’ so the line going through them must belong to both planes.
If point B’ on plane p’ is on the other side of plane p with respect to point A and not on line k, the line going through A and B’ must pierce plane p in some point O”. Note that points O, O’ and O” must be on the same straight line because if they were not, O, O’ and O” would constitute a plane p’ that would have to be identical with plane p i.e. point A would have to be on plane p and we know it is not.
Conclusion: if two planes intersect at all, they must intersect along a single straight line in 3D space.
A plane perpendicular to a straight line
Let’s take a closer look at two intersecting planes p and p’ described in the previous section. We know they intersect along a straight line that will be denoted here with letter f (planes p and p’ are omitted in the picture below together with points O and O”). Next, on plane p we construct line h that is perpendicular to line f in point O, and on plane p’ we construct line h’ that passes through O and is perpendicular to f.
Straight lines h and h’ constitute a third plane denoted in the picture with p”. On this plane we can draw yet another straight line t that crosses point O. We know that h and h’ are perpendicular to f. Is t perpendicular to f as well? Whether it is or isn’t, so what?
Let’s assume for the moment it is. Line t is arbitrary so what we really proved is that every line t that belongs to plane p” and goes through point O is perpendicular to straight line f. We can say that plane p” is perpendicular to line f in point O.
Now imagine on line f some point C. From all we have stated so far and from the Pythagorean theorem it follows that the segment CO must be the shortest route between point C and plane p” and this in turn lets us introduce coordinates in 3D. Can you see how?
Coordinates in 3D
Imagine we have a plane in space that is established on three points P0, P1 and P2 where the segments P0P1 and P0P2 are perpendicular.
Now imagine some random point P that is not on this plane. To find the coordinates of P in relation to P0, P1 and P2 we find on our plane such point O that PO is the shortest possible. If P is in front of the plane, its third coordinate will be z=|PO| and if it is behind: z=-|PO|. As for the coordinates x and y of point O on the plane, we know how two find them from the previous chapter. As you can see, all of that hinges on whether there actually exists a shortest unique route between point P and the plane. In the previous section we found some hints as to what features such route should have. What’s left to do is to find it or prove that this approach cannot work.
The shortest route between a point and a plane
Let’s pick point P in space and some plane p. If P is on the plane, then the distance between them is 0. What if P is not on the plane? Is it possible to find on plane p such point O that PO is the shortest distance between this plane and point P?
Let’s draw on plane p any line (g in the picture below). From the previous chapter we know how to find the shortest route PA between point P and line g (the shortest route is perpendicular to g). Next, on plane p we construct line q that goes through point A and is perpendicular to g. The shortest route PO between point P and line q must be perpendicular to line q. Is segment PO perpendicular to every line on plane p that goes through point O?
To establish this, let’s choose on line g some point B≠A. Now the question is whether angle ∠BOP is right or not. From the Pythagorean theorem we know that:
|PO|2=|AP|2-|AO|2
|BO|2=|AO|2+|AB|2
|PB|2=|AP|2+|AB|2
so it must be true that:
|BO|2+|PO|2=|AP|2+|AB|2=|PB|2
and from this it follows that angle ∠BOP must be right. Note that the same reasoning can be repeated no matter where on line g our point B is, so in the situation depicted below it must be true that: ∠B0OP=∠B1OP=∠B2OP=90o.
In the next step, on plane p we construct line g’ that is perpendicular to the segment OB1. From the Pythagorean theorem it follows that |OD0| and |OD2| must be longer than |OB1|. Due to the fact ∠B0OP=∠B1OP=∠B2OP=90o we can also prove (again using Pythagorean theorem) that segments |PD0| and |PD2| must be longer than |PB1|. In the same fashion it can be demonstrated that for every point D between D0 and D2 segment |PD| is the shortest for D=B1 so angle ∠PB1D2 must be right.
Using the reasoning outlined above, we can prove that every line h on plane p that goes through point O must be perpendicular to segment PO and also that if some line f on the same plane is perpendicular to line h in some point E, segment PE must be perpendicular to line f.
All of this means that segment PO constitutes the shortest distance between plane p and point P. Now that we know how to find it, the coordinates in 3D can be introduced just as described in the previous section. We have quickly found what we wanted! We could close this chapter and move on, however, as usual there is always more to it than anticipated. Please read on.
Symmetries
As we have seen in the previous chapter, two triangles constructed in exactly the same way can have different internal distances depending on the properties of the space itself. In other words, if we constructed triangle PAB in two different locations, there is no guarantee that |PC| would have the same length in both them:
To determine whether this is the case or not, we can construct the same triangle in multiple places and see whether |PC| and other internal distances always yield the same results. Of course verifying this everywhere is impractical (there is an infinite number of situations to consider). For a finite number of examples it is true so let’s assume it is. If we encounter a situation in which our assumption fails, we will modify our model.
Just like in the previous chapter, we are going to formulate our assumption in terms of preservation of distances and shape under specific transformations of objects in space. This time we operate in three dimensions so instead of a triangle, we need a three-dimensional object. Let’s add to the triangle built with three vertices P0, P1, P2 the fourth vertex P3 that is not on the same plane as the triangle P0P1P2.
Note that even if the distances between all the vertices are preserved when we move this pyramid somewhere else, there is no guarantee that P3 won’t flip to P’3. For both versions the distances between vertices are the same, yet it is not exactly the same object.
To distinguish between them, we will use the concept of chirality. When we look at the triangle P0P1P2 from the point of view of P3, moving from P0 to P1 and then via P2 back to P0 is seen as counter-clockwise. On the other hand, from the point of view of P’3, it looks clockwise. Following the convention accepted in chemistry, we will call the P3 version cis, and the P’3 version trans.
From this direct experience we can tell that moving a pyramid in space doesn’t change the distances between every pair of points that belong to the pyramid, and its chirality doesn’t change either. What’s left to do is defining what exactly it means to move object in 3D.
Rotations in 3D
Let’s rotate some point P by angle α around axis a. Note that the rotation axis has a direction so it is clear which angles are positive i.e. counter-clockwise.
To rotate, first, we need to find a plane of rotation. To do that, we find the shortest route PO between P and axis “a” (PO is perpendicular to “a”). Then, we find line b that passes through O and is not on the same plane as P and line a. On this very plane (constituted by lines a and b) we find another line b’ that is perpendicular to line a. The segment PO and line b’ constitute another plane that is perpendicular to line a, and on this plane we perform a rotation just like described in the previous chapter.
Using this procedure we can rotate any collection of points in 3D space e.g. a pyramid. Let’s summarize the properties of rotation with the following:
Postulate 6-3D: Rotation:
- preserves the distance between every pair of points i.e. the distance before the rotation is the same as after
- preserves chirality i.e. a pyramid doesn’t get flipped on the other side of its base
One of the consequences of this postulate is that after the rotation, points that were on the same line are still on the same line (now rotated) i.e. rotation doesn’t deform straight lines. The argument for this is the same as in the section about rotations in 2d so I won’t repeat it here.
Note that the same must be true for a plane. Points that were on the same plane before the rotation, after the rotation are still on the same plane i.e. the plane was rotated without any deformation. This is a consequence of the fact that rotations do not deform lines.
Translations in 3D
Translation, only changing the position of an object, is defined in the same way for 3D as for 2D. However in 3D, a displacement d is a directed segment in 3D space. To apply this displacement to point P, first, we construct a plane containing d and point P, and then we move P to P’ in exactly the same way as on a 2D plane.
The properties of translation can by summarized as:
Postulate 7-3d: Translation:
- preserves the distances between points i.e. the distance before the translation is the same as after
- preserves chirality i.e. a translated pyramid is not flipped on the other side of its base
The properties of translation are the same as for rotation so translation doesn’t deform straight lines and planes i.e. points that were on the same line or plane before a translation are still on the same line or plane just translated to a new location.
Pyramid
Thus far a pyramid was introduced without specifying how to construct one. It’s time to rectify this and demonstrate 3d coordinates and postulates in action.
When we choose three points in space: P0, P1, P2 they constitute the base of our pyramid and also a plane with the front side defined by a counterclockwise appearance of the route P0→ P1→ P2→P0. The fourth vertex P3 can be located with its coordinates with respect to P0, P1 and P2. The first two coordinates: x, y give us the position of point O on our plane where OP3 is the shortest route between P3 and the plane. Coordinate z in turn specifies the distance between P3 and the plane. When z is positive, P3 is in front of the plane and when negative, then it is behind. The question is how to construct a line perpendicular to plane P0P1P2 in point O. We have not done it yet.
Let’s pick some random point Q above plane P0P1P2 (lines h, g and g’ below are on plane P0P1P2):
We already know how to construct line f that is perpendicular to our plane and passing through point Q. Let’s draw on our plane P0P1P2 line h that goes through points O and N (NQ is the shortest route between Q and the plane). Next, we add on plane P0P1P2 lines g and g’ perpendicular to h in points N and O. If we translate lines h, g and f along segment NO, then line f becomes f’ that passes through point O, line g coincides with g’, and h moves along itself. According to postulate 7-3D the distances between the points that moved didn’t change so f’ must be perpendicular to lines g’ and h just like it was before moving (in the previous chapter, angles are expressed in terms of relative distances). Now that we constructed line f’ that is perpendicular to plane P0P1P2, we can locate P3 on f’ using the coordinate z and we are pretty much done.
Summary
In this chapter we have established how to locate a point in space using three real numbers.
Also, we managed to grasp mathematically the experimental fact that a recipe to construct some object (e.g. a pyramid) produces the same result no matter where in space it is utilized, that is: rotation and translation of an object doesn’t change its shape and size.
We know it is possible to express with real numbers a distance of translation and angle of rotation. How about direction of translation and axis of rotation? Can they be described with numbers as well? We will find out in the next chapter.