We can navigate in 3-D to any point (x_R, y_R, z_R), in Cartesian Coordinates, by starting at the origin (0, 0, 0) and in the 2-D, x-y plane, walking in the x-direction (1, 0, 0) a distance x_R, then turning left 90 degrees and walking a distance y_R in the y-direction (0, 1, 0), still on the x-y plane.  This is the same as we did in 2-D except we put an extra ‘0’ in the third ‘z’ direction.  At this point we are still on the x-y plane at the point (x_R, y_R, 0).  The z-direction is straight up, so we would now need a ladder to climb up or a shovel to dig down, it is helpful to pretend we can fly (or dig).  From this point (x_R, y_R, 0) on the x-y “ground” plane, we fly straight up in the z-direction (0, 0, 1) a distance z_R, and we successfully arrive at the point (x_R, y_R, z_R).

We can express this journey with vectors, as we did in 2-D, by connecting (adding) lines together to form the path:

   (x_R, y_R, z_R) = (1, 0, 0)*x_R + (0, 1, 0)*y_R + (0, 0, 1)*z_R
                        = (x_R, 0, 0) + (0, y_R, 0) + (0, 0, z_R)
                        = (x_R, y_R, z_R)

        And using dot product notation, this can be written:

   (x_R, y_R, z_R) = (1, 0, 0)*x_R + (0, 1, 0)*y_R + (0, 0, 1)*z_R
                        = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]^.(x_R, y_R, z_R)

The [(1, 0, 0), (0, 1, 0), (0, 0, 1)] matrix is called the identity matrix.

We could also arrive faster by flying along a straight line connecting the two. Fly along a direction (D_x, D_y, D_z), from the origin directly to this point, (x_R, y_R, z_R) a distance “R” away:

   (x_R, y_R, z_R) = (D_x, D_y, D_z)*R

        Where the distance ‘R’ is found by:

    R² = x_R² + y_R² + z_R²

From the equation above for this line, we can easily find the direction vector from the origin to any point (x_R, y_R, z_R), similar to what we did in 2-D.

      (D_x, D_y, D_z) = (x_R/R, y_R/R, z_R/R)

Recall that the direction vector always has a distance of 1 unit, as seen in the diagram above.  In the diagram, we have attempted to show a 3-D path on a 2-D piece of paper.  Note that in the diagram, D_r is the distance from the origin to the point (D_x, D_y, 0) in the x-y plane. From the Pythagorean theorem, we find that the length:

      D_r² = D_x² + D_y²    and
      D_r² + D_z² = 1        thus combining these
      D_x² + D_y² + D_z² = 1.

The total length of the distance vector is 1 unit, as required.  Everything is consistent.  Now we use the definition of the ‘Sin’ and ‘Cos’ and can see, from the diagram above, that:

     Cos(A_xy) = D_x/D_r
     Sin(A_xy) = D_y/D_r

      Cos(A_z) = D_r/1
      Sin(A_z) = D_z/1

    Thus

       D_x = Cos(A_xy)*D_r = Cos(A_xy)*Cos(A_z)
       D_y = Sin(A_xy)*D_r = Sin(A_xy)*Cos(A_z)
       D_z = Sin(A_z)

And so, if we stand at the origin facing in the x-direction, and rotate our “head” an angle A_xy to the left, and then tilt our head an angle A_z off the x-y plane to look directly at the point (x_R, y_R, z_R), we can find the direction vector pointing to this point using these angles:

   (D_x, D_y, D_z) = (Cos(A_xy)*Cos(A_z), Sin(A_xy)*Cos(A_z), Sin(A_z))

       And then draw the line to this point:

    (x_R, y_R, z_R) = (D_x, D_y, D_z)*R

Now that we are staring at a point in the sky, with our nose pointed directly at the point, we can ask about “reference frame” of our head.  What direction is our left ear pointing, for example.  Our left ear is pointing in the direction, still in the x-y plane, that is 90 to the left of the A_xy angle we rotated our head (-D_y/D_r, D_x/D_r, 0). We divide by D_r because in 3-D we must make the length of the direction vector equal to 1.  If this seems vague, look back and review how to find 90 degree directions in the 2-D plane in the top diagram taken from previous posts.

When we tilt our head, we can ask what is the direction that the top of our head points, 90 degrees up from the angle A_z. It perhaps takes a bit more imagination to see this, since we are facing in the x-y plane (D_x/D_r, D_y/D_r, 0) and have tilted our head up to look at D_z, we find that the “shadow” of our head is pointed behind us, in the x-y plane (-D_x/D_r, -D_y/D_r, 0), then looking at the elevation D_r (when rotated 90 degrees) we find that (-D_x*D_z/D_r, -D_y*D_z/D_r, D_r) is the direction of the top of our head.  So, in the reference frame of our head:

     Direction of our Nose:     (D_x, D_y, D_z)
     Direction of Left ear:         (-D_y/D_r, D_x/D_r, 0)
     Direction of Top of head:  (-D_x*D_z/D_r, -D_y*D_z/D_r, D_r)

Let us verify that these direction vectors all have a length of 1.  The Nose direction we have verified.  For the Left ear direction:

    L^.L = (-D_y/D_r)² + (D_x/D_r)² + 0² = (D_y² + D_x²)/D_r² = 1

For the Top of head direction:

    T^.T = (-D_x*D_z/D_r)² + (-D_y*D_z/D_r)² + D_r² = (D_y² + D_x²)*D_z²/D_r² + D_r²
                                                                    = D_z² + D_r² = 1

Now let’s find what happens when we find the dot product of these direction vectors, we will call them the N, L, and T directions respectively.

N^.N = (D_x, D_y, D_z)^.(D_x, D_y, D_z) = D_x² + D_y² + D_z² = 1

N^.L = (D_x, D_y, D_z)^.(-D_y/D_r, D_x/D_r, 0) = -D_x*D_y/D_r + D_y*D_x/D_r + 0
                                                                  = 0

N^.T = (D_x, D_y, D_z)^.(-D_x*D_z/D_r, -D_y*D_z/D_r, D_r)
        = -D_x²*D_z/D_r + -D_y²*D_z/D_r + D_z*D_r
        = -(D_x² + D_y²)*D_z/D_r + D_z*D_r
        = -D_r²*D_z/D_r + D_z*D_r
        = 0

L^.T = (-D_y/D_r, D_x/D_r, 0)^.( -D_x*D_z/D_r, -D_y*D_z/D_r, D_r)
       = D_y*D_x*D_z/D_r² + -D_x*D_y*D_z/D_r² + 0
       = 0

We also verified that L^.L = 1 and T^.T = 1 above.  So, in general, if we dot any of these N, L, T direction vectors with themselves, we get ‘1’, but if we dot any of them with each other, we get ‘0’.  It turns out that this is a requirement for any three perpendicular direction vectors in a 3-D Cartesian reference frame.  I find the “head” reference frame to be more natural because we can relate to it better, so I use this reference frame for whatever direction I am looking.  It is also a great reference frame for video games.

We have already seen from perpendicular 2-D paths how we can convert between a reference frame to our “ground” x-y-z reference frame.  To understand this, let’s take a journey: Fly in the Nose direction a distance S_N, then fly in the Left ear direction a distance S_L, and then in the Top of head direction a distance S_T.  Then we will arrive at the point:

    (x, y, z) = N*S_N + L*S_L + T*S_T

       or in dot product notation:

     (x, y, z) = [N, L, T]^.(S_N, S_L, S_T)

This amazing equation lets you convert any point in your “head frame” of reference (S_N, S_L, S_T), when your Nose is pointing in a direction N:(D_x, D_y, D_z), to a point in the “ground” frame (x, y, z).  You might also want to do this the other way around.  Review how we did this in 2-D, by “dotting” each side by each of the direction vectors, we get:

    (S_N, S_L, S_T) = [N, L, T]^.(x, y, z)

This is how video games convert a 3-D world, stored in “ground” (x, y, z) coordinates, into the perspective of the viewer’s head (S_N, S_L, S_T) coordinates.  The details need to be worked out but using this equation we can map a 3-D world onto a 2-D screen in the viewer’s perspective, no matter the relative direction of the viewer’s head.  We will stop here, there is a lot to take in.

On the Cartesian plane, if we walk along the x-direction (1, 0) for a distance of 4 units, then turn left 90 degrees (perpendicular) and walk a distance of 3 units, we arrive at the point (4, 3), which is a distance of 5 units from the origin (4² + 3² = 5²), great.  Well, what happens if we do the same, but start off walking in any direction (D_x, D_y) 4 units and then turn left 90 degrees (-Dy, Dx) and walk 3 units; we are still 5 units from the origin.  We have just “rotated” the whole path to a new direction.  Where would we end up (x, y)?  We use our orienteering skills:

     (x, y) = (D_x, D_y)*4 + (-D_y, D_x)*3
     (x, y) = [(D_x, D_y), (-D_y, D_x)]^.(4, 3)  –  using the dot product

Now notice that the matrix [(D_x, D_y), (-D_y, D_x)] when dotted with the cartesian point (4, 3) rotates it to another point on the circle (radius 5).  See the diagram. This matrix is called the “rotation matrix”.

Now that we are getting a bit more comfortable with the dot product, let’s play around and see what happens when we dot direction vectors with each other?  We’ll start off by dotting a direction with itself:

     (D_x, D_y)^.(D_x, D_y) = D_x² + D_y² = 1

Any direction dotted with itself is just “1”.  What about dotting a direction with its perpendicular direction?

    (D_x, D_y)^.(-D_y, D_x) = D_x*(-D_y) + D_y*D_x = 0

It turns out that any direction dotted with its perpendicular direction is “0”.  Now how about dotting the position (x, y) above with the direction vector?

   (D_x, D_y)^.(x, y) = (D_x, D_y)^.((D_x, D_y)*4 + (-D_y, D_x)*3)
                             = (D_x, D_y)^.(D_x, D_y)*4 + (D_x, D_y)^.(-D_y, D_x)*3
                             = 1*4 + 0*3
                             = 4

And so, dotting a direction vector (D_x, D_y) with any position vector (x, y) gives you the distance (S_x) you would have to travel in that direction before making a 90 turn and traveling the distance (S_y) that would bring you to the position (x, y).  Now let’s dot the perpendicular direction with the position vector:

   (-D_y, D_x)^.(x, y) = (-D_y, D_x)^.((D_x, D_y)*4 + (-D_y, D_x)*3)
                             = (-D_y, D_x)^.(D_x, D_y)*4 + (-D_y, D_x)^.(-D_y, D_x)*3
                             = 0*4 + 1*3
                             = 3

You can see that dotting the perpendicular direction with a position also gives you the distance (S_y) you would have to travel in the perpendicular direction (-D_y, D_x) to get to the position (x, y).

And so can you see that in this case:

     [(D_x, D_y), (-D_y, D_x)]^.(x, y) = (4, 3)

So in general if

     (x, y) = [(D_x, D_y), (-D_y, D_x)]^.(S_x, S_y)

              then

     [(D_x, D_y), (-D_y, D_x)]^.(x, y) = (S_x, S_y)

What does this mean?  The rotation matrix […] when it “operates” on (is dotted with) any point (x, y), gives you the coordinates of the point relative to a new set of “x-y” axis that is pointing in a direction (D_x, D_y) relative to the old axis.  Or you can say that it rotates the point around the origin.  Either vantage point is valid, you can rotate the point around the origin or rotate the reference frame in the opposite way around the origin; either way, it is describing the same thing.

Remember the concept of “rotational symmetry”, that we can pick the direction of the x-y axis to be in any direction we want.  Now we have the tools to convert the cartesian coordinates of a set of points from any reference frame to a reference frame pointing in any direction we choose.

THIS IS A FUNDAMENTAL CONCEPT of the MATH Language: ROTATIONAL SYMMETRY means that the universe is the same no matter the direction of your frame of reference.  The Rotation matrix gives you a way of describing the same universe from any prospective of direction.  It is pure magic.

One more point, remember that

   (D_x, D_y) = (Cos(A), Sin(A))

Where “A” is the angle of the direction off x-axis.  Thus, the rotation matrix “R(A)” can be written as:

     R(A) = [(Cos(A), Sin(A)), (-Sin(A), Cos(A))]

And there you have it.

The diagram above (which looks a little like Homer Simpson) shows you how to build a 2-dimensional universe from 1-dimensional universes. There is a lot in the diagram so we will go over it a piece at a time then you can use your imagination to build another universe.

Start with the original 1-dimensional line, the one that passes through the points labeled O, and O+ on the diagram. Remember that we need only 2 points on the line, any 2 of them will generate the whole line. We then need to find another point, labeled P, that is not on the original line. The lines radiating out of the point P represent all the lines connecting P with all the points on the original line “O”.

Each one of the line segments between point P with a point on the line spans a different distance. We find a line of symmetry with respect to P by finding the segment with the shortest distance. This segment goes to the point labeled O. Point O is an important point of reflection, a point of symmetry on the line with respect to P, any point on the positive side of O and its reflection on the negative side of O has the same distance to Point P.

We use the label “r” to refer to the distance between point P and point O. In math language, the symbol “r” reminds us of a “radial” distance from the point P. This r distance is referred to as the “distance” between P and the line. If we put together all the points (on the radial lines) that are a distance r from point P, we form a circle of radius r around P. Notice that since r is the shortest distance between P and the line, this circle only touches the line at point O and cannot reach the line anywhere else. Can you find this circle in the diagram?

The line passing through O and P is a special line of symmetry in this case. It is often called the “normal” of the line that passes through point P. It is also called the perpendicular to the line passing through P. Any set of points on the “positive” side of a line can be reflected onto the “negative” side by finding the perpendiculars passing through the points and then reflecting the points along their perpendicular lines the same distance on the “negative” side of the line. In this diagram, this symmetry is obvious, every point on the right (+) side of the perpendicular (O P) is a reflection of points the left (-) side and vice versa.

In the diagram, we show the point O+ on the positive side of the original line at a distance r from the point O. Its reflection on the negative side is labeled O-. We can easily see that the line (O O+) is also the perpendicular of the line (O P). If we wanted to, we could reflect this whole diagram through the original line and we would get the diagram flipped upside down on the bottom half. For clarity’s sake, I didn’t do this in the diagram.

You might be starting to see the amazing symmetries represented in this diagram. It gets even more obvious when you look at the midpoint between O+ and P. This is labeled M+. Its reflection is labeled M-. If we draw a line segment between M+ and M-, as shown in the diagram, notice that it is also perpendicular to the line of symmetry. To show this better I drew a circle around point M+ and M- that touches the perpendicular line at one point. It also becomes obvious that M+ and M- are the same distance from the original line O O+, by symmetry. All distances are preserved upon reflection.

Alright, now we use our imagination to see what would happen if we drew another perpendicular to the original line through point M+ and another through M-. We can use the small circles drawn to visualize where the circle touches the line to draw these perpendiculars. We can now reflect everything in the middle of these perpendiculars on either side. Like two mirrors facing each other, the reflections go on forever in both directions.
From these symmetries, the line M- M+ can be shown to always be the same distance away from the original line forever in both directions. These lines will never cross.

We have finally found two parallel 1-dimensional spaces that do not intersect each other. Do you think you could find an infinite number of parallel lines in the vertical direction by using reflections?

This is a concept utilized by the mathematician Rene Descartes. He built what is known as the Cartesian plane, composed of infinitesimally close parallel vertical and horizontal lines all normal to each other. These concepts give us many ways to look at 2-dimensional space. One of the most important qualities of this space is that it is assumed to be exactly the same in all directions. Very flat indeed.

Let’s imagine that you have a universe with only two points in it. You now draw a line segment between the two points and choose one of them to be the reference point. The location of any point along the line segment is now determined by the distance it is from the reference point. Now make a third point on the line segment that is exactly half-way between the two points. This new point is located at exactly half of the distance between the two end points.

Can you see the symmetry that is formed by these three points? The distance from the center point to either of the two end points is exactly the same. If you now were to choose the center point to be the reference point, then both end points would be the same distance away, but in opposite directions.

You would now need more than just the distance measurement alone to determine the location of points along the line. With three points, now direction becomes important. Again, using the axiom of choice and can choose which direction from the center point is positive, and which direction is negative.

We can then, finally, determine the location of every point on the line by specifying both a distance and a direction (positive or negative) from the center point. This arrangement of determining the locations of points on a line is called a “number line”. We call the distance and direction of each point on the line the Point’s coordinates. A Point’s coordinates uniquely determine its location on the line. Finally, we have a way of determining the location of a Point.

The center point is a special point of symmetry between the two endpoints. We can “rotate” the direction (exchange negative for positive directions) and the location (distance and direction) of the two identical endpoints would be exactly the same. This is called “rotational symmetry”. It means you can switch reference directions and the list of all the point coordinates will be the same. Remember that since you can’t tell the difference between points, the order of the list of points is not important.

Please believe me, taking the time to understand these concepts makes math so much more fun and easier later on. Learning these concepts is like learning the basic vocabulary of the math language.

How can the exact location of the Point be described? This is a conceptual problem. It is also where some very advanced concepts of math (found in calculus and number theory) come to play. After some thought, it becomes obvious that if there was just one point in the universe, all by itself, it does not have an absolute location until you look at where it is relative to another point in the universe.

If there were only two points in the universe you could imagine a line segment drawn between the two points, and then there is some distance between the two points, you could pick one of the points as the reference point. Then you could describe the location of the second point as being a certain distance from the reference point. You only need one (1) distance measurement from the reference point to describe the location of the Point. This is why such a line is called 1-dimensional space.

You have now discovered quite a few interesting concepts, including the concept of the number “2”. Two identical objects in space that are distinguished by the distance between them. You have also discovered the concept of the number “1”, the concept of distance, and you are also starting to realize why Einstein’s theory is called the theory of relativity. Distances and times are only measurable relative to a reference point, and you can pick any of the points to be the reference point.

The idea that you can pick any of the points to be the reference point is called the axiom of choice. An axiom is something that is assumed to be true even though it has no definite proof. The axiom of choice forms the conceptual foundation for all mathematics.

Another concept that comes from using the distance from a reference point to describe location is the idea of symmetry. You can exchange the locations of two points, and you have not changed anything. It is impossible to tell the two points apart, and both points still have the same locations relative to every other point. This is called exchange symmetry. Symmetries makes math and physics much easier.