Euclidean Space

The real numbers are denoted by $\real$ . I'll assume you're familiar with the basic properties of $\real$ , but I'll mention less familiar things as they come up.

The standard 2-dimensional plane is denoted $\real^2$ . It consists of ordered pairs of real numbers:

$$\real^2 = \{(x, y) \mid x, y \in \real\}.$$

You're probably familiar with how a point $(x, y)$ is located in the plane:

$$\hbox{\epsfysize=1.5in \epsffile{euclidean-space-1.eps}}$$

The "x" and "y" on the axes label the positive x-axis and positive y-axis.

Similarly, 3-dimensional space is denoted $\real^3$ . It consists of ordered triples of real numbers:

$$\real^3 = \{(x, y, z) \mid x, y, z \in \real\}.$$

(There are also $\real^4$ , $\real^5$ , and so on, defined in similar fashion.)

Just as it's conventional to use x and y to denote the first and second coordinate variables in $\real^2$ , it is conventional to use x, y, and z to denote the first, second, and third coordinate variables in $\real^3$ . The picture shows how a typical point $(x, y, z)$ is located in space:

$$\hbox{\epsfysize=1.75in \epsffile{euclidean-space-2.eps}}$$

The "x", "y", and "z" on the axes label the positive x-axis, the positive y-axis, and the positive z-axis.

At this point, I should note a convention that we'll always follow. The labelling of the positive axes in $\real^2$ and $\real^3$ follow the "right-hand rule". In the x-y-plane, you curl the fingers of your right hand from the positive x-axis to the positive y-axis through a $90^\circ$ angle. Check for yourself with the picture above.

(Note that in some computer graphics applications --- for instance, in the SVG graphics language --- it's conventional to have the positive y-axis going "downward" rather than "upward".)

For $\real^3$ , curl the fingers of your right hand from the positive x-axis to the positive y-axis. As you do so, your thumb points in the direction of the positive z-axis. Check for yourself with the picture above.

Graphing things in 3 dimensions is obviously harder than graphing things in 2 dimensions. While you don't need to have great artistic skills, you should practice making diagrams in 3 dimensions as they are a huge aid to understanding. Short of taking a drawing class, the best approach might be to copy other peoples' pictures until you get the idea. For starters, you can practice plotting points in $\real^3$ .

Example. Plot the points $(2, 4, 3)$ , $(1, -2, -3)$ , and $(-2, -4, 2)$ . The pictures below aren't perfectly scaled; I just want to locate the points in approximately the right places.

$$\hbox{\epsfysize=1.75in \epsffile{euclidean-space-3.eps}} \hskip0.2in \hbox{\epsfysize=1.75in \epsffile{euclidean-space-4.eps}} \hskip0.2in \hbox{\epsfysize=1.75in \epsffile{euclidean-space-5.eps}} \quad\halmos$$

The axes and the coordinate planes. Consider the x-axis. It is perpendicular to the y-z plane, and passes through the origin $(y, z) = (0, 0)$ of the y-z-plane. Therefore, it is determined by the equations

$$y = 0, \quad z = 0.$$

$$\hbox{\epsfysize=2in \epsffile{euclidean-space-7.eps}}$$

I'll discuss lines later on, and in particular their equations in parametric form. The parametric equations for the x-axis are

$$x = t, \quad y = 0, \quad z = 0.$$

Similarly, the y-axis is

$$x = 0, \quad z = 0.$$

The z-axis is

$$x = 0, \quad y = 0.$$

Now consider the x-y plane. It consists of all the points at "z-level" zero --- that is $z = 0$ .

$$\hbox{\epsfysize=2in \epsffile{euclidean-space-8.eps}}$$

Likewise, the y-z plane is $x = 0$ , and the x-z plane is $y = 0$ .

Distance. The distance between points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by

$$\hbox{distance} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}.$$

(You can write "$(x_2 - x_1)^2$ ", and so on, instead; the squaring makes the order of subtraction irrelevant.)

Here's where the formula comes from.

$$\hbox{\epsfysize=2.25in \epsffile{euclidean-space-6.eps}}$$

The box has sides of lengths $|x_1 - x_2|$ , $|y_1 - y_2|$ , and $|z_1 - z_2|$ . By Pythagoras' Theorem, the diagonal s of the bottom of the box is

$$s = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.$$

I can drop the absolute values because I'm squaring the terms.

Again by Pythagoras' Theorem, the distance is

$$\hbox{distance} = \sqrt{s^2 + (z_1 - z_2)^2} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}.$$

Example. Find the distance between the points $(4, -1, 5)$ and $(2, 6, -3)$ .

$$\sqrt{(4 - 2)^2 + (-1 - 6)^2 + (5 - (-3))^2} = \sqrt{4 + 49 + 64} = \sqrt{117}.\quad\halmos$$

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