A lot of the following is going to be rather intuitively clear, but we need to build up a framework where we are all speaking the same language to develop the powerful tools that we are going to find over the coming sections. We will be dealing here specifically with three dimensional space but we will discuss along the way the extension of these concepts to higher dimensional spaces. The higher dimensional stuff is not examinable but I think that sometimes it helps to understand the things which are special about three dimensions, and the things which are not.

In particular, I can recommend having a look at the web page of John Baez who discusses the regular polytopes in different numbers of dimensions here.

It’s clear that to define where you are in three dimensional space you need to set up a few key ingredients first. What you need is first of all an origin – a place to call home from which you will relatively describe your position. Everything will be defined in terms of distances and directions from the origin. The other thing that you need to specify are the axes of your space. Generally we will call them the $x$, $y$ and $z$ axes. The important point about them is that we want them to be orthogonal – ie. all at right angles to each other. Clearly in four you would need four axes, and in 11 dimensions you would need 11 orthogonal axes. Think for a bit about why in particular we desire our axes to be orthogonal…

In three dimensions the three axes define for us three orthogonal planes. We can think of these as three 2-dimensional planes which are all at rightangles to each other. We denote these the $xy$-plane, the $xz$-plane and the $yz$-plane. You might have thought that if you have three of these in three dimensions then in some other number $d$ of dimensions you will have $d$ planes, but this isn’t the case. To define a plane we need to pick two axes. In three dimensions this is choosing two directions from three, or $_3C_2$, which is three. In four dimensions you pick two directions from four. ie. $_4C_2=6$. If we label the axes in four dimensions as $x,y,z,r$ for instance, then we have the $xy$, the $xz$, the $xr$, the $yz$, the $yr$ and the $zr$ planes. In 11 dimensions there are $_{11}C_2=55$ planes. We’re not going to enumerate them! In fact there is a simpler way of writing $_dC_2$ which is that this is always going to be given by $\frac{d(d-1)}{2}$. Actually 3 dimensional space is already special for this reason.

In two dimensions there are said to be four quadrants. where both the $x$ and $y$ values are positive, two when one is positive and one negative and one when they are both negative. In three dimensions there are eight of them. We can denote them by $+++$, $++-$, $+-+$, $+--$, $-++$, $-+-$, $--+$, $---$. We call these octants. This means that in each sector we have to choose either positive or negative for each of the directions. This gives $2^d$ sectors in $d$ dimensions.

Positions in three dimensions
A point in three dimensions is denoted $P(a,b,c)$ and is a distance $a$ from the $(y,z)$ plane, a distance $b$ from the $(x,z)$ plane and a distance $c$ from the $(x,y)$ plane (Note that sometimes we will call it the $(x,y)$ plane and sometimes the $xy$ plane. Learn to go seamlessly between these notations.

A point $P(3,4,2)$ and its distances from the various coordinate planes.

We can see from figure above that the point defines for us a rectangle with the origin as the extreme vertex. In particular we can project the point onto the coordinate planes. If we project the point $P(3,4,2)$ onto the $xy$ coordinate plane, this is like shining a light from above and seeing where the shadow of the original point lies on the $xy$ plane. The answer is another point (call it $Q$) which is at $Q(3,4,0)$. Similarly we can project onto the other planes and we get $R(3,0,2)$ and $S(0,4,2)$. These can similarly be projected onto the coordinate axes. If we take $Q$ and project it onto the $y$ axis we get a point at $(0,4,0)$, etc.

Projections can be very useful for looking at straight lines in 3 dimensions. If you are given the 3 points $(3,7,-2)$, $(2,4,2)$ and $(1,3,3)$ and asked whether they form a straight line, then it’s not clear by drawing them as a two dimensional picture (which all pictures are), whether or not they form a straight line. We also don’t quite know yet how to find gradients in higher dimensions. However, one thing we can be sure of is that a straight line in 3 dimensions will still look like a straight line when we project it onto the coordinate planes. In order to check that it does form a straight line we’ll have to project it onto all three planes. First, let’s project it onto the $xy$-plane. The points project to: $(3,7,0)$, $(2,4,0)$ and $(1,3,0)$. Now we can calculate the gradient between the first and the second point (call it $g_1$) and between the second and the third point (call it $g_2$): $g_1=\frac{4-7}{2-3}=3$ and $g_2=\frac{3-4}{1-2}=1$. So the gradients are different and thus this can’t be a straight line. If the projection isn’t a straight line then the original line can’t be straight either. Had we found that this projection was a straight line then we still would have had to check the other projections to see whether they gave straight lines. Only if all three projections give straight lines can we be sure that the line in three dimensions is straight.

Functions in three dimensions

A point in three dimensional space is given by three real numbers, thus the set of all points in three dimensional space is $\mathbb{R}\times \mathbb{R}\times \mathbb{R}=\mathbb{R}^3=\{(x,y,z)|x,y,z\in \mathbb{R}\}$. In two dimensions, a curve is a set of points in $\mathbb{R}^2$. In 3D, an equation involving $x$, $y$ and $z$ is a surface in $\mathbb{R}^3$. For instance, in 2D we could look at the set:

$\{(x,y)|y=x^2\}$

or

$\{(x,y)|y=3\}$

The first gives the set of points along the curve $y=x^2$ and the second gives the set of points on the straight line at $y=3$. How about in three dimensions? The set of points described by:

$\{(x,y,z)|z=3\}$

Involves all points such that $z=3$, so for instance $P(4,15,3)$ is such a point, and $P(-12,2,3)$ is such a point. In fact all points, independent of the $x$ and $y$ values but with $z=3$ satisfy the requirements to be in this set. This means that the set of points is just the flat surface at $z=3$. This is shown in the figure below:

The function $z=3$ plotted in three dimensions.

In fact anything of the form:

$\{(x,y,z)|ax+by+cz=d\}$

will correspond to a flat surface in 3-dimensions. ie. Any linear function corresponds to a flat surface. As soon as we have any higher powers or non-integer powers of the directions, or if we have any other functions, we will have curved surfaces. This is very important to remember – linear functions correspond to flat surfaces in any number of dimensions.

Try using Wolfram alpha to plot surfaces in 3 dimensions. For instance, if you want to plot $\{(x,y,z)|3x+2y+z=4\}$ you have to solve the equation for z which gives you $z=4-(3x+2y)$ and then you can write in Wolfram Alpha:

$Plot3D\left[4-(3x+2y),\{x,-5,5\},\{y,-5,5\}\right]$

Convince yourself that all linear functions will give flat surfaces.

Let’s look at a slightly less trivial function. How about:

$\{(x,y,z)|x^2+y^2=1\}$

Of course we can just write this as “the surface in three dimensions specified by the function $x^2+y^2=1$” rather than writing it in the language of the set of points, but they are equivalent. This says “I don’t care about what value $z$ is but I want all those points whose $x$ and $y$ values square and add to one. ie. for each $z$ values we have a circle of radius 1 centred at $(0,0,z)$. This surface is shown here:

The function $x^2+y^2=1$ plotted in three dimensions.

We can see that because $z$ is not constrained, this is just a cylinder. ie. a circle defined at all values of $z$. We could add another equation. For instance we could have:

$\{(x,y,z)|x^2+y^2=1,z=3\}$

which now is explicitly the $z=3$ slice through the previous surface. It is simply a circle at $z=3$ of radius 1 centred at $(0,0,3)$.

The function $x^2+y^2+z^2=1$ is of course a sphere of radius 1. We can define a sphere of any radius, centred at any point $(x_0,y_0,z_0)$ with: $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2$.

Regions in 3d

The definition of regions in three dimensions is also relatively intuitive. For instance: the region defined by $1\le x^2+y^2+z^2\le 4$ is the region in between the spheres of radius 1 and 2 centred on the origin.

The distance from the origin to the point $P(3,4,2)$ is given by by using Pythagoras with the red, green and blue lines. ie. $distance^2= (\sqrt{3^2+4^2})^2+(2)^2=\sqrt{3^2+4^2}^2+ 2^2$.

Distances in three dimensions

We can ask about distances in three dimensions as a simple extension of two dimensions. We already know that the equation for a sphere in 3 dimensions of radius $r$ centred about the point $(a,b,c)$ is $\{(x,y,z)|r=(x-a)^2+(y-b)^2+(z-c)^2\}$. This is thus the set of points (ie. the surface in this case) all of whom are a distance $r$ from the point $(a,b,c)$. This already tells us that to find the distance between two points at $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$ all we need is $\sqrt{(a_x-b_x)^2+(a_y-b_y)^2+(a_z-b_z)^2}$ as you would expect. The following diagram may not be intuitive straight away so figure out carefully what the different colours in the diagram correspond to. You can see the equation for the distance in 3d coming directly from Pythagoras by looking at the distances of the projections of these points onto the coordinate axes.

 How clear is this post?