Working in two and three dimensions

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We talk about basic geometry in higher dimensions.

The word geometry can be broken into geo meaning “world” and metry meaning “measure.” In this section we will tell you what our mathematical “world” is, and how we “measure” it.

Higher dimensions

In our previous courses, we studied functions where the input was a single real number and the output was a single real number. Note, the word “real” is being used in a technical sense:

A real number is a number that has a (possibly infinite) decimal representation. The set of all real numbers is denoted by $\R$ .

Which of the following are real numbers?

$\sqrt {7}$ $0$ $-3$ $\pi$ $e$ $x$ $\sqrt {-1}$ $\infty$

When we say a function maps a real number to a real number, we write: $f:\R \to \R$ When working in two dimensions, we need a way of talking about ordered pairs of numbers. We denote the set of all ordered pairs of real numbers by $\R ^2$ . When working in three dimensions we denote the set of all ordered triples of real numbers by $\R ^3$ . In three dimensions we have three coordinates axes, the $x$ -axis, $y$ -axis, and $z$ -axis:

The axes point according to the right-hand-rule:

Of course you will need to “spin” your hand around to align your pointer-finger with the $x$ -axis and your middle-finger with the $y$ -axis. Then your thumb will point in the $z$ -direction.

Which of the following axes are aligned according to the right-hand rule?

Point the “pointer finger” of your right hand in the positive direction of the $x$ -axis while simultaneously pointing your “middle finger” in the positive direction of the $y$ -axis. Your thumb will point in the positive direction of the $z$ -axis.

Basic plotting

To plot a point $(a,b,c)$ in $\R ^3$ , you move $a$ in the $x$ -direction, $b$ in the $y$ -direction and $c$ in the $z$ direction.

Of course, we’re going to be plotting many points. We typically described groups of points, as those that satisfy a given equation involving $x$ , $y$ , and $z$ . Here is a place where working in three dimensions is really different from working in two.

In $\R ^2$ , any equation involving $x$ and/or $y$ draws a curve.
In $\R ^3$ , any equation involving $x$ , $y$ , and/or $z$ draws a surface.

The most basic surface in $\R ^3$ is a plane.

Which equation describes the $(y,z)$ -plane?

$x=0$ $y=0$ $z=0$

For every point on the $(y,z)$ -plane, the $x$ -coordinate is zero.

Can you describes the solution set of $y=2$ in $\R ^3$ ?

It’s a horizontal line. It’s a vertical line. It’s a plane parallel to the $(x,y)$ -plane. It’s a plane parallel to the $(x,z)$ -plane. It’s a plane parallel to the $(y,z)$ -plane.

$y=2$ consists of all those points where $y=2$ , but $x$ and $z$ are allowed to be anything.

Another way to think of the point $(a,b,c)$ is as the intersection of the planes $x=a$ , $y=b$ , $z=c$ .

Move the point around below to see the planes that define it.

While three planes need not intersect at all, the intersection of two (nonparallel) planes is a line.

What is the intersection of the $(x,y)$ -plane and the $(y,z)$ -plane?

The $x$ -axis. The $y$ -axis. The $z$ -axis.

What is the intersection of the $(x,y)$ -plane and the $(x,z)$ -plane?

The $x$ -axis. The $y$ -axis. The $z$ -axis.

What is the intersection of the $(x,z)$ -plane and the $(y,z)$ -plane?

The $x$ -axis. The $y$ -axis. The $z$ -axis.

Distance and spheres

So the objects in our geometry are made of points, and now we must tell you how we plan to “measure” objects. To do this, we’ll use our old friend, the distance formula.

Given two points $P=(x,y)$ and $Q=(a,b)$ in $\R ^2$ , the distance between them is given by: $|PQ|=\sqrt {(x-a)^2 + (y-b)^2}$

This is nothing more than a corollary of the Pythagorean Theorem. Plot the points $P=(a,b)$ and $Q=(x,y)$ :

We may now construct a right triangle with horizontal side length $\answer [given]{(x-a)}$ and vertical side length $\answer [given]{(y-b)}$ whose hypotenuse is the shortest path between the two points:

By the Pythagorean Theorem, the length of this path is given by $|PQ|=\sqrt {(x-a)^{2}+(y-b)^{2}}.$

What is the distance between the points $P=(-17,379)$ and $Q=(14,-101)$ in $\R ^2$ ? $|PQ|=\answer {481}\unit {units}.$

On a completely related note, what’s the most famous theorem in mathematics? I’ll tell you: The Pythagorean Theorem. In essence, the distance formula is The Pythagorean Theorem. Let’s see if we can explain why the Pythagorean Theorem is true.

Given a right triangle,

we have that $a^2 + b^2 = c^2.$

Given a right triangle with legs $a$ and $b$ , and hypotenuse $c$ , we can make the following squares each with side length $(a+b)$ :

Since both squares above have side length $(a+b)$ , both large squares have the same area. Moreover, if we remove the triangles:

Both diagrams must still have the same area, since we removed an equal amount from each diagram. Hence $c^2=a^2+b^2$ .

The distance formula also extends to higher dimensions:

Given two points $P=(x,y,z)$ and $Q=(a,b,c)$ in $\R ^3$ , the distance between them is given by: $|PQ|=\sqrt {(x-a)^2 + (y-b)^2 + (z-c)^2}$

To see why this is true, draw a picture!

Ignoring the vertical components of the points, we can make a right-triangle whose legs have lengths $(x-a)$ and $(y-b)$ .

Thus by the Pythagorean Theorem, the length of the hypotenuse of the right-triangle above is: $\answer [given]{\sqrt {(x-a)^2+(y-b)^2}}$ Now, moving this segment up, we can form another right-triangle where the legs have lengths $\sqrt {(x-a)^2+(y-b)^2}$ and $(z-c)$ :

The length of the hypotenuse of this new triangle is the distance between $P$ and $Q$ . Again by the Pythagorean Theorem, we see $\begin{align*} |PQ|^2 &= \left(\sqrt{(x-a)^2+(y-b)^2}\right)^2 + (z-c)^2\\ |PQ| &= \sqrt{(x-a)^2+(y-b)^2 + (z-c)^2}. \end{align*}$ Thus the distance formula in $\R ^3$ is indeed what we claimed.

What is the distance between the points $P=(0,0,0)$ and $Q=(1,1,1)$ in $\R ^3$ ? $|PQ|=\answer {\sqrt {3}}\unit {units}.$

In general we can extend this notion of distance to $\R ^n$ :

Given two points $P=(x_1,x_2,\dots ,x_n)$ and $Q=(a_1,a_2,\dots ,a_n)$ in $\R ^n$ , the distance between them is given by: $|PQ|=\sqrt {(x_1-a_1)^2 + (x_2-a_2)^2 + \dots + (x_n-a_n)^2}$

What is the distance between the points $P=(0,0,0,0)$ and $Q=(1,1,1,1)$ in $\R ^4$ ? $|PQ|=\answer {2}\unit {units}.$

Circles and spheres, disks and balls

Let me remind you what the definition of a circle is:

A circle is the set of points $(x,y)$ in $\R ^2$ that are a fixed, nonzero, equal distance from a given point $C$ , where $C$ is the center of the circle.

Is the center point of a circle part of the circle?

yes no

Well, the distance from the center of a circle to the center of the circle is zero, and the radius of a circle is nonzero, so the center is not part of the circle.

From the definition of a circle, we see that it is intimately related to the distance formula. Indeed, it is also the case the equation of a sphere is related to the distance formula in $\R ^3$ :

The equation for a circle of radius $r$ centered at the point $(a,b)$ in $\R ^2$ is $r^2=(x-a)^2 + (y-b)^2$ The equation for a sphere of radius $r$ centered at the point $(a,b,c)$ in $\R ^3$ is $r^2 = (x-a)^2 + (y-b)^2 + (z-c)^2$ In general, the equation for a $n$ -dimensional “sphere” of radius $r$ centered at the point $(c_1,c_2,c_3,\dots ,c_n)$ in $\R ^n$ is $r^2 = \sum _{i=1}^n(x_i-c_i)^2$

This follows directly from the distance formula since a $n$ -sphere is the set of points that are equidistant from a given point in $\R ^n$ .

In general, a $n$ -dimensional “ball” of radius $r$ centered at the point $(a_1,a_2,a_3,\dots ,a_n)$ is given by the inequality $r^2 \ge \sum _{i=1}^n(x_i-a_i)^2$

Here the “ $\ge$ ” fills-in the $n$ -dimensional sphere.

The equation $(x-1)^2+y^2+(z+2)^2 = 4$ has a solution set in $\R ^3$ which forms a sphere. What is the center and radius of this sphere? $\text {Radius} = \answer {2}$ $\text {Center} = \left (\answer {1},\answer {0}, \answer {-2} \right )$

The expression $(x-1)^2+y^2+(z+2)^2$ is the square of the distance from $(1,0,-2)$ to $(x,y,z)$ . If the square of the distance is $4$ , then that distance is $2$ . Since the solution set of this equation is all points which are a distance of $2$ away from $(1,0,-2)$ , then this is a sphere of radius $2$ centered at $(1,0,-2)$ .

Things really get interesting when we have both spheres and planes around. Spheres can intersect planes at no points (if they are missing the plane), at one point (if they are “just” touching the plane), or at an infinite number of points (here the intersection is a circle).

Let’s see some examples.

How many points are on the intersection of $(x-3)^2 + (y+2)^2 + (z-1)^2 = 9$ and the $(y,z)$ -plane? Describe the intersection.

The first thing we should note, is that the surface defined by $(x-3)^2 + (y+2)^2 + (z-1)^2 = 9$ is a sphere, of radius $\answer [given]{3}$ , centered at the point $(\answer [given]{3},\answer [given]{-2},\answer [given]{1})$ . Moreover, the $(y,z)$ -plane is the plane $\answer [given]{x}=0$ . So, we have to solve the following system of equations $(x-3)^2 + (y+2)^2 + (z-1)^2 = 9$ $x=0$ By plugging $x=0$ into the first equation, we obtain that $(y+2)^2 + (z-1)^2 =0$ So, the sum of the two squares on the left is equal to $0$ . This implies that both terms on the left are equal to $0$ . Therefore, the system has a unique solution, namely $(\answer [given]{0},\answer [given]{-2},\answer [given]{1})$ .

That wasn’t too bad, let’s see another.

How many points are on the intersection of $(x-3)^2 + (y-1)^2 + (z+2)^2 = 8$ and the $(x,y)$ -plane? Describe the intersection.

The first thing we should note is that the surface defined by $(x-3)^2 + (y-1)^2 + (z+2)^2 = 8$ is a sphere, of radius $\answer [given]{\sqrt {8}}$ , centered at the point $(\answer [given]{3},\answer [given]{1},\answer [given]{-2})$ . Moreover, the $(x,y)$ -plane is the plane $\answer [given]{z}=0$ . So, all the points on the intersection satisfy the system of equations $(x-3)^2 + (y-1)^2 + (z+2)^2 = 8$ $z=0$ By plugging $z=0$ into the first equation, we obtain that $(x-3)^2 + (y-1)^2 = 4$ Hence, the intersection consists of infinitely many points and they form a circle of radius $r=\answer [given]{2}$ . The circle of intersection is centered at $(\answer [given]{3},\answer [given]{1},\answer [given]{0})$ , in the plane $z=\answer [given]{0}$ .

For our last example, we’ve left the easiest case of all.

How many points are on the intersection of $(x+1)^2 + (y+3)^2 + (z-2)^2 = 6$ and the $(x,z)$ -plane? Describe the intersection.

The first thing we should note, is that the surface defined by $(x+1)^2 + (y+3)^2 + (z-2)^2 = 6$ is a sphere, of radius $\answer [given]{\sqrt {6}}$ , centered at the point $(\answer [given]{-1},\answer [given]{-3},\answer [given]{2})$ . Moreover, the $(x,z)$ -plane is the plane $\answer [given]{y}=0$ . Since the plane is $\answer [given]{3}$ units away from the center of the sphere, and the radius of the sphere is $\answer [given]{\sqrt {6}}$ , there are no points on the intersection! We could reach the same conclusion by setting up a system of equations $(x+1)^2 + (y+3)^2 + (z-2)^2 = 6$ $y=0$ By plugging $y=0$ into the first equation, we get that $(x+1)^2 + (z-2)^2 = -3$ which implies that the system has no solutions.