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Mathematical Expression Editor
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.
1 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 \(\mathbb {R}\).
When we say a function maps a real number to a real number, we write:
\[ f:\mathbb {R} \to \mathbb {R} \]
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 \(\mathbb {R}^2\). When working in three
dimensions we denote the set of all ordered triples of real numbers by \(\mathbb {R}^3\). 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.
2 Basic plotting
To plot a point \((a,b,c)\) in \(\mathbb {R}^3\), 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 \(\mathbb {R}^2\), any equation involving \(x\) and/or \(y\) draws a curve.
In \(\mathbb {R}^3\), any equation involving \(x\), \(y\), and/or \(z\) draws a surface.
The most basic surface in \(\mathbb {R}^3\) 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 \(\mathbb {R}^3\)?
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.
3 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 \(\mathbb {R}^2\), 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 \(\mathbb {R}^2\)?
\[ |PQ|=\answer {481}\mathrm {units}. \]
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 \(\mathbb {R}^3\), 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 \(\mathbb {R}^3\) is indeed what we claimed.
What is the distance between the points \(P=(0,0,0)\) and \(Q=(1,1,1)\) in \(\mathbb {R}^3\)?
\[ |PQ|=\answer {\sqrt {3}}\mathrm {units}. \]
In general we can extend this notion of distance to \(\mathbb {R}^n\):
Given two points \(P=(x_1,x_2,\dots ,x_n)\) and \(Q=(a_1,a_2,\dots ,a_n)\) in \(\mathbb {R}^n\), the distance between them is given by:
What is the distance between the points \(P=(0,0,0,0)\) and \(Q=(1,1,1,1)\) in \(\mathbb {R}^4\)?
\[ |PQ|=\answer {2}\mathrm {units}. \]
3.1 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 \(\mathbb {R}^2\)
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 \(\mathbb {R}^3\):
The equation for a circle of radius \(r\) centered at the point \((a,b)\) in \(\mathbb {R}^2\) 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 \(\mathbb {R}^3\) 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 \(\mathbb {R}^n\) 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 \(\mathbb {R}^n\).
In general, the equation for a \(n\)-dimensional “ball” of radius \(r\) centered at the point \((a_1,a_2,a_3,\dots ,a_n)\) is
\[ r^2 \ge \sum _{i=1}^n(x_i-a_i)^2 \]
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 \(\mathbb {R}^3\) which forms a sphere. What is the center and
radius of this sphere?
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 one point (if they are “just” touching the plane), no points (if
they are missing the plane), or 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 \(\answer [given]{x}=0\).
Since the plane is \(\answer [given]{3}\) units away from the center of the sphere, and the radius
of the sphere is \(\answer [given]{3}\), there is exactly one point on the intersection, 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 \(\answer [given]{z}=0\). Since the
plane is \(\answer [given]{2}\) units away from the center of the sphere, and the radius of the
sphere is \(\answer [given]{\sqrt {8}}\), there are an infinite number of points on the intersection. The
intersection is in fact a circle. The center of the circle is the point \((\answer [given]{3},(\answer [given]{1},(\answer [given]{0})\). The radius of
the circle can be found by plugging \(z=\answer [given]{0}\) into
\[ (x-3)^2 + (y-1)^2 + (z+2)^2 = 8 \]
and rearranging to find
\[ (x-3)^2 + (y-1)^2 = \answer [given]{4}. \]
Hence,
the intersection is a circle of radius \(\answer [given]{2}\), centered at the point \((\answer [given]{3},(\answer [given]{1},(\answer [given]{0})\), 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!