The Coordinate System

Section 6.1 The Coordinate System

Throughout the text thus far we have focused investigating functions of the form

y = f (x),

$y=f(x)\text{,}$ with one independent and one dependent variable. Such functions can be represented in two dimensions, using two numerical axes that allow us to identify every point in the plane with two numbers. We now shift our focus to three-dimensional space; to identify every point in three dimensions we require three numerical values. The obvious way to make this association is to add one new axis, perpendicular to the

x

$x$ - and

y

$y$ -axes we already understand. We could, for example, add a third axis, the

z

$z$ -axis, with the positive

z

$z$ -axis coming straight out of the page, and the negative

z

$z$ -axis going out the back of the page. This is difficult to work with on a printed page, so more often we draw a view of the three axes from an angle:

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You must then imagine that the

z

$z$ -axis is perpendicular to the other two. Just as we have investigated functions of the form

y = f (x)

$y=f(x)$ in two dimensions, we will investigate three dimensions largely by considering functions; now the functions will (typically) have the form

z = f (x, y) .

$z=f(x,y)\text{.}$ Due to the fact that we are used to having the result of a function graphed in the vertical direction, it is somewhat easier to maintain that convention in three dimensions. To accomplish this, we normally rotate the axes so that

z

$z$ points up; investigate by rotating the axes below:

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Figure 6.1. The 3D coordinate axes.

Note that if you imagine looking down from above, along the

z

$z$ -axis, the positive

z

$z$ -axis will come straight toward you, the positive

y

$y$ -axis will point up, and the positive

x

$x$ -axis will point to your right, as usual. Any point in space is identified by providing the three coordinates of the point, as shown; naturally, we list the coordinates in the order

(x, y, z) .

$(x,y,z)\text{.}$ One useful way to think of this is to use the

x

$x$ and

y

$y$ coordinates to identify a point in the

x

$x$ -

y

$y$ -plane, then move straight up (or down) a distance given by the

z

$z$ coordinate.

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It is now fairly simple to understand some “shapes” in three dimensions that correspond to simple conditions on the coordinates. In two dimensions the equation

x = 1

$x=1$ describes the vertical line through

(1, 0) .

$(1,0)\text{.}$ In three dimensions, it still describes all points with

x

$x$ -coordinate 1, but this is now a plane, as in Figure 6.2.

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Figure 6.2. The plane

x = 1 .

$x=1\text{.}$

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Recall the very useful distance formula in two dimensions which comes directly from the Pythagorean Theorem: the distance between points

(x_{1}, y_{1})

$(x_1,y_1)$ and

(x_{2}, y_{2})

$(x_2,y_2)$ is

\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} .

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\text{.}$ What is the distance between two points

(x_{1}, y_{1}, z_{1})

$(x_1,y_1,z_1)$ and

(x_{2}, y_{2}, z_{2})

$(x_2,y_2,z_2)$ in three dimensions? Geometrically, we want the length of the long diagonal labelled

c

$c$ in the “box” in Figure 6.3. Since

a,

$a\text{,}$

b,

$b\text{,}$

c

$c$ form a right triangle,

a^{2} + b^{2} = c^{2} .

$a^2+b^2=c^2\text{.}$

b

$b$ is the vertical distance between

(x_{1}, y_{1}, z_{1})

$(x_1,y_1,z_1)$ and

(x_{2}, y_{2}, z_{2}),

$(x_2,y_2,z_2)\text{,}$ so

b = | z_{1} - z_{2} | .

$b=|z_1-z_2|\text{.}$ The length

a

$a$ runs parallel to the

x

$x$ -

y

$y$ -plane, so it is simply the distance between

(x_{1}, y_{1})

$(x_1,y_1)$ and

(x_{2}, y_{2}),

$(x_2,y_2)\text{,}$ that is,

a^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} .

$a^2=(x_1-x_2)^2+(y_1-y_2)^2\text{.}$ Now we see that

c^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}

$c^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$ and

c = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}} .

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

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It is sometimes useful to give names to points, for example we might let

P_{1} = (x_{1}, y_{1}, z_{1}),

$P_1=(x_1,y_1,z_1)\text{,}$ or more concisely we might refer to the point

P_{1} (x_{1}, y_{1}, z_{1}),

$P_1(x_1,y_1,z_1)\text{,}$ and subsequently use just

P_{1} .

$P_1\text{.}$ Distance between two points in either two or three dimensions is sometimes denoted by

d,

$d\text{,}$ so for example the formula for the distance between

P_{1} (x_{1}, y_{1}, z_{1})

$P_1(x_1,y_1,z_1)$ and

P_{2} (x_{2}, y_{2}, z_{2})

$P_2(x_2,y_2,z_2)$ might be expressed as

d (P_{1}, P_{2}) = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}} .

$\begin{equation*} d(P_1,P_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\text{.} \end{equation*}$

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Definition 6.1. Distance.

The distance between points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ in two dimensions is

d (P_{1}, P_{2}) = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}

$\begin{equation*} d(P_1,P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \end{equation*}$

The distance between points $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ in three dimensions is

d (P_{1}, P_{2}) = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}}

$\begin{equation*} d(P_1,P_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2} \end{equation*}$

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Figure 6.3. Distance in three dimensions.

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In two dimensions, the distance formula immediately gives us the equation of a circle: the circle of radius

r

$r$ and center at

(h, k)

$(h,k)$ consists of all points

(x, y)

$(x,y)$ at distance

r

$r$ from

(h, k),

$(h,k)\text{,}$ so the equation is

r = \sqrt{(x - h)^{2} + (y - k)^{2}}

$r=\sqrt{(x-h)^2+(y-k)^2}$ or

r^{2} = (x - h)^{2} + (y - k)^{2} .

$r^2=(x-h)^2+(y-k)^2\text{.}$ Now we can get the similar equation

r^{2} = (x - h)^{2} + (y - k)^{2} + (z - l)^{2},

$r^2=(x-h)^2+(y-k)^2+(z-l)^2\text{,}$ which describes all points

(x, y, z)

$(x,y,z)$ at distance

r

$r$ from

(h, k, l),

$(h,k,l)\text{,}$ namely, the sphere with radius

r

$r$ and center

(h, k, l) .

$(h,k,l)\text{.}$

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Definition 6.2. Equation of a Sphere.

A sphere with centre $(x_0,y_0,z_0)$ and radius $r$ is described by

r^{2} = (x - x_{0})^{2} + (y - y_{0})^{2} + (x - x_{0})^{2}

$\begin{equation*} r^2=(x-x_0)^2+(y-y_0)^2+(x-x_0)^2 \end{equation*}$

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Exercises for Section 6.1.

Exercise 6.1.1.

Sketch the location of the points $(1,1,0)\text{,}$ $(2,3,-1)\text{,}$ and $(-1,2,3)$ on a single set of axes.

Exercise 6.1.2.

Describe geometrically the set of points $(x,y,z)$ that satisfy $z=4\text{.}$

Exercise 6.1.3.

Describe geometrically the set of points $(x,y,z)$ that satisfy $y=-3\text{.}$

Exercise 6.1.4.

Describe geometrically the set of points $(x,y,z)$ that satisfy $x+y=2\text{.}$

Exercise 6.1.5.

The equation $x+y+z=1$ describes some collection of points in $\ds \R^3\text{.}$ Describe and sketch the points that satisfy $x+y+z=1$ and are in the $x$-$y$-plane, in the $x$-$z$-plane, and in the $y$-$z$-plane.

Exercise 6.1.6.

Find the lengths of the sides of the triangle with vertices $(1,0,1)\text{,}$ $(2,2,-1)\text{,}$ and $(-3,2,-2)\text{.}$

Answer

$3\text{,}$ $\sqrt{26}\text{,}$ $\sqrt{29}$

Exercise 6.1.7.

Find the lengths of the sides of the triangle with vertices $(2,2,3)\text{,}$ $(8,6,5)\text{,}$ and $(-1,0,2)\text{.}$ Why do the results tell you that this isn't really a triangle?

Answer

$\sqrt{14}\text{,}$ $2\sqrt{14}\text{,}$ $3\sqrt{14}\text{.}$

Exercise 6.1.8.

Find an equation of the sphere with center at $(1,1,1)$ and radius 2.

Answer

$(x-1)^2+(y-1)^2+(z-1)^2=4\text{.}$

Exercise 6.1.9.

Find an equation of the sphere with center at $(2,-1,3)$ and radius 5.

Answer

$(x-2)^2+(y+1)^2+(z-3)^2=25\text{.}$

Exercise 6.1.10.

Find an equation of the sphere with center $(3,-2,1)$ and that goes through the point $(4,2,5)\text{.}$

Exercise 6.1.11.

Find an equation of the sphere with center at $(2,1,-1)$ and radius 4. Find an equation for the intersection of this sphere with the $y$-$z$-plane; describe this intersection geometrically.

Answer

$(x-2)^2+(y-1)^2+(z+1)^2=16\text{,}$ $(y-1)^2+(z+1)^2=12$

Exercise 6.1.12.

Consider the sphere of radius 5 centered at $(2,3,4)\text{.}$ What is the intersection of this sphere with each of the coordinate planes?

Exercise 6.1.13.

Show that for all values of $\theta$ and $\phi\text{,}$ the point $(a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi)$ lies on the sphere given by $x^2+y^2+z^2=a^2\text{.}$

Exercise 6.1.14.

Prove that the midpoint of the line segment connecting $(x_1,y_1,z_1)$ to $(x_2,y_2,z_2)$ is at $\ds\left({x_1+x_2\over 2},{y_1+y_2\over 2},{z_1+z_2\over 2}\right)\text{.}$

Exercise 6.1.15.

Any three points $P_1(x_1,y_1,z_1)\text{,}$ $P_2(x_2,y_2,z_2)\text{,}$ $P_3(x_3,y_3,z_3)\text{,}$ lie in a plane and form a triangle. The triangle inequality says that $d(P_1,P_3)\le d(P_1,P_2)+d(P_2,P_3)\text{.}$ Prove the triangle inequality using either algebra (messy) or the law of cosines (less messy).

Exercise 6.1.16.

Is it possible for a plane to intersect a sphere in exactly two points? Exactly one point? Explain.