2.2 Vectors in Three Dimensions

Three-Dimensional Coordinate Systems

As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal x-axis and the vertical y-axis. We can add a third dimension, the z-axis, which is perpendicular to both the x-axis and the y-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions we encounter in real life.

In Figure 2.23(a), the positive z-axis is shown above the plane containing the x- and y-axes. The positive x-axis appears to the left and the positive y-axis is to the right. A natural question to ask is: How was arrangement determined? The system displayed follows the right-hand rule. If we take our right hand and align the fingers with the positive x-axis, then curl the fingers so they point in the direction of the positive y-axis, our thumb points in the direction of the positive z-axis. In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the standard representation.

Figure 2.23 (a) We can extend the two-dimensional rectangular coordinate system by adding a third axis, the z-axis, that is perpendicular to both the x-axis and the y-axis. (b) The right-hand rule is used to determine the placement of the coordinate axes in the standard Cartesian plane.

In two dimensions, we describe a point in the plane with the coordinates $\left(x,y\right).$ Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate, $z,$ is appended to indicate alignment with the z-axis: $\left(x,y,z\right).$ A point in space is identified by all three coordinates (Figure 2.24). To plot the point $\left(x,y,z\right),$ go x units along the x-axis, then $y$ units in the direction of the y-axis, then $z$ units in the direction of the z-axis.

Figure 2.24 To plot the point $\left(x,y,z\right)$ go $x$ units along the x-axis, then $y$ units in the direction of the y-axis, then $z$ units in the direction of the z-axis.

Example 2.11

Locating Points in Space

Sketch the point $\left(1,\mathrm{-2},3\right)$ in three-dimensional space.

Solution

To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive $x$ direction, $2$ units in the negative $y$ direction, and $3$ units in the positive $z$ direction. Complete the prism to plot the point (Figure 2.25).

Figure 2.25 Sketching the point $\left(1,\mathrm{-2},3\right).$

In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a coordinate plane: the xy-plane, the xz-plane, and the yz-plane (Figure 2.26). We define the xy-plane formally as the following set: $\left\{\left(x,y,0\right):x,y\in ℝ\right\}.$ Similarly, the xz-plane and the yz-plane are defined as $\left\{\left(x,0,z\right):x,z\in ℝ\right\}$ and $\left\{\left(0,y,z\right):y,z\in ℝ\right\},$ respectively.

To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the xy-plane, the wall to your right is the xz-plane, and the wall to your left is the yz-plane.

Figure 2.26 The plane containing the x- and y-axes is called the xy-plane. The plane containing the x- and z-axes is called the xz-plane, and the y- and z-axes define the yz-plane.

In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants. The octants fill ${ℝ}^3$ in the same way that quadrants fill ${ℝ}^2,$ as shown in Figure 2.27.

Figure 2.27 Points that lie in octants have three nonzero coordinates.

Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space.

If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We that the distance $d$ between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ in the xy-coordinate plane is given by the formula

The formula for the distance between two points in space is a natural extension of this formula.

The proof of this theorem is left as an exercise. (Hint: First find the distance $d_1$ between the points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_1\right)$ as shown in Figure 2.28.)

Figure 2.28 The distance between $P_1$ and $P_2$ is the length of the diagonal of the rectangular prism having $P_1$ and $P_2$ as opposite corners.

Example 2.12

Distance in Space

Find the distance between points $P_1=\left(3,\text{−}1,5\right)$ and $P_2=\left(2,1,\text{−}1\right).$

Figure 2.29 Find the distance between the two points.

Solution

Substitute values directly into the distance formula:

$$\begin{array}{cc}\hfill d\left(P_1,P_2\right)& =\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}\hfill \\ & =\sqrt{{\left(2-3\right)}^2+{\left(1-\left(\mathrm{-1}\right)\right)}^2+{\left(\mathrm{-1}-5\right)}^2}\hfill \\ & =\sqrt{(\mathrm{–1}{)}^2+2^2+{\left(\mathrm{-6}\right)}^2}\hfill \\ & =\sqrt{41}.\hfill \end{array}$$

Before moving on to the next section, let’s get a feel for how ${ℝ}^3$ differs from ${ℝ}^2.$ For example, in ${ℝ}^2,$ lines that are not parallel must always intersect. This is not the case in ${ℝ}^3.$ For example, consider the line shown in Figure 2.30. These two lines are not parallel, nor do they intersect.

Figure 2.30 These two lines are not parallel, but still do not intersect.

You can also have circles that are interconnected but have no points in common, as in Figure 2.31.

Figure 2.31 These circles are interconnected, but have no points in common.

We have a lot more flexibility working in three dimensions than we do if we stick with only two dimensions.

Writing Equations in ℝ³

Now that we can represent points in space and find the distance between them, we can learn how to write equations of geometric objects such as lines, planes, and curved surfaces in ${ℝ}^3.$ First, we start with a simple equation. Compare the graphs of the equation $x=0$ in $ℝ,{ℝ}^2,\text{and}{ℝ}^3$ (Figure 2.32). From these graphs, we can see the same equation can describe a point, a line, or a plane.

Figure 2.32 (a) In $ℝ,$ the equation $x=0$ describes a single point. (b) In ${ℝ}^2,$ the equation $x=0$ describes a line, the y-axis. (c) In ${ℝ}^3,$ the equation $x=0$ describes a plane, the yz-plane.

In space, the equation $x=0$ describes all points $\left(0,y,z\right).$ This equation defines the yz-plane. Similarly, the xy-plane contains all points of the form $\left(x,y,0\right).$ The equation $z=0$ defines the xy-plane and the equation $y=0$ describes the xz-plane (Figure 2.33).

Figure 2.33 (a) In space, the equation $z=0$ describes the xy-plane. (b) All points in the xz-plane satisfy the equation $y=0.$

Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one of the coordinate planes. When a plane is parallel to the xy-plane, for example, the z-coordinate of each point in the plane has the same constant value. Only the x- and y-coordinates of points in that plane vary from point to point.

As we have seen, in ${ℝ}^2$ the equation $x=5$ describes the vertical line passing through point $\left(5,0\right).$ This line is parallel to the y-axis. In a natural extension, the equation $x=5$ in ${ℝ}^3$ describes the plane passing through point $\left(5,0,0\right),$ which is parallel to the yz-plane. Another natural extension of a familiar equation is found in the equation of a sphere.

Figure 2.34 Each point $\left(x,y,z\right)$ on the surface of a sphere is $r$ units away from the center $\left(a,b,c\right).$

The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance.

Example 2.14

Finding an Equation of a Sphere

Find the standard equation of the sphere with center $\left(10,7,4\right)$ and point $\left(\mathrm{-1},3,\mathrm{-2}\right),$ as shown in Figure 2.35.

Figure 2.35 The sphere centered at $\left(10,7,4\right)$ containing point $\left(\mathrm{-1},3,\mathrm{-2}\right).$

Solution

Use the distance formula to find the radius $r$ of the sphere:

$$\begin{array}{cc}\hfill r& =\sqrt{{\left(\mathrm{-1}-10\right)}^2+{\left(3-7\right)}^2+{\left(\mathrm{-2}-4\right)}^2}\hfill \\ & =\sqrt{{(\mathrm{-11})}^2+{(\mathrm{-4})}^2+{(\mathrm{-6})}^2}\hfill \\ & =\sqrt{173}.\hfill \end{array}$$

The standard equation of the sphere is

$${\left(x-10\right)}^2+{\left(y-7\right)}^2+{\left(z-4\right)}^2=173.$$

Example 2.15

Finding the Equation of a Sphere

Let $P=\left(\mathrm{-5},2,3\right)$ and $Q=\left(3,4,\mathrm{-1}\right),$ and suppose line segment $PQ$ forms the diameter of a sphere (Figure 2.36). Find the equation of the sphere.

Figure 2.36 Line segment $PQ.$

Solution

Since $PQ$ is a diameter of the sphere, we know the center of the sphere is the midpoint of $PQ.$ Then,

$$\begin{array}{cc}\hfill C& =\left(\frac{\mathrm{-5}+3}{2},\frac{2+4}{2},\frac{3+\left(\mathrm{-1}\right)}{2}\right)\hfill \\ & =\left(\mathrm{-1},3,1\right).\hfill \end{array}$$

Furthermore, we know the radius of the sphere is half the length of the diameter. This gives

$$\begin{array}{cc}\hfill r& =\frac{1}{2}\sqrt{{\left(\mathrm{-5}-3\right)}^2+{\left(2-4\right)}^2+{\left(3-\left(\mathrm{-1}\right)\right)}^2}\hfill \\ & =\frac{1}{2}\sqrt{64+4+16}\hfill \\ & =\sqrt{21}.\hfill \end{array}$$

Then, the equation of the sphere is ${\left(x+1\right)}^2+{\left(y-3\right)}^2+{\left(z-1\right)}^2=21.$

Example 2.16

Graphing Other Equations in Three Dimensions

Describe the set of points that satisfies $\left(x-4\right)\left(z-2\right)=0,$ and graph the set.

Solution

We must have either $x-4=0$ or $z-2=0,$ so the set of points forms the two planes $x=4$ and $z=2$ (Figure 2.37).

Figure 2.37 The set of points satisfying $\left(x-4\right)\left(z-2\right)=0$ forms the two planes $x=4$ and $z=2.$

Example 2.17

Graphing Other Equations in Three Dimensions

Describe the set of points in three-dimensional space that satisfies ${\left(x-2\right)}^2+{\left(y-1\right)}^2=4,$ and graph the set.

Solution

The x- and y-coordinates form a circle in the xy-plane of radius $2,$ centered at $\left(2,1\right).$ Since there is no restriction on the z-coordinate, the three-dimensional result is a circular cylinder of radius $2$ centered on the line with $x=2\text{and}y=1.$ The cylinder extends indefinitely in the z-direction (Figure 2.38).

Figure 2.38 The set of points satisfying ${\left(x-2\right)}^2+{\left(y-1\right)}^2=4.$ This is a cylinder of radius $2$ centered on the line with $x=2\text{and}y=1.$

Working with Vectors in ℝ³

Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.

Three-dimensional vectors can also be represented in component form. The notation $\text{v}=\langle x,y,z\rangle$ is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, $\left(0,0,0\right),$ and terminal point $\left(x,y,z\right).$ The zero vector is $0=\langle 0,0,0\rangle .$ So, for example, the three dimensional vector $\text{v}=\langle 2,4,1\rangle$ is represented by a directed line segment from point $\left(0,0,0\right)$ to point $\left(2,4,1\right)$ (Figure 2.39).

Figure 2.39 Vector $\text{v}=\langle 2,4,1\rangle$ is represented by a directed line segment from point $\left(0,0,0\right)$ to point $\left(2,4,1\right).$

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If $\text{v}=\langle x_1,y_1,z_1\rangle$ and $\text{w}=\langle x_2,y_2,z_2\rangle$ are vectors, and $k$ is a scalar, then

If $k=\mathrm{-1},$ then $k\text{v}=\left(\mathrm{-1}\right)\text{v}$ is written as $\text{−}\text{v},$ and vector subtraction is defined by $\text{v}-w=v+\left(\text{−}\text{w}\right)=v+\left(\mathrm{-1}\right)\text{w}.$

The standard unit vectors extend easily into three dimensions as well—$\text{i}=\langle 1,0,0\rangle ,$ $\text{j}=\langle 0,1,0\rangle ,$ and $\text{k}=\langle 0,0,1\rangle$—and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in ${ℝ}^3$ in the following ways:

Example 2.18

Vector Representations

Let $\overrightarrow{PQ}$ be the vector with initial point $P=(3,12,6)$ and terminal point $Q=\left(\mathrm{-4},\mathrm{-3},2\right)$ as shown in Figure 2.40. Express $\overrightarrow{PQ}$ in both component form and using standard unit vectors.

Figure 2.40 The vector with initial point $P=(3,12,6)$ and terminal point $Q=\left(\mathrm{-4},\mathrm{-3},2\right).$

Solution

In component form,

$$\begin{array}{cc}\hfill \overrightarrow{PQ}& =\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle \hfill \\ & =\langle \mathrm{-4}-3,\mathrm{-3}-12,2-6\rangle =\langle \mathrm{-7},\mathrm{-15},\mathrm{-4}\rangle .\hfill \end{array}$$

In standard unit form,

$$\overrightarrow{PQ}=\mathrm{-7}\text{i}-15\text{j}-4\text{k}.$$

As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space (Figure 2.41).

Figure 2.41 To add vectors in three dimensions, we follow the same procedures we learned for two dimensions.

We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.

We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

Example 2.20

Throwing a Forward Pass

A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of $30\text{°}$ (see the following figure). Write the initial velocity vector of the ball, $\text{v},$ in component form.

Solution

The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector $\text{w}$ extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of $30\text{°}$ (see the following figure). This vector would have the same direction as $\text{v},$ but it may not have the right magnitude.

The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is

$$\text{Dist from QB to receiver}=\sqrt{{15}^2+{20}^2}=\sqrt{225+400}=\sqrt{625}=25\text{yd}.$$

We have $\frac{25}{\Vert \text{w}\Vert }=\text{cos}30\text{°}.$ Then the magnitude of $\text{w}$ is given by

$$\Vert \text{w}\Vert =\frac{25}{\text{cos}30\text{°}}=\frac{25·2}{\sqrt{3}}=\frac{50}{\sqrt{3}}\text{yd}$$

and the vertical distance from the receiver to the terminal point of $\text{w}$ is

$$\text{Vert dist from receiver to terminal point of}\text{w}=\Vert \text{w}\Vert \text{sin}30\text{°}=\frac{50}{\sqrt{3}}·\frac{1}{2}=\frac{25}{\sqrt{3}}\text{yd}.$$

Then $\text{w}=\langle 20,15,\frac{25}{\sqrt{3}}\rangle ,$ and has the same direction as $\text{v}.$

Recall, though, that we calculated the magnitude of $\text{w}$ to be $\Vert \text{w}\Vert =\frac{50}{\sqrt{3}},$ and $\text{v}$ has magnitude $60$ mph. So, we need to multiply vector $\text{w}$ by an appropriate constant, $k.$ We want to find a value of $k$ so that $\Vert k\text{w}\Vert =60$ mph. We have

$$\Vert k\text{w}\Vert =k\Vert \text{w}\Vert =k\frac{50}{\sqrt{3}}\text{mph,}$$

so we want

$$\begin{array}{}\\ \\ \hfill k\frac{50}{\sqrt{3}}& =\hfill & 60\hfill \\ \hfill k& =\hfill & \frac{60\sqrt{3}}{50}\hfill \\ \hfill k& =\hfill & \frac{6\sqrt{3}}{5}.\hfill \end{array}$$

Then

$$\text{v}=k\text{w}=k\langle 20,15,\frac{25}{\sqrt{3}}\rangle =\frac{6\sqrt{3}}{5}\langle 20,15,\frac{25}{\sqrt{3}}\rangle =\langle 24\sqrt{3},18\sqrt{3},30\rangle .$$

Let’s double-check that $\Vert \text{v}\Vert =60.$ We have

$$\Vert \text{v}\Vert =\sqrt{{\left(24\sqrt{3}\right)}^2+{\left(18\sqrt{3}\right)}^2+{\left(30\right)}^2}=\sqrt{1728+972+900}=\sqrt{3600}=60\text{mph}.$$

So, we have found the correct components for $\text{v}.$