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Mathematics LibreTexts

11.2: Vectors

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Learning Objectives
  • Describe three-dimensional space mathematically.
  • Locate points in space using coordinates.
  • Write the distance formula in three dimensions.
  • Write the equations for simple planes and spheres.
  • Perform vector operations in R3.

Vectors are useful tools for solving two-dimensional problems. Life, however, happens in three dimensions. To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases the topography of the land is important. Does your planned route go through the mountains? Do you have to cross a river? To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a natural extension of the two-dimensional Cartesian coordinate plane into 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.

Definition: Three-dimensional Rectangular Coordinate System

The three-dimensional rectangular coordinate system consists of three perpendicular axes: the x-axis, the y-axis, and the z-axis. Because each axis is a number line representing all real numbers in , the three-dimensional system is often denoted by 3.

In Figure 11.2.1a, 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 this 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 (Figure 11.2.1b). 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.

This figure has two images. The first is a 3-dimensional coordinate system. The x-axis is forward, the y-axis is horizontal to the left and right, and the z-axis is vertical. The second image is the 3-dimensional coordinate system axes with a right hand. The thumb is pointing towards positive z-axis, with the fingers wrapping around the z-axis from the positive x-axis to the positive y-axis.
Figure 11.2.1: (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 (x,y). 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: (x,y,z). A point in space is identified by all three coordinates (Figure 11.2.2). To plot the point (x,y,z), 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.

This figure is the positive octant of the 3-dimensional coordinate system. In the first octant there is a rectangular solid drawn with broken lines. One corner is labeled (x, y, z). The height of the box is labeled “z units,” the width is labeled “x units” and the length is labeled “y units.”
Figure 11.2.2: To plot the point (x,y,z) 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 11.2.1: Locating Points in Space

Sketch the point (1,2,3) 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 11.2.3).

This figure is the 3-dimensional coordinate system. In the fourth octant there is a rectangular solid drawn. One corner is labeled (1, -2, 3).
Figure 11.2.3: Sketching the point (1,2,3).
Exercise 11.2.1

Sketch the point (2,3,1) in three-dimensional space.

Hint

Start by sketching the coordinate axes. e.g., Figure 11.2.3. Then sketch a rectangular prism to help find the point in space.

Answer

This figure is the 3-dimensional coordinate system. In the first octant there is a rectangular solid drawn. One corner is labeled (-2, 3, -1).

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 11.2.4). We define the xy-plane formally as the following set: {(x,y,0):x,y}. Similarly, the xz-plane and the yz-plane are defined as {(x,0,z):x,z} and {(0,y,z):y,z}, 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.

This figure is the first octant of a 3-dimensional coordinate system. Also, there are the x y-plane represented with a rectangle with the x and y axes on the plane. There is also the x z-plane on the x and z axes and the y z-plane on the y and z axes.
Figure 11.2.4: 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 3 in the same way that quadrants fill 2, as shown in Figure 11.2.5.

This figure is the 3-dimensional coordinate system with the first octant labeled with a roman numeral I, I, II, III, IV, V, VI, VII, and VIII. Also, for each quadrant there are the signs of the values of x, y, and z. They are: I (+, +, +); 2nd (-, +, +); 3rd (-, -, +); 4th (+, -, +); 5th (+, +, -); 6th (-, +, -); 7th (-, -, -); and 8th (+, -, -).
Figure 11.2.5: 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 know that the distance d between two points (x1,y1) and (x2,y2) in the xy-coordinate plane is given by the formula

d=(x2x1)2+(y2y1)2.

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

The Distance between Two Points in Space

The distance d between points (x1,y1,z1) and (x2,y2,z2) is given by the formula

d=(x2x1)2+(y2y1)2+(z2z1)2.

The proof of this theorem is left as an exercise. (Hint: First find the distance d1 between the points (x1,y1,z1) and (x2,y2,z1) as shown in Figure 11.2.6.)

This figure is a rectangular prism. The lower, left back corner is labeled “P sub 1=(x sub 1,y sub 1,z sub 1). The lower front right corner is labeled “(x sub 2, y sub 2, z sub 1)”. There is a line between P sub 1 and P sub 2 and is labeled “d sub 1”. The upper front right corner is labeled “P sub 2=(x sub 2,y sub 2,z sub 2).” There is a line from P sub 1 to P sub 2 and is labeled “d (P sub 1,P sub 2).” The front right vertical side is labeled “|z sub 2-z sub 1|”.
Figure 11.2.6: The distance between P1 and P2 is the length of the diagonal of the rectangular prism having P1 and P2 as opposite corners.
Example 11.2.2: Distance in Space

Find the distance between points P1=(3,1,5) and P2=(2,1,1).

This figure is the 3-dimensional coordinate system. There are two points. The first is labeled “P sub 1(3, -1, 5)” and the second is labeled “P sub 2(2, 1, -1)”. There is a line segment between the two points.
Figure 11.2.7: Find the distance between the two points.
Solution

Substitute values directly into the distance formula (Equation ???):

d(P1,P2)=(x2x1)2+(y2y1)2+(z2z1)2=(23)2+(1(1))2+(15)2=(1)2+22+(6)2=41.

Exercise 11.2.2

Find the distance between points P1=(1,5,4) and P2=(4,1,1).

Hint

d=(x2x1)2+(y2y1)2+(z2z1)2

Answer

52

Before moving on to the next section, let’s get a feel for how 3 differs from 2. For example, in 2, lines that are not parallel must always intersect. This is not the case in 3. For example, consider the lines shown in Figure 11.2.8. These two lines are not parallel, nor do they intersect.

Figure 11.2.8: 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 11.2.9.

Figure 11.2.9: 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 stuck with only two dimensions.

Writing Equations in 3

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 3. First, we start with a simple equation. Compare the graphs of the equation x=0 in , 2,and 3 (Figure 11.2.10). From these graphs, we can see the same equation can describe a point, a line, or a plane.

This figure has three images. The first is a horizontal axis with a point drawn at 0. The second is the two dimensional Cartesian coordinate plane. The third is the 3-dimensional coordinate system. It is inside of a box and has a grid drawn at the y z-plane.
Figure 11.2.10: (a) In , the equation x=0 describes a single point. (b) In 2, the equation x=0 describes a line, the y-axis. (c) In 3, the equation x=0 describes a plane, the yz-plane.

In space, the equation x=0 describes all points (0,y,z). This equation defines the yz-plane. Similarly, the xy-plane contains all points of the form (x,y,0). The equation z=0 defines the xy-plane and the equation y=0 describes the xz-plane (Figure 11.2.11).

This figure has two images. The first is the 3-dimensional coordinate system. It is inside of a box and has a grid drawn at the x y-plane. The second is the 3-dimensional coordinate system. It is inside of a box and has a grid drawn at the x z-plane.
Figure 11.2.11: (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.

Equations of Planes Parallel to Coordinate Planes
  1. The plane in space that is parallel to the xy-plane and contains point (a,b,c) can be represented by the equation z=c.
  2. The plane in space that is parallel to the xz-plane and contains point (a,b,c) can be represented by the equation y=b.
  3. The plane in space that is parallel to the yz-plane and contains point (a,b,c) can be represented by the equation x=a.
Example 11.2.3: Writing Equations of Planes Parallel to Coordinate Planes
  1. Write an equation of the plane passing through point (3,11,7) that is parallel to the yz-plane.
  2. Find an equation of the plane passing through points (6,2,9),(0,2,4), and (1,2,3).
Solution
  1. When a plane is parallel to the yz-plane, only the y- and z-coordinates may vary. The x-coordinate has the same constant value for all points in this plane, so this plane can be represented by the equation x=3.
  2. Each of the points (6,2,9),(0,2,4), and (1,2,3) has the same y-coordinate. This plane can be represented by the equation y=2.
Exercise 11.2.3

Write an equation of the plane passing through point (1,6,4) that is parallel to the xy-plane.

Hint

If a plane is parallel to the xy-plane, the z-coordinates of the points in that plane do not vary.

Answer

z=4

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

Definition: Sphere

A sphere is the set of all points in space equidistant from a fixed point, the center of the sphere (Figure 11.2.12), just as the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, the distance from the center to a point on the sphere is called the radius.

This image is a sphere. It has center at (a, b, c) and has a radius represented with a broken line from the center point (a, b, c) to the edge of the sphere at (x, y, z). The radius is labeled “r.”
Figure 11.2.12: Each point (x,y,z) on the surface of a sphere is r units away from the center (a,b,c).

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.

Standard Equation of a Sphere

The sphere with center (a,b,c) and radius r can be represented by the equation

(xa)2+(yb)2+(zc)2=r2.

This equation is known as the standard equation of a sphere.

Example 11.2.4: Finding an Equation of a Sphere

Find the standard equation of the sphere with center (10,7,4) and point (1,3,2), as shown in Figure 11.2.13.

Figure 11.2.13: The sphere centered at (10,7,4) containing point (1,3,2).

Solution

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

r=(110)2+(37)2+(24)2=(11)2+(4)2+(6)2=173

The standard equation of the sphere is

(x10)2+(y7)2+(z4)2=173.

Exercise 11.2.4

Find the standard equation of the sphere with center (2,4,5) containing point (4,4,1).

Hint

First use the distance formula to find the radius of the sphere.

Answer

(x+2)2+(y4)2+(z+5)2=52

Example 11.2.5: Finding the Equation of a Sphere

Let P=(5,2,3) and Q=(3,4,1), and suppose line segment ¯PQ forms the diameter of a sphere (Figure 11.2.14). Find the equation of the sphere.

This figure is the 3-dimensional coordinate system. There are two points labeled. The first point is P = (-5, 2, 3). The second point is Q = (3, 4, -1). There is a line segment drawn between the two points.
Figure 11.2.14: 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,

C=(5+32,2+42,3+(1)2)=(1,3,1).

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

r=12(53)2+(24)2+(3(1))2=1264+4+16=21

Then, the equation of the sphere is (x+1)2+(y3)2+(z1)2=21.

Exercise 11.2.5

Find the equation of the sphere with diameter ¯PQ, where P=(2,1,3) and Q=(2,5,1).

Hint

Find the midpoint of the diameter first.

Answer

x2+(y2)2+(z+2)2=14

Example 11.2.6: Graphing Other Equations in Three Dimensions

Describe the set of points that satisfies (x4)(z2)=0, and graph the set.

Solution

We must have either x4=0 or z2=0, so the set of points forms the two planes x=4 and z=2 (Figure 11.2.15).

This figure is the 3-dimensional coordinate system. It has two intersecting planes drawn. The first is the x y-plane. The second is the y z-plane. They are perpendicular to each other.
Figure 11.2.15: The set of points satisfying (x4)(z2)=0 forms the two planes x=4 and z=2.
Exercise 11.2.6

Describe the set of points that satisfies (y+2)(z3)=0, and graph the set.

Hint

One of the factors must be zero.

Answer

The set of points forms the two planes y=2 and z=3.

This figure is the 3-dimensional coordinate system. It has two intersecting planes drawn. The first is the x z-plane. The second is parallel to the y z-plane at the value of z = 3. They are perpendicular to each other.

Example 11.2.7: Graphing Other Equations in Three Dimensions

Describe the set of points in three-dimensional space that satisfies (x2)2+(y1)2=4, and graph the set.

Solution

The x- and y-coordinates form a circle in the xy-plane of radius 2, centered at (2,1). 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 and y=1. The cylinder extends indefinitely in the z-direction (Figure 11.2.16).

This figure is the 3-dimensional coordinate system. It has a vertical cylinder parallel to the z-axis and centered around line parallel to the z-axis with x = 2 and y = 1.
Figure 11.2.16: The set of points satisfying (x2)2+(y1)2=4. This is a cylinder of radius 2 centered on the line with x=2 and y=1.
Exercise 11.2.7

Describe the set of points in three dimensional space that satisfies x2+(z2)2=16, and graph the surface.

Hint

Think about what happens if you plot this equation in two dimensions in the xz-plane.

Answer

A cylinder of radius 4 centered on the line with x=0 and z=2.

This figure is the 3-dimensional coordinate system. It has a cylinder parallel to the y-axis and centered around the y-axis.

Working with Vectors in 3

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 v=x,y,z is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, (0,0,0), and terminal point (x,y,z). The zero vector is 0=0,0,0. So, for example, the three dimensional vector v=2,4,1 is represented by a directed line segment from point (0,0,0) to point (2,4,1) (Figure 11.2.17).

This figure is the 3-dimensional coordinate system. It has a vector drawn. The initial point of the vector is the origin. The terminal point of the vector is (2, 4, 1). The vector is labeled “v = <2, 4, 1>.”
Figure 11.2.17: Vector v=2,4,1 is represented by a directed line segment from point (0,0,0) to point (2,4,1).

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If v=x1,y1,z1 and w=x2,y2,z2 are vectors, and k is a scalar, then

v+w=x1+x2,y1+y2,z1+z2

and

kv=kx1,ky1,kz1.

If k=1, then kv=(1)v is written as v, and vector subtraction is defined by vw=v+(w)=v+(1)w.

The standard unit vectors extend easily into three dimensions as well, ˆi=1,0,0, ˆj=0,1,0, and ˆk=0,0,1, and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in 3 in the following ways:

v=x,y,z=xˆi+yˆj+zˆk

.

Example 11.2.8: Vector Representations

Let aPQ be the vector with initial point P=(3,12,6) and terminal point Q=(4,3,2) as shown in Figure 11.2.18. Express aPQ in both component form and using standard unit vectors.

This figure is the 3-dimensional coordinate system. It has two points labeled. The first point is P = (3, 12, 6). The second point is Q = (-4, -3, 2). There is a vector from P to Q.
Figure 11.2.18: The vector with initial point P=(3,12,6) and terminal point Q=(4,3,2).
Solution

In component form,

aPQ=x2x1,y2y1,z2z1=43,312,26=7,15,4.

In standard unit form,

aPQ=7ˆi15ˆj4ˆk.

Exercise 11.2.8

Let S=(3,8,2) and T=(2,1,3). Express ST in component form and in standard unit form.

Hint

Write aST in component form first. T is the terminal point of aST.

Answer

aST=1,9,1=ˆi9ˆj+ˆk

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 11.2.19).

This figure is the first octant of the 3-dimensional coordinate system. It has has three vectors in standard position. The first vector is labeled “A.” The second vector is labeled “B.” The third vector is labeled “A + B.” This vector is in between vectors A and B.
Figure 11.2.19: 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.

Properties of Vectors in Space

Let v=x1,y1,z1 and w=x2,y2,z2 be vectors, and let k be a scalar.

  • Scalar multiplication: kv=kx1,ky1,kz1
  • Vector addition: v+w=x1,y1,z1+x2,y2,z2=x1+x2,y1+y2,z1+z2
  • Vector subtraction: vw=x1,y1,z1x2,y2,z2=x1x2,y1y2,z1z2
  • Vector magnitude: v=x21+y21+z21
  • Unit vector in the direction of v: 1vv=1vx1,y1,z1=x1v,y1v,z1v,ifv0

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 11.2.9: Vector Operations in Three Dimensions

Let v=2,9,5 and w=1,1,0 (Figure 11.2.20). Find the following vectors.

  1. 3v2w
  2. 5w
  3. 5w
  4. A unit vector in the direction of v
This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled “v = <-2, 9, 5>.” The second vector is labeled “w = <1, -1, 0>.”
Figure 11.2.20: The vectors v=2,9,5 and w=1,1,0.
Solution

a. First, use scalar multiplication of each vector, then subtract:

3v2w=32,9,521,1,0=6,27,152,2,0=62,27(2),150=8,29,15.

b. Write the equation for the magnitude of the vector, then use scalar multiplication:

5w=512+(1)2+02=52.

c. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:

5w=∥5,5,0∥=52+(5)2+02=50=52

d. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:

vv=1v2,9,5=1(2)2+92+522,9,5=11102,9,5=2110,9110,5110.

Exercise 11.2.9:

Let v=1,1,1 and w=2,0,1. Find a unit vector in the direction of 5v+3w.

Hint

Start by writing 5v+3w in component form.

Answer

1310,5310,8310

Example 11.2.10: 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° (see the following figure). Write the initial velocity vector of the ball, v, in component form.

This figure is an image of two football players with the first one throwing the football to the second one. There is a line segment from each player to the bottom of the image. The distance from the first player to the bottom of the image is 20 yards. The distance from the second player to the same point on the bottom of the image is 15 yards. The two line segments are perpendicular. There is a broken line segment from the first player to the second player. There is a vector from the first player. The angle between the broken line and the vector is 30 degrees.

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 w extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of 30° (see the following figure). This vector would have the same direction as v, but it may not have the right magnitude.

This figure is the image of two football players with the first player throwing the football to the second player. The distance between the two players is represented with a broken line segment. There is a vector from the first player. The angle between the vector and the broken line segment is 30 degrees. There is a vertical broken line segment from the second player. Also, there is a right triangle formed from the two broken line segments and the vector from the first player is labeled “w” and is the hypotenuse.

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

Dist from QB to receiver =152+202=225+400=625=25 yd.

We have 25w=cos30°. Then the magnitude of w is given by

w=25cos30°=2523=503 yd

and the vertical distance from the receiver to the terminal point of w is

Vert dist from receiver to terminal point of w=wsin30°=50312=253 yd.

Then w=20,15,253, and has the same direction as v.

Recall, though, that we calculated the magnitude of w to be w=503 yd, and v has magnitude 60 mph. So, we need to multiply vector w by an appropriate constant, k. We want to find a value of k so that kw∥=60 mph*. We have

kw=kw=k503 yd,

so we want

k(503 yd)=60 mph

k=60350 mph / yd

k=635 mph / yd.

Then

v=kw=k20,15,253=63520,15,253=243,183,30.

Let’s double-check that v=60 mph. We have

v=(243)2+(183)2+(30)2=1728+972+900=3600=60 mph.

So, we have found the correct components for v.

Note *

Readers who have been watching the units of measurement may be wondering what exactly is going on at this point: haven't we just mixed yards and miles per hour? We haven't, but the reason is subtle. One way to understand it is to realize that there are really two parallel coordinate systems in this problem: one gives positions down the field, across the field, and up into the air in units of yards; the other gives speeds down the field, across the field, and up into the air in units of miles per hour. The vector w is calculated in the position coordinate system; vector v will be in the speed system. Because corresponding axes in each system are parallel, directions in the two systems are also parallel, so the claim that w and v point in the same direction is correct. The constant k that we're looking for is a conversion factor between the magnitudes of these two vectors, converting from the position system to the speed one in the process. And as seen above, our calculation of k produces the right units for such a conversion, namely miles per hour per yard.

Exercise 11.2.10

Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of 40 mph and an angle of 45°. Write the initial velocity vector of the ball, v, in component form.

Hint

Follow the process used in the previous example.

Answer

v=162,122,202

Key Concepts

  • The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples (x,y,z) are used to describe the location of a point in space.
  • The distance d between points (x1,y1,z1) and (x2,y2,z2) is given by the formula d=(x2x1)2+(y2y1)2+(z2z1)2.
  • In three dimensions, the equations x=a,y=b, and z=c describe planes that are parallel to the coordinate planes.
  • The standard equation of a sphere with center (a,b,c) and radius r is (xa)2+(yb)2+(zc)2=r2.
  • In three dimensions, as in two, vectors are commonly expressed in component form, v=x,y,z, or in terms of the standard unit vectors, v=xˆi+yˆj+zˆk.
  • Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let v=x1,y1,z1 and w=x2,y2,z2 be vectors, and let k be a scalar.

Scalar multiplication:

kv=kx1,ky1,kz1

Vector addition:

v+w=x1,y1,z1+x2,y2,z2=x1+x2,y1+y2,z1+z2

Vector subtraction:

vw=x1,y1,z1x2,y2,z2=x1x2,y1y2,z1z2

Vector magnitude:

v=x21+y21+z21

Unit vector in the direction of v:

vv=1vx1,y1,z1=x1v,y1v,z1v,v0

Key Equations

Distance between two points in space:

d=(x2x1)2+(y2y1)2+(z2z1)2

Sphere with center (a,b,c) and radius r:

(xa)2+(yb)2+(zc)2=r2

Glossary

coordinate plane
a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the xy-plane, xz-plane, or the yz-plane
right-hand rule
a common way to define the orientation of the three-dimensional coordinate system; when the right hand is curved around the z-axis in such a way that the fingers curl from the positive x-axis to the positive y-axis, the thumb points in the direction of the positive z-axis
octants
the eight regions of space created by the coordinate planes
sphere
the set of all points equidistant from a given point known as the center
standard equation of a sphere
(xa)2+(yb)2+(zc)2=r2 describes a sphere with center (a,b,c) and radius r
three-dimensional rectangular coordinate system
a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple (x,y,z) that plots its location relative to the defining axes

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

Example 11.2.10 has been modified by Doug Baldwin and Paul Seeburger to clarify the units of measurement that it uses and how it uses them.

Paul Seeburger also created dynamic versions of Figures 11.2.8,11.2.9 and 11.2.13 using CalcPlot3D.


11.2: Vectors is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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