Chapter 07 - Vectors-And-Dot Products

Objectives

1 Use magnitude and direction

to show vectors are equal.

2 Visualize scalar multiplication,

vector addition, and vector

subtraction as geometric

vectors.

3 Represent vectors in the

rectangular coordinate

system.

4 Perform operations with

vectors in terms of i and j.

5 Find the unit vector in the

direction of v.

6 Write a vector in terms of its

magnitude and direction.

7 Solve applied problems

involving vectors.

VectorsS e c t i o n 7.6

718 Chapter 7 Additional Topics in Trigonometry

107. I multiplied two complex numbers in polar form by firstmultiplying the moduli and then multiplying the arguments.

108. The proof of the formula for the product of two complexnumbers in polar form uses the sum formulas for cosinesand sines that I studied in the previous chapter.

109. My work with complex numbers verified that the onlypossible cube root of 8 is 2.

110. Prove the rule for finding the quotient of two complexnumbers in polar form. Begin the proof as follows, using theconjugate of the denominator’s second factor:

Perform the indicated multiplications. Then use thedifference formulas for sine and cosine.

111. Plot each of the complex fourth roots of 1.

r11cos u1 + i sin u12r21cos u2 + i sin u22 =

r11cos u1 + i sin u12r21cos u2 + i sin u22 #

1cos u2 - i sin u221cos u2 - i sin u22 .

Group Exercise112. Group members should prepare and present a seminar on

mathematical chaos. Include one or more of the followingtopics in your presentation: fractal images, the role ofcomplex numbers in generating fractal images, algorithms,iterations, iteration number, and fractals in nature. Be sureto include visual images that will intrigue your audience.

Preview Exercises

Exercises 113–115 will help you prepare for the material covered

in the next section.

113. Use the distance formula to determine if the line segmentwith endpoints and (0, 3) has the same length asthe line segment with endpoints (0, 0) and (3, 6).

114. Use slope to determine if the line through and (0, 3) is parallel to the line through (0, 0) and (3, 6).

115. Simplify: 4 15x + 4y2 - 2 16x - 9y2.1-3, -32

1-3, -32

f

f

F1

F1

F2

F2

F3

F3

Fg

Fg

It’s been a dynamic lecture, but now that it’s over it’s obvious that my professor is exhausted.She’s slouching motionless against the board and—what’s that? The forces acting against herbody, including the pull of gravity, are appearing as arrows. I know that mathematics revealsthe hidden patterns of the universe, but this is ridiculous. Does the arrangement of the arrowson the right have anything to do with the fact that my wiped-out professor is not sliding downthe wall?

Ours is a world of pushes and pulls. For example, suppose you are pulling a cart upa 30° incline, requiring an effort of 100 pounds. This quantity is described by

giving its magnitude (a number indicating size, including a unit of measure) and alsoits direction. The magnitude is 100 pounds and the direction is 30° from thehorizontal. Quantities that involve both a magnitude and a direction are called vector

quantities, or vectors for short. Here is another example of a vector:

You are driving due north at 50 miles per hour.The magnitude is the speed, 50 milesper hour. The direction of motion is due north.

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 718

Section 7.6 Vectors 719

Some quantities can be completely described by giving only their magnitudes.For example, the temperature of the lecture room that you just left is 75°. Thistemperature has magnitude, 75°, but no direction. Quantities that involve magni-tude, but no direction, are called scalar quantities, or scalars for short. Thus, a scalarhas only a numerical value. Another example of a scalar is your professor’s height,which you estimate to be 5.5 feet.

In this section and the next, we introduce the world of vectors, which literallysurround your every move. Because vectors have nonnegative magnitude as well asdirection, we begin our discussion with directed line segments.

Directed Line Segments and Geometric Vectors

A line segment to which a direction has been assigned is called a directed line segment.Figure 7.48 shows a directed line segment from to We call the initial point and

the terminal point.We denote this directed line segment by

The magnitude of the directed line segment is its length. We denote this by

Thus, is the distance from point to point Because distance isnonnegative, vectors do not have negative magnitudes.

Geometrically, a vector is a directed line segment. Vectors are often denotedby boldface letters, such as v. If a vector v has the same magnitude and the samedirection as the directed line segment we write

Because it is difficult to write boldface on paper, use an arrow over a single letter,such as to denote v, the vector v.

Figure 7.49 shows four possible relationships between vectors v and w. InFigure 7.49(a), the vectors have the same magnitude and the same direction, andare said to be equal. In general, vectors v and w are equal if they have the same

magnitude and the same direction. We write this as v = w.

v!,

v = PQ!.

PQ!,

Q.P7PQ! 77PQ

! 7 .PQ

!PQ

!.

QPQ.P

This sign shows a distance anddirection for each city. Thus, the signdefines a vector for eachdestination.

1 Use magnitude and direction

to show vectors are equal.

Terminal point

Initial point

P

Q

Figure 7.48 A directed linesegment from to QP

v w v

wv w

vw

(a) v 5 w becausethe vectors have thesame magnitude andsame direction.

(b) Vectors v and whave the samemagnitude, butdifferent directions.

(c) Vectors v and whave the samemagnitude, butopposite directions.

(d) Vectors v and whave the samedirection, butdifferent magnitudes.

Figure 7.49 Relationships between vectors

Showing That Two Vectors Are Equal

Use Figure 7.50 to show that

Solution Equal vectors have the same magnitude and the same direction. Usethe distance formula to show that u and v have the same magnitude.

Thus, u and v have the same magnitude: 7u 7 = 7v 7 .

ivi= Ï(x2-x1)2+(y2-y1)

2=Ï(3-0)2+(6-0)2Magnitudeof v

= Ï32+62=Ï9+36=Ï45 (or 3Ï5)

iui= Ï(x2-x1)2+(y2-y1)

2=Ï[0-(–3)]2+[3-(–3)]2Magnitudeof u

= Ï32+62=Ï9+36=Ï45 (or 3Ï5)

u = v.

EXAMPLE 1

−1

1

345

−2−3−4−5

1 2 3 4 5−3−4−5

y

x

v

u

Terminal point:(0, 3)

Initial point:(−3, −3)

Terminal point:(3, 6)

Initial point:(0, 0)

Figure 7.50

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 719

One way to show that u and v have the same direction is to find the slopes ofthe lines on which they lie. We continue to use Figure 7.50 on the previous page.

u lies on a line passing through

v lies on a line passing through

Because u and v are both directed toward the upper right on lines having the sameslope, 2, they have the same direction.

Thus, u and v have the same magnitude and direction, and

Check Point 1 Use Figure 7.51 to show that

A vector can be multiplied by a real number. Figure 7.52 shows three suchmultiplications: and Multiplying a vector by any positive real number

(except for 1) changes the magnitude of the vector, but not its direction. This can beseen by the blue and green vectors in Figure 7.52. Compare the black and blue vec-tors. Can you see that 2v has the same direction as v but is twice the magnitude of v?Now, compare the black and green vectors: has the same direction as v but is halfthe magnitude of v.

12 v

- 32 v.2v, 12 v,

u = v.

u = v.

(0, 0) and (3, 6).Line on which

v liesy2-y1

x2-x1

6-0

3-0

6

3m= = =2=

(23,23) and (0, 3).Line on which

u liesy2-y1

x2-x1

3-(–3)

0-(–3)

6

3m= = =2=


2 Visualize scalar multiplication,

vector addition, and vector

subtraction as geometric

vectors.

(−2, 6) (5, 6)

(−5, 2) (2, 2)

−1

1234567

−2−3

1 2 3 4 5−1−2−3−4−5

y

x

vu

Figure 7.51

v 2v qv −wv

Figure 7.52 Multiplying vector v byreal numbers

Now compare the black and red vectors in Figure 7.52. Multiplying a vector by

a negative number reverses the direction of the vector. Notice that has theopposite direction as v and is the magnitude of v.

The multiplication of a real number and a vector v is called scalar multiplication.We write this product as kv.

k

32

- 32 v

Scalar Multiplication

If is a real number and v a vector, the vector is called a scalar multiple of thevector v. The magnitude and direction of are given as follows:

The vector has a magnitude of We describe this as the absolutevalue of times the magnitude of vector v.

The vector has a direction that is

• the same as the direction of v if and

• opposite the direction of v if k 6 0.

k 7 0,

kv

kƒk ƒ 7v 7 .kv

kv

kvk

A geometric method for adding two vectors is shown in Figure 7.53 at the top ofthe next page.The sum of u and v, denoted by is called the resultant vector. Hereis how we find this vector:

1. Position u and v, so that the terminal point of u coincides with the initialpoint of v.

2. The resultant vector, extends from the initial point of u to the terminalpoint of v.

u + v,

u + v

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 720


The difference of two vectors, is defined as whereis the scalar multiplication of u and The difference is shown

geometrically in Figure 7.54.v - u-1: -1u.-u

v - u = v + 1-u2,v - u,

Terminal point of v

Resultant vector

Initial point of u

v

u

u + v

Figure 7.53 Vector addition the terminal point of u coincides withthe initial point of v.

u + v;

Wiped Out, ButNot Sliding Downthe Wall

The figure shows the sum of fivevectors:

Notice how the terminal point ofeach vector coincides with the ini-tial point of the vector that’sbeing added to it. The vector sum,from the initial point of to theterminal point of f, is a singlepoint. The magnitude of a singlepoint is zero. These forces add upto a net force of zero, allowing theprofessor to be motionless.

F1

F1 + F2 + F3 + Fg + f.

f

F1

F2

F3

Fg

v

u−u

−u

v − u

Figure 7.54 Vector subtraction the terminalpoint of v coincides with the initial point of -u.

v - u;

Vectors in the Rectangular Coordinate System

As you saw in Example 1, vectors can be shown in the rectangular coordinatesystem. Now let’s see how we can use the rectangular coordinate system torepresent vectors. We begin with two vectors that both have a magnitude of 1. Suchvectors are called unit vectors.

3 Represent vectors in the

rectangular coordinate system.

The i and j Unit Vectors

Vector i is the unit vector whose direction is along thepositive Vector j is the unit vector whosedirection is along the positive y-axis.

x-axis.

1

j

iO

y

1x

Why are the unit vectors i and j important? Vectors in the rectangular coordi-nate system can be represented in terms of i and j. For example, consider vector vwith initial point at the origin, (0, 0), and terminal point at The vector vis shown in Figure 7.55. We can represent v using i and j as v = ai + bj.

P = 1a, b2.

O

y

x

v is the vectoraddition ofai and bj.

bj is a scalarmultiple of j.

ai is a scalarmultiple of i.

a

P = (a, b)b

bjv = ai + bj

ai

Figure 7.55 Using vector addition, vector vis represented as v = ai + bj.

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 721


Representing a Vector in RectangularCoordinates and Finding Its Magnitude

Sketch the vector and find its magnitude.

Solution For the given vector and The vector canbe represented with its initial point at the origin, (0, 0), as shown in Figure 7.56. Thevector’s terminal point is then We sketch the vector by drawing anarrow from (0, 0) to We determine the magnitude of the vector by using thedistance formula. Thus, the magnitude is

Check Point 2 Sketch the vector and find its magnitude.

The vector in Example 2 was represented with its initial point at the origin. Avector whose initial point is at the origin is called a position vector. Any vector inrectangular coordinates whose initial point is not at the origin can be shown to beequal to a position vector. As shown in the following box, this gives us a way torepresent vectors between any two points.

v = 3i - 3j

7v 7 = 3a2 + b2 = 41-322 + 42 = 29 + 16 = 225 = 5.

1-3, 42.1a, b2 = 1-3, 42.

b = 4.v = -3i + 4j, a = -3

v = -3i + 4j

EXAMPLE 2

Representing Vectors in Rectangular Coordinates

Vector v, from (0, 0) to is represented as

The real numbers and are called the scalar components of v. Note that

• is the horizontal component of v, and

• is the vertical component of v.

The vector sum is called a linear combination of the vectors i and j.The magnitude of is given by

7v 7 = 3a2 + b2 .

v = ai + bj

ai + bj

b

a

ba

v = ai + bj.

1a, b2,

(−3, 4)

(0, 0)

2

4

−2

−4−5

5

−1

−3

2 4−2 5−1−4 1 3−3−5

y

x

Initial point

Terminal point

v = −3i + 4j

Figure 7.56 Sketchingin the rectangular

coordinate systemv = -3i + 4j

Representing Vectors in Rectangular Coordinates

Vector v with initial point and terminal point is equalto the position vector

v = 1x2 - x12i + 1y2 - y12j.P2 = 1x2 , y22P1 = 1x1 , y12

We can use congruent triangles, triangles with the same size and shape, toderive this formula. Begin with the right triangle in Figure 7.57(a). This triangleshows vector v from to In Figure 7.57(b), we movevector v, without changing its magnitude or its direction, so that its initial point is atthe origin. Using this position vector in Figure 7.57(b), we see that

where and are the components of v. The equal vectors and the right angles in theright triangles in Figures 7.57(a) and (b) result in congruent triangles.The correspondingsides of these congruent triangles are equal, so that and Thismeans that v may be expressed as

v=ai+bj=(x2-x1)i+(y2-y1)j.

Horizontal component:x-coordinate of terminalpoint minus x-coordinate

of initial point

Vertical component:y-coordinate of terminalpoint minus y-coordinate

of initial point

b = y2 - y1 .a = x2 - x1

ba

v = ai + bj,

P2 = 1x2 , y22.P1 = 1x1 , y12

b

P = (a, b)

v = ai + bj

(0, 0)

v

a

y

x

Figure 7.57(b)

y

x

P2 = (x2, y2)

P1 = (x1, y1)

v y2 − y1

x2 − x1

Figure 7.57(a)

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 722

Study Tip

When finding the distance from to the order in which thesubtractions are performed makes no difference:

When writing the vector from to must be the terminal pointand the order in the subtractions is important:

v=(x2-x1)i+(y2-y1)j.

(x2, y2), the terminal point, isused first in each subtraction.

P2 = 1x2 , y22, P2P1 = 1x1 , y12d = 41x2 - x122 + 1y2 - y122 or d = 41x1 - x222 + 1y1 - y222 .

P2 = 1x2 , y22,P1 = 1x1 , y12


Thus, any vector between two points in rectangular coordinates can be expressed interms of i and j. In rectangular coordinates, the term vector refers to the positionvector expressed in terms of i and j that is equal to it.

Representing a Vector in Rectangular Coordinates

Let v be the vector from initial point to terminal point Write v in terms of i and j.

Solution We identify the values for the variables in the formula.

Using these values, we write v in terms of i and j as follows:

Figure 7.58 shows the vector from to represented in

terms of i and j and as a position vector.

P2 = 1-2, 52P1 = 13, -12v = 1x2 - x12i + 1y2 - y12j = 1-2 - 32i + 35 - 1-124j = -5i + 6j.

P2=(–2, 5)

x2 y2

P1=(3, –1)

x1 y1

P2 = 1-2, 52.P1 = 13, -12EXAMPLE 3

(−5, 6)

P1 = (3, −1)

P2 = (−2, 5)

2

4

−2

−4−5

5

−1

−3

2 4−2 5−1−4 1−3−5

y

x

v = −5i + 6j

Figure 7.58 Representing thevector from to as aposition vector

1-2, 5213, -12

4 Perform operations with vectors

in terms of i and j.

Check Point 3 Let v be the vector from initial point to terminalpoint Write v in terms of i and j.

Operations with Vectors in Terms of i and j

If vectors are expressed in terms of i and j, we can easily carry out operations suchas vector addition, vector subtraction, and scalar multiplication. Recall thegeometric definitions of these operations given earlier. Based on these ideas, we canadd and subtract vectors using the following procedure:

P2 = 12, 72. P1 = 1-1, 32

Adding and Subtracting Vectors in Terms of i and j

If then

v - w = 1a1 - a22i + 1b1 - b22j. v + w = 1a1 + a22i + 1b1 + b22j

v = a1 i + b1 j and w = a2 i + b2

j,

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 723


Adding and Subtracting Vectors

If find each of the following vectors:

a. b.

Solution

a. These are the given vectors.

Add the horizontal components.

Add the vertical components.

Simplify.

b. These are the given vectors.

Subtract the horizontal components.

Subtract the vertical components.

Simplify.

Check Point 4 If and find each of the followingvectors:

a. b.

How do we perform scalar multiplication if vectors are expressed in terms of iand j? We use the following procedure to multiply the vector v by the scalar k:

v - w.v + w

w = 4i - 5j,v = 7i + 3j

= - i + 13j

= 15 - 62i + 34 - 1-924j v - w = 15i + 4j2 - 16i - 9j2

= 11i - 5j

= 15 + 62i + 34 + 1-924j v + w = 15i + 4j2 + 16i - 9j2

v - w.v + w

v = 5i + 4j and w = 6i - 9j,

EXAMPLE 4

Scalar Multiplication with a Vector in Terms of i and j

If and is a real number, then the scalar multiplication of the vectorv and the scalar is

kv = 1ka2i + 1kb2j.k

kv = ai + bj

Scalar Multiplication

If find each of the following vectors:

a. 6v b.

Solution

a. The scalar multiplication is expressed with

the given vector.

Multiply each component by 6.

Simplify.

b. The scalar multiplication is expressed with

the given vector.

Multiply each component by

Simplify.

Check Point 5 If find each of the following vectors:

a. 8v b. -5v.

v = 7i + 10j,

= -15i - 12j

-3. = 1-3 # 52i + 1-3 # 42j -3v = -315i + 4j2

= 30i + 24j

= 16 # 52i + 16 # 42j 6v = 615i + 4j2

-3v.

v = 5i + 4j,

EXAMPLE 5

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 724


Vector Operations

If and find

Solution

Operations are expressed with the

given vectors.

Perform each scalar multiplication.

Add horizontal and vertical components

to perform the vector addition.

Simplify.

Check Point 6 If and find

Properties involving vector operations resemble familiar properties of realnumbers. For example, the order in which vectors are added makes no difference:

Does this remind you of the commutative property Just as 0 plays an important role in the properties of real numbers, the zero

vector 0 plays exactly the same role in the properties of vectors.

a + b = b + a?

u + v = v + u.

6v - 3w.w = 4i - 5j,v = 7i + 3j

= 8i + 34j

= 120 - 122i + 116 + 182j = 20i + 16j - 12i + 18j

4v - 2w = 415i + 4j2 - 216i - 9j2

4v - 2w.w = 6i - 9j,v = 5i + 4j

EXAMPLE 6

The Zero Vector

The vector whose magnitude is 0 is called the zero vector, 0. The zero vector isassigned no direction. It can be expressed in terms of i and j using

0 = 0i + 0j.

5 Find the unit vector in the

direction of v.

Properties of Vector Addition and Scalar Multiplication are given as follows:

Properties of Vector Addition and Scalar Multiplication

If u, v, and w are vectors, and and are scalars, then the following propertiesare true.

Vector Addition Properties

1. Commutative property

2. Associative property

3. Additive identity

4. Additive inverse

Scalar Multiplication Properties

1. Associative property

2. Distributive property

3. Distributive property

4. Multiplicative identity

5. Multiplication property of zero

6. Magnitude property7cv 7 = ƒ c ƒ 7v 70u = 0

1u = u

1c + d2u = cu + du

c1u + v2 = cu + cv

1cd2u = c1du2

u + 1-u2 = 1-u2 + u = 0

u + 0 = 0 + u = u

1u + v2 + w = u + 1v + w2u + v = v + u

dc

Unit Vectors

A unit vector is defined to be a vector whose magnitude is one. In many applications ofvectors, it is helpful to find the unit vector that has the same direction as a given vector.

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 725


Finding a Unit Vector

Find the unit vector in the same direction as Then verify that thevector has magnitude 1.

Solution We find the unit vector in the same direction as v by dividing v by itsmagnitude. We first find the magnitude of v.

The unit vector in the same direction as v is

This is the scalar multiplication of v and

Now we must verify that the magnitude of this vector is 1. Recall that the magnitude

of is Thus, the magnitude of is

Check Point 7 Find the unit vector in the same direction as Thenverify that the vector has magnitude 1.

Writing a Vector in Terms of Its Magnitude and Direction

Consider the vector The components and can be expressed in termsof the magnitude of v and the angle that v makes with the positive Thisangle is called the direction angle of v and is shown in Figure 7.59. By the definitionsof sine and cosine, we have

Thus,

v = ai + bj = 7v 7 cos ui + 7v 7 sin uj.

a = 7v 7 cos u b = 7v 7 sin u.

cos u =a

7v 7 and sin u =b

7v 7

x-axis.u

bav = ai + bj.

v = 4i - 3j.

B a5

13b2

+ a -

12

13b2

= A25

169+

144

169= A

169

169= 21 = 1.

513 i - 12

13 j3a2 + b2 .ai + bj

113 .

v

7v 7 =5i - 12j

13=

5

13 i -

12

13 j.

7v 7 = 3a2 + b2 = 452 + 1-1222 = 225 + 144 = 2169 = 13

v = 5i - 12j.

EXAMPLE 7

Finding the Unit Vector that Has the Same Direction

as a Given Nonzero Vector v

For any nonzero vector v, the vector

is the unit vector that has the same direction as v. To find this vector, divide v byits magnitude.

v

7v 7

Discovery

To find out why the procedure in thebox produces a unit vector, workExercise 112 in Exercise Set 7.6.

y

x

v = ai + bj

(a, b)

u

||v||

Figure 7.59 Expressing a vectorin terms of its magnitude, andits direction angle, u

7v 7 ,

Writing a Vector in Terms of Its Magnitude and Direction

Let v be a nonzero vector. If is the direction angle measured from the positiveto v, then the vector can be expressed in terms of its magnitude and

direction angle as

v = 7v 7 cos ui + 7v 7 sin uj.

x-axisu

6 Write a vector in terms of its

magnitude and direction.

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 726

Figure 7.60 Vector v represents

a wind blowing at 20 miles per hour

in the direction N30°W.


A vector that represents the direction and speed of an object in motion is

called a velocity vector. In Example 8, we express a wind’s velocity vector in terms

of the wind’s magnitude and direction.

Writing a Vector Whose Magnitude and DirectionAre Given

The wind is blowing at 20 miles per hour in the direction N30°W. Express its velocity

as a vector v in terms of i and j.

Solution The vector v is shown in Figure 7.60. The vector’s direction angle, from

the positive to v, is

Because the wind is blowing at 20 miles per hour, the magnitude of v is 20 miles per

hour: Thus,

Use the formula for a vector in terms

of magnitude and direction.

and

and

Simplify.

The wind’s velocity can be expressed in terms of i and j as

Check Point 8 The jet stream is blowing at 60 miles per hour in the direction

N45°E. Express its velocity as a vector v in terms of i and j.

Application

Many physical concepts can be represented by vectors. A vector that represents a

pull or push of some type is called a force vector. If you are holding a 10-pound

package, two force vectors are involved. The force of gravity is exerting a force of

magnitude 10 pounds directly downward. This force is shown by vector in Figure

7.61. Assuming there is no upward or downward movement of the package, you are

exerting a force of magnitude 10 pounds directly upward. This force is shown by

vector in Figure 7.61. It has the same magnitude as the force exerted on your

package by gravity, but it acts in the opposite direction.

If and are two forces acting on an object, the net effect is the same as if

just the resultant force, acted on the object. If the object is not moving, as

is the case with your 10-pound package, the vector sum of all forces is the zero

vector.

Finding the Resultant Force

Two forces, and of magnitude 10 and 30 pounds, respectively, act on an

object. The direction of is N20°E and the direction of is N65°E. Find the

magnitude and the direction of the resultant force. Express the magnitude to the

nearest hundredth of a pound and the direction angle to the nearest tenth of a

degree.

Solution The vectors and are shown in Figure 7.62. The direction angle

for from the positive to the vector, is or 70°. We express

using the formula for a vector in terms of its magnitude and direction.

F190° - 20°,x-axisF1 ,

F2F1

F2F1

F2 ,F1

EXAMPLE 9

F1 + F2 ,

F2F1

F2

F1

v = -10i + 1023j.

= -10i + 1023j

sin 120° =23

2.cos 120° = -

1

2 = 20a -

1

2b i + 20¢23

2≤ j

u = 120°.7v 7 = 20 = 20 cos 120°i + 20 sin 120°j

v = 7v 7 cos ui + 7v 7 sin uj

7v 7 = 20.

u = 90° + 30° = 120°.

x-axis

EXAMPLE 8

v

30 8

20u

y

x

7 Solve applied problems

involving vectors.

F1

F2

Force ofgravity

Figure 7.61 Force vectors

u

Resultantforce, F

F1

10 pounds

F2

30 pounds

Direction angle ofthe resultant force

y

x

65°

20°

Figure 7.62

A-BLTZMC07_663-746-hr1 6-10-2008 14:56 Page 727


and

Use a calculator.

Figure 7.62 illustrates that the direction angle for from the positive to the vector, is or 25°. We express using the formula for a vector interms of its magnitude and direction.

and

Use a calculator.

The resultant force, F, is Thus,

Use and found above.

Add the horizontal components.

Add the vertical components.

Simplify.

Figure 7.63 shows the resultant force, F, without showing and Now that we have the resultant force vector, F, we can find its magnitude.

The magnitude of the resultant force is approximately 37.74 pounds.To find the direction angle of the resultant force, we can use

These ratios are illustrated for the right triangle in Figure 7.63.Using the first formula, we obtain

Thus,

Use a calculator.

The direction angle of the resultant force is approximately 35.8°.In summary, the two given forces are equivalent to a single force of approximately

37.74 pounds with a direction angle of approximately 35.8°.

Check Point 9 Two forces, and of magnitude 30 and 60 pounds,respectively, act on an object. The direction of is N10°E and the direction of is N60°E. Find the magnitude, to the nearest hundredth of a pound, and thedirection angle, to the nearest tenth of a degree, of the resultant force.

We have seen that velocity vectors represent the direction and speed of movingobjects. Boats moving in currents and airplanes flying in winds are situations inwhich two velocity vectors act simultaneously. For example, suppose v representsthe velocity of a plane in still air. Further suppose that w represents the velocity ofthe wind. The actual speed and direction of the plane is given by the vector This resultant vector describes the plane’s speed and direction relative to theground. Problems involving the resultant velocity of a boat or plane are solvedusing the same method that we used in Example 9 to find a single resultant forceequivalent to two given forces.

v + w.

F2F1

F2 ,F1

u = cos-1a30.61

37.74b L 35.8°.

cos u =a

7F 7 L30.61

37.74.

cos u =a

7F 7 or sin u =b

7F 7 .u,

7F 7 = 3a2 + b2 = 4130.6122 + 122.0822 L 37.74

F2 .F1

= 30.61i + 22.08j.

= 13.42 + 27.192i + 19.40 + 12.682jF2 ,F1 L 13.42i + 9.40j2 + 127.19i + 12.68j2

F = F1 + F2

F1 + F2 .

L 27.19i + 12.68j

u = 25°.7F2 7 = 30 = 30 cos 25°i + 30 sin 25°j

F2 = 7F2 7 cos ui + 7F2 7 sin uj

F290° - 65°,x-axisF2 ,

L 3.42i + 9.40j

u = 70°.7F1 7 = 10 = 10 cos 70°i + 10 sin 70°j

F1 = 7F1 7 cos ui + 7F1 7 sin uj

Study Tip

If the direction angle,of F can also be found using

tan u =b

a.

u,F = ai + bj,

u

y

x

b = 22.08

a = 30.61

(30.61, 22.08)Resultant force:F = 30.61i + 22.08j

iFi

Figure 7.63

u

Resultantforce, F

F1

10 pounds

F2

30 pounds

Direction angle ofthe resultant force

y

x

65°

20°

FIgure 7.62 (repeated)

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 728


Exercise Set 7.6

Practice Exercises

In Exercises 1–4, u and v have the same direction. In each exercise:

a. Find b. Find c. Is Explain.

1.

2.

3.

4.

In Exercises 5–12, sketch each vector as a position vector and find

its magnitude.

5. 6.

7. 8.

9. 10.

11. 12. v = -5jv = -4i

v = 5i - 2jv = -6i - 2j

v = - i - jv = i - j

v = 2i + 3jv = 3i + j

21

4

−2

−4

3

−1

−3

21 4−2 −1−4

y

x

(3, −4)

(−3, 3)

(−3, −2)

(3, 1)

v

u

2

4

−2

−4

3

−3

2 3 51−3 −1

y

x

(−2, −1)

(−1, 1) (5, 1)

(4, −1)v

u

2

456

−2

431−4 −3

y

x

(−4, 6)(−2, 5)

(2, −1)(0, 0)

vu

1

3456

−2

2 4 51 3−3

y

x

(4, 6)

(5, 4)

(0, 0)

v

u

(−1, 2)

u = v?7v 7 .7u 7 .

In Exercises 13–20, let v be the vector from initial point to terminal

point Write v in terms of i and j.

13.

14.

15.

16.

17.

18.

19.

20.

In Exercises 21–38, let

Find each specified vector or scalar.

21. 22.

23. 24.

25. 26.

27. 5v 28. 6v

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

In Exercises 39–46, find the unit vector that has the same direction

as the vector v.

39. 40.

41. 42.

43. 44.

45. 46.

In Exercises 47–52, write the vector v in terms of i and j whose

magnitude and direction angle are given.

47. 48.

49. 50.

51. 52.

Practice Plus


Find each specified vector or scalar.

53. 54.

55. 56. 7v + w 72 - 7v - w 727u + v 72 - 7u - v 723u - 14v - w24u - 12v - w2

u = -2i + 3j, v = 6i - j, w = -3i.

7v 7 = 14 , u = 200°7v 7 = 1

2 , u = 113°

7v 7 = 10, u = 330°7v 7 = 12, u = 225°

7v 7 = 8, u = 45°7v 7 = 6, u = 30°

u7v 7

v = i - jv = i + j

v = 4i - 2jv = 3i - 2j

v = 8i - 6jv = 3i - 4j

v = -5jv = 6i

7u - w 77w - u 77 -2u 772u 74w - 3v3v - 4w

3u + 4v3w + 2v

-7w-4w

w - vv - u

v - wu - v

v + wu + v

u = 2i - 5j, v = -3i + 7j, and w = - i - 6j.

P1 = 14, -52, P2 = 14, 32P1 = 1-3, 42, P2 = 16, 42P1 = 1-1, 62, P2 = 17, -52P1 = 1-1, 72, P2 = 1-7, -72P1 = 1-7, -42, P2 = 10, -22P1 = 1-8, 62, P2 = 1-2, 32P1 = 12, -52, P2 = 1-6, 62P1 = 1-4, -42, P2 = 16, 22

P2 .P1

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 729



Prove each property by obtaining the vector on each side of the

equation. Have you proved a distributive, associative, or

commutative property of vectors?

57.

58.

59.

60.

In Exercises 61–64, find the magnitude to the nearest hundredth,

and the direction angle to the nearest tenth of a degree, for each

given vector v.

61. 62.

63.

64.

Application Exercises

In Exercises 65–68, a vector is described. Express the vector in

terms of i and j. If exact values are not possible, round components

to the nearest tenth.

65. A quarterback releases a football with a speed of 44 feet persecond at an angle of 30° with the horizontal.

66. A child pulls a sled along level ground by exerting a forceof 30 pounds on a handle that makes an angle of 45° withthe ground.

67. A plane approaches a runway at 150 miles per hour at anangle of 8° with the runway.

68. A plane with an airspeed of 450 miles per hour is flying in thedirection N35°W.

Vectors are used in computer graphics to determine lengths of

shadows over flat surfaces. The length of the shadow for v in the

figure shown is the absolute value of the vector’s horizontal

component. In Exercises 69–70, the magnitude and direction

angle of v are given. Write v in terms of i and j. Then find the

length of the shadow to the nearest tenth of an inch.

69.

70. 7v 7 = 1.8 inches, u = 40°

7v 7 = 1.5 inches, u = 25°

v

v = 17i - 3j2 - 110i - 3j2v = 14i - 2j2 - 14i - 8j2

v = 2i - 8jv = -10i + 15j

u,7v 7 ,

1c + d2u = cu + du

c1u + v2 = cu + cv

1u + v2 + w = u + 1v + w2u + v = v + u

w = a3 i + b3 j.

v = a2 i + b2 j

u = a1i + b1 j

71. The magnitude and direction of two forces acting on an objectare 70 pounds, S56°E, and 50 pounds, N72°E, respectively.Find the magnitude, to the nearest hundredth of a pound, andthe direction angle, to the nearest tenth of a degree, of theresultant force.

72. The magnitude and direction exerted by two tugboats towinga ship are 4200 pounds, N65°E, and 3000 pounds, S58°E,respectively. Find the magnitude, to the nearest pound, andthe direction angle, to the nearest tenth of a degree, of theresultant force.

73. The magnitude and direction exerted by two tugboats towinga ship are 1610 kilograms, N35°W, and 1250 kilograms,S55°W, respectively. Find the magnitude, to the nearestkilogram, and the direction angle, to the nearest tenth of adegree, of the resultant force.

74. The magnitude and direction of two forces acting on anobject are 64 kilograms, N39°W, and 48 kilograms, S59°W,respectively. Find the magnitude, to the nearest hundredth ofa kilogram, and the direction angle, to the nearest tenth of adegree, of the resultant force.

The figure shows a box being pulled up a ramp inclined at 18°

from the horizontal.

Use the following information to solve Exercises 75–76.

75. If the box weighs 100 pounds, find the magnitude of the forceneeded to pull it up the ramp.

76. If a force of 30 pounds is needed to pull the box up the ramp,find the weight of the box.

In Exercises 77–78, round answers to the nearest pound.

77. a. Find the magnitude of the force required to keep a3500-pound car from sliding down a hill inclined at 5.5°from the horizontal.

b. Find the magnitude of the force of the car against the hill.

78. a. Find the magnitude of the force required to keep a280-pound barrel from sliding down a ramp inclined at12.5° from the horizontal.

b. Find the magnitude of the force of the barrel againstthe ramp.

ƒBC!ƒ = magnitude of the force of the box against the ramp

to pull the box up the ramp

7AC! 7 = magnitude of the force needed

7BA! 7 = weight of the box

BA!= force of gravity

18°18°

FB

DE

A

C

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 730


The forces acting on an object are in equilibrium

if the resultant force is the zero vector:

In Exercises 79–82, the given forces are acting on an object.

a. Find the resultant force.

b. What additional force is required for the given forces to be

in equilibrium?

79.

80.

81.

82.

83. The figure shows a small plane flying at a speed of 180 milesper hour on a bearing of N50°E. The wind is blowing fromwest to east at 40 miles per hour. The figure indicates that v represents the velocity of the plane in still air and w

represents the velocity of the wind.

a. Express v and w in terms of their magnitudes anddirection angles.

b. Find the resultant vector,

c. The magnitude of called the ground speed of theplane, gives its speed relative to the ground.Approximatethe ground speed to the nearest mile per hour.

d. The direction angle of gives the plane’s truecourse relative to the ground. Approximate the truecourse to the nearest tenth of a degree. What is theplane’s true bearing?

v + w

v + w,

v + w.

v + wv

w

Wind's speed:40 miles per hour

Plane's speed:180 miles per hour

y

x

50°

y

x

70°40°

20° 0

8

6

4

F2

F3

F1

y

x

F2

F3

F4

F1

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

F1 = -2i + 3j, F2 = i - j, F3 = 5i - 12j

F1 = 3i - 5j, F2 = 6i + 2j

F1 + F2 + F3 + Á + Fn = 0.

F1 , F2 , F3 , Á , Fn 84. Use the procedure outlined in Exercise 83 to solve thisexercise. A plane is flying at a speed of 400 miles per houron a bearing of N50°W. The wind is blowing at 30 miles perhour on a bearing of N25°E.

a. Approximate the plane’s ground speed to the nearestmile per hour.

b. Approximate the plane’s true course to the nearesttenth of a degree. What is its true bearing?

85. A plane is flying at a speed of 320 miles per hour on abearing of N70°E. Its ground speed is 370 miles per hour andits true course is 30°. Find the speed, to the nearest mile perhour, and the direction angle, to the nearest tenth of adegree, of the wind.

86. A plane is flying at a speed of 540 miles per hour on abearing of S36°E. Its ground speed is 500 miles per hour andits true bearing is S44°E. Find the speed, to the nearest mileper hour, and the direction angle, to the nearest tenth of adegree, of the wind.

Writing in Mathematics87. What is a directed line segment?

88. What are equal vectors?

89. If vector v is represented by an arrow,how is represented?

90. If vectors u and v are represented by arrows, describe howthe vector sum is represented.

91. What is the vector i?

92. What is the vector j?

93. What is a position vector? How is a position vectorrepresented using i and j?

94. If v is a vector between any two points in the rectangularcoordinate system, explain how to write v in terms of i and j.

95. If two vectors are expressed in terms of i and j, explain howto find their sum.

96. If two vectors are expressed in terms of i and j, explain howto find their difference.

97. If a vector is expressed in terms of i and j, explain how to findthe scalar multiplication of the vector and a given scalar

98. What is the zero vector?

99. Describe one similarity between the zero vector and thenumber 0.

100. Explain how to find the unit vector in the direction of anygiven vector v.

101. Explain how to write a vector in terms of its magnitudeand direction.

102. You are on an airplane. The pilot announces the plane’sspeed over the intercom. Which speed do you think is beingreported: the speed of the plane in still air or the speed afterthe effect of the wind has been accounted for? Explain youranswer.

103. Use vectors to explain why it is difficult to hold a heavystack of books perfectly still for a long period of time. Asyou become exhausted, what eventually happens? Whatdoes this mean in terms of the forces acting on the books?

k.

u + v

-3v

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 731


Critical Thinking ExercisesMake Sense? In Exercises 104–107, determine whether

each statement makes sense or does not make sense, and explain

your reasoning.

104. I used a vector to represent a wind velocity of 13 milesper hour from the west.

105. I used a vector to represent the average yearly rate ofchange in a man’s height between ages 13 and 18.

106. Once I’ve found a unit vector u, the vector must also bea unit vector.

107. The resultant force of two forces that each have a magnitudeof one pound is a vector whose magnitude is two pounds.

In Exercises 108–111, use the figure shown to determine whether

each statement is true or false. If the statement is false, make the

necessary change(s) to produce a true statement.

108.

109.

110.

111.

112. Let Show that is a unit vector in thedirection of v.

In Exercises 113–114, refer to the navigational compass shown in

the figure. The compass is marked clockwise in degrees that start

at north 0°.

N, 0°

W, 270° E, 90°

S, 180°

v

7v 7v = ai + bj.

7A 7 = 7C 7B - E = G - F

D + A + B + C = 0

A + B = E

A

B

C

D

EF

G

-u

113. An airplane has an air speed of 240 miles per hour and acompass heading of 280°.A steady wind of 30 miles per houris blowing in the direction of 265°. What is the plane’s truespeed relative to the ground? What is its compass headingrelative to the ground?

114. Two tugboats are pulling on a large ship that has goneaground. One tug pulls with a force of 2500 pounds in acompass direction of 55°. The second tug pulls with a forceof 2000 pounds in a compass direction of 95°. Find themagnitude and the compass direction of the resultant force.

115. You want to fly your small plane due north, but there is a75 kilometer wind blowing from west to east.

a. Find the direction angle for where you should head theplane if your speed relative to the ground is 310 kilometersper hour.

b. If you increase your air speed, should the directionangle in part (a) increase or decrease? Explain youranswer.

Preview Exercises

Exercises 116–118 will help you prepare for the material covered

in the next section.

116. Find the obtuse angle rounded to the nearest tenth of adegree, satisfying

where and

117. If find the following vector:

118. Consider the triangle formed by vectors u, v, and w.

a. Use the magnitudes of the three vectors to write the Lawof Cosines for the triangle shown in the figure:

b. Use the coordinates of the points shown in the figure towrite algebraic expressions for and 7w 72.

7u 7 , 7u 72, 7v 7 , 7v 72, 7w 7 , 7u 72 = ?.

(0, 0)

(a1, b1)

(a2, b2)u

y

x

vu

w

21-22 + 41-627w 72 w.

w = -2i + 6j,

w = - i + 4j.v = 3i - 2j

cos u =31-12 + 1-22142

7v 7 7w 7 ,

u,

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 732

Talk about hard work! Ican see the weightlifter’s

muscles quivering from theexertion of holding the bar-bell in a stationary position

above her head. Still, I’m not sure ifshe’s doing as much work as I am, sitting at my

desk with my brain quivering from studyingtrigonometric functions and their applications.

Would it surprise you to know that neitheryou nor the weightlifter are doing any work at all?The definition of work in physics and mathematicsis not the same as what we mean by “work” in

everyday use. To understand what is involved in real work, we turn to a new vectoroperation called the dot product.

The Dot Product of Two Vectors

The operations of vector addition and scalar multiplication result in vectors. Bycontrast, the dot product of two vectors results in a scalar (a real number), ratherthan a vector.

Section 7.7 The Dot Product 733

Objectives

1 Find the dot product

of two vectors.

2 Find the angle between

two vectors.

3 Use the dot product to

determine if two vectors

are orthogonal.

4 Find the projection of a

vector onto another vector.

5 Express a vector as the sum

of two orthogonal vectors.

6 Compute work.

The Dot ProductS e c t i o n 7.7

Definition of the Dot Product

If and are vectors, the dot product is definedas follows:

The dot product of two vectors is the sum of the products of their horizontalcomponents and their vertical components.

v # w = a1 a2 + b1 b2 .

v # ww = a2 i + b2 jv = a1 i + b1 j

1 Find the dot product

of two vectors.

Finding Dot Products

If and find each of the following dot products:

a. b. c.

Solution To find each dot product, multiply the two horizontal components, andthen multiply the two vertical components. Finally, add the two products.

c. v ? v=5(5)+(–2)(–2)=25+4=29

Multiply the horizontal componentsand multiply the vertical components of

v = 5i − 2j and v = 5i − 2j.

b. w ? v=–3(5)+4(–2)=–15-8=–23


w = −3i + 4j and v = 5i − 2j.

a. v ? w=5(–3)+(–2)(4)=–15-8=–23


v = 5i − 2j and w = −3i + 4j.

v # v.w # vv # w

w = -3i + 4j,v = 5i - 2j

EXAMPLE 1

A-BLTZMC07_663-746-hr2 13-10-2008 16:28 Page 733


Check Point 1 If and find each of the followingdot products:

a. b. c.

In Example 1 and Check Point 1, did you notice that and producedthe same scalar? The fact that follows from the definition of the dotproduct. Properties of the dot product are given in the following box. Proofs for someof these properties are given in the appendix.

v # w = w # vw # vv # w

w # w.w # vv # w

w = 2i - j,v = 7i - 4j

(0, 0)

(a1, b1)

(a2, b2)u

y

x

vu

w

Figure 7.64

Alternative Formula for the Dot Product

If v and w are two nonzero vectors and is the smallest nonnegative anglebetween them, then

v # w = 7v 7 7w 7 cos u.

u

Properties of the Dot Product

If u, v, and w are vectors, and is a scalar, then

1.

2.

3.

4.

5. 1cu2 # v = c1u # v2 = u # 1cv2v # v = 7v 720 # v = 0

u # 1v + w2 = u # v + u # wu # v = v # u

c

The Angle between Two Vectors

The Law of Cosines can be used to derive another formula for the dot product. Thisformula will give us a way to find the angle between two vectors.

Figure 7.64 shows vectors and By the definitionof the dot product, we know that Our new formula for the dotproduct involves the angle between the vectors, shown as in the figure. Apply theLaw of Cosines to the triangle shown in the figure.

iui2=ivi2+iwi2-2ivi iwi cos u

iui = Ï(a1 − a2)2 + (b1 − b2)

2

u = (a1 − a2)i + (b1 − b2)j

ivi = Ïa12 + b1

2

v = a1i + b1j

iwi = Ïa22 + b2

2

w = a2i + b2j

u

v # w = a1 a2 + b1 b2 .w = a2 i + b2 j.v = a1 i + b1 j

a12 - 2a1 a2 + a2

2 + b12 - 2b1 b2 + b2

2 = a12 + b1

2 + a22 + b2

2 - 2 7v 7 7w 7 cos u

1a1 - a222 + 1b1 - b222 = 1a12 + b1

22 + 1a22 + b2

22 - 2 7v 7 7w 7 cos u Substitute the squares of the

magnitudes of vectors u, v, and w

into the Law of Cosines.

Use the Law of Cosines.

Square the binomials using

1A - B22 = A2 - 2AB + B2.

-2a1 a2 - 2b1 b2 = -2 7v 7 7w 7 cos u Subtract and from

both sides of the equation.

b22a1

2 , a22 , b1

2 ,

Divide both sides by -2.

By definition,v ? w = a1a2 + b1b2.

a1a2+b1b2=ivi iwi cos u

v # w = 7v 7 7w 7 cos u Substitute for the

expression on the left

side of the equation.

v # w

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 734


2 Find the angle between

two vectors.

Solving the formula in the box for gives us a formula for finding theangle between two vectors:

cos u

Formula for the Angle between Two Vectors

If v and w are two nonzero vectors and is the smallest nonnegative anglebetween v and w, then

cos u =v # w7v 7 7w 7 and u = cos-1¢ v # w

7v 7 7w 7 ≤ .

u

Finding the Angle between Two Vectors

Find the angle between the vectors and shown inFigure 7.65. Round to the nearest tenth of a degree.

Solution Use the formula for the angle between two vectors.

cos u =v # w7v 7 7w 7

w = - i + 4j,v = 3i - 2ju

EXAMPLE 2

Begin with the formula for the cosine

of the angle between two vectors.

=13i - 2j2 # 1- i + 4j2

432 + 1-22241-122 + 42

Substitute the given vectors in the

numerator. Find the magnitude of

each vector in the denominator.

=31-12 + 1-22142213217

Find the dot product in the numerator.

Simplify in the denominator.

= -

11

2221Perform the indicated operations.

The angle between the vectors is

Use a calculator.

Check Point 2 Find the angle between the vectors and Round to the nearest tenth of a degree.

w = i + 2j.v = 4i - 3j

u = cos-1¢ -

11

2221≤ L 137.7°.

u

u

(−1, 4)

(3, −2)

4

−2

5

−1

−4−3

−5

2 4−2−4−5 3 5−1−3

y

x

w = −i + 4j

v = 3i − 2j

Figure 7.65 Finding the anglebetween two vectors

Parallel and Orthogonal Vectors

Two vectors are parallel when the angle between the vectors is 0° or 180°.If the vectors point in the same direction. If the vectors point inopposite directions. Figure 7.66 shows parallel vectors.

u = 180°,u = 0°,u

v

wv w

u

Figure 7.66 Parallel vectors

Two vectors are orthogonal when the angle between the vectors is 90°, shownin Figure 7.67. (The word orthogonal, rather than perpendicular, is used to describevectors that meet at right angles.) We know that If v and w areorthogonal, then

Conversely, if v and w are vectors such that then or orIf then so v and w are orthogonal.u = 90°,cos u = 0,cos u = 0.

7w 7 = 07v 7 = 0v # w = 0,

v # w = 7v 7 7w 7 cos 90° = 7v 7 7w 7 102 = 0.

v # w = 7v 7 7w 7 cos u.

v

w

u

Figure 7.67Orthogonal vectors:

andcos u = 0u = 90°

3 Use the dot product to determine

if two vectors are orthogonal.

and Vectors point in the same direction.

cos u = 1.u = 0° and Vectorspoint in opposite directions.

cos u = -1.u = 180°

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 735


The Dot Product and Orthogonal Vectors

Two nonzero vectors v and w are orthogonal if and only if Becausethe zero vector is orthogonal to every vector v.0 # v = 0,

v # w = 0.

Determining Whether Vectors Are Orthogonal

Are the vectors and orthogonal?

Solution The vectors are orthogonal if their dot product is 0. Begin by finding

The dot product is 0. Thus, the given vectors are orthogonal. They are shown inFigure 7.68.

Check Point 3 Are the vectors and orthogonal?

Projection of a Vector Onto Another Vector

You know how to add two vectors to obtain a resultant vector. We now reverse thisprocess by expressing a vector as the sum of two orthogonal vectors. By doing this, youcan determine how much force is applied in a particular direction. For example, Figure

7.69 shows a boat on a tilted ramp. The force due to gravity, F, is pulling straight downon the boat. Part of this force, is pushing the boat down the ramp. Another part ofthis force, is pressing the boat against the ramp, at a right angle to the incline.Thesetwo orthogonal vectors, and are called the vector components of F. Notice that

A method for finding and involves projecting a vector onto another vector.Figure 7.70 shows two nonzero vectors, v and w, with the same initial point.The

angle between the vectors, is acute in Figure 7.70(a) and obtuse in Figure 7.70(b).A third vector, called the vector projection of v onto w, is also shown in each figure,denoted by projw v.

u,

F2F1

F = F1 + F2 .F2 ,F1

F2 ,F1 ,

w = 6i - 4jv = 2i + 3j

v # w = 16i - 3j2 # 1i + 2j2 = 6112 + 1-32122 = 6 - 6 = 0

v # w.

w = i + 2jv = 6i - 3j

EXAMPLE 3

v

w

projw v

u

Figure 7.70(b)

v

w

projw v

u

Figure 7.70(a)

u

4

−2

5

−1

−4

321

−3

−5

2 4−2−4 3 5 6−1−3x

w = i + 2j

v = 6i − 3j

y

Figure 7.68 Orthogonal vectors

F

F1

F2

Figure 7.69

4 Find the projection of a vector

onto another vector.

How is the vector projection of v onto w formed? Draw the line segment fromthe terminal point of v that forms a right angle with a line through w, shown in red.The projection of v onto w lies on a line through w, and is parallel to vector w. Thisvector begins at the common initial point of v and w. It ends at the point where thedashed red line segment intersects the line through w.

Our goal is to determine an expression for We begin with itsmagnitude. By the definition of the cosine function,

Multiply both sides by

Reverse the two sides.

We can rewrite the right side of this equation and obtain another expression for themagnitude of the vector projection of v onto w. To do so, use the alternate formulafor the dot product, v # w = 7v 7 7w 7 cos u.

7projw v 7 = 7v 7 cos u.

7v 7 . 7v 7 cos u = 7projw v 7

iprojwvi

ivicos u=

This is the magnitude of thevector projection of v onto w.

projw v.

The discussion at the bottom of the previous page is summarized as follows:

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 736

Divide both sides of by

The expression on the right side of this equation, is the same expressionthat appears in the formula for Thus,

We use the formula for the magnitude of to find the vector itself. Thisis done by finding the scalar product of the magnitude and the unit vector in thedirection of w.

v ? w

iwiv ? w

iwi2projwv= w=

This is the magnitude ofthe vector projection of

v onto w.

This is the unitvector in the

direction of w.

ba w

iwiba

projw v

7projw v 7 = 7v 7 cos u =v # w7w 7 .

7projw v 7 .7v 7 cos u,

v # w7w 7 = 7v 7 cos u.

7w 7 :v # w = 7v 7 7w 7 cos u


The Vector Projection of v Onto w

If v and w are two nonzero vectors, the vector projection of v onto w is

projw v =v # w7w 72 w.

Finding the Vector Projection of One Vector Onto Another

If and find the vector projection of v onto w.

Solution The vector projection of v onto w is found using the formula for

The three vectors, v, w, and are shown in Figure 7.71.

Check Point 4 If and find the vector projection of v onto w.

We use the vector projection of v onto w, to express v as the sum of twoorthogonal vectors.

projw v,

w = i - j,v = 2i - 5j

projw v,

=21-22 + 4162A240 B2 w =

20

40 w = 1

2 1-2i + 6j2 = - i + 3j

projw v =v # w7w 72 w =

12i + 4j2 # 1-2i + 6j2A41-222 + 62 B2 w

projw v.

w = -2i + 6j,v = 2i + 4j

EXAMPLE 4

The Vector Components of v

Let v and w be two nonzero vectors. Vector v can be expressed as the sum of twoorthogonal vectors, and where is parallel to w and is orthogonal to w.

Thus, The vectors and are called the vector components of v.Theprocess of expressing v as is called the decomposition of v into and v2 .v1v1 + v2

v2v1v = v1 + v2 .

v1 = projw v =v # w7w 72 w, v2 = v - v1

v2v1v2 ,v1

5 Express a vector as the sum of

two orthogonal vectors.

4567

−2−1 21 4−2−4 3−1−3

x

y

projw vv

w

Figure 7.71 The vectorprojection of v onto w

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 737


Decomposing a Vector into Two Orthogonal Vectors

Let and Decompose v into two vectors, and whereis parallel to w and is orthogonal to w.

Solution These are the vectors we worked with in Example 4. We use theformulas in the box on the previous page.

We obtained this vector in Example 4.

Check Point 5 Let and (These are the vectors fromCheck Point 4.) Decompose v into two vectors, and where is parallel to w and is orthogonal to w.

Work: An Application of the Dot Product

The bad news:Your car just died.The good news: It died on a level road just 200 feetfrom a gas station. Exerting a constant force of 90 pounds, and not necessarilywhistling as you work, you manage to push the car to the gas station.

Although you did not whistle, you certainly did work pushing the car 200 feetfrom point to point How much work did you do? If a constant force F isapplied to an object, moving it from point to point in the direction of the force,the work, done is

You pushed with a force of 90 pounds for a distance of 200 feet. The work done byyour force is

or 18,000 foot-pounds.Work is often measured in foot-pounds or in newton-meters.The photo on the left shows an adult pulling a small child in a wagon. Work is

being done. However, the situation is not quite the same as pushing your car.Pushing the car, the force you applied was along the line of motion. By contrast, theforce of the adult pulling the wagon is not applied along the line of the wagon’s motion.In this case, the dot product is used to determine the work done by the force.

W = 190 pounds21200 feet2

W = 1magnitude of force21distance from A to B2.W,

BAB.A

200 feet

Force: 90 pounds

A B

v2

v1v2 ,v1

w = i - j.v = 2i - 5j

v2 = v - v1 = 12i + 4j2 - 1- i + 3j2 = 3i + j

v1 = projw v = - i + 3j

v2v1

v2 ,v1w = -2i + 6j.v = 2i + 4j

EXAMPLE 5

6 Compute work.

F

Definition of Work

The work, done by a force F moving an object from to is

W = F # AB!.

BAW,

When computing work, it is often easier to use the alternative formula for thedot product. Thus,

It is correct to refer to as either the work done or the work done by the force.W

W=F ? AB=iFi iABi cos u.

iFi is themagnitude

of the force.

iABi is thedistance over

which theconstant force

is applied.

u is the anglebetween theforce and thedirection of

motion.

A-BLTZMC07_663-746-hr 26-09-2008 17:00 Page 738


Computing Work

A child pulls a sled along level ground by exerting a force of 30 pounds on a ropethat makes an angle of 35° with the horizontal. How much work is done pulling thesled 200 feet?

Solution The situation is illustrated in Figure 7.72. The work done is

Thus, the work done is approximately 4915 foot-pounds.

Check Point 6 A child pulls a wagon along level ground by exerting a force of20 pounds on a handle that makes an angle of 30° with the horizontal. How muchwork is done pulling the wagon 150 feet?

Magnitudeof the force

is 30 pounds.

Distanceis

200 feet.

The anglebetween theforce and thesled’s motion

is 35°.

W=iFi iABi cos u=(30)(200) cos 358≠4915.

EXAMPLE 6

30 pounds

35°

Figure 7.72 Computing work done pullingthe sled 200 feet

Exercise Set 7.7

Practice Exercises

In Exercises 1–8, use the given vectors to find and

1. 2.

3. 4.

5.

6.

7. 8.


Find each specified scalar.

9. 10.

11. 12.

13. 14.

15. 16.

In Exercises 17–22, find the angle between v and w. Round to the

nearest tenth of a degree.

17.

18.

19. 20.

21. 22.

In Exercises 23–32, use the dot product to determine whether v and

w are orthogonal.

23. 24.

25.

26.

27.

28.

29. 30.

31. 32.

In Exercises 33–38, find Then decompose v into two vectors,

and where is parallel to w and is orthogonal to w.

33. 34. v = 3i - 2j, w = 2i + jv = 3i - 2j, w = i - j

v2v1v2 ,v1

projw v.

v = 5i, w = -6jv = 3i, w = -4j

v = 5i, w = -6iv = 3i, w = -4i

v = 5i - 5j, w = i - j

v = 2i - 2j, w = - i + j

v = 8i - 4j, w = -6i - 12j

v = 2i + 8j, w = 4i - j

v = i + j, w = - i + jv = i + j, w = i - j

v = 3j, w = 4i + 5jv = 6i, w = 5i + 4j

v = i + 2j, w = 4i - 3jv = -3i + 2j, w = 4i - j

v = -2i + 5j, w = 3i + 6j

v = 2i - j, w = 3i + 4j

51v # w241u # v215v2 # w14u2 # vv # u + v # wu # v + u # wv # 1u + w2u # 1v + w2

u = 2i - j, v = 3i + j, and w = i + 4j.

v = i, w = -5jv = 5i, w = j

v = -8i - 3j, w = -10i - 5j

v = -6i - 5j, w = -10i - 8j

v = 7i - 2j, w = -3i - jv = 5i - 4j, w = -2i - j

v = 3i + 3j, w = i + 4jv = 3i + j, w = i + 3j

v # v.v # w

35.

36.

37. 38.

Practice PlusIn Exercises 39–42, let

Find each specified scalar or vector.

39. 40.

41. 42.

In Exercises 43–44, find the angle, in degrees, between v and w.

43.

44.

In Exercises 45–50, determine whether v and w are parallel,

orthogonal, or neither.

45.

46.

47.

48.

49.

50.

Application Exercises51. The components of represent the respective

number of gallons of regular and premium gas sold at astation. The components of represent therespective prices per gallon for each kind of gas. Find anddescribe what the answer means in practical terms.

52. The components of represent the respectivenumber of one-day and three-day videos rented from a videostore.The components of represent the prices torent the one-day and three-day videos, respectively. Find and describe what the answer means in practical terms.

v # ww = 3i + 2j

v = 180i + 450j

v # ww = 2.90i + 3.07j

v = 240i + 300j

v = -2i + 3j, w = -6i - 4j

v = 3i - 5j, w = 6i +18

5 j

v = -2i + 3j, w = -6i - 9j

v = 3i - 5j, w = 6i + 10j

v = -2i + 3j, w = -6i + 9j

v = 3i - 5j, w = 6i - 10j

v = 3 cos 5p

3 i + 3 sin

5p

3 j, w = 2 cos pi + 2 sin pj

v = 2 cos 4p

3 i + 2 sin

4p

3 j, w = 3 cos

3p

2 i + 3 sin

3p

2 j

proju1v - w2proju1v + w24u # 15v - 3w25u # 13v - 4w2

u = - i + j, v = 3i - 2j, and w = -5j.

v = 2i + j, w = 6i + 3jv = i + 2j, w = 3i + 6j

v = 2i + 4j, w = -3i + 6j

v = i + 3j, w = -2i + 5j

A-BLTZMC07_663-746-hr 26-09-2008 17:01 Page 739


53. Find the work done in pushing a car along a level road frompoint to point 80 feet from while exerting a constantforce of 95 pounds. Round to the nearest foot-pound.

54. Find the work done when a crane lifts a 6000-pound boulderthrough a vertical distance of 12 feet. Round to the nearestfoot-pound.

55. A wagon is pulled along level ground by exerting a force of40 pounds on a handle that makes an angle of 32° with thehorizontal. How much work is done pulling the wagon100 feet? Round to the nearest foot-pound.

56. A wagon is pulled along level ground by exerting a force of25 pounds on a handle that makes an angle of 38° with thehorizontal. How much work is done pulling the wagon100 feet? Round to the nearest foot-pound.

57. A force of 60 pounds on a rope is used to pull a box up aramp inclined at 12° from the horizontal. The figure showsthat the rope forms an angle of 38° with the horizontal. Howmuch work is done pulling the box 20 feet along the ramp?

58. A force of 80 pounds on a rope is used to pull a box up aramp inclined at 10° from the horizontal. The rope forms anangle of 33° with the horizontal. How much work is donepulling the box 25 feet along the ramp?

59. A force is given by the vector The force movesan object along a straight line from the point (4, 9) to thepoint (10, 20). Find the work done if the distance is measuredin feet and the force is measured in pounds.

60. A force is given by the vector The force movesan object along a straight line from the point (8, 11) to thepoint (18, 20). Find the work done if the distance is measuredin meters and the force is measured in newtons.

61. A force of 4 pounds acts in the direction of 50° to thehorizontal. The force moves an object along a straight linefrom the point (3, 7) to the point (8, 10), with distancemeasured in feet. Find the work done by the force.

62. A force of 6 pounds acts in the direction of 40° to thehorizontal. The force moves an object along a straight linefrom the point (5, 9) to the point (8, 20), with the distancemeasured in feet. Find the work done by the force.

63. Refer to Figure 7.69 on page 736. Suppose that the boatweighs 700 pounds and is on a ramp inclined at 30°.Represent the force due to gravity, F, using

a. Write a unit vector along the ramp in the upward direction.

b. Find the vector projection of F onto the unit vectorfrom part (a).

c. What is the magnitude of the vector projection in part(b)? What does this represent?

64. Refer to Figure 7.69 on page 736. Suppose that the boatweighs 650 pounds and is on a ramp inclined at 30°.Represent the force due to gravity, F, using

a. Write a unit vector along the ramp in the upward direction.

b. Find the vector projection of F onto the unit vectorfrom part (a).

c. What is the magnitude of the vector projection in part(b)? What does this represent?

F = -650j.

F = -700j.

F = 5i + 7j.

F = 3i + 2j.

60 pounds

38°

12°

A,B,A

Writing in Mathematics65. Explain how to find the dot product of two vectors.

66. Using words and no symbols, describe how to find the dotproduct of two vectors with the alternative formula

67. Describe how to find the angle between two vectors.

68. What are parallel vectors?

69. What are orthogonal vectors?

70. How do you determine if two vectors are orthogonal?

71. Draw two vectors, v and w, with the same initial point. Showthe vector projection of v onto w in your diagram. Thendescribe how you identified this vector.

72. How do you determine the work done by a force F in movingan object from to when the direction of the force is notalong the line of motion?

73. A weightlifter is holding a barbell perfectly still above hishead, his body shaking from the effort. How much work isthe weightlifter doing? Explain your answer.

74. Describe one way in which the everyday use of the word work

is different from the definition of work given in this section.

Critical Thinking ExercisesMake Sense? In Exercises 75–78, determine whether each

statement makes sense or does not make sense, and explain

your reasoning.

75. Although I expected vector operations to produce anothervector, the dot product of two vectors is not a vector, but areal number.

76. I’ve noticed that whenever the dot product is negative, theangle between the two vectors is obtuse.

77. I’m working with a unit vector, so its dot product withitself must be 1.

78. The weightlifter does more work in raising 300 kilogramsabove her head than Atlas, who is supporting the entire world.

BA

v # w = 7v 7 7w 7 cos u.

In Exercises 79–81, use the vectors

to prove the given property.

79. 80.

81.

82. If find a vector orthogonal to v.

83. Find a value of so that and are orthogonal.

84. Prove that the projection of v onto i is

85. Find two vectors v and w such that the projection of v onto w is v.

Group Exercise86. Group members should research and present a report on

unusual and interesting applications of vectors.

1v # i2i.

-4i + bj15i - 3jb

v = -2i + 5j,

u # 1v + w2 = u # v + u # w

1cu2 # v = c1u # v2u # v = v # u

u = a1 i + b1 j, v = a2 i + b2 j, and w = a3 i + b3 j,

A-BLTZMC07_663-746-hr1 3-10-2008 9:05 Page 740

Date post:	24-Oct-2014
Category:	Documents
Upload:	pete-rodrigue
View:	1,171 times
Download:	4 times

Chapter 07 - Vectors-And-Dot Products

Documents