Chapter 1 Dot Product

7/31/2019 Chapter 1 Dot Product

1/18

Chapter 1: Vectors and theGeometry of Space

The Dot Product of Two Vectors


2/18

In our first two lessons on vectors, you have studied:

Properties of vectors;

Notation associated with vectors;

Vector Addition;

Multiplication by a Scalar.

In this lesson you will study the dot product of two vectors. The dot productof two vectors generates a scalar as described below.

The Definition of Dot Product

The dot product of two vectors

.

,,

2211

2121

vuvuvu

isvvvanduuu

The dot product of two vectors

.

,,,,

332211

321321

vuvuvuvu

isvvvvanduuuu


3/18

Properties of the Dot Product

2vvv5.

0v04.

vcuvuc)vuc(3.

wuvu)wv(u:holdspropertyvedistributiThe2.

.uvvu:holdspropertyecommutativThe1.

scalar.abecletand

spaceinorplanein thevectorsbewand,v,uLet

The proof of property 5.

2

2

3

2

2

2

1

22

3

2

2

2

1

2

2

3

2

2

2

1321321

321

Therefore

,,v,,v

.,,vvSuppose

vvv

vvvvvvv

vvvvvvvvv

vv

Proofs of otherproperties aresimilar.


4/18

Consider two non-zero vectors. We can use their dotproduct and their magnitudes to calculate the angle

between the two vectors. We begin with the sketch.

u

v

uv

From the Law of Cosines where c is the

side opposite the angle theta:

cos2

cos2

222

222

vuvuuv

abbac


5/18

22

2

2

)()(

)()(

uvuv

uuvuuvvv

uvuuvv

uvuvuv

:thatfollowsIt.wwwthathavewe

productdotfordevelopedjustthat wepropertiestheFrom

2

cos2222

vuvuuv

From the previous slide

Substituting from above

vu

vu

vuvu

vuvu

cos

cos

cos22

cos22

2222

vuvuuvuv

Simplifying


6/18

You have just witnessed the proof of the followingtheorem:

.vu

vucosthenvandu

vectorsnonzeroobetween twangletheisIf

Example 1Find the angle between the vectors:

.14,vand5,3

u

Solution

117.0or04.2

......4537.0cos

1714

7

11659

512

1,45,3

1,45,3cos

vu

vu


7/18

Example 2

.6-3,-3,zand2-1,3,-w

:ifzandwctorsbetween veangletheFind

Solution

2

03699419

1239

6,3,32,1,3

6,3,32,1,3cos

vu

vu

True or False? Whenever two non-zero vectors areperpendicular, their dot product is 0.

Think before you click.

Congratulate yourself if you choseTrue!


8/18

002cosand02cos

vuvu

vu

vu

vu

True or False? Whenever two non-zero vectors areperpendicular, their dot product is 0.

This is true. Since the two vectors are perpendicular,

the angle between them will be .2

or90

True or False? Whenever you find the angle between

two non-zero vectors the formula

will generate angles in the interval

vu

vu

cos

.2

0


9/18




vu

vu

cos

.2

0

This is False. For example, consider the vectors:.2,1-vand1,4

u

vu

vucos

Finish this on your own then click for the answer.

2 2 4

1

1

2

x

y


10/18




vu

vu

cos

.2

0

This is False. For example, consider the vectors:

.2,1-vand1,4

u

5.167or9.2

...9762.085

9

517

18

1,21,4

1,21,4

vu

vucos

When the cosine is negative the angle between the twovectors is obtuse.


11/18

An Application of the Dot Product: Projection

The Tractor Problem: Consider the familiar example ofa heavy box being dragged across the floor by a rope.If the box weighs 250 pounds and the angle betweenthe rope and the horizontal is 25 degrees, how muchforce does the tractor have to exert to move the box?

Discussion: the force being exerted by the tractorcan be interpreted as a vector with direction of 25

degrees. Our job is to find the magnitude.

25

v


12/18

The Tractor Problem, Slide 2: we are going to look at theforce vector as the sum of its vertical component vectorand its horizontal component. The work of moving the

box across the floor is done by the horizontal component.

x

v

1v

2v

y

25

lbsv

v

vv

v

v

8.275......9063.0

250

25cos

25cos

25cos

1

1

1

Hint: for maximum accuracy, dont round off until theend of the problem. In this case, we left the value of

the cosine in the calculator and did not round off untilthe end.

Conclusion: the tractor has to exert

a force of 275.8 lbs before the boxwill move.


13/18

The problem gets more complicated when the directionin which the object moves is not horizontal or vertical.

u

v1

w

2w

x

y

.wtoorthogonalisand

vtoparalleliswthatso

wusketch,In the

12

1

21

w

w

.wcomponentthe

bydonebeingisworkthe

ofallthenvofdirection

in theoriginat the

objectanmovingisuby

drepresenteforceaIf

1

.vtoorthogonalu

ofcomponentvectorthecalledis

.projasdenotedisandvontouofprojectionthecalledis

2

v1

1

w

uw

w


14/18

A Vocabulary Tip:

When two vectors are perpendicular we say that they are orthogonal.

When a vector is perpendicular to a line or a plane we say that the vector isnormal to the line or plane.

u

v

1w

2w

x

y

The following theorem willprove very useful in the

remainder of this course.

vv

vuuprojv

2

:followsascalculated

becanvontouofprojectionthe

thenvectors,nonzeroarevanduIf

.vvectorofmultipleaisv

ontouofprojectionthesoscalarais

sparenthesitheinsidequantityThe


15/18

.vtoorthogonaluofcomponent

vectorthefindanduprojfind6,3isvand5,9isuif:Example v

1

2

22

8.3,6.7

3,645

573,6

936

2730

3,63,6

3,69,5

w

vv

vuuprojv

2.5,6.28.3,6.79,5

9,58.3,6.7w

thatnoticesketch,theFrom

2

2

21

w

wuw

2 2 4

2

4

(5,9)

(6,3)


16/18

Work is traditionally defined as follows: W = FD where F is the constantforce acting along the line of motion and D is the distance traveled alongthe line of motion.

Example: an object is pulled 12 feet across the floor using a force of 100pounds. Find the work done if the force is applied at an angle of 50 degreesabove the horizontal.

50

100 lb

12 ft

Solution A: using W = FD we use theprojection of F in the x direction.

pounds-foot35.771)12)(50cos100( FDW

pounds.-foot35.771)12(50cos10012,050sin100,100cos50

thatalsoNote.12,0isvectordirectiontheand

50sin100,100cos50isformcoordinateinvectorforcethat thenote:BSolution


17/18

Two Ways to Calculate Work

v

vprojv

uW:BMethod

uW:AMethod:BMethodbyorAMethodbycalculatedbecanvvectoralong

napplicatioofpointitsmovinguforceconstantabydoneWork W

Example: Find the work done by a force of 20 lb acting in the directionN50W while moving an object 4 ft due west.

lbsft28.61W

)4)(40)(cos20(projW:SolutionA v

f tlbvu

lbsft28.61)0(140sin20)4(140cos20

0,4and140sin20,140cos20u:SolutionB

vuW

v

u

v

50

4vand20

u


18/18

axes.-zandy-,x-,positivewith themakesvthat][0,intervalin the

and,,anglesthearevvectornonzeroaofanglesdirectionThe

v

.vvectorofcosinesdirectionthecalledare

,cosand,cos,cosangles,directiontheseofcosinesThe

We can calculate the direction cosines byusing the unit vectors along each positiveaxis and the dot product.

i j

k

The last topic of this lesson concerns Direction Cosines.

v

v

v

vvv

kv

kv

v

v

v

vvv

jv

jv

v

v

v

vvv

iv

iv

3321

2321

1321

1

1,0,0,,cos

1

0,1,0,,cos

1

0,0,1,,cos

Chapter 1 Dot Product

Documents

Transcript of Chapter 1 Dot Product