Vectors and the Geometry of Space

The Dot Product of Two Vectors

Written by Richard Gill

Associate Professor of Mathematics

Tidewater Community College, Norfolk Campus, Norfolk, VA

With Assistance from a VCCS LearningWare Grant

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 product of two vectors generates a scalar as described below.

The Definition of Dot ProductThe dot product of two vectors

.

,,

2211

2121

vuvuvu

isvvvanduuu

The dot product of two vectors

.

,,,,

332211

321321

vuvuvuvu

isvvvvanduuuu

Properties of the Dot Product

2vvv 5.

0v0 4.

vcuvuc)vuc( 3.

wuvu)wv(u :holdsproperty vedistributi The 2.

.uvvu :holdsproperty ecommutativ The 1.

scalar. a be clet and

spacein or plane in the vectorsbe w and,v,uLet

The proof of property 5.

2

23

22

21

22

32

22

1

2

23

22

21321321

321

Therefore

,,v,,v

.,,vv Suppose

vvv

vvvvvvv

vvvvvvvvv

vv

Proofs of other properties are similar.

Consider two non-zero vectors. We can use their dot product 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

cos2222

222

vuvuuv

abbac

22

2

2

)()(

)()(

uvuv

uuvuuvvv

uvuuvv

uvuvuv

: thatfollowsIt .www that have we

productdot for developedjust that weproperties theFrom2

cos2222

vuvuuv

From the previous slide

Substituting from above

vu

vu

vuvu

vuvu

cos

cos

cos22

cos222222

vuvuuvuv

Simplifying

You have just witnessed the proof of the following theorem:

.vu

vucos then v and u

vectorsnonzero obetween tw angle theis If

Example 1

Find the angle between the vectors:

.14,v and 5,3

u

Solution

106.9or 87.1

......2911.0cos

1734

7

116259

512

1,45,3

1,45,3cos

vu

vu

Example 2

.6-3,-3,z and 2-1,3,-w

:if z and w ctorsbetween ve angle theFind

Solution

2

03699419

1239

6,3,32,1,3

6,3,32,1,3cos

vu

vu

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

Think before you click.Congratulate yourself if you chose True!

002

cos and 02

cos

vuvu

vu

vu

vu

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

This is true. Since the two vectors are perpendicular,

the angle between them will be .

2or 90

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

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

This is False. For example, consider the vectors:

.2,1-v and 1,4

u

vu

vucos

Finish this on your own then click for the answer.

x

y

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

This is False. For example, consider the vectors:

.2,1-v and 1,4

u

4.139or 43.2

...7592.085

7

517

18

1,21,4

1,21,4

vu

vucos

When the cosine is negative the angle between the two vectors is obtuse.

An Application of the Dot Product: Projection

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

Discussion: the force being exerted by the tractor can be interpreted as a vector with direction of 25 degrees. Our job is to find the magnitude.

25

v

The Tractor Problem, Slide 2: we are going to look at the force vector as the sum of its vertical component vector and 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, don’t round off until the end of the problem. In this case, we left the value of the cosine in the calculator and did not round off until the end.

Conclusion: the tractor has to exert a force of 275.8 lbs before the box will move.

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

u

v

1w

2w

x

y

.w toorthogonal is and

v toparallel is w that so

wu sketch, In the

12

1

21

w

w

.wcomponent the

by done being is work the

of all then v ofdirection

in theorigin at the

object an moving is uby

drepresente force a If

1

.v toorthogonal u

ofcomponent vector thecalled is

.proj as denoted is and v

onto u of projection thecalled is

2

v1

1

w

uw

w

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 is normal to the line or plane.

u

v

1w

2w

x

y

The following theorem will prove very useful in the remainder of this course.

vv

vuuprojv

2

:follows as calculated

becan v onto u of projection the

then vectors,nonzero are v and u If

.v vector of multiple a is v

onto u of projection thesoscalar a is

sparenthesi theinsidequantity The

.v toorthogonal u ofcomponent

vector thefind and uproj find 6,3 is v and 5,9 is u if :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

9,58.3,6.7

w

that notice sketch, theFrom

2

2

21

w

w

uw

(5,9)

(6,3)

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

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

50

100 lb

12 ft

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

pounds-foot 35.771)12)(50cos100( FDW

pounds.-foot 35.771)12(50cos10012,050sin100,100cos50

thatalso Note .12,0 isvector direction theand

50sin100,100cos50 is form coordinatein vector force that thenote :BSolution

Two Ways to Calculate Work

v

vprojv

u W:B Method

u W:A Method

:B Methodby or A Methodby calculated becan v vector along

n applicatio ofpoint its moving u forceconstant aby doneWork W

Example: Find the work done by a force of 20 lb acting in the direction N50W while moving an object 4 ft due west.

lbsft 28.61W

)4)(40)(cos20(proj W:SolutionA v

ftlbvu

lbsft 28.61)0(140sin20)4(140cos20

0,4 and 140sin20,140cos20u :SolutionB

vuW

v

u

v

50

4v and 20

u

axes.-z and y-, x-,positive with themakes v that ][0, interval in the

and , , angles theare v vector nonzero a of anglesdirection The

v.v vector of cosinesdirection thecalled are

,cos and ,cos ,cos angles,direction theseof cosines The

We can calculate the direction cosines by using the unit vectors along each positive axis 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

For comments on this presentation you may email the author Professor Richard Gill rgill@tcc.edu or the publisher of the VML Dr. Julia Arnold atjarnold@tcc.edu