CHAPTER 13Vector ApplicationsContents:A. Problems involving vector operationsB. Lines in 2-D and 3-DC. The angle between two linesD. Constant velocity problemsE. The shortest distance from a line to a pointF. Intersecting linesG. Relationships between lines

A. PROBLEMS INVOLVING VECTOR OPERATIONSThe sum of vectors is called the resultant vector.

Example 1In still water, Jacques can swim at 1.5m/s. Jacques is at point A on the edge of a canal, and considers point B directly opposite. A current is flowing from the left at a constant speed of 0.5m/s.

a. If Jacques dives in straight towards B, and swims without allowing for the current, what will his actual speed and direction be?

b. Jacques wants to swim directly across the canal to point B.

i. At what angle should Jacques aim to swim in order that the current will correct his direction?

ii What will Jacques’ actual speed be?

Suppose c is the current’s velocity vector, s is the velocity

vector Jacques would have if the water was still, and

f = c + s is Jacques’ resultant velocity vector.

a. Jacques aims directly across the river, but the current takes him downstream to the right.

Applying the Pythagorean Theorem and a little trigonometry.

Jacques has an actual speed of approximately 1.58m/s and his direction of motion is approximately 18.4o to the right of his intended line.

b. Jacques needs to aim to the left of B so the current will correct his direction.

Exercise 13A

1. An athlete can normally run with constant speed 6 m/s. Using a vector diagram to illustrate each situation, find the athlete’s speed if:

a. he is assisted by a wind of 1 m/s from directly behind him

b. he runs into a head wind of 1 m/s.

a.

b.

6 m/s 1 m/s

7 m/s

6 m/s

1 m/s

5 m/s

5 An airplane needs to fly due east from one city to another at a speed of 400 km/h. However, a 50 km/h wind blows constantly from the north-east.

a. How does the wind affect the speed of the airplane?

b. In what direction must the airplane head to compensate for the wind?

a.

Using the Law of Cosine, compute the value for x

The calm airspeed of the plane 437 km/h which is slowed to 400 km/h by the northeasterly wind.

400 km/h

50 km/h

x135o

437

)135cos()50)(400(250400

cos2

22

222

x

x

Cabbac

b.

Determine q: use the Law of Sine.

north of due east.

400 km/h

50 km/h

x135oq

65.4

437

)135sin(50sin

437

)135sin(

50

sin

1

LINES IN 2-D AND 3-D

In both 2-D and 3-D geometry we can determine the equation of a line using its direction and any fixed point on the line.

Suppose a line passes through a fixed point A with position

vector a, and that the line is parallel to the vector b.

Vector Equation of a line

A line passes through the point A(1, 5) and has direction

vector . Describe the line using:

a. A vector equation,

b. Parametric equations,

c. and a Cartesian equation.

2

3

Parametric equation

tybottom

txtop

ty

x

25:

31:

2

3

5

1

Cartesian equation:

Solve both equation for t. Set them equal to one another and simplify.

1332

153222

5

3

1

2

5

3

1

yx

yx

yx

yt

xt

LINES IN 3-D

Find a vector equation and the corresponding parametric

equations of the line through (1,-2, 3) in the direction

4i + 5j – 6k.

NON-UNIQUENESS OF THE VECTOR EQUATION OF A LINE

Going from (5, 4) to (7, 3), go2 to the right and 1 down.

Going from (7, 3) to (5,4), go2 to the left and 1 up.

Find parametric equations of the line through A(2,-1, 4) and

B(-1, 0, 2).

Find the directional vector AB or BA

2

1

3

42

10

21

AB

Rttztytx

t

z

y

x

,24,1,32

2

1

3

4

1

2

Using point A Using point B

Rsszsysx

s

z

y

x

,22,,31

2

1

3

2

0

1

2

1

3

24

01

12

BA

Exercise 13B

1644124

4

43.1

44,3.1

,4

1

4

3.1

yxyx

yxaiii

tytxaii

Rtty

xai

292542255

5

2

2

5.1

52,25.1

,5

2

2

5.1

yxyx

yxbiii

tytxbii

Rtty

xbi

42373427

73

6.1

7,36.1

,7

3

0

6.1

yxyx

yxciii

tytxcii

Rtty

xci

2122221

112

1.1

11,21.1

,1

2

11

1.1

yxyx

yx

diii

tytxdii

Rtty

xdi

yesthereforetsame

tt

tt

Rtta

1312

123

,3

1

1

2

2

33

Solve for t, set equal and simplify

5128

2)4(21

221

12

1

12

1.3

k

k

yx

yx

tytx

b

Sub in (k, 4)

THE ANGLE BETWEEN TWO LINES

Exercise 13C

Find the angle between the lines

L1: x = -4 + 12t, y = 3 + 5t

L2: x = 3s, y = -6 – 4s

7.7565

16cos

65

16

25169

|)20(36|cos

4

3,

5

12

1

21

bb

6.2830772

154cos

30772

154

98314

|)35()128(9|cos

5

8

3

,

7

16

3

1

21

bb

7

48

0748

0731603

03

0

7

16

3

.

x

x

x

x

b

Find the measure of the acute angle between the lines

2x + y = 5 and 3x – 2y = 8.

3.6065

4cos

65

4

135

|)6(2|cos

3

2,

2

1

3

2

2

3823

2

1

1

252

1

21

bb

vectordirectionslopeyx

vectordirectionslopeyx

Find the measure of the angle between the lines:

y = 2 – x and x – 2y = 7

Answer: 71.6o

CONSTANT VELOCITY PROBLEMS

a

a. The initial position of the object occurs at t = 0

)9,1(9,1

9

1

4

30

9

1,0

4

3

9

1

oryxtherefore

y

x

y

xtat

ty

x

3

10

4

33

9

1,3

1

7

4

32

9

1,2

5

4

4

31

9

1,1

9

1

4

30

9

1,0

4

3

9

1

y

x

y

xtat

y

x

y

xtat

y

x

y

xtat

y

x

y

xtat

ty

x

4

3

9

1t

y

x

velocity vector

43

Is the velocity vector

sm

bspeed

/5

)4(3

4

3

||

22

To find the position of the object at time t, we need to find the parametric equationfor x and y. The coordinates of (x, y) will give us our position.

Step 1: set up the vector equation of the object.Step 2: find the parametric equation from the vector equationStep 3: write the coordinate as an ordered pair of parametric equations.

b. To determine the speed of the object, we need to find the magnitude or length of the vector.

c. Plug in the value for t into our ordered pair.

d. Due east of (0, 0) means when y = 0 (aka x-intercept)

22. EXERCISE 13d

1a. The initial position occurs at t = 0.

x(0) = 1 + 2(0) = 1

y(0) = 2 – 5(0) = 2

the initial position (1, 2)

1b. x = 1 + 2t y = 2 - 5t (x, y)

0 1 2 (1, 2)

1 3 -3 (3, -3)

2 5 -8 (5, -8)

3 7 -13 (7, -13)

0 1 2 3 4 5 6 7 8

-14

-12

-10

-8

-6

-4

-2

0

2

4

Series1

29

52.1

5

2

5

2

2

1.1

22

d

velocity

ty

xc

THE SHORTEST DISTANCE FROMA LINE TO A POINT

Perpendicular/orthogonal

EXERCISE 13E

)0,18(

18362:int

)12,0(

12363:int

3632.

B

xxx

A

yyy

yxa

)123

2,(

123

2

3632.

xxR

xy

yxb

123

24

0123

24

12

18

012

180.

x

x

x

xPR

ABc

13

84

313108

236

3

236

13

108

21626

081447218

03

23612)4(18

012

18

3

2364

0.

xy

x

x

xx

xx

xx

ABPRd

25

21

2

2

12

2

131

2

121

2

1

01694

1

3

2

1

23

2

1

3

2

1

23

2

32

131

121

2

31

21

.

R

t

ttt

t

t

t

PR

t

t

t

t

t

t

PR

t

t

t

Ra

2

6

1

1

2

2

1

21

21

1

12

1

22

13

2

12

1

23

2

t

t

t

PR

INTERSECTING LINES

EXERCISE 13F

)0,5()6,8()3,2(1

21

2

0

2)1(22

1

252

,2

52

2

1

1

5

0

1

2

2

0.

CBy

xA

rt

r

rr

tforrsub

tr

tr

tra

BCAB

BC

AC

AB

53)06()58(

3)03()52(

53)63()82(

22

22

22

RELATIONSHIPS BETWEEN LINES

LINE CLASSIFICATION IN 3 DIMENSION