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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