Chapter 13

Report
CHAPTER 13
Vector Applications
Contents:
A. Problems involving vector operations
B. Lines in 2-D and 3-D
C. The angle between two lines
D. Constant velocity problems
E. The shortest distance from a line to a point
F. Intersecting lines
G. Relationships between lines
A. PROBLEMS INVOLVING VECTOR
OPERATIONS
The sum of vectors is called the resultant vector.
Example 1
In 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.
6 m/s
1 m/s
7 m/s
6 m/s
b.
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.
x
135o
50 km/h
400 km/h
Using the Law of Cosine, compute the value for x
c 2  a 2  b 2  2ab cos C
x  400 2  50 2  2(400)(50) cos(135)
x  437
The calm airspeed of the plane 437 km/h which is slowed
to 400 km/h by the northeasterly wind.
b.
x
q
135o
50 km/h
400 km/h
Determine q: use the Law of Sine.
sin q sin(135)

50
437
  sin(135)  
q  sin 1  50
 
  437  
q  4.65
north of due east.
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
 3
vector   . Describe the line using:
 2
a. A vector equation,
b. Parametric equations,
c. and a Cartesian equation.
Parametric equation
 x  1  3
      t  
 y   5  2
top : x  1  3t
bottom : y  5  2t
Cartesian equation:
Solve both equation for t. Set them equal to one another
and simplify.
x 1
y 5
t
3
2
x 1 y  5

3
2
2 x  2  3 y  15
t
2 x  3 y  13
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), go
2 to the right and 1 down.
Going from (7, 3) to (5,4), go
2 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
 1 2    3

  
AB   0   1   1 
 2  4    2

  
Using point A
 2   1  3 

  
BA    1  0     1
 42   2 

  
Using point B
 x   2    3
     
  y     1  t  1 
 z   4    2
     
 x    1  3 
     
  y    0   s  1
z  2   2 
     
x  2  3t , y  1  t , z  4  2t , t  R
x  1  3s, y   s, z  2  2 s, s  R
Exercise 13B
 x   3  1
1ai.       t  , t  R
 y    4  4
1aii. x  3  t , y  4  4t
y4
1aiii. x  3 
4
4 x  12  y  4
4 x  y  16
 x   5   2
1bi.       t  , t  R
 y   2  5 
1bii. x  5  2t , y  2  5t
x 5 y 2
1biii.

2
5
5 x  25  2 y  4
5 x  2 y  29
 x    6  3
1ci.       t  , t  R
 y  0  7
1cii. x  6  3t , y  7t
x6 y
1ciii.

3
7
7 x  42  3 y
7 x  3 y  42
 x    1   2 
1di.       t  , t  R
 y   11   1 
1dii. x  1  2t , y  11  t
x 1
1diii.
 y  11
2
x  1  2 y  22
x  2 y  21
 3   2  1 
3a       t  , t  R
  2   1    3
3 2t
 2  1  3t
t 1
t 1
same t therefore yes
Solve for t, set equal and simplify
x 1
3b.
t
2
x 1
 y 1
2
x  1  2 y  2
k  1  2(4)  2
k  8  2  1  5
y 1  t
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
12 
 3 
b1   ,
b2   
5
  4
| 36  (20) | 16
cos q 

65
169 25
1  16 
q  cos    75.7
 65 
 3 


b1    16 ,
 7 


 3 
 
b2   8 
  5
 
| 9  (128)  (35) |
154
cos q 

314 98
30772
 154 
q  cos 
  28.6
 30772 
1
b.
 3   0 

  
  16     3   0
 7   x 

  
3  0   16    3  7 x  0
48  7 x  0
48
x
7
Find the measure of the acute angle between the lines
2x + y = 5 and 3x – 2y = 8.
2x  y  5
2
slope 
1
3x  2 y  8
3
slope 
2
 1 
 2
b1   ,
b2   
  2
 3
| 2  (6) |
4
cos q 

5 13
65
 4 
q  cos 1 
  60.3
 65 
 1 
direction vector   
  2
 2
direction vector   
 3
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
 x  1  3 
      t  
 y  9   4
 x  1  3 
      0 
at t  0,
 y  9   4
therefore x  1, y  9 or (1, 9)
 x  1
    
 y  9
 x  1  3 
      t  
 y  9   4
 x  1  3 
      0 
at t  0,
 y  9   4
at t  1,
at t  2,
at t  3,
 x  1  3 
      1 
 y  9   4
 x  1  3 
      2 
 y  9   4
 x  1  3 
      3 
 y  9   4
 x  1
    
 y  9
 x   4
    
 y  5
 x  7
    
 y 1
 x   10 
    
 y    3
 x  1  3 
      t  
 y  9   4
velocity vector
 3 
 
  4
Is the velocity vector
speed | b |
 3 
  
  4
 32  (4) 2
 5 m/s
To find the position of the object at time t, we need to find the parametric equation
for 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 equation
Step 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)
4
2
0
0
2
4
6
8
-2
-4
Series1
-6
-8
-10
-12
-14
 x  1  2 
1c .      t  
 y   2   5
 2 
velocity   
  5
1d .
2
2
29
2


 5
THE SHORTEST DISTANCE FROM
A LINE TO A POINT
Perpendicular/orthogonal
EXERCISE 13E
a. 2 x  3 y  36
y  int : 3 y  36
y  12
A  (0, 12)
x  int : 2 x  36
B  (18, 0)
x  18
b. 2 x  3 y  36
2
y   x  12
3
2
R  ( x,  x  12)
3
 0  18    18 
  

c. AB  
12  0   12 
x4

  x4 
 2

PR   2
  x  12  0    x  12 
 3
  3

d .PR  AB  0
 x  4    18 
 36  2 x   
  0


  12 
 3 
 36  2 x 
 18( x  4)  12
0
 3 
 18 x  72  144  8 x  0
 26 x  216
108
x
13
 108 
36  2

36  2 x
13  84

y


3
3
13
 1  2t 


a. R    1  3t 
 2t 


 1  2t  1   2t 

 

PR    1  3t  1   3t  2 
 2  t  3   t 1 

 

 2   2t   2 
  
  
PR   3    3t  2    3   4t  9t  6  t  1  0
 1   t 1   1 
  
  
1
2

1 
 1  2  
2   2 



 1  
R    1  3     1 2 
 2  


5
2


 21 
  

2 

t
 1 
 2  
2

   1 

2
t


2

  1  
 1 
PR   3t  2    3   2     1 2     1
 t  1    2     1 2  2   1

  1






1
 2

  

6

2
INTERSECTING LINES
EXERCISE 13F
 0  2 0  1 
a.    r       t  
 2   1   5    1
2r  t
2 r  5t
sub 2r for t ,
2  r  5  2r
r 1
t  2r  2(1)  2
 x  0
 2
A        1   (2, 3)
 y   2
1
B  (8, 6)
C  (5, 0)
AB  (2  8)  (3  6)  3 5
2
2
AC  (2  5) 2  (3  0) 2  3
BC  (8  5) 2  (6  0) 2  3 5
AB  BC
RELATIONSHIPS BETWEEN LINES
LINE CLASSIFICATION IN 3 DIMENSION

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