where i points the right and j points up. In other words, i = 0
,1 and j = 1,
0 .
In this notation, the vector u = ,
3 4
− would be written as u = i
3 − 4 j since both
indicate a displacement of 3 units to the right, and 4 units down.
While it can be convenient to think of the vector u = x, y as an arrow from the origin to the point ( x, y), be sure to remember that most vectors can be situated anywhere in the
plane, and simply indicate a displacement (change in position) rather than a position.
Section 8.4 Vectors 495
It is common to need to convert from a magnitude and angle to the component form of
the vector and vice versa. Happily, this process is identical to converting from polar
coordinates to Cartesian coordinates, or from the polar form of complex numbers to the
a+bi form.
Example 5
Find the component form of a vector with length 7 at an angle of 135 degrees.
Using the conversion formulas x = r cos(θ) and y = r sin(θ), we can find the components
7 2
x = 7co 135
s(
)
° = −
2
7 2
y = 7sin 135
(
)
° =
2
This vector can be written in component form as
7 2 7 2
−
,
.
2
2
Example 6
Find the magnitude and angle θ representing the vector u = ,
3 2
− .
First we can find the magnitude by remembering the relationship between x, y and r: 2
r = 32 + (− )
2 2 = 13
r = 13
Second we can find the angle. Using the tangent,
− 2
tan(θ ) =
3
2
=
−
tan 1
θ
− ≈ −
69
.
33 ° , or written as a coterminal positive angle, 326.31°, because
3
we know our point lies in the 4th quadrant.
Try it Now
3. Using vector v from Try it Now #1, the vector that travels from the origin to the
point (3, 5), find the components, magnitude and angle θ that represent this vector.
496 Chapter 8
In addition to representing distance movements, vectors are commonly used in physics
and engineering to represent any quantity that has both direction and magnitude,
including velocities and forces.
Example 7
An object is launched with initial velocity 200 meters per second at an angle of 30
degrees. Find the initial horizontal and vertical velocities.
By viewing the initial velocity as a vector, we can resolve the vector into horizontal and
vertical components.
3
x = 200co 30
s(
)
° = 200⋅
≈
205
.
173
m/sec
2
200 m/s
100 m/s
1
30°
y = 200sin 30
(
)
° = 200 ⋅ = 100 m/sec
173 m/s
2
This tells us that, absent wind resistance, the object will travel horizontally at about 173
meters each second. Gravity will cause the vertical velocity to change over time – we’ll
leave a discussion of that to physics or calculus classes.
Adding and Scaling Vectors in Component Form
To add vectors in component form, we can simply add the corresponding components.
To scale a vector by a constant, we scale each component by that constant.
Combining Vectors in Component Form
To add, subtract, or scale vectors in component form
If u = u , u , v = v , v , and c is any constant, then 1
2
1
2
u + v = u + v , u + v
1
1
2
2
u − v = u − v , u − v
1
1
2
2
u
c = cu , cu
1
2
Example 8
Given u = ,
3 2
− and v = − ,14 , find a new vector w = u
3 − v
2
Using the vectors given,
w = u
3 − v
2
= 3 ,
3 2
− − 2 − ,14
Scale each vector
= ,
9 6
− − − 8
,
2
Subtract corresponding components
=
,
11 14
−
Section 8.4 Vectors 497
By representing vectors in component form, we can find the resulting displacement
vector after a multitude of movements without needing to draw a lot of complicated non-
right triangles. For a simple example, we revisit the problem from the opening of the
section. The general procedure we will follow is:
1) Convert vectors to component form
2) Add the components of the vectors
3) Convert back to length and direction if needed to suit the context of the question
Example 9
A woman leaves home, walks 3 miles north, then 2 miles southeast. How far is she
from home, and what direction would she need to walk to return home? How far has
she walked by the time she gets home?
Let’s begin by understanding the question in a little more depth.
When we use vectors to describe a traveling direction, we often
N
position things so north points in the upward direction, east
NW
NE
points to the right, and so on, as pictured here.
W
E
Consequently, travelling NW, SW, NE or SE, means we are
SW
SE
travelling through the quadrant bordered by the given directions
S
at a 45 degree angle.
With this in mind, we begin by converting each vector to components.
A walk 3 miles north would, in components, be 3
,
0 .
A walk of 2 miles southeast would be at an angle of 45° South of East, or measuring
from standard position the angle would be 315°.
Converting to components, we choose to use the standard position angle so that we do
not have to worry about whether the signs are negative or positive; they will work out
automatically.
2
− 2
2co 315
s(
),
° 2sin 315
(
)
° = 2 ⋅
, 2 ⋅
≈ .
1
,
414 − 414
.
1
2
2
Adding these vectors gives the sum of the movements in component
2
form
3
3
,
0 +
,
414
.
1
− 414
.
1
=
586
.
1,
414
.
1
To find how far she is from home and the direction she would need to walk to return
home, we could find the magnitude and angle of this vector.
Length = 1 414
.
2 + 586
.
1
2 = 125
.
2
498 Chapter 8
To find the angle, we can use the tangent
586
.
1
tan(θ ) =
414
.
1
1 586
.
=
−
tan 1
θ
=
.
48 273° north of east
1 414
.
Of course, this is the angle from her starting point to her ending point. To return home,
she would need to head the opposite direction, which we could either describe as
180°+48.273° = 228.273° measured in standard position, or as 48.273° south of west (or
41.727° west of south).
She has walked a total distance of 3 + 2 + 2.125 = 7.125 miles.
Keep in mind that total distance travelled is not the same as the length of the resulting
displacement vector or the “return” vector.
Try it Now
4. In a scavenger hunt, directions are given to find a buried treasure. From a starting
point at a flag pole you must walk 30 feet east, turn 30 degrees to the north and
travel 50 feet, and then turn due south and travel 75 feet. Sketch a picture of these
vectors, find their components, and calculate how far and in what direction you
must travel to go directly to the treasure from the flag pole without following the
map.
While using vectors is not much faster than using law of cosines with only two
movements, when combining three or more movements, forces, or other vector
quantities, using vectors quickly becomes much more efficient than trying to use
triangles.
Example 10
Three forces are acting on an object as shown below, each measured in Newtons (N).
What force must be exerted to keep the object in equilibrium (where the sum of the
forces is zero)?
6 N
7 N
30°
300°
4 N
We start by resolving each vector into components.
Section 8.4 Vectors 499
The first vector with magnitude 6 Newtons at an angle of 30 degrees will have
components
3
1
6co 30
s(
),
° 6sin 30
(
)
° = 6 ⋅
,6 ⋅
= 3 3,3
2
2
The second vector is only in the horizontal direction, so can be written as − ,
7 0 .
The third vector with magnitude 4 Newtons at an angle of 300 degrees will have
components
1
− 3
4co 300
s(
),
° 4sin 300
(
)
° = 4 ⋅ ,4 ⋅
= ,
2 − 2 3
2
2
To keep the object in equilibrium, we need to find a force vector x, y so the sum of
the four vectors is the zero vector, 0
,
0 .
3 3, 3 + 7
− , 0 + 2, − 2 3 + x, y = 0, 0
Add component-wise
3 3 − 7 + 2, 3+ 0 − 2 3 + x, y = 0, 0
Simplify
3 3 − 5, 3− 2 3 + x, y = 0, 0
Solve
x, y = 0, 0 − 3 3 − 5, 3− 2 3
x, y = 3
− 3 + 5, − 3+ 2 3 ≈ 0.196
−
, 0.464
This vector gives in components the force that would need to be applied to keep the
object in equilibrium. If desired, we could find the magnitude of this force and
direction it would need to be applied in.
Magnitude = ( 0
− .
)
196 2 + .
0 4642 = 0 504
.
N
Angle:
464
.
0
tan(θ ) =
− 196
.
0
0 464
.
=
−
tan 1
θ
= −67 089
.
° .
− 196
.
0
This is in the wrong quadrant, so we adjust by finding the next angle with the same
tangent value by adding a full period of tangent:
θ = − .
67
°
089 +
°
180 =
°
911
.
112
To keep the object in equilibrium, a force of 0.504 Newtons would need to be applied at
an angle of 112.911°.
500 Chapter 8
Important Topics of This Section
Vectors, magnitude (length) & direction
Addition of vectors
Scaling of vectors
Components of vectors
Vectors as velocity
Vectors as forces
Adding & Scaling vectors in component form
Total distance travelled vs. total displacement
Try it Now Answers
v
2
− v
1
2.
3.
−1 5
v = ,
3 5
magnitude = 34
θ = tan =
°
04
.
59
3
4.
50 ft
30 ft
75 ft
v= v= ° ° v= −
1
,
30 0
2
50co 30
s(
),50sin(30 )
3
,
0 75
v =
f
30 + 50co 30
s(
),
° 50sin 30
(
)
° − 75 = 73
,
301
.
− 50
Magnitude = 88.73 feet at an angle of 34.3° south of east.
Section 8.4 Vectors 501
Section 8.4 Exercises
Write the vector shown in component form.
1.
2.
Given the vectors shown, sketch u+ v, u− v, and 2 u.
3.
4.
Write each vector below as a combination of the vectors u and v from question #3.
5.
6.
From the given magnitude and direction in standard position, write the vector in
component form.
7. Magnitude: 6, Direction: 45°
8. Magnitude: 10, Direction: 120°
9. Magnitude: 8, Direction: 220°
10. Magnitude: 7, Direction: 305°
Find the magnitude and direction of the vector.
11. ,
0 4
12. − ,
3 0
13. ,
6 5
14. ,
3 7
15. − ,
2 1
16. − ,
10 13
17. ,
2 − 5
18. ,
8 − 4
19. − ,
4 − 6
20. − ,19
Using the vectors given, compute u+ v, u− v, and 2 u−3 v.
21. u = ,
2 − 3 , v = ,15
22. u = − ,
3 4 , v = − ,
2 1
502 Chapter 8
23. A woman leaves home and walks 3 miles west, then 2 miles southwest. How far
from home is she, and in what direction must she walk to head directly home?
24. A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far
from the marina is the boat, and in what direction must it sail to head directly back to
the marina?
25. A person starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles
south, 4 miles southwest, and 2 miles east. How far have they walked? If they
walked straight home, how far would they have to walk?
26. A person starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles
south, 5 miles southwest, and 3 miles east. How far have they walked? If they
walked straight home, how far would they have to walk?
27. Three forces act on an object: F = − −
F =
F =
− . Find the net
1
,
8 5 , 2
,
0 1 , 3
,
4 7
force acting on the object.
28. Three forces act on an object: F =
F =
F =
− . Find the net force
1
,
2 5 , 2
,
8 3 , 3
,
0 7
acting on the object.
29. A person starts walking from home and walks 3 miles at 20° north of west, then 5
miles at 10° west of south, then 4 miles at 15° north of east. If they walked straight
home, how far would they have to walk, and in what direction?
30. A person starts walking from home and walks 6 miles at 40° north of east, then 2
miles at 15° east of south, then 5 miles at 30° south of west. If they walked straight
home, how far would they have to walk, and in what direction?
31. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing
from the southwest at 80 km/hr. How many degrees off course will the plane end up
flying, and what is the plane’s speed relative to the ground?
32. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing
from the northwest at 50 km/hr. How many degrees off course will the plane end up
flying, and what is the plane’s speed relative to the ground?
33. An airplane needs to head due north, but there is a wind blowing from the southwest
at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due
north, the pilot will need to fly the plane how many degrees west of north?
Section 8.4 Vectors 503
34. An airplane needs to head due north, but there is a wind blowing from the northwest
at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due
north, the pilot will need to fly the plane how many degrees west of north?
35. As part of a video game, the point (5, 7) is rotated counterclockwise about the origin
through an angle of 35 degrees. Find the new coordinates of this point.
36. As part of a video game, the point (7, 3) is rotated counterclockwise about the origin
through an angle of 40 degrees. Find the new coordinates of this point.
37. Two children are throwing a ball back and forth straight across the back seat of a car.
The ball is being thrown 10 mph relative to the car, and the car is travelling 25 mph
down the road. If one child doesn't catch the ball and it flies out the window, in what
direction does the ball fly (ignoring wind resistance)?
38. Two children are throwing a ball back and forth straight across the back seat of a car.
The ball is being thrown 8 mph relative to the car, and the car is travelling 45 mph
down the road. If one child doesn't catch the ball and it flies out the window, in what
direction does the ball fly (ignoring wind resistance)?
504 Chapter 8
Section 8.5 Parametric Equations
Many shapes, even ones as simple as circles, cannot be represented as an equation where