6.3
Vectors in the Plane
Day 1 2015
A vector is a quantity with both a magnitude and a
direction.
A ball flies through the air at a certain speed and in a
particular direction. The speed and direction are the
velocity of the ball. The velocity is a vector quantity
since it has both a magnitude and a direction.
Vectors are used to represent velocity, force, tension,
and many other quantities.
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2
A quantity with magnitude and direction is represented by
a directed line segment PQ with initial point P and
terminal point Q.
P
Q
The vector v = PQ is the set of all directed line segments
of length ||PQ|| which are parallel to PQ.
Two vectors, u and v, are equal if the line segments
representing them are parallel and have the same length
or magnitude.
v
u
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3
Vector Representation by Directed Line Segments
Let u be represented by the directed line segment from
P = (0,0) to Q = (3,2), and let v be represented by the directed
line segment from R = (1,2) to S = (4,4). Show that u = v.
S
4
Using the distance formula, show
that u and v have the same length.
Show that their slopes are equal.
v
3
Q
R
2
u 
3  02  2  02
 13
v 
4  12  4  22
 13
u
1
Slopes of u and v are both
P
1
2
3
4
2
3
Scalar multiplication is the product of a scalar, or
real number, times a vector.
For example, the scalar 3 times v results in the vector 3v,
three times as long and in the same direction as v.
v
3v
The product of -
and v gives a vector half as long
as and in the opposite direction to v.
v
-
v
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5
Vector Addition
To add vectors u and v:
v
u
1. Place the initial point of u at the terminal point of v.
2. Draw the vector with the same initial point as v and
the same terminal point as u.
v
u
u
v
v+u
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6
Vector Subtraction as adding the opposite.
v
To subtract vectors u and v: u  v
Add the opposite of v to u: u +( v)
u
1. Place the initial point of  v at the terminal point of u.
2. Draw the vector u  v from the initial point of u to the
terminal point of v.
u
u
uv
-v
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-v
7
Vector Subtraction
v
u
To subtract vectors u and v:
1. Place the initial point of v at the initial point of u.
2. Draw the vector u  v from the terminal point of v
to the terminal point of u.
v
v
uv
u
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u
8
A vector with initial point (0, 0) is in standard position and is
represented uniquely by its terminal point (v1, v2). This is the
component form of a vector v, written as v1 , v2 .
y
(v1, v2)
x
If v is a vector with initial point P = (p1 , p2) and terminal point
Q = (q1 , q2), then
1. The component form of v is
v = q1  p1, q2  p2
2. The magnitude (or length) of v is
y
P (p1, p2) Q (q1, q2)
x
||v|| = ( q 1  p 1 ) 2  ( q 2  p 2 ) 2
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Remember that to write a vector in
component form: v = q1  p1, q2  p2
Use terminal point – initial point.
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10
Example:
Find the component form and magnitude of the vector v with
initial point P = (3, 2) and terminal point Q = (1, 1).
Component form:
v  4,3
The magnitude of v is
||v|| =
(  4 ) 2  ( 3 ) 2 = 2 5 = 5.
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11
You Try: Find the component form and length of the vector v
that has initial point (4,-7) and terminal point (-1,5)
Let P = (4, -7) = (p1, p2) and Q = (-1, 5) = (q1, q2).
Then, the components of v = are given by
v1 ,v2
Thus, v = 5, 12
The length of v is
v  (5)  12  169  13
2
2
Two vectors u = u1, u2 and v = v1, v2 are equal if and only if
u1 = v1 and u2 = v2 .
Example: If u = PQ, v = RS, and w = TU with P = (1, 2),
Q = (4, 3), R = (1, 1), S = (3, 2), T = (-1, -2), and U = (1, -1),
determine which of u, v, and w are equal. (solution follows.)
Calculate the component form for each vector:
u = 4  1, 3  2 = 3, 1
v = 3  1, 2  1 = 2, 1
w = 1  (-1), 1  (-2) = 2, 1
Therefore v = w but v =/ u and w =/ u.
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13
Operations on Vectors in the Coordinate Plane
Let u = x1 , y1 , v = x2 , y2 , and let c be a scalar.
1. Scalar multiplication cu = cx1 , cy1
2. Addition
3. Subtraction
u + v = x1  x2 , y1  y2
u  v = x1  x2 , y1  y2
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14
Examples: Given vectors u = 4, 2 and v = 2,5
Find -2u, u+v, and u-v
Try these on
y your own.
-2u = 2 4, 2  8, 4
(4, 2)
u
(-8, -4)
u + v = 4, 2 + 2,5 = 6, 7
y
(2, 5)
v
-2u
u  v = 4, 2  2,5 = 2, 3
y
(2, 5)
(6, 7)
v
(4, 2)
u
x
x
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(4, 2)
u
uv
(2, -3)
x
15
Vector Operations
Ex. Let v =  2,5 and w = 3,4 . Find the following vectors.
a. 2v
b. w – v
w  v  5, 1
2v   4,10
10
4
8
2v
6
3
w
4
2
-v
v
2
-4
-2
1
2
-2
-1
w-v
1
2
3
4
5
A unit vector is a vector whose magnitude = 1. In many vector
applications it is useful to find a unit vector that has the same
direction as a given vector v. To do this, divide v by its length to
obtain:
v  1 
  v
u = unit vector 
v  v 
Ex: Find a unit vector in the direction of v =
v

v
2,5 
2  5
2
2
1

2,5

29

2,5
2 5
,
29 29
You try:
a) Find a unit vector in the direction of
3, 4
3 4
,
5 5
b) Find the magnitude of the unit vector you just found.
1
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18
Homework 6.3 Day 1
• Pg.417 1-35 odd
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19
Digital Lesson
Vectors in the Plane
Day 2 2015
Standard unit vectors
The unit vectors i  1, 0 and j  0,1 are called
the standard unit vectors. i represents one unit of
horizontal movement and j represents one unit of
vertical movement. Any vector can be represented
by what is called a linear combination of unit vectors.
Example: Vector v  2, 6
can be
represented as a linear combination of
unit vectors by rewriting it as v= 2i-6j.
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22
Writing a Linear Combination of Unit Vectors
Let u be the vector with initial point (2, -5) and terminal point
(-1, 3). Write u as a linear combination of the standard unit
vectors of i and j.
6
Solution
(-1, 3)
10
4
u  3,8
2
-2
-2
u 2
4
-4
(2, -5)
8
6
 3i  8 j
4
Graphically,
it looks like…
8j
2
-3i
-6
-4
-8
-2
2
-2
Writing a Linear Combination of Unit Vectors
Let u be the vector with initial point (1, -7) and terminal point
(-1, 2).Write u as a linear combination of the standard unit
vectors i and j.
Begin by writing the component form of the vector u.
u  2,9
u  2i  9 j
Example:
Find the vector v with the given magnitude
and the same direction as u.
v  3,
u  4, 4
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3 2 3 2
v
,
2
2
25
You try:
Find the vector v with the given magnitude
and the same direction as u.
v  10,
u  2i  3 j
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v
20 13
i  30 13
13
13
j
26
Vector Operations
Let u = -3i + 8j and let v = 2i - j. Find 2u - 3v.
2u - 3v = 2(-3i + 8j) - 3(2i - j)
= -6i + 16j - 6i + 3j
= -12i + 19 j
Finding Direction Angles
The direction angle  of a vector v is the angle formed by
the positive half of the x-axis and the ray along which v lies.
y
y
θ
x
v
θ
x
If v = x, y , then tan  =
If v = 3, 4 ,
then tan  =
v
y
.
x
and  = 53.13.
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y
(x, y)
v
x
y
x
29
• Find the magnitude and direction angle of
the vector:
u  3i  3 j
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u  3 2,   45
30
• Find the magnitude and direction angle of
the vector:
v  3i  4 j
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v  5,   306.87
31
• Find the direction angle of the vector:
v  3i  4 j
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v  5,
  126.87
32
• Find the direction angle of the vector:
v  3i  4 j
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v  5,
  233.13
33
Writing Vectors using Direction Angles
If u is a unit vector such that  is the angle (measured
counter-clockwise) from the positive x-axis to u, the terminal
point of u lies on the unit circle and you have:
u  x, y  cos ,sin   cos i  sin 
j
To indicate a length other than 1, multiply the unit
vector by a magnitude. v  v cos ,sin 
The vector v  5 cos30,sin 30 has a magnitude
of 5 in the direction of 30 degrees.
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34
• Find the component form of v given its
magnitude and the angle it makes with the
positive x-axis:
v  3 and v is in the direction of 3i+4j
v  3 cos53.13,sin 53.13
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35
Homework 6.3 Day 2
• Pg.418 37-61 odd
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36
• Find the direction angle and write the
component form using it:
v  4 and v is in the direction of -5i-2j
v  4 cos 201.8,sin 201.8
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37
Vectors in the Plane
Day 3 2015
(use class notes handout)
HWQ
Add vectors u and v.
u= 2i-j v = -i+j
Explain how you found the sum.
u+v = 1i+0j = I
Add two vectors by adding the horizontal components (x) ,
and the vertical (y) components.
The result is a new vector called the resultant of u and v.
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39
Resultant Vectors (sum of vectors)
• The sum of two or more vectors is called the
resultant of the vectors.
• Find the resultant of u and v.
u  3,
u  30
v  4,
v  60
resultant vector: r = 4.6, 4.96

E. Find the magnitude of the vertical and horizontal components of this distance vector.
1.A ship leaving port sails for 75 miles in a bearing N 55
Applications
A ship leaving port sails for 75 miles at a bearing N
55 E. Find the magnitude of the vertical and
horizontal components of this distance vector.
Distance Vector: 75 cos35,sin 35
75cos35  61.44 miles
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75sin 35  43.02 miles
41

1.Find
of vertical
the vector
represents
the velocity
of distance
a plane descending
at a speed of 100 mph. @ an angle o
E. Findthe
thecomponent
magnitudeform
of the
andthat
horizontal
components
of this
vector.
1.A ship leaving port sails for 75 miles in a bearing N 55
Applications
Find the component form of the vector that
represents the velocity of a plane
descending at a speed of 100 mph.
@ an angle 30 degrees below horizontal.
v  100 cos 210,sin 210
3 1
v  100 
,
2
2
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v  50 3, 50
v 
 50 3 
2
  50   100
2
42

E. Find the magnitude of the vertical and horizontal components of this distance vector.
1.A ship leaving port sails for 75 miles in a bearing N 55
Applications
An airplane is traveling at a speed of 500 mph. with an air
bearing of 330 degrees, at a fixed altitude, with negligible wind
velocity.
As the airplane reaches a certain point, it encounters a wind
blowing with a velocity of 70 mph. in the direction
N 45E.
Find :
Resultant Speed
a. The resultant speed.
v  522.5 mph
b. The direction of the airplane.
Direction of the Plane
  112.6
True Direction = 337.4
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43

E. Find the magnitude of the vertical and horizontal components of this distance vector.
1.A ship leaving port sails for 75 miles in a bearing N 55
Applications
A piling for a high-rise building is pushed by two bulldozers at
exactly the same time. One bulldozer exerts a force of 1550
pounds in a westerly direction. The other bulldozer pushes
the piling with a force of 3050 pounds in a northerly direction.
What is the magnitude of the resultant force upon the piling?
r 
 1550 
2
  3050   3421.25
2
What is the direction of the resulting force upon the piling?
 3050 
tan 
  116.94
 1550 
1
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44

E. Find the magnitude of the vertical and horizontal components of this distance vector.
1.A ship leaving port sails for 75 miles in a bearing N 55
You Try
The Shanghai World Finance Center building in Shanghai,
China, is 1508 feet tall. Suppose that a piling for building is
pushed by two bulldozers at exactly the same time. One
bulldozer exerts a force of 900 pounds in an easterly
direction. The other bulldozer pushes the piling with a force
of 2150 pounds in a northerly direction.
What is the magnitude of the resultant force upon the
piling?
2
2
r   900    2150   2330.77
What is the direction of the resulting force upon the piling?
 2150 
tan 
  67.29
 900 
1
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45
Homework 6.3 Day 3
• Pg.418 63-65 odd, 69-75odd ,81,82
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46