Transcript Slide 1

Chapter 7: Vectors and the
Geometry of Space
Section 7.1
Vectors in the Plane
Written by Dr. Julia Arnold
Associate Professor of Mathematics
Tidewater Community College, Norfolk Campus, Norfolk, VA
With Assistance from a VCCS LearningWare Grant
In this first lesson on vectors, you will learn:
• Component Form of a Vector
• Vector Operations;
• Standard Unit Vectors;
• Applications of Vectors.
What is a vector?
Many quantities in geometry and physics can be characterized
by a single real number: area, volume, temperature, mass and
time. These are defined as scalar quantities.
Quantities such as force, velocity, and acceleration involve both
magnitude and direction and cannot be characterized by a single
real number.
To represent the above quantities we use a directed line
segment.
What is a directed line segment?
First let us look at a directed line segment:
Q
P
This line segment has a beginning, (the
dot) and an ending (the arrow point).
We call the beginning point the “initial point” . Here we have
called it P. The ending point (arrow point) is called the
“terminal point” and here we have called it Q.
The vector is the directed line segment and is denoted by
PQ
In some text books vectors will be denoted by bold type letters
such as u, v, or w.
However, we will denote vectors the same way you will denote
vectors by writing them with an arrow above the letter.
v  PQ
It doesn’t matter where a vector is positioned. All of the
following vectors are considered equivalent.
because they are pointing in the same direction and the
line segments have the same length.
How can we show that two vectors u and v are equivalent?
Suppose v is the vector with initial point (0,0) and terminal
point (6,4), and u is the vector with initial point (1,2) and
terminal point (7,6) .
Since a directed line
segment is made up of its
magnitude (or length) and its
direction, we will need to
show that both vectors have
the same magnitude and are
going in the same direction.
Looks verify but are not
proof.
y
u
6
v
4
2
2
4
6
x
8
How can we show that two vectors u and v are equivalent?
The symbol we use to denote the magnitude of a vector
is what looks like double absolute value bars.
Thus v
represents the magnitude or length of the vector v .
v 
6  02  4  02
 36  1 6  52  2 26
u 
7  12  6  22
 36  1 6  52  2 26
Both vectors have the same length, verified by using the
distance formula.
To show that the two vectors have the same direction we
compute the slope of the lines.
40 4 2
(0,0) and (6,4)
 
(1,2) and (7,6) .
60 6 3
62 4 2
 
7 1 6 3
Since they are also equal we
Conclude the vectors are equal
v u
What is standard position for a vector in the plane?
Since all vectors of the same magnitude and direction are
considered equal, we can position all vectors so that their
initial point is at the origin of the Cartesian coordinate
system. Thus the terminal point would represent the vector.
v  v1, v 2
Would be the vector whose terminal point
would be (v1,v2) and initial point (0,0)
The notation v  v1 , v 2
form of v.
is referred to as the component
v1 and v2 are called the components of v.
If the initial point and terminal point are both (0,0) then we
call this the zero vector denoted as 0 .
Here is the formula for putting a vector in standard position:
If P(p1,p2) and Q(q1,q2) represent the initial point and terminal
point respectively of a vector, then the component form of the
vector PQ is given by:
PQ  q1  p1, q 2  p2  v1, v2
And the length is given by:
PQ 
q1  p1 2  q 2  p 2 2
 v12  v 2 2
Special Vectors
If v  v1 , v 2
represents the vector v in standard
position from P(0,0) to Q(v1,v2) and if the length of v,
v  1 Then v is called a unit vector.
The length of a vector v may also be called the norm of v.
If v  0 then v is the zero vector
0.
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
u
4
3
1
2
1
1
1
2
3
4
y
2
3
4
x
5
Move the first
vector into
standard
position.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
u
4
3
v
1
2
1
1
1
2
3
4
y
2
3
4
x
5
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
u
4
3
v
1
2
1
1
1
2
3
4
y
2
3
4
x
5
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
We will start with
addition
and look at the
geometric
interpretation.
3
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Move the second
vector so that its
initial point is at the
terminal point of
the first vector.
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
The result or the
resultant vector
3
is the one with
initial point the
origin and the
terminal point at
the endpoint of
vector v.
2
1
1
1
2
3
4
x
5
1
2
3
4
y
See that the resultant vector can be found by adding
the components of the vectors, u and v.
2
Is written in standard
position.
3
u  v  1, 2
v
u
4
u  2 ,1
v  1  2 , 2  1   1,1
v
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
3
v
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Notice, that if
vector v is moved
to standard
position. The
resultant vector
becomes the
diagonal of a
parallelogram.
Vector Operations:
Now we need to define vector addition and scalar multiplication.
4
3
v
2
1
3
2
u
4
1
1
1
2
3
4
y
2
3
4
x
5
Notice, that if
vector v is moved
to standard
position. The
resultant vector
becomes the
diagonal of a
parallelogram.
If u  u1 , u 2
and v  v1 , v 2
then the vector sum of u and v is
u  v  u1  v1, u 2  v2
Next we look at a scalar multiple of a vector,
Example: Suppose we have a vector
ku  ku1, ku 2
2, 3 that we double.
Geometrically, that would mean it would be twice as long, but
the direction would stay the same. Thus only the length is
affected.
If k u  ku1 , k u 2
then
ku 
ku1 2  ku2 2
 k 2 u 12  k 2 u 2 2
k 2 ( u 12  u 2 2 )  k u 12  u 2 2  k u
If u  u1 , u 2
and v  v1 , v 2
then the vector sum of u and v is
u  v  u1  v1, u 2  v2
If u  u1 , u 2 and k is a scalar then ku  ku1 , ku 2
Since -1 is a scalar, the negative of a vector is the same as
multiplying by the scalar -1. So,  u   u ,u
1
2
Example: The negative of the vector 2, 3
would become
 2,  3 Making the terminal point in the opposite direction of
the original terminal point.
y
4
3
2, 3
2
1
4
3
2
1
1
1
2
 2,  3
3
4
2
3
4
x
5
If u  u1 , u 2
and v  v1 , v 2
then the vector sum of u and v is
u  v  u1  v1, u 2  v2
If u  u1 , u 2 and k is a scalar then ku  ku1 , ku 2
The negative of v  v1 , v 2
is  v   v1 ,  v 2
Lastly, we examine the difference of two vectors:
u  v  u  (v)  u1  v1, u 2  v2
using the definition of the sum of two vectors and the negative
of a vector.
Geometrically, what is the difference? Let u and v be the vectors below.
What is u – v?
y
4
vectors u
and v are
in
standard
position.
Now,
create
the
vector -v
u  3, 3
3
v   2, 2
2
1
4
3
2
1
1
1
2
3
4
2
3
4
x
5
Geometrically, what is the difference? Let u and v be the vectors below.
What is u – v?
y
4
vectors u
and v are
in
standard
position.
Now,
create
the
vector -v
u  3, 3
3
v   2, 2
2
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
Geometrically, what is the difference?
y
4
Use the
parallelogram
principle to
draw the sum
of u - v
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
How do u  v and u  v relate to our parallelogram?
u  v  1, 5
y
4
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
How do u  v and u  v relate to our parallelogram?
u  v  1, 5
y
4
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
How do u  v and u  v relate to our parallelogram?
u  v  1, 5
y
4
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
How do u  v and u  v relate to our parallelogram?
u  v  1, 5
y
4
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
How do u  v and u  v relate to our parallelogram?
u  v  1, 5
y
4
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
How do u  v and u  v relate to our parallelogram?
u  v  1, 5
y
4
u  3, 3
3
v   2, 2
2
u  v  5,1
1
4
3
2
1
1
2
3
4
1
2
 v  2,2
3
4
x
5
They are
both
diagonals of
the
parallelogram.
If u  u1 , u 2
and v  v1 , v 2
then the vector sum of u and v is
u  v  u1  v1, u 2  v2
If u  u1 , u 2 and k is a scalar then
The negative of v  v1 , v 2
If u  u , u
1 2
and
ku  ku1, ku 2
is  v   v1 ,  v 2
v  v1, v 2
then the vector difference of u and v is
u  v  u1  v1, u 2  v2
Vector Properties of Operations
Let
u, v and w be vectors in the plane and let c, and d be scalars.
The commutative property:
uv  vu

(u  v)  w  u  v  w
The associative property:
Additive Identity Property:
u 0  0u  u
Additive Inverse Property:
u u  0
 
cdu   cd(u)
Associative Property with scalars:
Distributive Property:
(c  d)u  cu  du
Distributive Property:
c(u  v)  cu  cv
Also
1(u)  u, 0u  0

The length of a scalar multiple of a vector is the length of the
vector times the scalar as was shown earlier and here again.
If
then
k u  ku1 , k u 2
ku 
ku1 2  ku2 2
 k 2 u 12  k 2 u 2 2
k 2 ( u 12  u 2 2 )  k u 12  u 2 2  k u
Every non-zero vector can be made into a unit vector:
Proof: First we will show that
 
 1 
1
u   v 
 v 
v
 
u
v 1
has length 1.
u
v
v

1
v, v  0
v
Since u is just a scalar multiple of v ,
they are both going in the same direction.
The process of making a non-zero vector v into a unit vector u in the direction
of v is called the normalization of v .
Thus, to normalize the vector v , multiply v by the scalar
4,  2  4 2   22  16  4  20  2 5
4,  2
by
1
2 5

4
,
2
2 5 2 5

2
5
,
1
5
Now we will show that the normalized vector has length 1.
2 1
,
5 5

v
.
4,  2 and show that the new vector has length 1.
Example: Normalize the vector
Multiply
1
4 1
5
 
 1 1
5 5
5
Standard Unit Vectors
The unit vectors <1,0> and <0,1> are called the standard unit vectors in the
plane and are denoted by the symbols i and j respectively.
i  1, 0 and j  0,1
Using this notation, we can write a vector in the plane in terms of the vectors
i and j as follows:
v  v1 , v 2  v1 1, 0  v 2 0,1  v1 i  v 2 j
v1i  v2 j
The scalars
v
is called a linear combination of i and j .
v1 and v 2
respectively.
are called the horizontal and vertical components of
Writing a vector in terms of sin and cos .
Let u be a unit vector in standard position that makes an angle  with
the x axis.
y
Thus
2
1
u
2
(cos , sin )

cos
1
1
2
sin 
1
2
x
u  cos , sin
Writing a vector in terms of sin and cos  continued.
Let v be a non-zero vector in standard position that makes an angle 
with the x axis.
Since we can make the vector v a unit vector by multiplying by the
reciprocal of its length it follows that
v
 cos , sin  cos i  sin j where  is theanglev makes
v
with the x  axis
v  v cos , sin  v cos i  v sin j
Example: Suppose vector v has length 4 and makes a 30o angle with the

positive x-axis. First we use the radian measure for  
6




v  4 cos , sin
 4 cos i  4 sin j 
6
6
6
6
4 3
4 1
i
j  2 3i  2 j
2
2
Sample Problems
Example 1:
Find the component form of the vector v and sketch the vector
in standard position with the initial point at the origin.
(-1,4)
y
4
3
2
(3,1)
1
4
3
2
1
1
1
2
3
4
2
3
4
x
5
Sample Problems
Example 1:
Find the component form of the vector v and sketch the vector
in standard position with the initial point at the origin.
(-1,4)
y
1  3, 4 1   4, 3
4
<-4,3>
3
2
(3,1)
1
4
3
2
1
1
1
2
3
4
2
3
4
x
5
Example 2: Given the initial point <1,5> and terminal point <-3,6>, sketch
the given directed line segment and write the vector in component form
and finally sketch the vector in standard position.
Example 2: Given the initial point <1,5> and terminal point <-3,6>, sketch
the given directed line segment and write the vector in component form
and finally sketch the vector in standard position.
Solution: <-3-1,6-5>=<-4,1>
y
6
5
4
3
2
1
7
6
5
4
3
2
1
1
1
2
3
4
5
6
2
3
4
5
6
x
7
Example 3: Use the graph below to sketch
u  2v
y
4
3
2
v
u
1
4
3
2
1
1
1
2
3
4
2
3
4
x
5
Example 3: Use the graph below to sketch
u  2v
y
4
First double the length of
v
3
Next move
2
v
u
into standard position.
u
Now move 2v
1
into standard position
4
3
2
1
1
1
2
3
4
2
3
4
x
5
Complete the
parallelogram and draw
the diagonal.
Example 4: Compute
a  b, a  2b, 3b, 3b  2a
For a  3,  1 , and b   4, 5
Example 4: Compute
a  b, a  2b, 3b, 3b  2a
For a  3,  1 , and b   4, 5
Solution:
a  b  3,  1   4, 5  3  4,  1  5   1, 4
a  2b  3,  1  2  4, 5  3,  1  8,  10  11,  11
3b  3  4, 5   12,15
3b  2a   12,15  2 3,  1   12,15   6, 2   18,17
 18,17 
 182  172
 613
Example 5: Compute
For
a  b, a  2b, 3b, 3b  2a
a  i  2 j, and b  3i  j
Example 5: Compute
For
a  b, a  2b, 3b, 3b  2a
a  i  2 j, and b  3i  j
Solution:
a  b  i  2 j  3i  j  4i  3 j
a  2b  i  2 j  2(3i  j)  i  2 j  6i  2 j  5i
3b  3(3i  j)  9i  3 j
3b  2a  9i  3 j  2(i  2 j)  9i  3 j  2i  4 j  7i  j
7i  j 
7 2  12
 50  5 2
Example 6: For each of the following vectors,
a) find a unit vector in the same direction
b) write the vector in polar coordinates i.e. v
1.
3, 6
2.
2i  4 j
3.
4i
4.
from 2,1 to 5, 2
 v cos , sin
Example 6: For each of the following vectors,
a) find a unit vector in the same direction
b) write the vector in polar coordinates v  v
1.
3, 6
a)
9  36  45  3 5
3
so
,
6

1
3 5 3 5
b) 3 5
2.
,
5
2
5 2 5
,
5
5

5
5 2 5
,
5
5
2i  4 j
a)
1
4  16  20  2 5 , so
2
 5
2 5
b) 2 5 
i
 5
5

3.
cos , sin
4i
a)
16  0  4

2i  4 j 
5
1
i
5
2
5
2 5
i
j
5
5
j
5

j



1
4i  i
4
b) 4 i
4.
from 2,1 to 5, 2
a ) 5  2, 2  1  3,1 ,
9  1  10, so
1
10
b)
10
3 10 10
,
10 10
3,1 
3
10
,
1
10

3 10 10
,
10 10
Example 7: Suppose there are two forces acting on a skydiver: gravity
at 150 lbs down and air resistance at 140 lbs up and 20 lbs to the right.
What is the net force acting on the skydiver?
150
20
140
Example 7: Suppose there are two forces acting on a skydiver: gravity
at 150 lbs down and air resistance at 140 lbs up and 20 lbs to the right.
What is the net force acting on the skydiver?
150
20
140
The net force is the sum of the three forces acting on
the skydiver.
Gravity would be -150j
Air Resistance would be 140j
The force to the right would be 20i
The sum would be 20i 10j
which would be 10 pounds
down and 20 pounds to the right.
Note: The 150 lbs represents the length of the vector. A unit vector pointing
in the same direction is 0, 1 or -j. Thus in polar coordinates the vector
would be 150(-j)=-150j
Example 8: Suppose two ropes are attached to a large crate. Suppose that
rope A exerts a force of 164,115 pounds on the crate and rope B exerts a
force of 177,177 . If the crate weighs 275 lbs., what is the net force acting
on the crate? Based on your answer, which way will the crate move.
A
B
Example 8: Suppose two ropes are attached to a large crate. Suppose that
rope A exerts a force of  164,115pounds on the crate and rope B exerts a
force of 177,177 . If the crate weighs 275 lbs., what is the net force acting
on the crate? Based on your answer, which way will the crate move.
A
B
Solution: The weight of the crate combined with gravity creates a force of
-275j or<0,-275>.
Adding the 3 vectors we get <13, 17
> 13 lbs right and 17 lbs up
Example 9: Find the horizontal and vertical components of the vector described.
A jet airplane approaches a runway at an angle of 7.5o with the horizontal, traveling
at a velocity of 160 mph.
Example 9: Find the horizontal and vertical components of the vector described.
A jet airplane approaches a runway at an angle of 7.5o with the horizontal, traveling
at a velocity of 160 mph.
7.5o
Solution: Remembering that speed is length of vector, we know that this vector
is 160 miles in length. Using polar coordinates the vector is
160 cos( 7.5o ), sin(7.5o )  160 .9914, .1305  158.63, 20 .88
Example 10: A woman walks due west on the deck of a ship at 3 miles per
hour. The ship is moving north at a speed of 22 miles per hour. Find the
speed and direction of the woman relative to the surface of the water.
N
W
E
S
Example 10: A woman walks due west on the deck of a ship at 3 miles per
hour. The ship is moving north at a speed of 22 miles per hour. Find the
speed and direction of the woman relative to the surface of the water.
N
Ship 22mph
W
E
S
Woman
3 mph
What angle is made by the woman relative to polar coordinates?

radians
In unit vector terms this would be <-1,0>
What angle is made by the ship relative to polar coordinates?

radians
2
In unit vector terms this would be <0,1>
Woman vector = 3  1, 0
Ship vector= 22 0,1
3  1, 0  22 0,1   3, 0  0, 22   3, 22
Adding the two vectors
Example 10: A woman walks due west on the deck of a ship at 3 miles per
hour. The ship is moving north at a speed of 22 miles per hour. Find the
speed and direction of the woman relative to the surface of the water.
N
W
Ship 22mph
E
Woman
3 mph
S
Woman vector = 3  1, 0
Ship vector= 22 0,1
3  1, 0  22 0,1   3, 0  0, 22   3, 22
Adding the two vectors
Notice this does not answer yet the question of speed and direction. Speed is
vector magnitude and direction should be in degrees with a compass direction so
how do we get that?
Example 10: A woman walks due west on the deck of a ship at 3 miles per
hour. The ship is moving north at a speed of 22 miles per hour. Find the
speed and direction of the woman relative to the surface of the water.
Ship 22mph

Resultant Vector  3, 22
This angle
Woman
3 mph
Magnitude:  3, 22  9  484  493  22 .2mph
To find direction we need an angle: tan   sin  22  7.33333...
cos  3

This would be our reference
  tan 1  7.33333...  82.23o
angle in Q3
When giving directions such as NW or SE you always begin with North or South
and the angle is measured from either North or South. So  is not the angle we
would use to give the direction. We would use its complement which is 7.77o and
say the woman is walking 22.2 mph in the direction N7.770W.
Exercise 11. Given the vector with magnitude and direction following. Write
the vector in component form.
0
2 , N30 W
Exercise 12. Given the vector in component form write the magnitude and
direction of the vector with respect to N, NE, NW, S, SE, or SW direction.
3i – 4j.
Exercise 11. Given the vector with magnitude and direction following. Write
the vector in component form.
0
N300W
2 , N30 W
Is in quadrant II with reference angle 60 degrees and from the
positive x axis 120 degrees, Thus the vector is
2 cos120, sin120  2
1 3
,
  1, 3
2 2
Exercise 12. Given the vector in component form write the magnitude and
direction of the vector with respect to N, NE, NW, S, SE, or SW direction.
3i – 4j.
Magnitude is:
9  16  25  5
This vector is in Quadrant IV, (+,-) and
4
3
 4
0
tan 1 
  53.13
 3 
tan  
Since 53.13 degrees would also be the reference angel
between the vector and the positive x – axis, we would
need to subtract from 90 degrees to find the angle
between the vertical South and the vector for giving the
direction of S(900  53.130 )E  S36.870 E
Your Homework for this section is in Blackboard under Assignments
Button. Click on Assignment 7.1