Vectors in the Plane
When I use the word scalar I mean a number where only the
quantity is important (ex: temperature) . When I use the word
vector, I am referring to something a little more complicated.
A vector has a quantity and a direction (ex's: velocity or force).
The displacement vector from one point to another is an arrow
with its tail at the 1st point and its tip at the end point. The
magnitude of the vector is the distance between the 2 points.
The direction of the vector is the direction of the arrow.
Example: The displacement vector from Phila. to Buffalo
Buffalo
Phila.
Vectors have both magnitude and a direction.
Def: Two vectors are equal if they have the same magnitude
and direction.
These two displacement vectors represent the same
vector even if the start and end point are different!
Def: The zero vector is the displacement vector with zero
length. There is no direction.
We can place a vector on a Rectangular Coordinate system.
The standard position is when the vector initiates from the
origin.
y
v
x
If a vector is in standard position then it can be specified by
giving the coordinates a  b of the tip of the vector.
y
(5,3)
v
v = (5,3)
x
Def: If   a  b then the length of this vector is È    
The length of the vector is called its magnitude and is denoted
by ll
Example: The magnitude of the vector   a   b is
Éa  b    È    È   
Def (Scalar multiplication): If   a  b then a  b
=a  b
If   a  b then
1. ll  kk ll
2. If    then  and  have the same direction.
3. If    then  and  have opposite directions.
Def (sum of 2 vectors): The sum of the vectors  and  is the
displacement vector which results from 1st applying a
displacement of  and then applying a displacement of 
Example:
v
w
The sum of  and  is illustrated below:
w
v
v+ w
Example: Find the sum of the vectors   a b and
  a b
Subtraction of 2 vectors:       a  b
Ex: Find    where   a b and   a b
y
w = (1,3)
v = (6,2)
x
w = (1,3)
v = (6,2)
-w
-w
v-w
(5,-1)
Of course a b  a b  a  b
   is the vector that starts at the tip of  and ends at
tip of 
Def: A unit vector is a vector of length 1. If   a  b then a

unit vector in the same direction as  is

ll
Example: Find a unit vector in the same direction as a  b

a  b
a  b



a  b or
È  
È
È



ŽÈ
•
È


Def: The unit vector a b is denoted by  and the unit vector
a b is denoted by 
y
1
j
i 1
x
If   a   b then      
In general, if   a b then     
Example: To add    and     just combine "like
terms".
    a    b     
Example: Find the magnitude and angle of the resultant of the 2 vectors
shown below.
y
141.95°
27.75°
60
40
x
The 2 vectors can be written as
  °     ° 
and
  °     ° 
The resultant is a  b   a  b 
      
The magnitude is È     
y
Find this angle 1st
60
40
x
To find the angle, 1st find 

 ° 

Therefore the resultant makes an angle of °  °  °
wrt positive x-axis.
Application Problem
Forces with magnitude 500 pounds and 200 pounds act on a
machine part at angles of 30° and  45°, respectively, with
the x-axis (see figure). Find the direction and magnitude of the
resultant force. Note that the resultant is the sum of the
forces.
500 lb
o
30
45 o
200 lb
x
There are 2 approaches.
1. Make a parallelogram and find the diagonal.
This angle is 180-75=105 deg
500
75°
200
R
200
Use the Law of Cosines.
      abab ° 
    È   pounds.
We can find the angle  makes with the x axis.
??
500
o 200
105
R
x
Find the angle shown above (call it )
Use the Law of Sines.
 
 °
  °

   
 



     °
This means that  makes an angle of     °
with the x-axis.
2.
Find the  and  components of the 2 forces.
The 500 pound force 
  °    °    
The 200 pound force 
 a  °b   a  °b 
  °    °    
The sum of the 2 forces is
a  b  a  b 
a  b  a  b      
which has a magnitude of È   
 pounds.
The angle vertor  (call it ) makes with the x-axis
satisfies
 

  

   °
 
