Recall: Two vectors in the plane are equal if they have the same length and same
direction, thus unless stated explicitly, we can consider a general vector A to have initial
point the origin (0,0) and terminal point ( Ax , Ay ) .
( Ax , Ay )
Ay
Ax
Therefore we need only give the terminal point to specify the vector. In order to
distinguish a vector from a point in the plane we use the notation  Ax , Ay  .
The vector A =  Ax , Ay  is the vector in the plane with initial point the origin and
terminal point ( Ax , Ay ) . The notation  Ax , Ay  is called the component form of the
vector and with x-component Ax and y-component Ay .
The norm of a vector A, denoted || A || is the length of the vector. If a vector is given in
component form, A =  Ax , Ay  , then || A || is the distance from the origin to the point
( Ax , Ay ) , i.e. || A || =
Ax2  Ay2 .
Consider now the vector PQ with initial point P  ( x1 , y1 ) and terminal point
Q  ( x2 , y2 )
Q  ( x2 , y2 )
y2  y1
P  ( x1 , y1 )
x2  x1
The component form of PQ is  x2  x1 , y2  y1  and || PQ || ( x2  x1 ) 2  ( y2  y1 ) 2
Example: Find the component form and the norm of the vector with initial point P=(-3,5)
and terminal point Q = (9,0)
Solution: PQ = <9-(-3), 0-5> = <12, -5>; || PQ || 12 2  (5) 2  169  13
There are two ways to represent a vector in terms of its direction and length.
Method I. Let  be the angle, 0 <  < 90, between the vector (with initial point at the
origin) and the x-axis.




Then a vector A with norm ||A|| and angle  has component form  Ax , Ay  =
<||A||cos, ||A||sin>, where sign depends on the quadrant which contains the terminal
point of the vector..
||A||
Ay

Ax
cos = Ax /||A|| and sin = Ay /||A||.
The quadrant that contains the terminal point of the vector can be specified by giving the
angle  in terms of directions
N
WN
EN
W
E
WS
ES
S
Example: Find the component form of the vector A of length 4 and direction W 30 N.
Solution: Since the vector has terminal point in the second quandrant, Ax < 0 and
Ay > 0.  Ax , Ay  = <-||A||cos, ||A||sin> = <-4cos30 , 4sin30 > =  2 3 ,2  .
We can reverse this process to find the the length and direction of a vector A if we are
given the component form  Ax , Ay  . We have already seen that || A || =
The direction  is given by  = arctan| Ay / Ax | if Ax  0, 90 if Ax = 0.
Ax2  Ay2 .
Example: Find the length and the direction of the vector A =  2,2 3  .
Solution: || A || = (2) 2  (2 3) 2  16 = 4;  = arctan|(-2 3 )/(-2)| = arctan 3 =
60. Since the terminal point of A is in quandrant IV, the direction is E 60 S.
Method II.  is the angle measured from the positive x-axis to the vector (from the
horizontal). If measured counterclockwise  is positive, clockwise  is negative. There
are no restrictions on .


Then  Ax , Ay  = <||A||cos, ||A||sin>, where the sign is now specified by the angle .
Example: Find the component form of the vector A of length 10 which makes an angle
with the horizontal of
i)
135
ii)
-90
Solution:
i)  Ax , Ay  = <||A||cos, ||A||sin> = <10cos135,10sin135> =  5 2 ,5 2 
ii)  Ax , Ay  = <||A||cos, ||A||sin> = <10cos(-90),10sin(-90)> = <0,-10>
As before we can reverse this process to find the the length and direction of a vector A if
we are given the component form  Ax , Ay  . || A || =
Ax2  Ay2 .
The direction  is given by  = arctan( Ay / Ax ) if Ax  0, 90 if Ax = 0, where if
necessary we may need to adjust  to the corresponding angle in the second or third
quadrant since the value of arctan is an angle in the first or fourth quadrant. This is done
by adding 180 to the value of arctan.
Example: Find the length and direction of the vector A=  5 2 ,5 2  .
 5 2 
 = arctan(-1) =
(5 2 ) 2  (5 2 ) 2 = 100 = 10; arctan 

5 2 
-45. Since the terminal point of A is in quandrant II, we must add 180 to this value to
obtain  = 135.
Solution: || A || =