Some terms to know:
Vectors:
Lines with length (magnitude) and direction
A
O
a
~
Vector
OA
OA  a~
Magnitude of OA is represented by OA
Or magnitude of a
is
represented
by
a
|
|
~
x
Direction is represented by column vector  
 y
Zero vector or Null vector
0
O   
•No length
•No direction
0
x
PQ   
 y
Q
y = +4
P
x = +2
Movement parallel
to the x-axis
Movement parallel
to the y-axis
 2
PQ   
 4
 x    2
QP      
 4
y

 
Q
y=-4
P
x=-2
PQ  RS
Equal vectors or equivalent vectors
Q
S
P
R
Same length
Same direction
PQ  SR
Negative vectors
Q
S
P
R
PQ  SR
Same length
opposite direction
NOTE: Negative sign
reverses the direction
Let’s do up Exercise 5A in your
foolscap paper
Q. 1, 3, 5
Vector addition
Vector subtraction
1   3 
    
 5   2
1   3 
    
 5   2
1  3 

 
 5  (2) 
1  3 

 
 5  (2) 
 4
  
3
  2
  
7 
Triangle law of vector addition
R
Q Ending point
P
Starting point
PQ 
PR  RQ
Parallelogram law of vector addition
R
Q Ending point
P
Starting point S
PQ  PR  PS
Let’s do up Exercise 5B in your
foolscap paper
Q. 3, 4, 6, 7
Scalar multiplication
x
PQ   
 y
x
3 PQ  3 
 y
 3x 
  
3y 
Let’s do up Exercise 5C in your
foolscap paper
Q. 1, 5, 10
 x  2 
PQ      
 5
y

 
P
Length (magnitude) of PQ  PQ
 2   5
2
y=-5
 29 units
x= 2
Q
2
Parallel vectors
Q
S
P
R
PQ  k RS
Same gradient
Gradient of vector
Q
y = +4
P
x = +2
gradient of PQ
height

base
4
 2
2
Collinear vectors
Lie on the same straight line
P, Q and R are collinear
=> P, Q and R lie on the same straight line
R
PQ  k PR
P
 3
Eg. Given c =   , find
~  4
(a) |2c|
(b) |-2c|
 3
(a) 2c = 2  
 4
6
  
8
(c) 2|c|
|2c| =
6 8
2
2
 36  64
 100
 10 units
 3
Eg. Given c =   , find
~  4
(a) |2c|
(b) |-2c|
 3
(b) -2c = -2  
 4
  6
  
  8
(c) 2|c|
|-2c| = (6)  (8)
2
 36  64
 100
 10 units
2
 3
Eg. Given c =   , find
~  4
(a) |2c|
(b) |-2c|
(c) 2|c| = 2 (3)  (4)
2
(c) 2|c|
2
 2 9  16
 2 25
 2(5)  10 units
Notice:
|2c| = 10 units
|-2c| = 10 units
2|c| = 10 units
Therefore, |2c| = |-2c| = 2|c|
In conclusion:
|kc| = |-kc| = k|c|, where k is any positive number
  3
1
Eg. Given a =   and b =   , find
~  2
~  5 
(a) |a| + |b|
(b) |a+b|
|a| + |b|
a+b
 1  2  (3)  5
2
2
 5  34
2
2
  2
  
 7 
|a+b| = (2) 2  7 2
 53
 8.07 units
 7.28 units
Note: |a| + |b| ≠ |a + b|
  3
1
Eg. Given a =   and b =   , find
~  2
~  5 
(c) |2b-a|
(d) 2|b| + |a|
2b-a
 7
  
 8 
2|b| + |a|
|2b-a| = (7)  8
2
 113
 10 .6 units
2
 2 (3)  5  1  2
2
2
 2 34  5
 9.43 units
Note: |2b -a| ≠ 2|b| - |a|
2
2
Let’s do up Exercise 5D in your
foolscap paper
Q. 1, 6, 8
Position vectors: always with respect to origin
Can you tell me what is the position vector of P?
y
O
P   2 , 4
 2
OP   
 4
x
In general, position vector of any point, say P,
is given by OP   x 
 y
y
O
P   x , y
x
PQ  ?
Position vectors
y
P
OP  PQ  OQ
PQ  OQ  OP
Q
O
x
Let’s do up Exercise 5E in your
foolscap paper
Q. 3, 6, 12