Vectors
OCR Stage 10
A VECTOR?
□ Describes the motion of an object
□ A Vector comprises
□ Direction
□ Magnitude
Size
□ We will consider
□ Column Vectors
□ General Vectors
□ Vector Geometry
Column Vectors
NOTE!
Vector a
a
Label is in BOLD.
2 up
When handwritten,
draw a wavy line
under the label
i.e.
4 RIGHT
4
 
2
COLUMN Vector
a
~
Column Vectors
Vector b
2 up
b
3 LEFT
 3 


2


COLUMN Vector?
Column Vectors
Vector u
2 down
n
4 LEFT
 4 



2


COLUMN Vector?
Describe these vectors
4
 
1
a
1
 
3
b
c
 2 


3


d
 4 



3


Alternative labelling
B
D
F
EF
E
AB
CD
G
C
A
GH
H
General Vectors
A Vector has BOTH a Length & a Direction
All 4 Vectors here are EQUAL in Length and
Travel in SAME Direction. All called k
k
k
k
k
k can be in any position
General Vectors
Line CD is Parallel to AB
B
A
CD is TWICE length of AB
k
D
2k
Line EF is Parallel to AB
E
C
-k
F
EF is equal in length to AB
EF is opposite direction to AB
Write these Vectors in terms of k
B
k
D
2k
½k
1½k
F
G
E
A
C
-2k
H
Combining Column Vectors
B
AB
A
k
C
AB
D
2
k   
1
CB
D  3k
2k
A
2
A B  3  2
C D  2 1 
 1
 
 64 
AB
C
D   
 32 
Simple combinations
C
4 
AB   
1 
1 
BC  
3 

 
B
A
5 
AC =  
4 
a  c   a  c
      

b  d  b  d 
Vector Geometry
Consider this parallelogram
Q
OP  a  RQ
OR  b  PQ
P
R
a
b
Opposite sides are Parallel
OQ  OP  PQ
 a+b
OQ  OR  RQ
 b+a
O
a+b  b+a
OQ is known as the resultant of a and b
Resultant of Two Vectors
□ Is the same, no matter which route is
followed
□ Use this to find vectors in geometrical
figures
Example
S is the Midpoint of PQ.
.
Q
S
P
OS
OS  OP  ½ PQ
R
a
b
O
Work out the vector
= a + ½b
Alternatively
S is the Midpoint of PQ.
.
Q
S
P
OS
OS  OR  RQ  QS
R
a
b
O
Work out the vector
= b + a - ½b
= ½b + a
= a + ½b
C
Example
AC= p, AB = q
p
A
M
q
Find BC
M is the Midpoint of BC
B
BC = BA + AC
= -q + p
=p-q
C
Example
AC= p, AB = q
p
A
M
q
Find BM
BM = ½BC
= ½(p – q)
M is the Midpoint of BC
B
C
Example
AC= p, AB = q
p
A
M is the Midpoint of BC
M
q
Find AM
AM = AB
B
+ ½BC
= q + ½(p – q)
= q +½p - ½q
= ½q +½p
= ½(q + p)
= ½(p + q)
Alternatively
C
AC= p, AB = q
p
A
M is the Midpoint of BC
M
q
Find AM
B
AM = AC + ½CB
= p + ½(q – p)
= p +½q - ½p
= ½p +½q
= ½(p + q)