Higher Mathematics
Unit 3.1
Vectors
1.
Introduction
A vector is a quantity with both
magnitude and direction.
It can be represented
using a direct line
segment
AB
A
This vector is named
A B or u or u
u
B
2.
Vectors in 3 - Dimensions
3
v
5
2
 ........
2 


v  ........
 5 
 ........ 
 3 
P
Z
 ........
-2 


O P  ........
 4 
 ........ 
 3 
Y
Q
2
O
3
-3
3
-2
R
X
Z
Y
O
 ........
3 


O Q  ........
 2 
 ........ 
 0 
Q
X
-3
3
-2
R
Z
Y
O
X
 ........
3 


O R  ........
 -3 
 ........ 
R
 -2 
3.
Finding the components of a Vector
from Coordinates
y
P (1, 2)
Q (6, 3)
Q
3
2
P
1
6
x
 ..........
6 - 1   .......
5 
PQ  


3 - 2   .......
1 
 ..........
y
S (-2, 1)
T (5, 3)
T
3
S
-2
1
5
x
 ..........
5 - -2   .......
7 
PQ  


3 - 1   .......
2 
 ..........
y
A (-2, -1)
B
1
B (4, 1)
x
A
-2
-1
4
 ..........
4 - -2   .......
6 
PQ  


1- - 1   .......
2 
 ..........
4.
Magnitude
4
4
AB  u   
-3
···
 
A
u
-3
B
AB  u 
2
2
4 + (-3)
···············

16  9

25  5
5.
Adding Vectors
7 
BC   
1
 ···
C
B
2 
AB   
4
 ···
1 
CD   
··· 
-6
A
D
 B C C D
A D Add
 A Bvectors
 2   7   1   10 
“ Nose-to-tail”
  
  
 4   1   6   1 
3
u  
1 
u+v
2
v  
4
v
u
Add vectors
“ Nose-to-tail”
3
u v   
1 
5
 
5
2
 
4
4 
AB  u  

 3 
A
 4 
B A  u  

3 
A
-u
u
B
B
B A is the negative of A B
 u is the negative of u
v
2
v  
4
 -2
v   
-4
 ...
3
u  
1 
u
u + -v
u-v
-v
u  v  u  v
Add the negative of the vector
“ Nose-to-tail”
3
 
1 
1 


 3 
 2 


 4 
The Zero Vector
2
v  
4
 2 
v  

 4 
v v
 v  v
-v
 2   2 
 

 4   4 
v
Back to the
start.
Gone nowhere
0
 
0
7.
Multiplication by a Scalar
1 
v  
2
v
2v
21 
2v  

2 2
2
 
4
2v has TWICE the MAGNITUDE of v,
but v and 2v have the SAME DIRECTION.
i.e. They are PARALLEL
8.
Position Vectors
y
 4 
p  OP  

2
 .... 
P (4, 2)
p
O
x
The position vector of a point P is the vector from the origin O, to P.
The position vector O P is denoted by p
x

p

y
then the components of the position vector of P are

z

If P has coordinates (x , y , z)





9.
Collinear points
NOT collinear
A
E
D
AB
B
BC
C
Collinear
If
AB  k BC
w h e re k  0
then the vectors are parallel and have
a point in common - namely B - ,
this makes them collinear
10.
Dividing lines in given ratios
“Section Formula”
Give up John, they are getting bored!!
11.
Unit Vectors
A unit vector is any vector whose
length (magnitude) is one
The vector



u  





2 
3 

2 
3 

1 

3 
is a unit vector
since
2
u 
u 1
2
2
2
1
     
3
3
3
2
There are three special unit vectors:
1
 
i 0
 
0
 
0
 
j 1
 
0
 
0
 
k  0
 
1
 
z
y
 0, 0,1 
k
j
i
 0,1, 0 
 1, 0, 0 
x
All vectors can be represented using
a sum of these unit vectors
P
Z
 ........
-2 


O P  ........
 4 
 ........ 
 3 
Y
O
X
OP 
-2 i +4 j +3 k
12.
Scalar Product
The scalar product (or “dot” product) is a kind of
vector “multiplication”. It is quite different from any
kind of multiplication we’ve met before.
 x1 
 x2 
 
 
a

y
The scalar product of the vectors
 1  and b   y 2 
z 
z 
 1
 2
is defined as:
a  b  a b co s 
where  is the angle
between the vectors,
pointing out from the
vertex
a
or
a  b  x1 x 2  y 1 y 2  z 1 z 2

b
Calculating the angle between two vectors
We have already seen that
Rearranging gives
a  b  a b co s 
co s  
ab
a b
And hence we can find the angle between two vectors
Some important results using the scalar product
1. The scalar product is a number not a vector
2. If either
3.
a 0
or
then a  b  0
b 0
Perpendicular vectors:
 Provided a and
b
are non zero then if
a b co s   0
then
co s   0
so
  90
ie a and
4.
b
0
are perpendiculiar
a  (b  c )  a  b  a  c