1.
1.1
Vector Algebra
Review of Basic Ideas
In engineering and science, physical quantities which are completely specified by
their magnitude (size) are known as scalars. Examples are: mass, temperature,
volume, resistance, charge, voltage, current, etc.
Other quantities possess both magnitude and direction and may be represented
geometrically by directed line segments known as vectors. The length of the line is
known as the magnitude of the vector and its direction is the direction of the vector.
Examples of vector quantities are: velocity, acceleration, force, electric field,
magnetic field etc and will be denoted by v, a, F, E, B, etc.
1.
Two vectors a and b are equal if they have the same magnitude and direction
irrespective of their initial points. We write
ab
2.
3.
a
b
A vector having the same magnitude as a but the opposite direction is
denoted by –a.
a
-a
Geometrically the sum of two vectors is given by the parallelogram law
b
a
4.
b
a
c
a
c
b
c  ab
The difference of two vectors a and b, represented by
c  a b
is defined as c = a + (-b).
a
a
5.
b
c
b
If a  b then a  b is the zero vector denoted by 0.
This has magnitude 0 but no direction.
1
c
-b
a
6.
Multiplication of a by a scalar, m, produces a vector ma with magnitude
m times that of a and direction the same as or opposite to that of a
according to whether m is positive or negative respectively. If
m = 0 then ma = 0.
a
7.
2a
a

1
a
2
Unit vectors are vectors with magnitude 1. If a is any vector then we
usually denote its magnitude by a . A unit vector with the same
direction as a will be
1.2
a
.
a
Components of a Vector
In a rectangular coordinate system in 3- D Euclidean space R 3 , orthogonal unit
vectors in the directions of the positive x, y and z axis are denoted by i, j and k
respectively.
z
z
k
O
i
y
O
j
r
xi
x
P(x,y,z)
y
zk
yj
x
The vector from the origin 0 to a point P is known as the position vector of P.
If P has Cartesian coordinates (x, y, z), then the position vector of P, r, may
be written as,
r = xi, + yj + zk
or
r = (x, y, z).
x, y, z are know as the components or coordinates of r with respect to the vectors i, j,
and k.
If P( x1 , y1 , z1 ) and Qx2 , y2 , z 2  are two points, the vector from P to Q will be
a  a1 i  a 2 j  a3 k  (a1 , a 2 , a3 ) where the components of a are
a1  x2  x1 , a2  y 2  y1 and a3  z 2  z1 .
2
Note that the ordered triple of components of a vector is unique with respect to a given
coordinate system.
If a  (a1 , a2 , a3 ) and b  (b1 , b2 , b3 ) , in terms of components we have:
Equality : a = b
Addition: a + b
iff a1  b1 , a2  b2 , a3  b3
 (a1  b1 , a2  b2 , a3  b3 )
 (a1  b1 )i  (a 2  b2 ) j  (a3  b3 )k
a
i
a

a

Scalar Multiplication: ma  (ma1 , ma2 , ma3 )
1

b
1


2
b
2
j

3

b
3
 ma1 i  ma2 j  ma3 k
Zero Vector : 0 = (0, 0, 0)
Magnitude : a 
Unit Vector:
a

a
a1  a 2  a3
2
2
1
a1  a 2  a3
2

2
3
(Pythagoras)
(a1 , a 2 , a3 )
2
a1 i  a 2 j  a3 k
a1  a 2  a3
2
2
2
Example
The vector a from P: (3, -2, 1) to Q: (1, 2, -4) has components
a1  1  3  2, a2  2  (2)  4, a3  4  1  5
Hence
a  (2,4,5)  2i  4 j  5k
a  (2) 2  4 2  (5) 2  45
And a unit vector in the direction of a is (2,4,5) / 45
If a has the initial point R : (1,2,3), then its terminal point is S : (-1,6,-2)
Direction Cosines
3
If r = xi + yj + zk, the direction of r may be specified by the cosines of the angles
made by r with the 3 coordinate axes.
z
N
P
r
O



y
M
L
x
l  cos  
OL x

OP r
m  cos  
n  cos  
OM
y

OP
r
ON
z

OP
r
l, m, and n are known as the direction cosines of r and li +mj +nk is a unit vector along OP
r  r (l i  m j  nk )
Example
If P is (3,2,6) = 3i + 2j +6k, r  7, l  3 / 7, m  2 / 7, n  6 / 7 and
  cos 1 (3 / 7),   cos 1 (2 / 7),   cos -1 (6 / 7).
1.3
An Introduction to Vector Spaces
If a, b, c are vectors and m, n are scalars, then
1. a + b = b + a
Commutative law of vector addition
2. a + (b + c) = (a + b) + c
Associative law of vector addition
3. a + 0 = a
Existence of 0 as an additive vector identity
4. a + (-a) = 0
Existence of additive inverses
5. m(a + b) = ma + mb
6. (m + n)a = ma + na
7. (mn)a = m(na)
8. 1a = a
Scalar distribution over vector addition
Vector distribution over scalar addition
Associative law for scalar multiplication
Multiplicative scalar identity
So far we have considered a, b, c etc to be vectors in R 3 and m, n etc to be real numbers
(  R ). However these ideas generalize to give an algebraic structure which is known as a
vector space.
4
A real vector space V is any non-empty set of elements a, b, c, … on which there are defined
the two algebraic operations of vector addition and scalar multiplication such that:
(a) for any a, b V , a + b is uniquely defined in V,
(b) for any a V and scalar m, ma is uniquely defined in V,
(c) properties 1-8 above hold.
Examples
1. V  R 3  {( a1 , a 2 , a3 ) : a1 , a 2 , a3  R}
2. V  R n  {( a1 , a 2 ,..., a n ) : ai  R for i  1,2,..., n}
3. V  P2 = {the set of all polynomials of degree at most 2}
4. V  {a cos x  b sin x : a, b  R}
However the following are not vector spaces – why?
5. {( a1 , a 2 , a3 ) : a1 , a 2 , a3  R  } R   {a : a  R, a  0}
6.
{all polynomials of degree 2}
7.
{all forces of the form f  mi  j , m  R }
In the following we shall normally restrict ourselves to the vector space R 3 , which is
probably most useful in practice, although generalizations will be mentioned. P2 , and more
generally Pn , above is useful for the approximation of functions and vector space 4, above
occurs in the solution of differential equations.
1.4
Vector Products
If a and b are two vectors, their dot product or scalar product, written a  b , is defined as,
 a b cos
ab  
0
if a  0, b  0
otherwise
where  is the angle between a and b.
Properties of the dot product
(i) The result is a scalar
(ii) a  b is zero if a = 0 or b = 0 or a and b are perpendicular (orthogonal)
(iii) a  a  a
5
(iv) a  b  b  a (symmetry)
(v) In terms of components, if a  a1 i  a 2 j  a3 k and b  b1 i  b2 j  b3 k
Then
a  b  (a1 i  a 2 j  a 3 k )  (b1 i  b2 j  b3 k )
 a1b1 i  i  a1b2 i  j  a1b3 i  k  a 2 b1 j  i  a 2 b2 j  j  a 2 b3 j  k  a3 b1 k  i  a 3 b2 k  j  a3 b3 k 
 a  b  a1b1  a 2 b2  a3 b3
since i  i  j  j  k  k  1 and
i j  j k  k i  0
This result is normally used to evaluate the dot product rather than the original
definition.
Example
If a = 5i + 4j + 2k and b = 4i – 5j + 3k, find a  b and the angle between the vectors.
a  b  (5  4)  (4  (5))  (2  3)  6
But a  45 , b  50 , hence
cos 
ab

ab
6
45  50

2
5 10
 2 
 arccos 

 5 10 
(vi)
B
a  b  a b cos 
 OA  OB cos
 OA  OX
b
O

X
a
A
Hence the projection or component of b in the direction of a is
1
b cos  a  b
a
 b  (unit vect or in a direction)
6
Example
A force F = 2i +3j +k acts on a particle which is displaced though d = i – j + 2k. Find the
component of F in the direction of d and the work done by the force.
d 23 2
1
Component of F is F  

d
6
6
Work done by F  (component of F in direction of d)  d

d
  F 
d

 F d

d


1
( vii)
ab  a b
(Schwartz inequality )
( viii) a  b  a  b
(ix)
(Triangle inequality )
a  b  a  b  2( a  b )
2
2
2
2
An Introduction to Inner Product Spaces
The concept of the dot product may be generalized to other vector spaces.
A real vector space V is called a real inner product space if with every pair of vectors a and b
in V there is associated a real number denoted by a, b and called the inner product of a and
b with the following properties
1. ma  nb, c  m a, c  n b, c
a, b V and m, n  R - Linearity
2. a , b  b, a
a, b V
3. a, a  0 and
a, a  0
iff
a 0
- Symmetry
- Positive - definitene ss
Example
1. R 3 is an inner product space with
3
a, b  a  b   ai bi
(the dot product)
i 1
2. R n is an inner product space with
n
a, b   a i bi
i 1
3.
Pn  {set of all polynomial s of degree  n} is an inner product space on an interval
7
  x   with

pn , qn   pn ( x)qn ( x)dx

3
In R we are familiar with the ideas of length and angle which are related to the vector dot
product by
Length of a  a  a  a
and
Angle,  , between a and b is given by
ab
cos 
a b
In an inner product space we use these relationships to define length and angle in terms of the
inner product.
The length of a, known as the norm of a and denoted by a
a 
is defined as,
a, a
The angle,  , between a and b is defined by
cos  
a, b
a
b
Two vectors a and b in an inner product space are said to be orthogonal if
a, b  0 .
Example
In the inner product space V of all continuous functions on 0,   with inner product

a, b   a( x)b( x)dx , we have
0
Length of f(x) = x is x 
Length of f(x) = sinx is


0
x 2 dx 
sin x 


0
3
3
sin 2 xdx 

x, s i nx   x s i nx d x
0
8

2
 x, sin x
Angle between x and sin x  cos 1 x
 x sin x

 6

 cos 1 

  




The vector product or cross product of two vectors a and b, written a  b , is defined as
 a b sin  v
ab  
0
if a  0, b  0
otherwise
Where v is a unit vector such that a, b, v form a right-handed triple, and  is the angle
between a and b.
V
b

a
Properties of the cross product
(i) The result is a vector
a  b is zero iff a = 0 or b = 0 or a and b are parallel.
(ii)
(iii)
aa  0
a  b  b  a
(iv)
a  b  a b sin  = area of parallelogram with sides a and b
(v)
i
j
a  b  a1
b1
a2
b2
k
a3  i (a2 b3-a 3b2 )  j (a3b1-a1b3 )  k (a1b2 -a 2 b1 )
b3
(Prove using results (vii) and (viii) below)
(vi) a  (b  c)  (a  b)  (a  c)
(vii) m (a  b)  (ma  b)  (a  mb)  (a  b)m
(viii)
i i  j  j  k k  0,
i  j  k , j  k  i, k  i  j
Example
If a = 5i + 4j + 2k and b = 4i – 5j + 3k, find a x b and a unit vector perpendicular to both a
and b.
9
i
j
k
ab  5
4
2 i
4 5 3
4
2
5 3
j
5 2
4 3
k
5
4
4 5
 22i  7 j  41k
A unit vector is 
ab
1

(22i  7 j  41k )
ab
2214
Example – Moment of a force
In mechanics the moment, m, of a force F about a point 0 is defined as the magnitude of F
times the perpendicular distance (d) from 0 to the line of action, L, of F. Let r be the vector
from 0 to any point on L.
O
d
m  F r sin 
r
F


m  rF
The vector m = r x F is called the vector moment of F about 0. Its direction is along the axis
about which F has a tendency to produce a rotation.
Example – Magnetic Field
The force F experienced by a point charge q moving with velocity v in a magnetic field of
flux density B is given by
F=qvxB
Example – Rotation
Consider a rigid body rotating with angular speed w about an axis. Let w be the vector with
magnitude w and direction along the axis such that the rotation of the body appears clockwise
looking along this direction.
Let P be any point in the body and d its distance from the axis.
Then P has speed wd
Let P have position vector r with respect to some point 0 on the axis.
Then
10
d  r sin 
wd  w r sin   w  r
And the velocity v of P is given by v  w  r
Products of three or more vectors follow naturally:
The triple scalar product is
- a scalar
a  (b  c)
Properties
i
j
k
a1
a2
a3
b3  b1
c3 c1
b2
c2
b3
c3
(i)
a  (b  c)  (a1 i  a2 j  a3 k )  b1 b2
c1 c2
(ii)
a  (b  c)  (a  b)  c - property of determinant and the triple scalar product is
usually written (a, b, c)
(iii)
(a, b, c) = (b, a, c)
(a, b, c) = (b, c, a) = (c, a, b)
(iv)
Geometrically the magnitude of (a, b, c) equals the volume of the parallelepiped
with a, b and c as adjacent edges.
v
a
c
b
Area of base  b  c
Perpendicular height = a  v , where v = unit vector in direction of b  c
Volume  a  v b  c
 abc
c
(v)
Three vectors are coplanar iff their triple scalar product is zero.
The triple vector product is
-a vector
a  (b  c)
11
Properties
Note that a  (b  c)  (a  b)  c in general
(i)
e.g. i  ( j  j )  0 whereas (i  j )  j  i
a  (b  c)  (a  c)b  (a  b)c (prove by expanding both sides in components –
(ii)
straightforward but tedious)
Some Vector Identities
(a) (a  b)  (c  d )  (a  c)(b  d )  (a  d )(b  c)
(b) (a  b)  (c  d )  (a, b, d )c  (a, b, c)d
(a  b)  (b  c)  (c  a)  (a, b, c) 2
(c)
Example
Prove identity (a) above.
(a  b)  (c  d )  a  b  (c  d )
triple scalar product of a, b, c  d
property (ii)
 a  (b  d )c  (b  c)d 
 (a  c)(b  d )  (a  d )(b  c)
1.5
Linear Dependence and Independence
If a1 , a 2 ,...., a k are any k vectors, then an expression of the form
k
m a
i
i
, ( m1 , m2 ,....., mk are any k scalars) is called a linear combination of
i 1
a1 , a 2 ,...., a k .
a1, a2 ,...., ak  is a linearly dependent set if at least one of the vectors can be represented
as a linear combination of the others. Otherwise the set is linearly independent.
Examples
The vectors a = 3i + 5j – 2k, b = 4j + 2k and c = i + j – k are linearly dependent since
a
1
b  3c
2
Hence vector a lies in the plane of the vectors b and c.
The vectors i, j and k are linearly independent.
An equivalent definition is : A set of k vectors is linearly independent iff
k
m a
i 1
i
i
 0 implies m1  m2    mk  0
12
Proof of equivalence :
Assume m p  0 for some 1  p  k , then
k
m a
i
i 1
i
0
k
iff m p a p    mi a i
i 1
i p
k
iff a p   
i 1
i p
mi
ai
mp
iff {a1 , a 2 ,..., a k } is linearly dependent.
If two vectors is 3-D space are linearly dependent they must be colinear. If three vectors in
3-D space are linearly dependent they must either be colinear of coplanar. Hence three
vectors form a linearly independent set iff their triple scalar product is not zero.
Four or more vectors is 3-D space will always be linearly dependent.
Example
If a = 3i + 5j – 2k, b = 4j + 2k, c = i + j – k, c = i + j – k
3 5 2
(a, b, c)  0 4 2  0
1 1
1
And hence a, b and c are linearly dependent.
The ideas of linear dependence and linear independence generalize to all vector spaces V. The
above definitions apply if a i  V for i = 1, 2, …, k.
If a vector space V contains a linearly independent set of n vectors whereas any set of n + 1 or
more vectors is linearly dependent then we say that V is n-dimensional or has dimension n. In
an n-dimensional vector space any set of n linearly independent vectors is called a basis of V.
Any vector v may be expressed uniquely as a linear combination of the vectors in the basis.
Note, however that a basis is not unique.
In an inner product space, vectors which are orthogonal are also linearly independent (can
13
you prove this?) but the converse is not true (give an example).
A basis {v1 , v 2 ,..., v n } in an n-dimensional inner product space V is said to be orthogonal if
vi , v j  0
if i  j
An orthogonal basis is orthonormal if
vi
2
 vi , vi  1
for i  1,2,...., n
Example
If V  R 3 , the dimension of R 3 is 3, as expected, and the 3 vectors i, j and k,
which are linearly independent, form an orthonormal basis for R 3 . Any vector v may
(i)
be
written as a linear combination of i, j and k.
(ii) The vectors i + j, 2i – j and k also form a basis for R 3 since they are linearly
independent and any vector in R 3 may be expressed as a linear combination
of these vectors, eg. –4i + 5j + 6k = 2(i + j) – 3(2i – j) + 6k. However they are
not useful in practice since they are not orthogonal.
(iii) The vectors a = 3i + 5j – 2k, b = 4j + 2k and c = i + j – k of the previous example do not
form a basis for R 3 since they are linearly dependent. The vector d = i + j + k, for
example, cannot be expressed as a linear combination of a, b and c.
Example
If V  P2 = {all polynomials in x with degree  2 } then l, x, x 2 form a basis for P 2
since they are linearly independent
iff a0  a1 x  a2 x 2  0
for all x
a0  a1  a2  0
And adding a 4th quadratic polynomial to the set would produce a linearly dependent set.
Thus P 2 has dimension 3. Clearly every polynomial of degree 2 can be written uniquely in
terms of 1, x, x 2
However with the usual inner product on [0, 1]
1
p2 , q2   p2 ( x)q2 ( x)dx
0
These basis polynomials are not orthogonal, since
1
1
 1  xdx  2  0
0
14
The set P0 ( x)  1, P1 ( x)  x 
1
1
and P2 ( x)  x 2  x  do form a linearly independent set
2
6
and are orthogonal to one another, i.e. they form an orthonormal basis.
Example
If V  R n , the n vectors
(1, 0, 0, …,0), (0, 1, 0,…,0), …(0, 0, …, 0, 1) form an orthonormal basis with respect to the
inner product
n
a, b   a i bi
i 1
and are known as the standard basis.
Example
The vector space of all continuous functions on an interval [ ,  ] has a basis
1, x, x 2 ,...., x n ,
Therefore the vector space is of infinite dimension.
15