REVIEW
OF
MATHEMATICS
Review of Vectors Analysis


T
T
a

[
a
,
a
,
a
]
,
b

[
b
,
b
,
b
]
Given
x
y
z
x
y
z
  T 
 
Dot product: a  b  a b  axbx  a y by  az bz  a b cos( )
 is the angle between the two vectors.

T
a
Example:   [1,3,5] a  b   1(2)  3(4)  5(6)  44
T
b  [2,4,6] 

a  35 
44 
1 


  cos 
b  56 
 35 56 

 
2
2
2
a

a

a

a

a

a
Magnitude of vector:
x
y
z
Example: a  [1,2,3]T  a  12  22  32  14
Review of Vectors Analysis


Vectors a and b are said to be perpendicular or
orthogonal if  
a b  0
Example:


T
T
a  1 0 0 ,
b  0 1 0
 
 
a b  0  a  b
Note that the above vectors represent the unit vectors for the Xaxis and Y-axis. They are definitely perpendicular or orthogonal.
Review of Vectors Analysis
 
Cross product: a  b  a y bz  a z by  a z bx  a x bz 
a b
x y
 a y bx 
 eˆ1 eˆ2 eˆ3 
 


a  b  det a x a y a z 
bx by bz 
 

a  b  a b sin( )
 is the angle between the two vectors.
Example:

 
T
a  [1,3,5]  a  b   3(6)  5(4) eˆ1  5(2)  1(6) eˆ2  1(4)  3(2) eˆ3

T
T
b  [2,4,6] 
 (2) (4)  2

a  35 


24 
1 
b  56   sin 

 35 56 
 

a  b  24 
Review of Vectors Analysis


The cross product of a and b provides
 us with a vector

which is perpendicular to both a and b


T
T
Example:
a  1 0 0 ,
b  0 1 0
 iˆ ˆj kˆ 
0 
 


a  b  det 1 0 0   0 iˆ  (0) ˆj  (1)kˆ  0
 
0 1 0 
1


Note that the above vectors represent the unit vectors for the Xaxis and Y-axis respectively. Their cross product is the unit
vector for the Z-axis, which is definitely perpendicular to both
the X-axis and the Y-axis.
Review of Vectors Analysis
Note that the unit vectors for the right handed Cartesian reference
frame are orthonormal basis vectors, i.e.
iˆ  ˆj  kˆ,
ˆj  kˆ  iˆ
kˆ  iˆ  ˆj
iˆ  ˆj  ˆj  kˆ  kˆ  iˆ  0
iˆ  ˆj  kˆ  1

   
a  b  a b cos( )  0      2n ,
n  1,2,
2
 
 
a  b  a b sin( )  0    2n ,
n  0,1,2,
Review of Vectors Analysis
        
Vector triple product: a  b  c   b a  c   c a  b 
Example:

T
a  0 1 0 
0   0 

     
T 
b  1 0 0 a  b  c   1   1  0 0 0
   

T
c  0 0 1 
0  0 
  
T
b a  c   1 0 0 (0  0  0)  0 0 0
  
T
c a  b   0 0 1 (0  0  0)  0 0 0
Review of Vectors Analysis
    
  
Scalar triple product: ab c   a  b  c   a  b  c
a x

 det bx
 cx
ay
by
cy
az 

bz 
c z 
Example:

T
a  0 1 0 
0 1 0 


T  
b  1 0 0  ab c  det 1 0 0  0  1(1  0)  0  1



T
c  0 0 1 
0 0 1
  
T
T
a  b  c   0 1 0  0  1 0  1
  
a  b  c  0 0  1T  0 0 1T  1
 
Review of Vectors Analysis
d 
a  a x
dt

T
a

[
a
,
a
,
a
]
Given
x
y
z
Example:
Given
a z 
T
T

2
a  t  t 2t  1
d 
T
a  2t  1 2
dt
d

 (a )   a x
dt

a  [ax , a y , az ]T
where  is a any constant
Example:
a y

a  t 2  t
d 
3a  32t
dt
2t  1
T
 1 2
T
a y
a z 
T
Review of Vectors Analysis


T
T
a

[
a
,
a
,
a
]
,
b

[
b
,
b
,
b
]
Given
x
y
z
x
y
z
d      
a  b   a  b  a  b
dt
Example:
2

 2
2
t
2
t

3
t




T

2
a  t  t 2t  1  d     
   t  0 


a

b


1

1

T 
  
  
 
b  2t  3 1  t   dt
 2    t  2t  1  1
d  
a  b   {(4t 2  6t )  1  (2t )}  {2t 2  2t  1}  6t 2  10t  2
dt
Review of Vectors Analysis


T
T
a

[
a
,
a
,
a
]
,
b

[
b
,
b
,
b
]
Given
x
y
z
x
y
z
d      
a  b   a  b  a  b
dt
Example:
 5t 2 
10t
sin( t )

 

cos(t )


d 
t t
 1    cos(t )   t    sin( t ) 
a

b



T

 
   

b  sin( t )  cos(t ) 0  dt
2
3


 3t   0   t   0 
ˆj
ˆj
 iˆ
 iˆ
kˆ 
kˆ 
d  




a  b  det  10t
1
 3t 2   det  5t 2
t
 t3 
dt
sin( t )  cos(t )
cos(t ) sin( t ) 0 
0 





a  5t 2


d  
a b
dt





d  
a b
dt


3 T

 

 3t 2 cos(t )
t 3 sin( t )

 

3

 3t 2 sin( t )


t
cos(
t
)
 

2
 {10t cos(t )  sin( t )} 5t sin( t )  t cos(t )

 



t 3 sin( t )  3t 2 cos(t )


   {t 3 cos(t )  3t 2 sin( t )} 
5t 2 sin( t )  11t cos(t )  sin( t )


Review of Vectors Analysis


d 
Ab   A b  Ab
dt
where A is a matrix of dimension comparable to the vector
being multiplied

Given b  [bx , by , bz ]T
Example:


b  t2

d 
T
t 1  b  2t 1 0
dt
2t 1 0 
 2 0 0
d
A   0 t 3   A  0 1 0

 dt


 2 0 3t 
0 0 3
T
2
2
2
2
2 0 0 t  2t 1 0  2t  2t  4t  1 6t  1
d  
 
  
 

Ab  0 1 0  t    0 t 3   1    t    t    2t 



 
dt
0 0 3  1   2 0 3t   0   3   4t   4t  3 
 
Eigenvalues and Eigenvectors

Let A be an nn matrix. If there exists a  and a nonzero n1 vector x
such that


Ax  x

then  is called an eigenvalue of A and x is called an eigenvector
of A corresponding to the eigenvalue 
Let In be a nn identity matrix. The eigenvalues of nn matrix A
can be obtained from:
det( A  I n )  0
A nn matrix A has at least one and at most “n” distinct eigenvalues
Example 1: Eigenvalues and Eigenvectors
Find the eigenvalues of
Solution:
1 2 3
A  0  4 2 


0 0 7
2
3 
1 2 3 
1 0 0 1  
A  I n  0  4 2   0 1 0   0
4
2 



 

0
7   
0 0 7 
0 0 1  0
2
3 
1  
det( A  I n )  0  det  0
4
2 0


0
7   
 0
1    4   7     0
  1,  4, 7
Example 2: Eigenvalues and Eigenvectors
What is the eigenvector of
 4  36 33 at =1?
1 
A
48
9
4

49 
 9 32 36



Ax  x   A  I x  0
  4  36 33 1 0 0   x1  0
1
   48
9
4   10 1 0   x2   0
 
    
49 
  9 32 36 0 0 1   x3  0
 36
33   x1  0
(4  49)
1
  x   0 
  48
(9  49)
4
 2   
49 
32
(36  49)  x3  0
  9
 45  36 33   x1  0
1
  48  40 4   x2   0
   
49 
32  13  x3  0
  9
Example 2: Eigenvalues and Eigenvectors
 45x1  36 x2  33x3  0
48x1  40 x2  4 x3  0
 9 x1  32 x2  13x3  0
Multiply 3rd eqn by -5 and add it to 1st eqn to eliminate x1
 45x1  36 x2  33x3  0
45x1  160x2  65x3  0
  196x2  98x3  0
x3
 2
x2
Example 2: Eigenvalues and Eigenvectors
 45x1  36 x2  33x3  0
48x1  40 x2  4 x3  0
 9 x1  32 x2  13x3  0
Divide 2nd eqn by x2 and simplify using the known result:
48
x
x1
 40  4 3  0
x2
x2
x1
48  40  4(2)  0
x2
x1 32 2
 

x2 48 3
Example 2: Eigenvalues and Eigenvectors
Story so far:
x3
 2,
x2

x  x1
x2
x1 2

x2 3
x3 
T
2

 x2  1 2 
3

T
We can obtain a normalized eigenvector using:
 2 1 2
x

2

x

3

xn   
2
x
2
x2    12  2 2
 3
T
1
 3 2
T

xn   1 2  2 3 6
7 3
7

Trigonometric Functions
sin 2 ( )  cos 2 ( )  1
sin(  )   sin( )
cos(  )  cos( )
tan(  )   tan( )
sin(1   2 )  sin(1 ) cos( 2 )  cos(1 ) sin( 2 )
cos(1   2 )  cos(1 ) cos( 2 )  sin(1 ) sin( 2 )
tan(1 )  tan( 2 )
tan(1   2 ) 
1  tan(1 ) tan( 2 )
Trigonometric Functions
sin( )
tan( ) 
cos( )
d
sin( )  cos( )
d
d
cos( )   sin( )
d
d
d
sin( )  cos( )   cos( )
dt
dt
d
d
cos( )   sin( )    sin( )
dt
dt
 sin( )d   cos( )
 cos( )d  sin( )