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Multiplication of vectors
 Two different interactions (what’s the difference?)
Scalar or dot product :
A  B | A || B | cos   B  A
 the calculation giving the work done by a force during a
displacement
 work and hence energy are scalar quantities which arise from the
multiplication of two vectors
 if A  B  0
• The vector A is zero
• Or The vector B is zero
• Or  = 90°
A  Axiˆ  Ay ˆj  Az kˆ
B  Bxiˆ  By ˆj  Bz kˆ
A  B  Ax Bx  Ay By  Az Bz
A

B
Vector or cross product :
A  B | A || B | sin  nˆ
 n is the unit vector along the normal to the plane containing A
and B and its positive direction is determined as the right-hand
screw rule
A  B  B  A
 the magnitude of the vector product of A and B is equal to the
area of the parallelogram formed by A and B
 if there is a force F acting at a point P with position vector r
relative to an origin O, the moment of a force F about O is
defined by :
L  r F
 If
A B  0
• The vector A is zero
• Or The vector B is zero
• Or  = 0°
A

B
Commutative law :
A B  B  A
A  B  B  A
Distribution law :
A  (B  C)  A  B  A  C
A (B  C)  A B  A C
Associative law :
A  BC  D  ( A  B)(C  D)
A  BC  ( A  B)C
A  B  C  ( A  B)  C
A  ( B  C )  ( A  B)  C
Unit vector relationships
 It is frequently useful to resolve vectors into components
along the axial directions in terms of the unit vectors i, j,
and k.
i× j = j×k = k×i = 0
i×i = j× j = k×k = 1
i×i = j× j = k×k = 0
i× j = k
j×k = i
k×i = j
A  Axiˆ  Ay ˆj  Az kˆ
B  Bxiˆ  By ˆj  Bz kˆ
A  B  Ax Bx  Ay By  Az Bz
iˆ
A  B  Ax
Bx
ˆj
Ay
By
kˆ
Az
Bz
A B  C
Scalar triple product
The magnitude of A B  C is the volume of the parallelepiped with edges parallel to
A, B, and C.
AB
C
B
A
A  B  C  A  B  C  B  C  A  B  C  A  C  A  B  [ A, B, C ]
A B  C
Vector triple product
The vector A B is perpendicular to the plane of A and B. When the further vector
product with C is taken, the resulting vector must be perpendicular to A B and
hence in the plane of A and B :
( A  B)  C  mA  nB
where m and n are scalar constants to be determined.
C  ( A  B)  C  mC  A  nC  B  0
m  C  B
n  C  A
( A  B)  C   (C  B) A  (C  A) B
Since this equation is valid
for any vectors A, B, and C
Let A = i, B = C = j:
  1
( A  B)  C  ( A  C ) B  ( B  C ) A
A  ( B  C )  ( A  C ) B  ( A  B)C
AB
C
B
A
Polar coordinate system:
x  r cos  ;
y  r sin  ;
r  x2  y 2 ;
  atan 2(y, x)
r  rrˆ
unit vectors:
rˆ  cos  xˆ  sin  yˆ ;
ˆ   sin  xˆ  cos  yˆ;
Remember to separate variables for integration calculations:
r and  should be integrated independently.
. Move anything related to r into r’s integration,
and move anything related to  r into ’s integration…
Please carefully observe the length of each small “cube” highlighted above.
Vector integration
Linear integrals
Vector area and surface integrals
Volume integrals
An arbitrary path of integration can be specified by defining a variable
position vector r such that its end point sweeps out the curve between P and Q
Q
dr
r
P
A vector A can be integrated between two fixed points along the curve r :

Q
P
Q
A  dr   ( Ax dx  Ay dy  Az dz )
P
Scalar product
if A·B = 0
The vector A is zero
or The vector B is zero
or  = 90°
Work, force and displacement
If the vector field is a force field and a particle at a point r experiences a force f,
then the work done in moving the particle a distance r from r is defined
as the displacement times the component of force opposing the displacement :
W  F  r
The total work done in moving the particle from P to Q is the sum of the increments
along the path. As the increments tends to zero:
Q
W   F  dr
P
When this work done is independent of the path, the force field is “conservative”.
Such a force field can be represented by the gradient of a scalar function :
F  W
A  
When a scalar point function is used to represent a vector field, it is called a
“potential” function :
gravitational potential function (potential energy)……………….gravitational force field
electric potential function ………………………………………..electrostatic force field
magnetic potential function……………………………………….magnetic force field
Two-dimensional system E  E x i  E y j
dr  dxi  dyj
Three dimensional system ndS  dxdyk

Q
P
Q
E  dr   ( E x dx  E y dy  Ez dz )
P
P
Q
Considering a surface S having element dS and curve C denotes the curve :
We also have : The surface integral of the velocity vector u gives
the net volumetric flow across the surface, which is called Flux.
 E  dS   E  ndS
 u  dS   u  ndS
S
Surface : a vector by reference to its boundary
area : the maximum projected area of the element
direction : normal to this plane of projection (right-hand screw rule)
n
dS  ndS
n is the normal direction of the surface.
The surface integral is then :
A
 E  dS   E  ndS
If E is a force field, the surface integral gives the total force acting on the surface.
If E is the velocity vector, the surface integral gives the net volumetric flow
across the surface. It is called “Flux”.
n
Flux:
 E  dS   EA
 E  dS   EA cos 
 is the angle
between
the Vector
and the normal
direction of the
surface
When the
Vector is
“coming in”,
Angle > 90
Flux <0
When the
vector Is
“going out”,
Angle <90
Flux >0
When the vector
is perpendicular
to the surface,
Angle =0. Flux is
maximized.
n
E

En


E  n = E cos

cos    cos(180   )   cos 
E  n ds = - E ds cos 
Carefully go over the solutions of
the 1st take home Math Quiz.
Question 4-7