1
Vector Calculus
The vector is the basic tool in the formalism of mechanics because it brings together in one concept two fundamental ideas, that is the size of the used parameter or the studied phenomenon and the direction in which it must be considered or in which it applies. The calculus rules that describe it are continuously exploited in the mathematical expression of the motion of bodies. This chapter lists them and develops them for the ease of use.
1.1. Vector space
1.1.1. Definition
The vector space E is a set with two operating laws: an internal law, from E E→ , which confers an Abelian group structure (commutative), and an external law, the multiplication by a scalar. The elements of a vector space are called vectors and, in the formalism of mechanics, are generally represented by an alphabetical symbol topped with an arrow: u .
1.1.1.1. Properties of the internal composition law
The internal composition law in the formalism of mechanics is the vector addition, denoted as +, and that has the following properties:
– if ,u v E∈ , so, u v E+ ∈ ;
COPYRIG
HTED M
ATERIAL
2 Movement Equations 2
– it is commutative: , ,u v E u v v u∀ ∈ + = + ;
– it is associative: ( ) ( ), , ,u v w E u v w u v w∀ ∈ + + = + + ;
– it has a neutral element denoted as 0 so that:, 0 0u E u u u∀ ∈ + = + = ;
– any element has an inverse (or opposite), that is to say: , ,∀ ∈ ∃ − ∈u E u E such as: ( ) 0u u u u+ − = − = .
1.1.1.2. Properties of the external composition law
The external composition law is identified in the mechanical formalism as the multiplication by a scalar λ ∈ of a vector ,u E∈ such that:
u u Eλ λ× = ∈ .
This law has the following properties:
– there is a neutral element, the scalar 1 as: 1 u u× = ;
– the law is distributive with respect to addition and in relation to the addition and multiplication in , as:
( )( )
( ) ( )
, , :
, , :
, , :
u v E u v u v
u E u u u
u E u u
λ λ λ λ
λ μ λ μ λ μ
λ μ λ μ λμ
⎧∀ ∈ ∀ ∈ × + = +⎪⎪∀ ∈ ∀ ∈ + × = +⎨⎪
∀ ∈ ∀ ∈ × × =⎪⎩
1.1.2. Vector space – dimension – basis
We say that n vectors 1 2, , , nu u u E∈… are linearly independent if the relation:
1 1 2 2 0n nu u u uα αλ λ λ λ+ + + = =…
Vector Calculus 3
has only the following solution:
0, 1, ,nαλ α= ∀ = … .
We say that a vector space E is of dimension n when it holds at most n linearly independent vectors; any other vector element of this space is then expressed as a linear combination with coefficients
αλ ∈ of these n vectors. Any other vector of the vector space E of dimension n can be expressed from only these n linearly independent vectors, which constitute the basis of what we call space.
Thus, if this basis consists of n linearly independent vectors
1 2, , , nu u u E∈… , any vector V of E can be written as:
1 1 2 2 n nV u u uμ μ μ= + + +… ,
where the coefficients 1 2, , , nμ μ μ ∈… are the components of V in the considered basis.
1.1.3. Affine space
The first area to be considered as the frame of the movement of physical bodies is terrestrial space. This can be represented by geometrical parameters and its properties fit well with the concept of vector space, a mathematical entity to which it is advised to give a reality; hence the notion of affine space of vector space.
This notion follows a precise mathematical definition that is not necessary to repeat here; it merely gives the physical space the mathematical specificity, which is necessary to make it the mechanical workplace, and thus the characteristics of the vector space rules applied to it.
4
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6 Movement Equations 2
The sign of the scalar product that is an algebraic quantity depends on whether θ is acute ( )cos 0θ > or obtuse ( )cos 0θ < .
1.3.1. Properties of the scalar product
Vis-à-vis the vector operations, specific to vector spaces, scalar product has the following properties:
– it is commutative:
a b b a⋅ = ⋅ ,
– it is distributive right and left with respect to the vector addition:
( )1 2 1 2a b b a b a b⋅ + = ⋅ + ⋅ and ( )1 2 1 2a a b a b a b+ ⋅ = ⋅ + ⋅ ,
– its multiplication by a scalar gives:
( ) ( ) ( )a b a b a bλ λ λ⋅ = ⋅ = ⋅ ,
– the scalar product of two linear combinations of the vectors unfolds as follows:
( )1 1 1 1
m n m n
i i j j i j i ji j i j
a b a bλ μ λ μ= = = =
⎛ ⎞ ⎛ ⎞⋅ = ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑ .
1.3.2. Scalar square – unit vector
As ( )cos , 1a a = , we obtain the scalar square of a by the
operation:
2 2a a a a⋅ = = so
2a a= .
If we consider a group of three linearly independent vectors ( )1 2 3, ,U U U of the affine space, we can define the three vectors:
u
which and thaThis b
( 1 2, ,u u
directio
(
1.3.3.
Con
b to th
The
which iunit vec
11
1
,UuU
=
are vectors wat can be theasis ( )u is
)3,u , becaus
ons. This bas
( ) ( 1 2u u u≡
Geometric
nsider two ve
he straight su
Figure 1.3
algebraic m
is the orthogoctor u , is giv
22
2
UuU
=
whose norme basis ( )us more usefse they offer
sis will be de
)3u .
c interpreta
ectors a and
upport of vect
3. Orthogonal
measurement o
onal projectioven by:
33
3
, UuU
=
ms are equal of the vectofully represer a same me
enoted as:
ation of the
d b , and proj
tor a whose
projection of a
of vector pro
on of the vec
Ve
,
to one, calleor affine spaented by theasurement r
scalar pro
ject orthogon
e unit vector
a vector on an
aoj b , denote
ctor b on the
ector Calculus
ed unit vectoace considerehe unit vectoreference in
oduct
nally the vec
is aua
= .
n axis
ed as aproj b
direction of t
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ors ed. ors all
tor
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the
8
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an
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Movement Eq
aproj b
The project
aproj b
Under thes
a b⋅ =
We conside
Figu
This vector
(aproj b− = −
nd its algebra
aproj b b− =
quations 2
cosb b θ=
tion vector p
( )b u b u= ⋅ =
e conditions
a proj= ×
er the projec
ure 1.4. Orthog
r is expressed
)( )u b u− ⋅ −
aic measure h
(cosb π θ−
a bu ba
θ ⋅= ⋅ =
aproj b is th
( )2
a b a
a
⋅= .
, we can also
aj b a pro= ⋅
tion b on −
gonal projectio
d as:
( )u b u= ⋅ =
has the value
) cosbθ = −
b .
herefore writ
o write:
aoj b .
a− according
on of a vector
( )2
a b ap
a
⋅=
e:
s u bθ = − ⋅ =
tten as:
g to Figure 1
r on an axis
aproj b ,
a b pa⋅− = −
1.4.
aproj b
Vector Calculus 9
1.3.4. Solving the equation a x⋅ = 0
Consider a given vector a and the equation 0a x⋅ = . It accepts two types of solution:
so and
0
, cos 02 2
x
x a a x π π
⎧ =⎪⎨⎪ ⊥ = =⎩
.
Therefore, the scalar product of two vectors is zero if one of the two vectors, at least, is zero, or if the two vectors are orthogonal.
1.4. Vector product ∧a b
1.4.1. Definition
The vector product of the two vectors a and b of affine space 3E of dimension 3 is represented by the operation a b∧ . Its result is the vector c a b= ∧ , which has the following properties:
– it is orthogonal to the plane formed by the vectors a and b : ( ),c a b⊥ Π ;
– it is oriented so that the trihedron ( ), ,a b c is direct (see below);
– sinc a b θ= × × with π θ π− ≤ ≤ .
NOTE.– According to the rule of the corkscrew by Maxwell, a corkscrew planted perpendicular to the plane ( ),a bΠ progresses,
when rotated from a toward b , in the direction of the vector c .
10
1.
ac
0 Movement E
.4.2. Geom
We considccording to F
Figur
As such, th
( OAire
Equations 2
Figure 1
etric interp
der the triangFigure 1.6.
re 1.6. Geome
he area of its
) 1OAB2
= ×
1.5. Maxwell’s
pretation of
gle OAB bu
etric interpreta
surface is eq
sina b θ×
corkscrew rul
f the vecto
uilt on the tw
ation of the ve
qual to 1 O2
×
12
a bθ = ∧
le
or product
wo vectors a
ctor product
OA BH× , so:
.
and b
:
Vector Calculus 11
1.4.3. Properties of vector product
The vector product has properties that are either its own or combined with the other operations of vectors:
– the vector product is anticommutative, that is to say:
b a a b∧ = − ∧ .
This property results from the fact that the trihedrons ( ), ,a b a b∧
and ( ), ,b a b a∧ must be both direct;
– multiplication by a scalar obeys the rule:
( ) ( ) ( )a b a b a bλ λ λ∧ = ∧ = ∧
– the vector product is distributive with respect to the addition of vectors:
( )( )
1 2 1 2
1 2 1 2
a b b a b a b
a a b a b a b
∧ + = ∧ + ∧
+ ∧ = ∧ + ∧
– the vector product of two linear combinations of vectors is developed:
( )1 1 1 1
m n m n
i i j j i j i ji j i j
a b a bλ μ λ μ= = = =
⎛ ⎞ ⎛ ⎞∧ = ∧⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑ .
1.4.4. Solving the equation a x∧ = 0
Consider a given vector a and equation 0a x∧ = . It accepts two types of solution:
and, as a result, and
0
, 0 sin 0 0
x
x a a xλ
⎧ =⎪⎨⎪ = = =⎩
.
12
tw
1.
1.
( a
1.
in
2 Movement E
Therefore, wo vectors, at
.5. Mixed p
.5.1. Defini
The mixed ), ,a b c , whic
( ), ,a b c
.5.2. Geom
Consider th
Figur
Consider tnterpretation
(Area O
Equations 2
the vector pt least, is zer
product ( a ,
tion
product of tch represents
) (a b c= ⋅ ∧
etric interp
he trihedron
re 1.7. Geome
the vector pof this opera
)OBDC 2= ×
product of twro, or if the tw
),b ,c
three vectorss the operatio
) .
pretation of
formed by th
etric interpreta
product b ∧ation, we wri
Area (OBC×
wo vectors iwo vectors a
s is the scalaron:
f the mixed
he three vect
ation of the mix
c ; accordinite:
C) b c= ∧ .
s zero if oneare co-linear.
r product den
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ixed product
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nd c .
eometric
Vector Calculus 13
The segment OH, orthogonal projection of a on the vector b c∧which is orthogonal to the plane ( ),b cΠ , is the height of the
parallelepiped constructed on the three vectors a , b , c . The volume of the parallelepiped is equal to:
( )OBDC OH×Aire , with OH cosa θ= × .
We deduce that
( ) ( )[ ]
( ) ( )parallelepiped , , 6 tetrahedron OABC
" " , ,
a b c
a b c a b c
⎡ ⎤ = ×⎣ ⎦
= ⋅ ∧ =
Vol Vol.
If we express this result as:
( ) ( )parallelepiped , ,a b c a b c⎡ ⎤ =⎣ ⎦Vol ,
this algebraic volume is positive when the trihedron ( ), ,a b c is direct,
negative when it is indirect.
1.5.3. Properties of the mixed product
1.5.3.1. Cases of nullity of the mixed product
The relationship ( ), , 0a b c = corresponds to the following cases:
– 0a = ;
– 0b c∧ = , which means one of these 0
0
such as
b
c
b cλ λ
⎧ =⎪⎪ =⎨⎪∃ ∈ =⎪⎩
;
– a b c⊥ ∧ , which means , such as a b cλ μ λ μ∃ ∈ = + .
14 Movement Equations 2
The mixed product of three vectors is zero when one of the vectors is zero or when two of them are at least linked.
1.5.3.2. Circular permutation of terms
If we consider the parallelepiped constructed on the three vectors a , b , c , the calculation of its volume does not depend on the order in which one considers them. We can thus calculate at first the vector product of a and b , then project c onto the vector a b∧ , or the vector product c a∧ and project b onto this vector; the main thing is to conserve the order of vectors, since the algebraic sign of the volume depends on the direct or indirect order of the three vectors. We can either write the equalities:
( ) ( ) ( )a b c b c a c a b⋅ ∧ = ⋅ ∧ = ⋅ ∧ ,
that is to say we can invert the operations, such that:
( ) ( )a b c a b c⋅ ∧ = ∧ ⋅ .
Symbolically therefore we write:
( ) ( ) ( ), , , , , ,a b c b c a c a b= = .
1.5.3.3. Permutation of two terms
For the same reasons related to the direct or indirect nature of the trihedron formed by the three vectors of a mixed product, in the order they are listed, the permutation of two vectors changes the sign of the result of the operation. So
( ) ( ), , , ,a c b a b c= − .
1.5.3.4. Multiplication by a scalar
Multiplying a mixed product by a scalar amounts to multiplying one of the vectors by this scalar:
( ) ( ) ( ) ( ), , , , , , , ,a b c a b c a b c a b cλ λ λ λ× = = = .
Vector Calculus 15
1.5.3.5. Distributivity
The operation is distributive with respect to the addition of vectors on each member of the mixed product, as, for example:
( ) ( ) ( )1 2 1 2, , , , , ,a a b c a b c a b c+ = + .
1.5.3.6. Mixed product of a combination of vectors
( )1 1 1 1 1 1
, , , ,l m n l m n
i i j j k k i j k i j ki j k i j k
a b c a b cλ μ ν λ μ ν= = = = = =
⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ ∑ ∑ ∑ ∑ .
1.6. Vector calculus in the affine space of dimension 3
1.6.1. Orthonormal basis
The orthogonal projection of vectors applies practically trigonometric functions. A trirectangular trihedron vector corresponds to linearly independent vectors since the orthogonal projection of one of them, on the plane formed by the two others, is always zero. The mechanical formalism is thereby greatly simplified. So it is worthwhile to resort, if another need does not justify it, to the orthogonal bases and, moreover, normed.
The vectors that constitute the orthonormal basis ( ) ( )1 2 3u u u u=
therefore verify all the relations:
2 2 21 2 3
1 2 2 3 3 1
1 2 3 2 3 1 3 1 2
1 1 1
0 0 0
u u u
u u u u u u
u u u u u u u u u
= = =
⋅ = ⋅ = ⋅ =
∧ = ∧ = ∧ =
.
16 Movement Equations 2
1.6.2. Analytical expression of the scalar product
Consider the orthogonal basis ( ) ( )1 2 3u u u u= , the two vectors
a and b , and their respective components ( )1 2 3, ,a a a and ( )1 2 3, ,b b bon it. These vectors are thus written:
1 1 2 2 3 3a a u a u a u= + + and 1 1 2 2 3 3b b u b u b u= + + .
The scalar product:
( ) ( )1 1 2 2 3 3 1 1 2 2 3 3a b a u a u a u b u b u b u⋅ = + + ⋅ + + ,
thus is developed, taking into account the properties of the unit vectors of the orthonormal basis:
1 1 2 2 3 3a b a b a b a b⋅ = + + .
If the basis was not orthonormal, we should take it into account, in the development, the expression of the different scalar products i ju u⋅ that are involved.
In the analytical form, the scalar square of the vector a is written as:
2 2 2 21 2 3a a a a a a= ⋅ = + + .
1.6.3. Analytical expression of the vector product
Similarly, the vector product of the two vectors:
1 1 2 2 3 3a a u a u a u= + + and 1 1 2 2 3 3b b u b u b u= + +
Vector Calculus 17
is written, taking into account the properties of the orthonormal basis:
( ) ( )( ) ( ) ( )
1 1 2 2 3 3 1 1 2 2 3 3
2 3 3 2 1 3 1 1 3 2 1 2 2 1 3"
a b a u a u a u b u b u b u
a b a b u a b a b u a b a b u
∧ = + + ∧ + +
= − + − + −.
We can calculate, using a practical method, this vector product by using the following determinant that is effectively developed with respect to the first line:
( ) ( ) ( )1 2 3
1 2 3 2 3 3 2 1 1 3 3 1 2 1 2 2 1 3
1 2 3
u u u
a a a a b a b u a b a b u a b a b u
b b b
= − − − + −
in order to obtain the expression.
1.6.4. Analytical expression of the mixed product
Consider the three vectors:
1 1 2 2 3 3 1 1 2 2 3 3, ,a a u a u a u b b u b u b u= + + = + +
1 1 2 2 3 3,c c u c u c u= + +
and their mixed product ( ), ,a b c :
,
( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( )
1 1 2 2 3 3
2 3 3 2 1 3 1 1 3 2 1 2 2 1 3
1 2 3 3 2 2 3 1 1 3 3 1 2 2 1
, ,
"
a b c a b c a u a u a u
b c b c u b c b c u b c b c u
a b c b c a b c b c a b c b c
= ⋅ ∧ = + +
⎡ ⎤⋅ − + − + −⎣ ⎦= − + − + −
…
…
( ) ( ) ( ) ( )1 2 3 3 2 2 3 1 1 3 3 1 2 2 1, ,a b c a b c b c a b c b c a b c b c⇒ = − + − + −
18
re
1.
1.
1.
Th
vepl
th
8 Movement E
esults that ma
, ,a b c
.7. Applicat
.7.1. Doubl
.7.1.1. Defin
Consider th
he vector b
ector d a=lane accordin
Fi
So we can
In the dir
hree vectors:
Equations 2
ay also be ca
1 1
2 2
3 3
a ba ba b
=
tions of ve
le vector pr
nition
he three vect
c∧ is ortho
( )b c∧ ∧ thng to the foll
igure 1.8. Illus
write ,λ μ∃ ∈
ect orthonor
alculated by u
1
2
3
ccc
.
ector calcul
roduct
tors a , b , c
ogonal to th
hat is orthogowing diagra
stration of the
∈ such as
rmal basis
using the det
lus
, and the ope
he plane ( bΠ
gonal to b ∧am.
double vector
d b cλ μ= +
( ) ( 1 2u u u=
terminant:
eration (a ∧
),b c ; theref
c∧ belongs
r product
c .
)2 3u , consi
( ).b c∧
fore, the
s to this
ider the
Vector Calculus 19
1 1 2 2 3 3
1 1 2 2 3 3
1 1 2 2 3 3
a a u a u a u
b b u b u b u
c c u c u c u
⎧ = + +⎪⎪ = + +⎨⎪ = + +⎪⎩
,
where ( ) ( ) ( )2 3 3 2 1 3 1 1 3 2 1 2 2 1 3b c b c b c u b c b c u b c b c u∧ = − + − + − .
If we consider the component 1u of the vector ( )d a b c= ∧ ∧ ,
which is of the form d b cλ μ= + , we can write:
( ) ( )( ) ( )( ) ( )
1 1 2 1 2 2 1 3 3 1 1 3
1 1 2 2 3 3 1 1 1 2 2 3 3 1
1 1
"
"
b c a b c b c a b c b ca c a c a c b a b a b a b c
a c b a b c
λ μ+ = − − −= + + − + +
= ⋅ − ⋅
.
In the same way, we obtain:
( ) ( ) ( ) ( )2 2 2 2 3 3 3 3,b c a c b a b c b c a c b a b cλ μ λ μ+ = ⋅ − ⋅ + = ⋅ − ⋅ ,
and we can identify:
,a c a bλ μ= ⋅ = ⋅ .
This gives the development of the double vector product
( ) ( ) ( )a b c a c b a b c∧ ∧ = ⋅ − ⋅ .
1.7.1.2. Particular characteristics of the double vector product
( ) ( ) ( ) ( )a b c c b a c a b c b a∧ ∧ = ∧ ∧ = ⋅ − ⋅ .
The difference, memberwise, of this relationship and the previous yields:
( ) ( ) ( ) ( )a b c a b c c b a a b c∧ ∧ = ∧ ∧ + ⋅ − ⋅ .
20 Movement Equations 2
In particular, when c a= , we have:
( ) ( )a b a a b a∧ ∧ = ∧ ∧ ,
and if λ∃ ∈ such as c aλ= :
( ) ( ) ( ) ( ) 0a b a a b a a b a a b aλ λ λ λ∧ ∧ − ∧ ∧ = ⋅ − ⋅ = ,
with the result that if a and c are collinear:
( ) ( )a b c a b c∧ ∧ = ∧ ∧ .
1.7.1.3. The case for nullity with the double vector product
Consider the relationship ( ) 0a b c∧ ∧ = . It is checked if:
– at least one of the three vectors, a , b or 0c = ;
– the vector 0b c∧ = λ⇒ ∃ ∈ such as b cλ= ;
– the vector a is collinear with b c∧ ;
λ⇒ ∃ ∈ such as ( )a b cλ= ∧ or else ( ),a b c⊥ Π .
1.7.1.4. Projection of a vector on a plane
Consider the plane ( )aΠ whose normal vector is a , and the unit
vector aua
= ; consider the vector b and its orthogonal projection on
the plane (parallel to a ).
The vector b is expressed in terms of its two projections:
( )a ab proj b proj bΠ= + with ( ) ( )2a
a bproj b u b u a
a
⋅= ⋅ = .
We
p
From
u
we obtvector axis:
b
The
the dir
given b
p
Figu
deduce that:
( )aproj bΠ =
m the formul
( )u b u∧ ∧ =
tain the expon an axis o
( )b u b u= ⋅ +
erefore, the p
ection defin
by:
( )aproj bΠ =
ure 1.9. Projec
:
ab proj b−
la of the dou
(2u b u b= − ⋅
pression of tof unit vector
( )u b u+ ∧ ∧
projection of
ned by vecto
( )u b u∧ ∧
ction of a vect
( )2
a bb
a
⋅= −
uble vector pr
)b u ,
the three-dimr u and on t
) .
f a vector b
or a , and th
(2
a b a
a
∧ ∧=
Ve
or on a plane
a .
roduct:
mensional pthe plane orth
on an orthog
he unit vect
)a.
ctor Calculus
projection ofhogonal to th
gonal plane,
or aua
= ,
21
f a his
in
is
22 Movement Equations 2
1.7.2. Resolving the equation a x b⋅ =
Consider two data, a vector 1 1 2 2 3 3a a u a u a u= + + and a scalar b .
This is about determining the vector 1 1 2 2 3 3x x u x u x u= + + ,
satisfying the equation a x b⋅ = , that is to say the relationship 1 1 2 2 3 3a x a x a x b+ + = that admits, as an equation, a double infinity of
solutions.
Indeed, if there are infinitely many solutions to choose 1x , for example, this value taken, there are still an infinite number of values to choose for 2x ; 3x then depends on these two choices.
Suppose that we know a particular solution 0x that satisfies the
equation 0a x b⋅ = ; it also satisfies the relationship:
( )0 0a x x⋅ − = ,
that is to say that the vector 0x x− is orthogonal to a .
We only needs to know one of these particular solutions in order to then express the general form of the equation’s solution.
Choosing one, among all the possible solutions, which is collinear with a , so the form 0x aλ= .
2
0 2ba x a a a b
aλ λ λ⋅ = ⋅ = = ⇒ = .
By stating:
OA a= , 0 0OM x= and OM x= ,
the ext( 0MΠ
1.7.3.
Suchthat is t
a
If wwrite:
a
that is t
x
Amorthogo
a
treme M of)a , passin
Figur
Resolving
h an equatioto say if:
0a b⋅ = so a
we know a pa
( )0a x x∧ −
to say that th
0x x aλ− =
ong all the ponal to a , th
0 0a x⋅ = w
f the vector g through M
re 1.10. Resol
the equati
on only has
1 1 2 2a b a b+ +
articular solu
0= ,
he vector x −
so 0x x=
possible vecthat is to say:
with 0a x∧ =
position OM
0M and orthog
lving the equa
ion a x b∧ =
a solution if
3 3 0a b = .
ution 0x sati
0x− is colline
aλ+ .
tor solutions,
b= .
Ve
M is situategonal to OA
ation a x b⋅ =
b
f a and b a
isfying a x∧
ear to a , and
, we choose,
ctor Calculus
ed in the plaA .
are orthogon
0x b= , we c
d is:
if there is, x
23
ane
nal,
can
0x
24
ob
a
thpa
4 Movement E
Multiplyingbtain:
( )0a a x∧ ∧
The genera
bxa
∧=
By stating,
OA a=
he extremityassing throug
F
Equations 2
g vectorially
( )0a x a= ⋅ −
al solution is
2a aλ∧ + .
as per the di
a , OB b= ,
M of the vegh 0M and p
Figure 1.11. R
y left by a
20a x a− = −
thus in the f
iagram below
0 0OM x=
ector OMparallel to O
Resolving the
the vector
20x a b= ∧
form of:
w:
and OM
is located oA .
equation a ∧
product abo
0x⇒ =
x= ,
on the straig
x b=
ove, we
2b a
a
∧= .
ght line
Vector Calculus 25
1.7.4. Equality of Lagrange
Consider two vectors a and b , and perform the scalar square of the two following terms, the one, the scalar a b⋅ , the other, the vector a b∧ .
( ) ( )
( ) ( ) ( ) ( )
2 21 1 2 2 3 3
2 2 2 22 3 3 2 3 1 1 3 1 2 2 1
a b a b a b a b
a b a b a b a b a b a b a b
⋅ = + +
∧ = − + − + −.
The sum memberwise of these two expressions yields, after development and simplification:
( ) ( ) ( )( ) ( )( )2 2 2 22 2 2 2 2 21 2 3 1 2 3a b a b a a a b b b a b∧ + ⋅ = + + + + = .
This equality of Lagrange is also written as:
2 2
1a b a ba b a b
⎛ ⎞ ⎛ ⎞∧ ⋅+ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
,
and reduces, in fact, to the trigonometric expression 2 2sin cos 1θ θ+ = .
1.7.5. Equations of planes
1.7.5.1. Plane normal to a vector and passing through a point
In the affine space that is associated with the frame 1 2 3O u u u ,
we consider a vector 1 1 2 2 3 3a a u a u a u= + + represented by the vector
position OA and a unit vector 1 1 2 2 3 3v v u v u v u= + + .
The question is to express the equation of the plane ( )A νΠ
passing through the point A and orthogonal to v .
26
eq
an
1.po
path
6 Movement E
Figure 1.
Consider th
OM x=
This point
AM v⋅ =
The plane
quation:
OM v⋅ =
nd has the Ca
1 1 2v x v+
.7.5.2. Planoint
Consider tassing throughe plane ( AΠ
AM ∈
Equations 2
12. Plane norm
he point M i
1 1 2 2x u x u+ +
M is such th
( OM OA= −
( )A vΠ i
OA v= ⋅ ,
artesian equa
2 2 3 3x v x+ =
ne defined
the two vectgh the point
)A ,b c is su
( )A ,b cΠ
rmal to a vecto
n this plane:
3 3x u+ .
hat:
)A 0v⋅ = .
is the locus
ation:
1 1 2 2v a v a+ +
by two vec
tors b and A of vector
uch that:
,λ μ⇒ ∃ ∈
or and passing
of points
3 3v a .
ctors and p
c , the planposition OA
such tha
g through a po
M that sati
passing thro
ne containingA . Any poin
at AM bλ=
oint
isfy the
rough a
g them, nt M of
b cμ+ .
Acc
O
In consdefinesthe prev
The
(
1.7.6.
Not
Figure 1.13.
cording to Fig
OM OA= +
sidering the s the orthogovious proble
e equation of
( ) OMb c∧ ⋅
Relations w
e the triangle
Fi
Plane defined
gure 1.13, w
AM OA+ = +
scalar produonal directioem.
f the plane is
( ) Ob c= ∧ ⋅
within the t
e represented
igure 1.14. Co
d by two vecto
we have:
b cλ μ+ + .
uct of this exn to the plan
given by the
OA .
triangle
d in the figur
onfiguration of
Ve
rs and through
xpression byne, one is br
e equation:
re below.
f a triangle
ctor Calculus
h a point
y b c∧ , whirought back
27
ich to
28 Movement Equations 2
Consider the relation of the Chasles BC BA AC= + and multiply its two members, scalarly, by BC . We obtain
2
2
BC BC BA BC AC BC BA CB CAcos cosa ac abβ γ
= ⋅ + ⋅ = ⋅ + ⋅⇒ = +
,
therefore cos cosa c bβ γ= + .
Considering now the scalar square of the Chasles relation:
( )2 2 2 2
2 2
BC BA AC BA AC 2 BA AC
AC AB 2 AC AB,
= + = + + × ⋅
= + − × ⋅
so 2 2 2 2 cosa b c bc α= + − .
According to the definition of the vector product, the area A of the triangle is given by:
2 sin sin sinbc ca abα β γ= = =A ,
and, by dividing each term of this expression by the product abc , we obtain a fundamental relationship of the triangle:
sin sin sina b c
α β γ= = .
1.8. Vectors and basis changes
1.8.1. Einstein’s convention
Given the size of mathematical expressions that will be developed in this work and formulas to be used therein, it quickly becomes essential to have a condensed notation system that facilitates writing and exploitation. The Einstein notation convention is a response to
Vector Calculus 29
that expectation. To illustrate the principle, consider the example of a system of linear equations of n variables ix for 1, ,i n= …
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
n n nn n n
a x a x a x ba x a x a x b
a x a x a x b
+ + + =⎧⎪ + + + =⎪⎨⎪⎪ + + + =⎩
……
……
.
Because of the Einstein notation convention, this system can be written in the following condensed form
, , 1, ,ij j ia x b i j n= ∀ = … ,
and is interpreted as follows:
– the index i that appears only once in each member of the equality indicates that there are as many relations as values of this index, so here n ; it is identified as free index. These are the n equations of the above system mixed into one relationship;
– the index j that appears only twice in the left section of the equality means that in each of the n equations of the given system, this member consists of the sum of n monomials obtained by giving to j its n values. This index is called summation index. The summation index only has meaning if the index appears only twice in a monomial convention.
NOTE.– Widely used in digital programming, this agreement is particularly interesting if the various indices used have the same range of variation; as will often be the case in the situations that will be discussed in this book.
For complete information on using the Einstein convention, take the example of the expression index:
ij i ja b c dα αβ β=
30 Movement Equations 2
– indices i and j appear only once in each of the two members. They are free indices. When their values are chosen, the expression of the relevant term ija outcomes;
– the indices α and β appear, however, twice in the right side of the index relationship; so these are the summation indices.
Consider, for the four indices, the values 1,2,3 . We obtain, for example, for 1, 2i j= = :
12 11 11 12 11 12 22 11 13 32
12 21 12 12 22 22 12 23 32
13 31 12 13 32 22 13 33 32
a b c d b c d b c db c d b c b b c db c d b c d b c d
= + ++ + ++ + +
…… ……
.
1.8.2. Transition table from basis ( )e to basis ( )E
Consider two direct orthonormal bases:
( ) ( ) ( ) ( )1 2 3 1 2 3,e x x x E X X X= = ,
and a vector f whose projection is expressed in the two bases.
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
1 1 2 2 3 3
1 1 2 2 3 3
f f x x f x x f x x f x x f x
f f X X f X X f X X f X X F X
α α α α
β β β β
⎧ = ⋅ + ⋅ + ⋅ = ⋅ =⎪⎨
= ⋅ + ⋅ + ⋅ = ⋅ =⎪⎩
The relationship between the two bases is expressed by the following expressions:
; ; ,p x X x p X X p xαβ α β α αβ β β αβ α= ⋅ = =
which can be summarized by the transition table ( ),p e E from the basis ( )e to basis ( )E where the reading is done by line to express the
Vector Calculus 31
vectors of basis ( )e according to those of basis ( )E or by column to determine those ( )E based on those ( )e .
( ) ( ) 1 2 3
1 11 12 13
2 21 22 23
3 31 32 33
e E X X X
x p p p
x p p p
x p p p
.
This table is also used to express the components of a vector of a basis in the other.
( ) ( )( ) ( )
f f x f p X p f X p F
F f X f p x p f x p f
α α αβ β αβ β αβ β
β β αβ α αβ α αβ α
= ⋅ = ⋅ = ⋅ =
= ⋅ = ⋅ = ⋅ =.
If we use a matrix representation of vectors and basis change operation, we can write:
1 11 12 13 1 1 11 21 31 1
2 21 22 23 2 2 12 22 32 2
3 31 32 33 3 3 13 23 33 3
orf p p p F F p p p ff p p p F F p p p ff p p p F F p p p f
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
.
From ( )[ ], ,p e E the matrix directly transcribed from the transition table from basis ( )e to basis ( )E , we have:
( )[ ]11 12 13
21 22 23
31 32 33
,p p p
p e E p p pp p p
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
,
and we see that the matrix that corresponds to a vector from the basis ( )E to a vector of basis ( )e is the transpose of the previous one, which can be expressed by:
( )[ ] ( )[ ], ,p E e p e E= .
32 Movement Equations 2
The basis change of the vector f between ( )e and ( )E is therefore written as a matrix:
[ ] ( ) ( )[ ][ ] ( )
[ ] ( ) ( )[ ][ ] ( ) ( )[ ][ ] ( )
,
, ,e E
E e e
f p e E f
f p E e f p e E f
=
= =.
1.8.3. Characterization of the transition table
The fact that the transition table takes place between two direct orthonormal bases means that its nine elements pαβ are not independent of each other. First note that the bases are orthonormal:
x xα β αβδ⋅ = and X Xμ ν μνδ⋅ = ,
,i i
x x p X p X p p X Xp p p p
α β αμ μ βν ν αμ βν μ ν
αμ βν μν α β αβδ δ⇒ ⋅ = ⋅ = ⋅
= = =
which gives six relations that should verify the elements of the transition table between two bases.
We get the same six relations between the elements of the transition table by writing:
i iX X p x p x p p x x p p p pμ ν αμ α βν β αμ βν α β αμ βν αβ μ ν μνδ δ⋅ = ⋅ = ⋅ = = = .
In addition, the orientation of the bases introduces new relationships, whether they are direct or indirect. If the basis ( )e is direct, as it is orthonormal, the relationship below is enough to express its orientation.
1 2 3x x x= ∧ ,
Vector Calculus 33
since the three vectors are orthogonal and this relationship enlightens unequivocally on the basis orientation. In the change of basis, the relation is written as:
1 2 3 2 3p X p X p X p p X Xα α β β γ γ β γ β γ= ∧ = ∧ .
The terms of this expression exist only if the values of the three indices , ,α β γ come within a circular permutation 1,2,3 or 1,3,2 , which gives another three relationships that should cross-check with the terms of the table. Since these are six distinct terms, it is clear that these relations are redundant.
When a transition table or a matrix change in basis has been established, we should check that:
– non-diagonal terms are anti-symmetric;
– vectors are unitary, that is to say, for each row or each column, the sum of squares of the three elements is equal to 1.
– vectors are orthogonal, that is to say the sum of the products of the relevant terms of two lines or two columns is zero, in checking it in the three combinations of lines or columns (1+2, 1+3, 2+3).
This control is required when drawing up transition table.