Advanced Structured Materials

Sorin VlaseMarin MarinAndreas Öchsner

Eigenvalue and Eigenvector Problems in Applied Mechanics

Advanced Structured Materials

Volume 96

Series editors

Andreas Öchsner, Faculty of Mechanical Engineering, Esslingen Universityof Applied Sciences, Esslingen, GermanyLucas F. M. da Silva, Department of Mechanical Engineering, Faculty ofEngineering, University of Porto, Porto, PortugalHolm Altenbach, Otto-von-Guericke University, Magdeburg, Sachsen-Anhalt,Germany

Common engineering materials reach in many applications their limits and newdevelopments are required to fulfil increasing demands on engineering materials.The performance of materials can be increased by combining different materials toachieve better properties than a single constituent or by shaping the material orconstituents in a specific structure. The interaction between material and structuremay arise on different length scales, such as micro-, meso- or macroscale, and offerspossible applications in quite diverse fields.

This book series addresses the fundamental relationship between materials and theirstructure on the overall properties (e.g. mechanical, thermal, chemical or magneticetc.) and applications.

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• classical fibre-reinforced composites (e.g. class, carbon or Aramid reinforcedplastics)

• metal matrix composites (MMCs)• micro porous composites• micro channel materials• multilayered materials• cellular materials (e.g. metallic or polymer foams, sponges, hollow sphere

structures)• porous materials• truss structures• nanocomposite materials• biomaterials• nano porous metals• concrete• coated materials• smart materials

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Sorin Vlase • Marin MarinAndreas Öchsner

Eigenvalue and EigenvectorProblems in AppliedMechanics

123

Sorin VlaseDepartment of Mechanical EngineeringTransilvania University of BraşovBraşov, Romania

Marin MarinDepartment of Mathematicsand Computer Science

Transilvania University of BraşovBraşov, Romania

Andreas ÖchsnerFakultät MaschinenbauEsslingen University of Applied SciencesEsslingen, Germany

ISSN 1869-8433 ISSN 1869-8441 (electronic)Advanced Structured MaterialsISBN 978-3-030-00990-8 ISBN 978-3-030-00991-5 (eBook)https://doi.org/10.1007/978-3-030-00991-5

Library of Congress Control Number: 2018957055

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https://doi.org/10.1007/978-3-030-00991-5

Preface

This volume presents, in a unitary way, some problems of applied mechanics,analyzed using the matrix theory and the properties of eigenvalues and eigenvec-tors. Problems and situations of different nature are studied. Different problems andstudies in mechanical engineering lead to patterns that are treated in a similar way.The same mathematical apparatus allows the study of mathematical structures suchas the quadratic forms but also mechanical problems such as multibody rigidmechanics, continuum mechanics, vibrations, elastic and dynamic stability ordynamic systems. A substantial number of engineering applications illustrate thisvolume.

Braşov, Romania Sorin VlaseBraşov, Romania Marin MarinEsslingen, Germany Andreas Öchsner

v

Contents

1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Vectors. Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Vector Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Addition of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.2 Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.2.3 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.4 Scalar Triple Product . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.2.5 Vector Triple Product . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.2.6 Applications of Vector Calculus . . . . . . . . . . . . . . . . . . . 24

1.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.2 Basic Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.2.1 Addition ðþMm�n � Mm�n ! Mm�nÞ . . . . . . . . . . . . 452.2.2 Scalar Multiplication ð�R � Mm�n ! Mm�nÞ . . . . . . . . 462.2.3 Matrix Multiplication ð�Mm�p � Mp�n ! Mm�nÞ . . . . 472.2.4 Inverse Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.2.5 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.2.6 Transposed of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 512.2.7 Trace of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.2.8 Matrix Representation of the Cross Product . . . . . . . . . . . 52

2.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.4 Ortogonal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.5 Some Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . 552.6 Block Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

vii

2.7 Matrix Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582.7.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . 582.7.2 Diagonalization of Symmetric Matrices . . . . . . . . . . . . . . 59

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3 Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.2 Extreme Values of a Real Function of Two Variables . . . . . . . . . . 633.3 Conics and Quadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.4 Quadratic Forms in a n-Dimensional Space . . . . . . . . . . . . . . . . . 653.5 Eigenvalues and Eigenvectors for Quadratic Forms . . . . . . . . . . . . 67

3.5.1 The Conditions for a Quadratic Form to Be Positive . . . . 673.5.2 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.5.3 Fundamental Theorems of Quadratic Forms . . . . . . . . . . . 693.5.4 Schur’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.6 Orthogonal Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 723.7 Invariants of Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4 Rigid Body Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.1 Finite Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4.1.1 Defining the Position of a Rigid Body . . . . . . . . . . . . . . 874.1.2 Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924.1.3 Bryan (Cardan) Angles . . . . . . . . . . . . . . . . . . . . . . . . . . 954.1.4 Finite Rotations and Commutativity . . . . . . . . . . . . . . . . 97

4.2 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1004.2.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1004.2.2 Moment of Inertia; Definitions . . . . . . . . . . . . . . . . . . . . 1054.2.3 Rotation of the Coordinates System . . . . . . . . . . . . . . . . 1084.2.4 Moment of Inertia of a Body Around an Axis . . . . . . . . . 1104.2.5 Directions of Extremum for the Moments of Inertia . . . . . 1114.2.6 A Property of the Principal Direction of Inertia . . . . . . . . 1134.2.7 Inertia Ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1154.2.8 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.2.9 Geometrical Moments of Inertia . . . . . . . . . . . . . . . . . . . 1364.2.10 Moment of Inertia of Planar Plates . . . . . . . . . . . . . . . . . 137

Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

5 Strain and Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.1 Strain Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

5.1.1 Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.1.2 Lagrangian and Eulerian Description . . . . . . . . . . . . . . . . 1425.1.3 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

viii Contents

5.1.4 Infinitesimal Deformation . . . . . . . . . . . . . . . . . . . . . . . . 1495.1.5 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . 1505.1.6 The Physical Significance of the Components

of the Strain Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1525.1.7 Transformation Induced by the Strain Tensor . . . . . . . . . 1535.1.8 Local Rigid Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

5.2 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.2.1 Stress State in a Point . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.2.2 Transformation of the Stress Tensor to Axis

Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1565.2.3 Normal Stress Corresponding to an Arbitrary

Direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1575.2.4 Extremal Conditions for Normal Stress . . . . . . . . . . . . . . 1585.2.5 Invariants of the Reduced Stress . . . . . . . . . . . . . . . . . . . 1615.2.6 Conic of Normal Stress . . . . . . . . . . . . . . . . . . . . . . . . . 1625.2.7 Quadric of Normal Stress . . . . . . . . . . . . . . . . . . . . . . . . 1635.2.8 Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 164

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

6 Modal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1676.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1676.2 Modal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

6.2.1 Eigenvalues—Natural Frequencies . . . . . . . . . . . . . . . . . 1686.2.2 Properties of the Eigenvalues . . . . . . . . . . . . . . . . . . . . . 1706.2.3 Orthogonality Properties . . . . . . . . . . . . . . . . . . . . . . . . . 1716.2.4 Rayleigh’s Quotient . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1726.2.5 Generalized Orthogonality Relationships . . . . . . . . . . . . . 1726.2.6 Definition of Relationships for the Damping Matrix . . . . . 1746.2.7 Normalized Vibration Modes . . . . . . . . . . . . . . . . . . . . . 1746.2.8 Decoupling the Motion Equations . . . . . . . . . . . . . . . . . . 175

6.3 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1786.4 Vibration of Continuous Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

6.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2216.4.2 Transverse Vibration of a Bar . . . . . . . . . . . . . . . . . . . . . 2216.4.3 Eigenvalues and Eigenmodes in Transverse

Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2236.4.4 Axial Vibrations of Bars . . . . . . . . . . . . . . . . . . . . . . . . 2266.4.5 Eigenvalues and Eigenfunction in Axial Vibration . . . . . . 2276.4.6 Torsional Vibration of the Bar . . . . . . . . . . . . . . . . . . . . 2316.4.7 Eigenvalues and Eigenfunctions in Torsional

Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

Contents ix

7 Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2397.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2397.2 Linear Systems with Two Degrees of Freedom . . . . . . . . . . . . . . 2407.3 Free Vibration of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

x Contents

Chapter 1Vectors

1.1 Vectors. Fundamental Notions

Mechanics uses physical quantities that cannot be characterized only by their mag-nitude (how it happens with a scalar) but also need other attributes to define them,namely the direction, or point of application (fixed or mobile). They represent a mul-titude of quantities, frequently used such as forces acting on a material point or solid,speeds, accelerations, moment of force etc. These quantities are called vectors.

A vector contains the measure of the magnitude or length (arithmetic element) towhich the direction (geometric element) is added. The direction is defined with oneword as the vector orientation.

It follows that the vector is geometrically represented by an oriented segment. Thedirection on which the vector will act is given by the line segment, with a definitedirection (it is a binary dimension geometrically indicated by an arrow at the end ofthe segment) and the magnitude (positive numerical value) is given by the length ofthe segment, which is represented by a certain convenient scale. In Fig. 1.1 a vectoras an oriented segment is represented. The point A is called the initial point or origin,base, tail and the point B is the terminal (final) point, head, endpoint, or tip (seeLiesen and Mehrmann 2015; Springer 2013; Simionescu 1982).

The vector notations are different, depending on the geometric or algebraicapproaches as well as the authors. Several notations have come to terms and arepresented below. Because the vector is geometrically represented by a directed linesegment, it has an origin called, for example, A and one end called, for example, B.In this case, the vector is marked with an arrow. The first letter indicates the originand the other the end. If the extremities are not called, vectors can be denoted by an

arrow letter(�a, �b, �u, �r

)or by a bar letter on the top

(ā, b̄, ū, r̄

). Sometimes the bar

is dropped and the letter is written in lowercase boldface as: (a, b, u, r) or lowercaseitalic boldface as: (a, b, u, r), especially in the Anglo-Saxon literature, in more math-ematical works. If algebraic representations are used, the vectors are also written

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2 1 Vectors

Fig. 1.1 A vector AB

as column matrix: {a}, {b}, {u}, {r}. In contrast to vectors, matrices are denoted byuppercase letters.

The vector magnitude (length or intensity) is denoted as in the case of algebra.

Thus, the vector magnitude of AB is∣∣∣−→AB

∣∣∣, for the vectors �a, �b, �u, �r , the moduleswill be |�a|,

∣∣∣�b∣∣∣, |�u|, |�r | or simpler as a, b, u, r . Sometimes for a, b, u, r, the notation

‖a‖, ‖b‖, ‖u‖, ‖r‖, is used.Classification of vectors: in mechanics, to characterize a vector used in a certain

type of problem, additional data is required. Thus, the vectors were classified intothe following three classes:

– a vector with fixed origin and head is called a bound vector. A fixed vector is themoment of a force in a point.

– a vector is called a free vector when only the direction and magnitude matter andthe origin is of no importance; so it can be considered any point of space.

For example velocity or acceleration.

– sliding vectors are vectors whose point of application can be found anywhere onthe support line of the vector without the mechanical effect on the body changing.Sliding vectors are forces that act on a rigid or angular velocity vector.

Equality of vectors:

Definition Two vectors are equal if they have the same magnitude and direction,they can be placed on the same or parallel lines.

The equality is expressed by the algebraic sign “=”. For example: �a � �b or ā � b̄or {a} � {b}.

1.1 Vectors. Fundamental Notions 3

Fig. 1.2 Representation ofthe unit vector

Equal vectors located on parallel lines are called equipollent vectors and theoperation by which a vector is moved by a translation on a line parallel to the vectorsupport is called an equipollent operation.

If two vectors have the same magnitude but opposite directions, they are calledopposite vectors and this is expressed by: ā � −b̄.

Two vectors that do not have the same magnitude but have the same support linedirection are called collinear vectors, regardless of the direction.

Unit vector: If we give a vector ā, we construct another vector ū with the samedirection as ā, whose magnitude is equal to unit size: u=1 (see Fig. 1.2). The vectorū is called the unit vector. If this choice is made, any colinear vector with ā can beexpressed with b̄ � λū.

Axis: If on a straight line we chose a positive sense, an origin and a unit of length,then we have defined an axis. The vector of the axis is the unit vector located on it,the sense of which coincides with the positive direction of the axis.

1.2 Vector Operations

1.2.1 Addition of Vectors

1.2.1.1 Sum of Two Vectors

Two vectors ā and b̄ are considered, see Fig. 1.3. We move both vectors through anequipollency in a point O of the space. It obtains a parallelogram formed with thesetwo vectors. The diagonal of this parallelogram is called the sum of the vectors ā and

4 1 Vectors

Fig. 1.3 Addition of twovectors

Fig. 1.4 Sum of severalvectors

b̄. The vectors ā and b̄ are called components of the vector c̄ and the vector c̄ � ā + b̄is called the resultant vector of the given vectors.

The geometric operation by which the resultant vector is constructed is named theparallelogram rule. The rule of addition has an experimental basis and is consideredan axiom if the vectors are forces. From a geometric point of view, the vector c̄ canbe obtained by placing at the head of vector ā the origin point of vector b̄. Then,vector c̄ resulting as the third side of the triangle so formed. If the rule of addition isconsidered, then it is immediately seen that it can be written: ā + b̄ � b̄ + ā that is,the addition of two vectors is a commutative operation.

1.2.1.2 Sum of Several Vectors

The sum can be naturally generalized if we deal with multiple vectors. Considerthe vectors ā1, ā2, . . . , ān . With vectors ā1 and ā2 we construct the sum vector s̄2 �ā1 + ā2, transporting ā2 to the head of vector ā1.

Then we construct the sum vector s̄3 � s̄2 + ā1 + ā2 + ā3, transporting ā3 to theendpoint of s̄2 (see Fig. 1.4).

The vector s̄n can be obtained by mathematical induction as the sum:

s̄n � ā1 + ā2 + · · · + ān.

1.2 Vector Operations 5

Consequently, it results that the sumvector is obtained by constructing the polygonformed by the vectors ā1, ā2, . . . , ān .

The resultant vector s̄ � s̄n � ā1, ā2 + · · ·+ ān , is obtained as an oriented segmentwith the origin point as the origin point of ā1 and the end point as the end point ofān . This process is called the polygonal contour rule.

For n vectors to have a null sum, the polygon built with them must be closed. Inparticular, in the case of three vectors, the condition that they have a null sum is that,by placing each vector with the origin at the end of the previous one, a triangle isformed. Because the triangle is a plane figure, the three vectors must be coplanar.

In the case of the sum of two vectors c̄ � ā + b̄, since the three form a triangle,one can write the inequality of the triangle in its known form:

c ≤ a + b. (1.1)

In the case of a polygonal contour generated by n vectors, which are placed one atthe extremity of the other, this inequality can be generalized in the following form:

s � |ā1 + ā2 + · · · + ān| ≤ a1 + a2 + · · · + an. (1.2)

We have this equality only if all the vectors are collinear and have the samephysical meaning.

1.2.1.3 Properties of Vectors Addition

The sum of the vectors has the following properties:

• The vector addition is commutative. This was shown when the sum was defined.It follows that the order in which the summation of more vectors is ordered isindifferent, that is, we have (Fig. 1.5):

ā + b̄ + c̄ � b̄ + ā + c̄ � b̄ + c̄ + ā � c̄ + b̄ + ā � c̄ + b̄ + ā. (1.3)

Fig. 1.5 Associativity of thevector addition

6 1 Vectors

• The vector addition is associative. If this is represented graphically, we immedi-ately find that we have:

(ā + b̄

)+ c̄ � ā + (b̄ + c̄). (1.4)

• The vector addition is distributive over the multiplication with a scalar.We will consider the vectors ā, b̄, c̄, d̄ , their sum is s̄ � ā + b̄+ c̄+ d̄ (see Fig. 1.6)

and the sum of the vectors mā,mb̄,mc̄ and md̄ is s̄ ′ � mā + mb̄ + mc̄ + md̄. Thesum s̄ is obtained by constructing the polygonal lineOABCD with the given vectors,the sum being OD.

We build the vectors:

OA′ � mOA � mā, OB ′ � mOB � mb̄,OC ′ � mOC � mc̄, OD′ � mOD � md̄.

The polygons OABCD and O′A′B′C′D are similar, having the sides parallel andthe same proportion of the sides, so

OD′ � mOD � m(ā + b̄ + c̄ + d̄).

However, we have:

OD′ � OA′ + A′B ′ + B ′C ′ + C ′D′� mā + mb̄ + m�c + md̄

By comparing the two expressions for−−→OD′ it results:

m(ā + b̄ + c̄ + d̄) � mā + m←b + mc̄ + md̄ (1.5)

Fig. 1.6 Addition of threevectors

1.2 Vector Operations 7

1.2.1.4 Decomposing a Vector

(a) Two-directional decomposition into two directions that determine a plane par-allel to the vector

Let us consider two straight lines, i.e.�1 and�2. LetO be the origin of the vector�a and A its head. We move the two straight lines in the origin O using a paralleltranslation. In this case, based on the mentioned hypotheses, the two lines and thegiven vector will be in the same plane (see Fig. 1.7).

It follows that the problem will be reduced in this case to finding two vectors withthe origin in O, �a1 along the straight line and �a2 along the straight line (�2) so thattheir sum is the given vector �a.

Taking into account that the rule of addition of two vectors is given by the par-alleogram rule, we will have to construct a parallelogram, with the sides on the lines(�1) and (�2) which will have as diagonal the vector �a. For this through the pointA, representing the head of the vector �a, one needs to construct two straight linesparallel to (�1) and (�2). One obtains the parallelogram OA1AA2, whose OA1 andOA2 sides represent exactly the vectors we are looking for, i.e. �a1 and �a2.

If the two straight lines (�1) and (�2) are perpendicular then �a1 and �a2 representthe orthogonal projections of vector �a on the two straight lines in both directions.

The decomposition of a vector in more than two directions, located in the sameplan, is undetermined (that is, there is an infinite number of decompositions of a vectorin three directions in the plane). This can be simply demonstrated if one arbitrarycomponent is taken in one direction and the other component is decomposed in twodirections.

(b) Decomposing a vector in three non-parallel directions in space

We note with Ox, Oy and Oz the three concurrent directions in which we want todecompose the vector ā with the origin in O. If the three directions are somewherein the space, we can move the vectors through O, using a parallel translation.

To perform the decomposition we proceed as follows: the lines OA and OA3determine a plane whose intersection with the Oxy plane is the straight line Ot. Nowthe vector �a can be decomposed along the directions Oz and Ot directions so that:ā � OA3 +OB � ā3 +OB. Furthermore, the OB vector found in theOxy plane can

Fig. 1.7 Decomposition of avector

8 1 Vectors

Fig. 1.8 Decomposition of avector in three directions

be decomposed in the directions Ox and Oy so that: OB � OA1 + OA2 � ā1 + ā2.It follows that the vector ā can be uniquely decomposed in the three directions givenso that (Fig. 1.8):

ā � OA3 + OA1 + OA2 � ā3 + ā1 + ā2.

If the three directions are the coordinate axes of an orthogonal system, then thethree projections are called the orthogonal projections of the vector on the three axes.

If the components ā1, ā2, ā3 are considered the sides of a parallelepiped then thegiven vector is the diagonal to this parallelepiped.

The decomposition of a vector in more than three directions not founded in thesame plane is indetermined.

(c) Algebraic representation of vectors

In the following, the concepts of vectorial calculus will be presented in bothgeometric and algebraic representation, since applications can be approached some-times more easily in the geometric representation and sometimes in the algebraicrepresentation.

Thus, if three unit vectors ī, j̄, k̄ of the same origin are considered, perpendiculartwo by two, which indicate the directions of the axes Ox, Oy, Oz, they will form anorthogonal coordinate system (see Fig. 1.9).

If a certain vector ā, is considered this vector can be decomposed in the three axesand it can be written as:

ā � ax ī + ay j̄ + azk̄. (1.6)

Themagnitudesax , ay, az are namedby the components of vector ā in the orthogo-nal coordinate systemOxyz.Wementionhere that it is possible to use a not-orthogonalsystem, but in the following, because the orthogonal system is used almost exclu-sively in common applications, the reference will be made to it, unless otherwisespecified.

1.2 Vector Operations 9

Fig. 1.9 Components of avector in an orthogonalcoordinate system

In this case the given vector can also be written in the form ā(ax , ay, az

)or

{a} �

⎧⎪⎨⎪⎩

axayaz

⎫⎪⎬⎪⎭

, (1.7)

in order to indicate its components. In this case the sum of two vectors can be definedas follows: if c̄ � ā + b̄ where ā(ax , ay, az

)and b̄

(bx , by, bz

)then the vector c̄ will

be the vector having the components:

cx � ax + bx ; cy � ay + by ; cz � az + bz .

The same can be written in matrix form as:

{c} �

⎧⎪⎨⎪⎩

cxcycz

⎫⎪⎬⎪⎭

� {a} + {b} �

⎧⎪⎨⎪⎩

axayaz

⎫⎪⎬⎪⎭

+

⎧⎪⎨⎪⎩

bxby

bz

⎫⎪⎬⎪⎭

�

⎧⎪⎨⎪⎩

ax + bxay + by

az + bz

⎫⎪⎬⎪⎭

. (1.8)

With this definition of the sum of two vectors, which will be reduced to a sum ofreal numbers, all the properties of the sum shown above are easily demonstrated.

1.2.1.5 Subtraction of Two Vectors

Definition. To subtract the vector b̄ from the vector ā, the task is to find the vectorc̄ that added to b̄ gives ā : b̄ + c̄ � ā. Subtracting of vectors is denoted by a minussign “−”:

10 1 Vectors

c̄ � ā − b̄. (1.9)

Geometrically, the subtraction can be made in two ways:

– for the parallelogram with a diagonal a and a side b at which the angle between aand b is known, we will have to construct the side c of the parallelogram, whichis joining the extremities of the vectors b and a and then leading through the headof a parallel to b and leading through the origin parallel to c. The direction of c isobtained from the condition that b summed with c must give a. It can be writtenas follows: c̄ � BA � OA − OB, (the relation of Chasles), i.e. the differencevector has as origin the extremity of the first vector and as end point the end pointof the second vector.

– it can be written c̄ � ā − b̄ � ā + (−b̄), that is, it adds ā to the vector (−b̄) (seeFig. 1.10).

Algebraically, the subtraction can be written:

{c} �

⎧⎪⎨⎪⎩

cxcycz

⎫⎪⎬⎪⎭

� {a} − {b} �

⎧⎪⎨⎪⎩

axayaz

⎫⎪⎬⎪⎭

−

⎧⎪⎨⎪⎩

bxby

bz

⎫⎪⎬⎪⎭

�

⎧⎪⎨⎪⎩

ax − bxay − byaz − bz

⎫⎪⎬⎪⎭

, (1.10)

i.e. the components of the difference are equal to the difference of the componentsof the two vectors.

1.2.1.6 Some Properties of Addition and Subtraction

• If the equality ā � b̄ is given, then a vector c̄ can be added to both sides and theequality remains.

• Given the equality ā � b̄, the same vector c̄ can be subtracted from both sides andthe equality remains. From the two properties it results that in a vectorial equality

Fig. 1.10 Subtraction of twovectors

1.2 Vector Operations 11

some terms can be passed from one part of the equality to the other, with a changedsign, as in any algebraic equality.

• If we have the vector equalities ā1 � b̄1 and ā2 � b̄2, then they can be added orsubtracted, term by term, and the equality remains.

1.2.1.7 Decomposing a Vector in Two Vectors

To decompose a known vector ā after two vectors b̄ and c̄ returns to decompose thevector ā in the directions of b̄ and c̄; it can be written:

OA � OA1 + OA2. (1.11)

These components can be expressed according to the vectors b̄ and c̄ (see Fig. 1.11)after the relations: OA1 � mb̄ and OA2 � nc̄ since the components are vectorscollinear with vectors b̄ and c̄. It results:

ā � mb̄ + nc̄. (1.11′)

It follows that any vector can be decomposed in two coplanar vectors with it. Ifthe two vectors are not collinear, decomposition is unique. If the vectors are collineardecomposition is indefinite.

The written relationship can be brought in a homogeneous form if we take:

m � −βα; n � −γ

α.

In this case the relation (1.11) can be written as:

αā + βb̄ + γ c̄ � 0, (1.12)

Fig. 1.11 Decomposition ofa vector in two vectors

12 1 Vectors

which is the condition that three vectors lie in the same plane. The relationshipsshows that if three coplanar vectors are given, any one of them can be expressed asa linear combination of the other two.

Algebraically, the relation (1.12) can be written as:

αax + βbx + γ cx � 0,αay + βby + γ cy � 0,αay + βby + γ cy � 0. (1.13)

According to the theory of homogeneous linear systems, there are non-zeroα, β, γthat satisfy the system if and only if the system determinant is zero:

∣∣∣∣∣∣∣

ax bx cxay by cy

az bz cz

∣∣∣∣∣∣∣� 0, (1.14)

which is the condition that three vectors are in the same plane (one column is a linearcombination of the other two).

Theorem The decomposition of a given vector in two other vectors, coplanar withit, is unique.

If it is assumed that there would be two decompositions we would have: ā �mb̄ + nc̄ and ā � m ′b̄ + n′c̄ with m � m ′ and n � n′. If we subtract these tworelationships, we get

(m − m ′)b̄ + (n − n′)c̄ � 0 or b̄ � − (n−n′)

(m−m ′) c̄ � λc̄ withλ � − (n−n′)

(m−m ′) , that is, b̄ and c̄ are collinear, which contradicts the hypothesis. It isvery easy to show that the cases m � m ′; n � n′ and m � m ′; n � n′ also contradictthe hypothesis, and it only remains: m � m ′; n � n′ that is, the decomposition isunique.

Theorem The necessary and sufficient condition for three vectors to be coplanar isto have the relationship (1.11) or (1.12) between them.

If the coplanar vectors ā, b̄ and c̄ are considered then ā can be decomposed in theother two vectors and it can be written: ā � mb̄ + nc̄ (the condition is necessary).

Reciprocally, the previously written relationship shows that ā is the sum of twovectors collinear with b̄, respectively c̄, that is a coplanar vector with them (thecondition is sufficient).

1.2.1.8 Scalar Multiplication

Let us consider a vector ā and the scalar λ. By definition, the product of a vectorwith a scalar λ is a vector b̄, colinear with ā, has the same direction with b̄ if λ ispositive and opposite direction if λ is negative and has the magnitude equal to |λ|ā.It is written as:

1.2 Vector Operations 13

�b � λ�a. (1.15)

1.2.1.9 Linearly Independent and Linearly Dependent CoplanarVectors

Since between two non-collinear vectors ā and b̄ there cannot be a linear relationship(if ā � λb̄ or αā + βb̄ � 0 the vectors would be collinear, i.e. a contradiction to thehypothesis), it is said that two non-collinear vectors are linearly independent.

Let us now consider three vectors that are in the same plane. Since relations (1.11)or (1.12) hold between them, there is a linear dependence between them. For thisreason, they are called linearly dependent.

If two non-collinear vectors b̄ and c̄ are considered, any coplanar vector with themcan be expressed as (1.11) or (1.12). So using these two vectors can express any othercoplanar vector with them. These vectors are called fundamental vectors.

1.2.1.10 Decomposing a Vector in Three Non-coplanar Vectors

We consider a vector ā and we propose to write it as the sum of three collinearvectors respectively with b̄, c̄, d̄ (they are not coplanar). For this, the three vectors gothrough an equipollency into a commonO-point of the space, where are the tail of thevector ā. It decomposes ā along the b̄, c̄, d̄ directions and it contains the componentsas (Fig. 1.12):

ā � OA1 + OA2 + OA3 (1.16)

Since OA1, OA2, OA3 are collinear respectively with b̄, c̄, d̄ , it can be written:OA1 � mb̄, OA1 � nc̄, OA1 � pd̄

Then it follows:

ā � mb̄ + nc̄ + pd̄. (1.16’)

Fig. 1.12 Decomposition ofa vector after three vectors

14 1 Vectors

If we introduce the following notations:

m − β/α, n � −γ /α, p � −δ/α,

we obtain the following expression:

αā + βb̄ + γ c̄ + δd̄ � 0. (1.17)

Theorem The decomposition of a vector in three non-coplanar vectors is unique.

Proof It is assumed that there are two distinct decompositions:

ā � mb̄ + nc̄ + pd̄ and ā � m ′b̄ + n′c̄ + p′d̄.

By subtraction it follows:

0 � (m − m ′)b̄ + (n − n′)c̄ + (p − p′)d̄. (1.18)

If m � m ′, n � n′, p � p′ it follows, according to the previous relationship,that the three vectors b̄, c̄, d̄ are coplanar, in contradiction with the hypothesis. Ifm � m ′, n � n′, p � p′, according to Eq. (1.17), it will result:

0 � (n − n′)c̄ + (p − p′)d̄, (1.18′)

that means that c̄ and d̄ are collinear, which is also contrary to the hypothesis. Casesin which only n � n′ or only p � p′ are treated analogously and lead to a violationof the hypotheses. It follows that only the case m � m ′, n � n′, p � p′ and thedecomposition is unique.

1.2.1.11 Linearly Independent and Linearly Dependent Non-coplanarVectors

If we have three non-coplanar vectors b̄, c̄, d̄ , then there cannot be a linear relationof the form: βb̄ + γ c̄ + δd̄ � 0 because this relationship characterizes three coplanarvectors. It is said that the three non-coplanar vectors are linearly independent. If wehave a set of four vectors, there will be the relationship (1.15) or (1.16) betweenthem which shows that one of them is a linear combination of the other three. Forthis reason, they are called linearly dependent.

If a group of three non-coplanar (linearly independent) vectors are chosen, it ispossible to express any vector from the three-dimensional space with them. That iswhy these vectors are called the fundamental vectors.

1.2 Vector Operations 15

1.2.1.12 The Conditions in Which Three Vectors Have Heads on a Lineor Four Vectors Which Have the Heads in the Same Plane

We give, without demonstration, two theorems that define the conditions in whichthree vectors which have the heads (terminal points) on a straight line or four vectorshave the heads in the same plane.

Theorem I Thenecessary and sufficient condition for the terminal point of the vectorOC � c̄ to be on the straight line AB is that, in the relationship d̄ � mā + nb̄ + pc̄,we have m + n + p � 1.Theorem II The necessary and sufficient condition for the vector’s OD � d̄ headto be found in the plane ABC is that, in the relationship c̄ � mā + nb̄, we havem + n � 1.

1.2.2 Dot Product

1.2.2.1 Definition and Properties

Consider two vectors ā and b̄ and we define an operation between these two vectorsresulting in the space of the real numbers, as follows:

Definition The number resulting from the product of the magnitude of the twovectors multiplied by the cosine of the angle between them is called the dot productof the two vectors ā and b̄.

The dot product is marked with a point: c � ā · b̄ � ab cos(ā, b̄). The scalarproduct has the following properties:

• The dot product is null if: (a) one of the vectors is null or (b) if the two vectors areperpendicular. Demonstration of the first property is immediate. For the seconddemonstration, considering the formula of the cosine of the angle between the twovectors, since the vectors are perpendicular, this cosine is zero and therefore theresult is zero.

• The square of a vector relative to the dot product operation is equal to the squareof magnitude. If the definition formula is applied and the angle between the twofactors of the product is zero, then the result is immediately obtained:

ā2 � ā · ā � a · a cos 0 � a2 (� |a|2). (1.19)

• The dot product is commutative. The demonstration follows immediately from thescalar product definition formula.

• The dot product is distributive over the scalar multiplication. The demonstrationis simple if it is taken into account that ā and λā are collinear and b̄ together λb̄are also collinear. Thus,

16 1 Vectors

cos(ā, b) � cos(λā, b̄) � cos(a, λb̄).

Then, from the definition formulas, it simply follows:

λ(ā · b̄) � (λā · b̄) � (a · λb̄). (1.20)

• The dot product of two vectors is equal with the magnitude of one, multiplied bythe projection of the other in the direction of the first vector. The property resultsas outlined in Fig. 1.13. We have:

āb̄ � ab cos(ā, b̄) � a⌊b cos(ā, b̄)⌋ � a prāb̄, (1.21)

because: prāb̄ � b cos(ā, b̄

). Similarly:

āb̄ � b prb̄ā. (1.22)

• We give, without demonstration, the following result: the scalar product is dis-tributive over the vector addition. We have:

ā(b̄ + c̄

) � āb̄ + āc̄. (1.23)

• If the previous property is taken into account, it can be written immediately:(ā + b̄

) · (c̄ + d̄) � āc̄ + b̄c̄ + ād̄ + b̄d̄,

i.e. the rule of multiplication of two polynomials is preserved.• The dot product of two vectors does not change if to one of the vectors is added avector perpendicular to the other. Let us consider two vectors ā and b̄ and a thirdvector ū perpendicular to ā. So we have: āū � 0. Then: ā(b + ū) � āb̄ + āū � āb̄what demonstrates the property.

• Next, consider an orthogonal coordinate system Oxyz that has the unit vectorsī, j̄, k̄. There are the following relationships, particularly important in the algebraicrepresentation of vectors:

ī · j̄ � j̄ · ī � cos π2

� 0, j̄ · k̄ � k̄ · j̄ � 0, k̄ · ī � ī · k̄ � 0

Fig. 1.13 Projection of avector on another vector

1.2 Vector Operations 17

(ī)2 � ( j̄)2 � (k̄)2 � cos 0 � 1. (1.24)

1.2.2.2 Algebraic Representations

If the representation (1.6) of vectors ā and b̄ is considered, then it can be written:

ā · �b � (ax ī + ay j̄ + azk̄)(bx ī + by j̄ + bzk̄

)

� axbx(ī)2

+ axby(ī · j̄) + axbz

(ī · k̄) + aybx

(j̄ · ī) + ayby

(j̄)2

+ aybz(j̄ · k̄) + azbx

(k̄ · ī) + azby

(k̄ · j̄) + azbz

(k̄)2

� axbx + ayby + azbz, (1.25)

i.e. the dot product is given by the sum of the product of the corresponding compo-nents of the two vectors. In matrix notation we have:

ā · b̄ � {a}T {b} �[ax ay az

]⎧⎪⎨⎪⎩

bxby

bz

⎫⎪⎬⎪⎭

� axbx + ayby + azbz .

The condition that two vectors are perpendicular will be written, if this represen-tation is adopted, as:

axbx + ayby + azbz � 0. (1.26)

Themagnitude of a vector is obtained if it is considered the dot product of a vectorwith itself, from the relation:

ā · ā � (ā)2 � a2 � a2x + a2y + a2z , (1.27)

from where:

a �√a2x + a

2y + a

2z . (1.28)

In matrix form, we have:

a2 � {a}T {a} � a2x + a2y + a2z .

The angle between two vectors is obtained from the relationship:

cos(ā, b̄

) � ā · b̄ab

� axbx + ayby + azbz√a2x + a2y + a2z

√b2x + b2y + b2z

. (1.29)

18 1 Vectors

Fig. 1.14 Cross productdefinition

The unit vector associated with a direction. It is considered a straight line (�) thatmakes the angles α, β and γ with the three axes of an orthogonal coordinate system.

In addition, consider a vector oriented along this line. Its three components willbe equal to his projections on the three axes, namely:

ux � u cosα, uy � u cosβ, uz � u cos γ.

Let us now consider the unit vector of the straight line (�), denoted by ūo. Thisunit vector is obtained by dividing the vector by its length:

ūo � ūu

� ī cosα + j̄ cosβ + k̄ cos γ. (1.30)

In matrix form, we can write:

{uo} �

⎧⎪⎨⎪⎩

cosαcosβcos γ

⎫⎪⎬⎪⎭

. (1.31)

Because ūo is an unit vector, it results immediately:

cos2 α + cos2 β + cos2 γ � 1. (1.32)

1.2.3 Cross Product

1.2.3.1 Definition and Properties

Consider two vectors and define an operation between the two vectors having as resulta vector, called the cross product (or vector product or outer product or directed areaproduct) (Fig. 1.14).

Definition Given the vectors ā � −→OA, b̄ � −→OB we define an operation, called thecross product of the two vectors, which associates a vector c̄ � −→OC determined inthe following way:

1.2 Vector Operations 19

(a) it is perpendicular to the plane determined by the given vectors (this determinesits direction);

(b) the direction of the vector is given by the right-hand rule;(c) the magnitude of the vector c̄ � −→OC is equal to the area of the parallelogram

that vectors−→OA and

−→OB span.

So it follows c � ab sin(ā, b̄) � 2 area(OAB). The vector product is markedwith the “×” sign: c̄ � ā × b̄.

Here are some of the properties of the vector product:

• The cross product is null if one of the two vectors is null or if the two vectors havethe same direction. Indeed, if in the expression of the cross product magnitudeit is considered that the angle between the two vectors is zero, then the sine ofthis vector is zero, so the magnitude of the vector is zero. In particular, the crossproduct of a vector with itself is zero.Parallelism of two vectors. Two vectors will be parallel if their cross product iszero. This is obvious if it is taken into account that the angle between the twoparallel vectors is zero, so the sine of this angle, which appears in the formula ofthe magnitude of the cross product, is zero. This property allows the testing ofsituations where two vectors are parallel.

• The cross product is anticommutative. Let us consider the products ā×b̄ and b̄×ā.The resulting vector in the two cases has the same direction, being oriented by astraight line perpendicular to the two vectors, but the orientation will be different.If we take ā over b̄ over the shortest way, we get an orientation, and if we take b̄over ā over the shortest way we get an opposite orientation. Since the magnitudesof the two vectors are equal, it can be written that:

ā × b̄ � −b̄ × ā or ā × b̄ + b̄ × ā � 0. (1.33)

• The cross product is distributive in regards to a multiplication with a scalar. Thedemonstration is simple if it is taken into account that ā and λā are collinear and,in the same time, b̄ and λb̄ are collinear. In this case, the angle between ā and b̄as the angle between λ ā and b̄ or ā and λ b̄ is the same, so the sine of this anglewill be the same in all three cases. It results:

λ(ā × b̄) � (λā) × b̄ � ā × (λb̄). (1.34)

• The cross product remains unchanged when the head point of a vector moves on aparallel line to the other vector (see Fig. 1.15). Let us consider the parallelogramOABM generated by the vectors ā and b̄ and the parallelogramOApBpMpgeneratedby the vectors ā and b̄p. The vector b̄p was obtained by moving its end point alonga line parallel to ā. The two parallegrams have the same area because they have thesame base and the same height. It follows that the cross product magnitude in thetwo cases is the same. Since the direction and sense does not change by moving

20 1 Vectors

Fig. 1.15 Cross product isinvariant if B moves along aline parallel to ā

the head point of b̄ along a parallel line with ā it results that the cross productvector is the same in the two cases.

• The cross product is distributive in regards to vector addition. Because the demon-stration, without being difficult, is longer, we do not give it. Thus, it can be written:

ā × (b̄ + c̄) � ā × b̄ + ā × c̄. (1.35)

• Using the previous property, we can write the cross product of two sums of vectorsas:

(ā + b̄

) × (c̄ + d̄) � ā × c̄ + ā × d̄ + b̄ × c̄ + b̄ × d̄. (1.35′)

The rule ofmultiplication remains the same as for polynomials, with the differencethat the order of the factors in the product is not arbitrary, but in each term ofdevelopment the first factor must belong to the first brackets and the second factormust belong to the second brackets. It can be shown as:

ā × b̄ + c̄ × d̄ � (ā − c̄) × (b̄ − d̄) + ā × d̄ + c̄ × b̄. (1.35′′)

• The cross product does not change if we add to one of the vectors another vectorparallel to the second vector of the product. The proof is trivial.

• If an orthogonal trihedral with the unit vectors ī, j̄, k̄ is considered, taking intoaccount the angles between the coordinate axes and the direction of the resultantvectors, it can be written:

ī × ī � j̄ × j̄ � k̄ × k̄ � 0,

and:

ī × j̄ � k̄, j̄ × k̄ � ī, k̄ × ī � j̄ . (1.36)

A coordinate system for which the above relations are valid is called the rightorthogonal coordinate system. If on the third axis the direction is defined by therelationship ī × j̄ � −k̄ we say that we have a left orthogonal coordinate system.

1.2 Vector Operations 21

Generally, the right orthogonal coordinate system will still be used unless otherwisespecified.

1.2.3.2 Algebraic Representations

Let us now consider two vectors ā and b̄ defined by their components:

ā � ax ī + ay j̄ + azk̄, b̄ � bx ī + by j̄ + bzk̄.

Taking into account the rules ofmultiplication of the unit vectors, it can bewritten:

c̄ � ā × b̄ � (ax ī + ay j̄ + azk̄) × (bx ī + by j̄ + bzk̄

)

� axbx(ī × ī) + axby

(ī × j̄) + axbz

(ī × k̄) + aybx

(j̄ × ī) + ayby

(j̄ × j̄)

+ aybz(j̄ × k̄) + azbx

(k̄ × ī) + azby

(k̄ × j̄) + azbz

(k̄ × k̄)

� (aybz − azby)ī + (azbx − axbz) j̄ +

(axby − aybx

)k̄.

Thus, the components of the vector will be given by the expressions:

cx � aybz − azby ; cy � azbx − axbz ; cz � axby − aybx . (1.37)

Symbolically, the cross product can be represented by the determinant:

c̄ � ā × b̄ �

∣∣∣∣∣∣∣∣

ī j̄ k̄ax ay az

bx by bz

∣∣∣∣∣∣∣∣.

In matrix form, for the representation of the cross product, the skew symmetric3×3 matrix, associated with the vector ā is:

[a] �⎡⎢⎣

0 −az ayaz 0 −ax

−ay ax 0

⎤⎥⎦,

and then the cross product is represented by the matrix product:

[a]{b} �⎡⎢⎣

0 −az ayaz 0 −ax

−ay ax 0

⎤⎥⎦

⎧⎪⎨⎪⎩

bxby

bz

⎫⎪⎬⎪⎭

�

⎧⎪⎨⎪⎩

aybz − azbyazbx − axbzaxby − aybx

⎫⎪⎬⎪⎭

. (1.38)

22 1 Vectors

1.2.4 Scalar Triple Product

1.2.4.1 Definition and Properties

Three vectors are considered, i.e. ā, b̄ and c̄. The scalar triple product of these threevectors is a scalar, defined by the combination:

d � ā · (b̄ × c̄). (1.39)

The triple scalar product has an interesting geometric significance. Consider thethree vectors, ā, b̄ and c̄. They will form the parallelepiped OBDCA′B′D′C′, seeFig. 1.16.

The magnitude of the cross product b̄× c̄, as shown above, represents the area ofthe parallelogram OBDC, and the cross product is the vector OM, perpendicular tothe plane determined by the vectors b̄ and c̄. Let us now take the perpendicular fromA′ to OM. A′A′′ is perpendicular to OM and will result that OA′′ is the height of theparallelepiped. Then the dot product between

−−→OM and ā will have amagnitude equal

to the product of OM and its projection of ā onto OM, i.e., OA′′ (a value equal to theproduct between the parallelogram base area and the height of the parallelepiped), soit will be equal to the parallelepiped volume. The scalar triple productmay be positiveor negative depending on the orientation of the trihedral formed with the three givenvectors. If the orientation of this trihedral is positive, the scalar triple product ispositive and if the orientation is negative, the scalar triple product is negative.

As a result, the scalar triple product of three vectors represents, in absolute value,the volume of the parallelepiped built on these vectors considered as edges.

Scalar triple product properties

• The scalar triple product of three vectors remains unchanged if its factors arecircularly permutated, i.e. we have:

Fig. 1.16 Graphicalrepresentation of the scalartriple product

1.2 Vector Operations 23

ā(b̄ × c̄) � b̄(c̄ × ā) � c̄(ā × b̄). (1.40)

The result can be easily demonstrated if it is taken into account that the threeproducts represent the volume of the parallelepiped built with the three vectors asedges, and the sign is the same in the three cases as the orientation of the trihedral ismaintained.

• The scalar triple product does not change if the signs and x are allowed betweenthem, i.e.

ā(b̄ × c̄) � (ā × b̄)c. (1.41)

The result is simply demonstrated by taking into account that the scalar product iscommutative and whether the above-mentioned property (which says that the mixedproduct does not change if the circular permutations are made), is used. Since thevalue of the scalar triple product depends only on the three vectors and the orientationof the trihedral defined by them, regardless of the sign that is placed between themor x, it is agreed to note the mixed product as:

ā(b̄ × c̄) � (ā × b̄)c � (āb̄c̄) � [āb̄c̄]. (1.42)

• If the two factors of the triple scalar product change between them, it changes thesign of the result:

(āb̄c̄

) � −(b̄āc̄) � −(āc̄b̄) � −(c̄b̄ā). (1.43)

• The scalar triple product is null if one of the factors is null or if the three vectors arecoplanar. The first case is obvious. For the second case it will be noticed that if thethree vectors are coplanar, the parallelepiped built with them has a zero volume,so the mixed product is null. A particular case, commonly found in practice, iswhen two vectors are collinear, so the three vectors are coplanar. So, if any twovectors of the mixed product are collinear, the mixed product is null.

1.2.4.2 Algebraic Representations

If the vectors ā � ax ī + ay j̄ + azk̄, b̄ � bx ī + by j̄ + bzk̄ and c̄ � cx ī + cy j̄ + czk̄ areconsidered, and the calculus are made, for the expression of the mixed product, oneobtains:

(āb̄c̄

) � axbycz + aybzcx + azbxcy − azbycx − aybxcz − axbzcy . (1.44)

This expression is precisely the determinant of the matrix having the vector com-ponents ā, b̄ and c̄ as lines, and so it can be written:

24 1 Vectors

(āb̄c̄

) �

∣∣∣∣∣∣∣

ax ay az

bx by bxcx cy cz

∣∣∣∣∣∣∣. (1.45)

If we write the scalar triple product in this form, all of the above-mentionedproperties are immediately demonstrated, if the determinant properties are takeninto account.

1.2.5 Vector Triple Product

If the vectors ā, b̄ and c̄ are considered the vector triple product is the vector d̄:

d̄ � ā × (b̄ × c̄). (1.46)

In the following we give, without demonstration (which does not pose problemsbut is more laborious), the main result of the vector triple product. Thus, the vectortriple product d̄ can be expanded as:

d̄ � ā × (b̄ × c̄) � (āc̄)b̄ − (āb̄)c̄. (1.47)

The vector triple product is null if one of the factors is null, if the vectors b̄ and c̄are collinear or if the vector ā is perpendicular to the plane determined by the vectorsb̄ and c̄. The demonstration is immediate.

1.2.6 Applications of Vector Calculus

1.2.6.1 Position Vector of a Point

Let us consider a point M in space and choose a fixed point O as the origin of thespace. Let us build the vector �r � −−→OM . The point of M is completely determinedby the vector and thus constructed. The vector is called the position point of thevectorM. In applications, different coordinate systems (cartesian, cylindrical, polar,spherical, natural etc.) can be used to define using different scalar components. Thesecomponents unambiguously define the position vector, thus the position of the pointM.

1.2 Vector Operations 25

Fig. 1.17 Support line of aforce

1.2.6.2 Support Line of a Force

Let us consider the force �F and its moment �MO in the point O (see Fig. 1.17), (wehave �F �MO �0). We propose to determine the support line of the force. For this, weneed to determine the solution �r from the vector equation �r × �F � �MO . Writing ncomponent, the relationship will be:

yZ − zY � MOx ,zX − x Z � MOy,xY − yX � MOz,

or

⎡⎢⎣

0 Z −Y−X 0 XY −Z 0

⎤⎥⎦

⎧⎨⎩xyz

⎫⎬⎭ �

⎧⎪⎨⎪⎩

MOxMOy

MOz

⎫⎪⎬⎪⎭

.

It is easy to see that the system determinant is zero:∣∣∣∣∣∣∣

0 Z −Y−X 0 XY −Z 0

∣∣∣∣∣∣∣� 0. (1.48)

Thus, the system does not have a single solution. If we consider the first twoequations as the main equations, it is obtained that the characteristic determinant isequal to zero:

�c �

∣∣∣∣∣∣∣

0 Z MOx−X 0 MOyY −Z MOz

∣∣∣∣∣∣∣� Z(XMOx + YMOy + ZMOz

) � Z( �R �MO

)� 0,

26 1 Vectors

by virtue of the fact that moment and force are two perpendicular vectors. It followsthat we have an undetermined system of two equations with three unknowns. Geo-metrically seen, the two equations are planes whose intersection will give us a line,which is the support line of the force.

Another method of determining the force support line is a vector method. Thus,if we premultiply the vector equation:

�r × �F � �MO , (1.49)

to left, with �F , in a cross product, one obtains:�F ×

(�r × �F

)� �F × �MO ,

and expanding the vector triple product it results:

F2�r −( �F�r

) �F � �F × �MO ,

and:

�r � �F × �MO�F2 +( �F�r

)

F2�F . (1.50)

Since on the basis of previous considerations we have seen that not all componentsof the vector are independent, one can choose as a parameter the expression:

λ �( �F�r

)

F2,

and we obtain:

�r � �F × �MOF2

+ λ �F . (1.51)

The equation obtained is a line, which has the direction of the force and passesthrough the point defined by the vector:

�d � �F × �MOF2

. (1.52)

The vector �d represents the distance from the origin to the line (it is perpendicularto the force—it comes froma cross product—so too on the support line and in additionwhen λ�0 it will result that its end belongs to the line). So the force support linehas the equation:

1.2 Vector Operations 27

�r � �d + λ �F, (1.53)

or, in components:

x � dx + λX,y � dy + λY,z � dz + λZ . (1.54)

By removing the parameter λ, the line can also be expressed as:

x − dxX

� y − dyY

� z − dzZ

. (1.55)

1.3 Applications

1.3.1. If the vectors ā, b̄ and c̄ are given, demonstrate that the following relationshipexists:

ā × (b̄ × c̄) + b̄ × (c̄ × ā) + c̄ × (ā × b̄) � 0.

1.3.2. If we have four vectors ā, b̄, c̄ and d̄ show that we have the following rela-tionship:

(ā × b̄) · (c̄ × d̄) �

∣∣∣∣∣ā · c̄ ā · d̄b̄ · c̄ b̄ · d̄

∣∣∣∣∣.

1.3.3. If we have the vectors ā, b̄, c̄ and d̄ show that we have the following relation-ship:

(ā × b̄) × (c̄ × d̄) � (āb̄d̄)c̄ − (āb̄c̄)d̄ � (c̄d̄ā)b̄ − (c̄d̄ b̄)ā.

1.3.4. Either vectors ā, b̄ and c̄ are non-coplanar. Let it show that if:

(ā × b̄) · (c̄ × d̄) � (ā · b̄) · (b̄ · c̄),

then the vectors ā and c̄ are perpendicular.

1.3.5. Show that:

(ā × b̄) × (b̄ × c̄) � (āb̄c̄)b̄.

1.3.6. Show that:

28 1 Vectors

[(ā × b̄)(b̄ × c̄)(c̄ × ā)] � (āb̄c̄)2.

1.3.7. Show that we have the expansion:

(ā × b̄) × [(b̄ × c̄) × (c̄ × ā)] � (āb̄c̄)[(c̄ā)b̄ − (c̄b̄)ā].

1.3.8. If ai , bi , ci (i � 1, 2, 3) are real numbers, show that the vectors:

ū1 �(b̄1 − c̄1

)ī + (c1 − a1) j̄ + (a1 − b1)k̄

ū2 �(b̄2 − c̄2

)ī + (c2 − a2) j̄ + (a2 − b2)k̄

ū3 �(b̄3 − c̄3

)ī + (c3 − a3) j̄ + (a3 − b3)k̄,

are coplanar.Note: If the mixed product of the three vectors is written as a determinant, itis found to be zero, so the vectors are coplanar.

1.3.9. If the non-coplanar vectors ā, b̄ and c̄ are given, calculate the scalar tripleproduct of the vectors:

(ā + b̄

),(b̄ + c̄

), (c̄ + ā) and interpret geometrically

the result.

Solution: P � 2(āb̄c̄)

1.3.10. If the non-coplanar vectors ā, b̄ and c̄ are given, calculate the scalar tripleproduct of the vectors:

(ā − b̄), (b̄ − c̄), (c̄ − ā).

Solution: P � 0.

1.3.11. From the development in two different ways of the product(ā × b̄)(c̄ × ū)

deduce the components of ū along the directions of the vectors ā, b̄ and c̄,considered non-coplanar.

Solution: ū � 1(āb̄c̄

) [(b̄c̄ū)ā + (c̄āū)b̄ + (āb̄ū)c̄].

1.3.12. If the non-coplanar vectors ā, b̄ and c̄ are given, constructed with thesevectors ū, v̄ and w̄ as follows:

ū � ā × (b̄ + c̄), v̄ � b̄ × (c̄ + ā), w̄ � c̄ × (ā + b̄).

Show that:

(a) ū × v̄ � v̄ × w̄ � w̄ × ū � (āb̄c̄)(ā + b̄ + c̄).(b) P � (ūv̄w̄) � 0.1.3.13. Write the equation of a straight line for the following cases:

1.3 Applications 29

Fig. 1.18 Straight linepassing through a point Mo

(a) It passes through the point Mo(r̄o) and has the direction given by the vector ā;(b) It passes through the points A(r̄A) and B(r̄B);(c) It passes through the origin and has the direction given by the vector ā.

Solution: (a) M being a certain point on the line (�), we have (see Fig. 1.18):−−−→MoM � λ�a with λ ∈ R. The relation can be written:

r̄ − r̄o � λā, λ ∈ R,

or

r̄ � r̄o + λā, λ ∈ R.

If the vectors r̄(x, y, z), r̄o(xo, yo, zo) and ā(ax , ay, az

)are defined by their Carte-

sian coordinates, the equation of the straight line takes the form:

x � xo + λax ; y � yo + λay ; z � zo + λaz, λ ∈ R

or, removing the parameter λ:

x − xoax

� y − yoay

� z − zoaz

.

Premultipling in a cross product, at right with ā the vectorial relation r̄ � r̄o +λā,one obtains:

r̄ × ā � r̄o × a.

Note b̄ � r̄o × ā(b̄ is a vector perpendicular to ā that ā · b̄ � 0). Then, the vectorequation of the straight line can be written as:

r̄ × ā � b̄ (with ā · b̄ � 0).

30 1 Vectors

Fig. 1.19 Straight linethrough the points A and B

(b) The vector−→AB � �rB − �rA is collinear with the straight line (see Fig. 1.19),

so if M is a point on the line, it can be written:−−−→MoM � λ−→AB, λ ∈ R, or

r̄ − r̄o � λ(r̄B − r̄A) or else:

r̄ � r̄o + λ(r̄B − r̄A).

If we consider the vectors r̄A(xA, yA, zA), r̄B(xB, yB, zB) defined by their Carte-sian coordinates, the equation of the straight line takes the form:

x � xo + λ(xB − xA); y � yo + λ(yB − yA); z � zo + λ(zB − zA), λ ∈ R

or, if λ is removed:

x − xoxB − xA �

y − yoyB − yA �

z − zozB − zA .

If we start now from the vectorial relationship multiplied to the right in a crossproduct with r̄B − r̄A, we get:

r̄ × (r̄B − r̄A) � r̄A × (r̄B − r̄A) � r̄A × r̄B,

or

r̄ × (r̄B − r̄A) � r̄A × r̄B .

(c) Using the previously obtained result with r̄A � 0 (point A is the same withorigin O) we have:

r̄ × r̄B � 0.

In this case r̄B indicates the direction of the line so r̄B � λā and then:r̄ × ā � 0.

1.3 Applications 31

1.3.14. Write the equation of a straight line parallel to the Ox axis.

Solution : r̄ × ī � b̄ with b̄ � α j̄ + β k̄.

1.3.15. Write the condition that three points are collinear.

Solution: Let us consider the points A, B and C. The condition that they arecollinear can be written as:

AC � μAB,

or

r̄C − r̄A � λ(r̄B − r̄A).

If we premultiply to the left in a cross product with r̄B − r̄A, one obtains: (r̄B −r̄A) × (r̄C − r̄A) � 0 or in a symmetrical form:

r̄A × r̄B + r̄B × r̄C + r̄C × r̄A � 0.

1.3.16. (a) Write the equation of the plane π that passes throughthe points A(r̄A), B(r̄B),C(r̄C). Numeric application:r̄A(1, 3, 5), r̄B(3, 2, 1), r̄C (2, 1, 5);

(b) Write the equation of the plane π passing through point A and thenormal direction is given by the vector ā;

(c) What is the condition for the points A(r̄A), B(r̄B),C(r̄C), D(r̄D) to becoplanar?

Solution: We have: AB � r̄B − r̄A; AM � r̄ − r̄A. A vector perpendicular to theplane will be perpendicular to the vectors AB; AC ; AM (Fig. 1.20). If n̄ is normalto the plane.

n̄ � AB × AC � r̄A × r̄B + r̄B × r̄C + r̄C × r̄A,

then we must have AM × n̄ � 0 or:

(r̄ − r̄A)[(rB − r̄A) × (rC − r̄A)] � 0.

After developments, we get:

r̄ [(r̄B − r̄A) × (r̄C − r̄A)] � r̄A[(r̄B − r̄A) × (r̄C − r̄A)],

or

r̄(r̄A × r̄B + r̄B × r̄C + r̄C × r̄A) � [r̄Ar̄Br̄C ],

32 1 Vectors

Fig. 1.20 A plane passingthrough three points A, Band C

or

r̄n � [r̄Ar̄Br̄C ].

Numeric application:

n̄ �

∣∣∣∣∣∣∣

ī j̄ k

3 − 1 2 − 3 1 − 52 − 1 1 − 3 5 − 5

∣∣∣∣∣∣∣� −8ī − 4 j̄ + 3k̄,

[r̄Ar̄Br̄C ] �

∣∣∣∣∣∣∣

1 3 53 2 12 1 5

∣∣∣∣∣∣∣� −35,

so the plane equation will be:

r̄(−8ī − 4 j̄ + 3k̄) � −35,

or

8x + 4y − 3z � 35.

(b) If �n it is the normal vector to the plane, it is perpendicular to any vector in plane,hence: (r̄ − r̄A)n̄ � 0 or r̄ n̄ � r̄An̄. In our case

r ā � r̄Aā.

1.3 Applications 33

(c) Let n̄ � r̄A × r̄B + r̄B × r̄C + r̄C × r̄A be a normal vector to the plane defined byA, B, C (see point a). If D there is in plan, we must have:

(r̄D − r̄A)n̄ � 0,

or

r̄Dn̄ � [r̄Ar̄Br̄C ].

We have:

r̄D[r̄A × r̄B + r̄B × r̄C + r̄C × r̄A] � [r̄Ar̄Br̄C ],

or

[rAr̄Br̄C ] + [r̄Br̄C r̄D] + [r̄C r̄Dr̄A] + [r̄Dr̄Ar̄B] � 0.

1.3.17. Solve the vector equation r̄ × a � b̄.Solution: If we pre-multiply the vector equation in a cross product with �a it is

obtained:

ā × (r̄ × ā) � ā × b̄.

or, expanding the vector triple product:

a2r̄ − (ar)ā � ā × b̄,

from where:

r̄ � ā × b̄a2

+ λā, with λ� (ār̄)a2

.

1.3.18. Let us show that d �∣∣∣ ā×b̄a2

∣∣∣ � ba represents the distance from origin of anEuclidian system to the line r̄ × ā � b̄(āb̄ � 0).

Solution: The equation r̄ × ā � b̄ has the solution: r̄ � ā×b̄a2 + λā. If λ � 0 thevector:

r |λ�0 � d̄ � ā × b̄a2 ,

connects the origin of the coordinate system with a point on the line and is perpen-dicular to the line, since it represents the vector whose magnitude is the distancefrom the origin to the line.

34 1 Vectors

Fig. 1.21 A planecontaining a line and a pointMo

1.3.19. Determine the equation of the plane that passes through the origin of thecoordinate system and is perpendicular to the vector ā.

Solution: In problem 1.3.16(b) one gets r̄A � 0 and r̄ ā � 0. If the coordinateplanes yOz, zOx, xOy, are considered they have the equations: r̄ ī � 0; r̄ j̄ �0; r̄ k̄ � 0.1.3.20. Write the vectorial form of the equation of a plane that intersects the axes

in the points A( 1a , 0, 0), B(0,1b , 0), C(0, 0,

1c ).

Solution: The plan equation is in this case:

xa + yb + zc � 1.

If the vectors r̄ � xī + y j̄ + zk̄ and n̄ � aī + b j̄ + ck̄ are considered, it can bewritten: r̄ n̄ � 1.1.3.21. Write the equation of the plane containing the line r̄ × ā � b̄, (āb̄ � 0)

and the point r̄o (see Fig. 1.21).

Solution: By solving the equation of the line we obtain:

r̄ � āxb̄a2

+ λā.

Point D for which λ � 0 belongs to the following plane:

r̄D � ā × b̄a2

.

A normal vector to the plane will be given by:

1.3 Applications 35

n̄ � (r̄D − r̄o)xā � r̄D �(āxb̄

a2− r̄o

)xā � b̄ − r̄oxā.

If we write the plan equation in the form (see 1.1.15b)

r̄ n̄ � r̄on̄,

and introducing n̄, one obtains:

r̄(b̄ − r̄o × ā

) � (b̄r̄o).

1.3.22. Write the equation of a plane perpendicular to the Oz axis and passingthrough the point r̄A.

Solution: r̄ k̄ � r̄Ak̄ � zA.

1.3.23. Determine the point of intersection between the plane r̄ · ā1 � c and theline r̄ × ā2 � b̄,

(ā2b̄ � 0

).

Application:

ā1(1, 3, 1), ā2(2, 1,−1), b̄(1,−1, 1), c � 1.

Solution: We pre-multiply, to the left, in a cross product with ā1, the line equation.It is obtained:

ā1 × (r̄ × a2) � ā1 × b̄.

Considering the vector triple product expansion, one obtains:

(ā1ā2)r̄ − (ā1r̄)ā2 � ā1 × b̄,

from where:

r̄ � ā1 × b̄ + cā2(ā1ā2)

.

For the given values:

ā1 × b̄ � 4ī − 4k̄; (ā1a2) � 4; r̄ I � 32i +

1

4j̄ − 5

4k̄.

1.3.24. Determine the points of intersection of the straight line r̄ × ā �b̄, (āb̄ � 0) with the coordinate planes (x=0, y=0, z=0). Application:ā(1, 3,−2); b̄(2, 2, 4)

36 1 Vectors

Solution: For the intersection with the yOz(r̄ · ī � 0) plane, the equation of the

straight is multiplied by �i in a dot product. It is obtained:

ī × (r̄ × ā) � i × b̄.

Expanding the vector triple product, one obtains:

(āī

)r̄ − (r̄ ī)ā � ī × b̄,

r̄xOy � ī × b̄(āī

) +(r̄ ī

)(āī

) ā � ī × b̄(āī

) .

Analogously, the other two intersections are obtained. For given values:

ī × b̄ � −4 j̄ + 2k; j̄ × b̄ � 4ī − 2k̄; k̄ × b̄ � −2ī + 2;(āī

) � 1; (ā j̄) � 3; (āk̄) � −2.

It results:

r̄xOy � −4 j̄ + 2k̄; r̄yOz � 43ī − 2

3k̄; r̄zOx � ī − k̄.

1.3.25. Determine the line intersection of planes: r̄ · a1 � c1; r̄ · ā2 � c2 (seeFig. 1.22).

Solution: If we write the line as: r̄ × ā � b̄, (āb̄ � 0), the vector ā will belongto the two planes. If ā1 and ā2 are perpendicular to the two planes, one can write:ā � ā1 × ā2. It remains b̄ to be determined. The equation of the line is, if ā is known:r̄ × (ā1 × ā2) � b̄. By developing the vector triple product, it results:

Fig. 1.22 Intersection oftwo plane is a straight line

1.3 Applications 37

Fig. 1.23 Distance from apoint to a plane

(r̄ ā2)ā1 − (r̄ ā1)ā2 � b̄.

Taking into account the relations of the planes definition, we obtain:

b̄ � c2ā1 − c1ā2.

The equation of the intersection line follows in the vector triple product form:

r̄ × (ā1 × ā2) � c2ā1 − c1ā2.

1.3.26. Let us consider: r̄ ·ā1 � c1, r̄ ·ā2 � c2, r̄ ·ā3 � c3.Determine the intersectionof the planes r̄ · ā1 � c1, r̄ · ā2 � c2, r̄ · ā3 � c3.

Solution: According to the previous problem (1.3.25), the intersection line of thefirst two planes is:

r̄ × (ā1 × ā2) � c2ā1 − c1ā2which must be intersected with the plan:

r̄ · ā3 � c3.

We pre-multiply, in a cross product, the line equation with ā3. Considering theexpansion of the vector triple product, it is obtained:

[ā3(ā1 × ā2)]r̄ − (ā3r̄ )(ā1 × ā2) � c2(ā3 × ā1) − c1(ā3 × ā2),

from where the point of intersection results:

r̄ I � c1(ā2 × ā3) + c2(ā3 × ā1) + c3(ā1 × ā2)[ā1ā2ā3]

.

1.3.27. Determine the distance from a point B(r̄B) to the plane r̄ · ā � c (seeFig. 1.23).

38 1 Vectors

Fig. 1.24 Distance from apoint A to a line

Solution: Let A be the foot of the perpendicular from B to the plane. We have:AB � λā or r̄B − r̄A � λā. If this relationship is multiplied in a dot product with ā,we obtain:

(r̄B − r̄A)ā � λa2.

Hence:

λ � r̄B ā − r̄Aāa2

� r̄B ā − ca2

.

The distance is:

d �∣∣∣−→AB

∣∣∣ � λa � r̄B ā − ca

.

1.3.28. Determine the distance frompoint A(r̄A) to the line r̄×ā � b̄ (see Fig. 1.24).Solution: Let M(r̄) the perpendicular foot from A to the line. The vector is per-

pendicular to the unit vector of the line. Thus, it can be written:

AM · ā � 0,

or

(r̄ − r̄A) · ā � 0r̄ · ā � r̄A · ā.

If the vectorial form of the equation of the line is pre-multiplied by vector ā, it isobtained, after the vector triple product expansion:

a2r̄ − (ār̄)ā � ā × b̄,

or

1.3 Applications 39

Fig. 1.25 The points onwhich the moment is known(M̄

)are situated on a

straight line

r̄ � ā × b̄a2

+(ār̄)

a2ā � ā × b̄

a2+

(ār̄A)

a2ā � r̄M ,

which represents the position vector of point M. Then the distance vector betweenpoint A and line is:

δ̄ � AM � r̄ − r̄A � ā × b̄a2

+(ār̄A)

a2ā − r̄A,

and its magnitude is:

δ �∣∣ā × b̄ + (ār̄A)ā − a2r̄A

∣∣a2

�√∑[

aybz − azby + ax (xAax + yAay + zAaz) − (a2x + a2y + a2z )xA]2

a2.

1.3.29. Determine the distance between the lines r̄ × ā1 � b̄1 and r̄ × ā2 � b̄2 (seeFig. 1.25).

Solution: Let AB be the distance vector between the two lines. Then, we will havethe relationships: AB · ā1 � 0 and AB · ā2 � 0. It follows that we must have therelationship: AB � r̄B − r̄A � λ(ā1 × ā2) and r̄A × ā1 � b̄1, r̄B × ā2 � b̄2. If wescalar multiply the vector

−→AB we get:

AB · (ā1 × ā2) � λ(ā1 × ā2)2,

or

(r̄B − r̄A)(ā1 × ā2) � λ(ā1 × ā2)2r̄B(ā1 × ā2) − r̄A(ā1 × ā2) � λ(ā1 × ā2)2.

If the properties of the scalar triple product are taken into account, one can read:

40 1 Vectors

−ā1(r̄B × ā2) − ā2(r̄A × ā1) � λ(ā1 × ā2)2.

By introducing the equations of the lines in brackets, it follows:

−ā1b̄2 − ā2b̄1 � λ(ā1 × ā2)2,

from where we deduce:

λ � −ā1b̄2 − ā2b̄1(ā1 × ā2)2 .

The distance between the two straight lines is thus obtained:

AB � −ā1b̄2 − ā2b̄1(ā1 × ā2)2 (ā1 × ā2),

and its magnitude:

AB �∣∣−ā1b̄2 − ā2b̄1

∣∣|(ā1 × ā2)| .

1.3.30. Determine the angle between the line r̄ x ā � b̄, (āb̄ � 0) and the plane�r · �m � c.

Solution: Let A′ ∈ π be the perpendicular foot fromA, belonging to the line, to theplane.We have AA′⊥π , then−→AA′ � λm̄. It results r̄A′ −r̄A � λm̄ or r̄A′ � r̄A+λm̄. Ifwe pre-multiply the equation of the straight line with m̄, it results: m̄x(r̄ x ā) � m̄xb̄.If the vector triple product develops, there is obtained:

r̄ � m̄ × b − (m̄r̄)ā(m̄ā)

,

and if we impose the condition that the intersection point belongs to the plan(r̄ · m̄ � c), one obtains the position vector of the point of intersection of the straightline with the plane:

r̄ I � m̄xb̄ − cā(m̄ā)

.

Let us find the direction of the straight line A′ I . We have:

ā1 � r̄ I − r̄A � m̄ × b̄ − cā(m̄ā)

− ā × b̄a2

.

From the vectorial equation of the line:

1.3 Applications 41

r̄ � ā × b̄a2

+ λ1ā,

it is considered for the choice of point A the particular case when λ1 � 0, therefore:

r̄A � ā × b̄a2

.

We have

(āā1) � aa1 cos(< āā1),

from where

cos(< āā1) � (āā1)aa1

.

But

(āā1) � ā(m̄ × b̄) − ca2

(m̄ā)− ā

(ā × b̄)a2

� m̄(b̄ × ā) − ca2

(m̄ā).

In this case, it follows:

cos(< āā1) �m̄(b̄×ā)−ca2

( �m�a)a∣∣∣ m̄×b̄−cā(m̄ā) − ā×b̄a2

∣∣∣.

1.3.31. Given the line on which the sliding force is F̄ , determine the points in whichthe moment vector is M̄

(F̄ · M̄ � 0), (see Fig. 1.25).

Solution: Let P be a point on the force support (where the moment is null) and Da point where the moment vector is M̄ . We have:

M̄ � F̄ × r̄ ′.

If we pre-multiply vectorally with F̄ we get:

r̄ ′ � − F̄ × M̄F2

+ λ′ F̄ .

The position vector of the point D is:

r̄D � r̄ + r̄ ′ � ā × b̄a2

+ λ ā − F̄ × M̄F2

+ λ′ F̄ .

42 1 Vectors

If it is taken into account that the force is colinear with the vector ā(F̄ � γ · ā),

it results:

r̄D �ā ×

(b̄ − γ M̄

γ 2

)

a2+ (λ + λ′γ )ā � ā × c̄

a2+ λ′′ā,

where: c̄ � b̄ − γ M̄γ 2

and λ′′ � λ + λ′γ that is, a parallel line with the force.

References

Liesen, J., Mehrmann, V.: Linear Algebra. Springer (2015)Simionescu, G.D.: Vector Algebra and Applications in Geometry. Tehnica, Bucharest (1982)Springer, C.E.: Tensor and Vector Analysis: with Applications to Differential Geometry. DoverPublication Inc., Mineola, New York (2013)

Chapter 2Matrices

2.1 Fundamental Notions

We introduce in the following the basic notions about matrices and the basic proper-ties interesting for our presentation. A number of other properties will be presentedin the following chapters. A matrix of dimension m × n is represented by a tablewith m lines and n columns (in box bracket or parentheses):

⎡⎢⎢⎢⎢⎣

a11 a12 . . . a1na21 a22 . . . a2n...

am1 am2 . . . amn

⎤⎥⎥⎥⎥⎦

,

∥∥∥∥∥∥∥∥∥∥

a11 a12 . . . a1na21 a22 . . . a2n...

am1 am2 . . . amn

∥∥∥∥∥∥∥∥∥∥,

⎛⎜⎜⎜⎜⎝

a11 a12 . . . a1na21 a22 . . . a2n...

am1 am2 . . . amn

⎞⎟⎟⎟⎟⎠

.

There are three different ways found in literature to write these sizes, but in thisbook we will prefer the notation with box brackets as it is easy to use.

The numbers of rows and columns of a matrix define its size. If a matrix has mrows and n columns, it is called a m-by-n matrix (or m × n matrix). The number mand n are called the dimensions of the matrix.

Notation of Matrices

There are different notations for matrices found in literature. Some notationswhich have become more used are outlined below. The set of matrices with m linesand n columns is denoted by Mm×n . Uppercase letters, such as A …, are used todenote matrices. These uppercase letters are sometimes written in bold as: Am×n ,Bm×n ,Km×n . It is also customary to insert the symbol between parentheses: [A]m×n ,[B]m×n , [K]m×n or representations of the form: [ai j ],

∥∥ai j∥∥, (ai j

). For amatrix having

a single column, the notation with braces is used (these matrices represent vectors):{A}, {B}, {X}.

© Springer Nature Switzerland AG 2019S. Vlase et al., Eigenvalue and Eigenvector Problems in Applied Mechanics, AdvancedStructured Materials 96, https://doi.org/10.1007/978-3-030-00991-5_2

43

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44 2 Matrices

Amatrix with a single row is called a row vector and amatrix with a single columnis called a column vector.

Square matrix: A matrix at which the number of rows is equal to the number ofcolumns is called a square matrix. If the number of rows and columns is equal to n,we have a matrix of size n. For example:

[A] �

⎡⎢⎢⎢⎣

1 2 3 4−1 2 2 23 5 1 −10 3 −1 2

⎤⎥⎥⎥⎦.

The set of squares matrix of dimension n is denoted by Mn . A square matrix isdenoted by An or, if the dimension is understood, it is simply written, A. To define asquare matrix, it is necessary to know its n2 elements.

Symmetric matrix. A matrix whose elements have the property ai j � a ji is calleda symmetric matrix, for example:

[A] �

⎡⎢⎢⎢⎣

1 2 3 42 0 −2 53 −2 1 −14 5 −1 2

⎤⎥⎥⎥⎦.

Observe that the elements of the matrix are symmetrical with respect to the firstdiagonal. For the definition of a symmetric matrix it is necessary to know only theelements on the diagonal and above it, so only 1 + 2 + 3 + · · · + n � n(n+1)2 elementsversus n2 are needed in the case of any matrix.

Skew-symmetric matrix. A matrix whose elements have the property ai j � −a jiis called a skew-symmetric (antisymmetric or antimetric) matrix. If i � j then thedefinition relation becomes: aii � −aii or 2aii � 0 so the elements on the maindiagonal of an antisymmetric matrix are zero, for example:

[A] �

⎡⎢⎢⎢⎣

0 2 3 4−2 0 −2 5−3 2 0 −1−4 −5 1 0

⎤⎥⎥⎥⎦,

To define a skew-symmetric matrix, it is necessary to know only the elementsabove or below the diagonal, so n(n−1)2 elements.

Diagonal matrix. A matrix with non-zero elements only on the main diagonal iscalled a diagonal matrix. The relation of definition of a diagonal matrix is: ai j � 0if i �� j and ai j �� 0 if i � j .

For example:

2.1 Fundamental Notions 45

[A] �

⎡⎢⎢⎢⎣

1 0 0 00 2 0 00 0 3 00 0 0 5

⎤⎥⎥⎥⎦.

To define a diagonal matrix, we only need to know the elements on the diagonal,so n elements.

Triangular matrix. A matrix that has nonzero elements on the diagonal and aboveor below is called a triangular matrix. A matrix is an upper triangular matrix if theelements below the main diagonal are zero:

[TS] �

⎡⎢⎢⎢⎣

1 4 2 −50 2 3 20 0 3 −10 0 0 5

⎤⎥⎥⎥⎦,

and a lower triangular matrix if the elements above the main diagonal are zero:

[Ti ] �

⎡⎢⎢⎢⎣

1 0 0 04 2 0 02 3 3 0

−5 2 −1 5

⎤⎥⎥⎥⎦.

2.2 Basic Operation

2.2.1 Addition (+Mm×n × Mm×n → Mm×n)

Consider two matrices [A], [B] ∈ Mm×n . The sum of these two matrices, denotedby +, is a matrix [C] ∈ Mm×n that reads:

[C] � [A] + [B], (2.1)

where matrix [C] elements are defined by the relationships: ci j � ai j + bi j . Forexample, if:

[A] �[

1 2 0−3 0 −1

]; [B] �

[2 2 −21 3 4

],

one obtains:

46 2 Matrices

[C] � [A] + [B] �[

1 2 0−3 0 −1

]+

[2 2 −21 3 4

]

�[

1 + 2 2 + 2 0 − 2−3 + 1 0 + 3 −1 + 4

]�

[3 4 −2

−2 3 −3

].

For the sum there is the null element (additive identity), which we denote by[O] ∈ Mm×n which, if [A] ∈ Mm×n , has the property:

[A] + [O] � [O] + [A] � [A]. (2.2)

All the entries in the null matrix are zero:

[O]m×n �

⎡⎢⎢⎢⎢⎢⎢⎣

0 0 . . . 00 0 . . . 0...

0 0 . . . 0

⎤⎥⎥⎥⎥⎥⎥⎦

. (2.3)

If we look at the zero matrix as a linear transformation, all the vectors aftertransformation are assigned to the zero vector. The null matrix is idempotent (whenit is multiplied by itself the result is itself): The rank of a zero matrix is 0.

The addition is commutative, that is:

[A] + [B] � [B] + [A], (2.4)

and associative:

([A] + [B]) + [C] � [A] + ([B] + [C]). (2.5)

2.2.2 Scalar Multiplication (·R × Mm×n → Mm×n)

Consider a real number, λ ∈ R, and a matrix [A] ∈ Mm×n . The scalar multiplicationof the matrix [A] with a scalar λ is a matrix [C], which reads:

[C] � λ · [A], (2.6)

where the elements of [C] are defined by the relationships: ci j � λ ai j . The scalarmultiplication can be extended if λ belongs to the space of complex numbers. Forexample, if λ � 2 and:

2.2 Basic Operation 47

[A] �[

1 2 0−3 0 −1

],

then the scalar multiplication by 2 gets:

[C] � λ · [A] � 2[

1 2 0−3 0 −1

]�

[2 4 0

−6 0 −2

].

Scalar multiplication is distributive in regards to the addition:

λ · ([A] + [B]) � λ · [A] + λ · [B], (2.7)

also:

λ · (μ · [A]) � (λ · μ) · [A],

and:

(λ + μ) · [A] � λ · [A] + μ · [A].

2.2.3 Matrix Multiplication (·Mm× p × M p×n → Mm×n)

Consider [A] ∈ Mm×p and [B] ∈ Mp×n . The product of these two matrices is amatrix [C] ∈ Mm×n which we write in