curs algebra liniara

Lecture Noteson

Linear Algebra,Analytic and Di¤erential Geometry

Nicoleta Brînzei

Jan. 2008

ii

Contents

To Students v

1 Linear Algebra 11.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 De�nition and Examples . . . . . . . . . . . . . . . . . . . 11.1.2 Subspaces in a Vector Space . . . . . . . . . . . . . . . . . 3

1.2 Basis and Dimension of a Vector Space . . . . . . . . . . . . . . . 61.2.1 Subspace Spanned by a Subset of a Vector Space . . . . . 61.2.2 Linear Dependence and Independence . . . . . . . . . . . 61.2.3 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . 81.2.4 Changes of Bases . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . 121.3.1 De�nition and Examples . . . . . . . . . . . . . . . . . . . 121.3.2 Kernel and Image of a Linear Transformation . . . . . . . 121.3.3 Linear Transformations on Finite-dimensional Vector Spaces 141.3.4 All Vector Spaces of Dimension n over the Same Field Are

Isomorphic . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4 Eigenvectors and Eigenvalues . . . . . . . . . . . . . . . . . . . . 161.5 Quadratic forms; Canonical Form . . . . . . . . . . . . . . . . . . 20

2 Analytic Geometry 232.1 Free (Spatial) Vectors . . . . . . . . . . . . . . . . . . . . . . . . 23

2.1.1 The Vector Space of Free Vectors . . . . . . . . . . . . . . 232.1.2 Dot (Scalar) Product of Free Vectors . . . . . . . . . . . . 262.1.3 The Cross (Vector) Product of Two Vectors . . . . . . . . 272.1.4 Mixed Triple Product . . . . . . . . . . . . . . . . . . . . 28

2.2 Equation of Lines and Planes in Space . . . . . . . . . . . . . . . 292.2.1 The Plane Containing a Point and Two Directions . . . . 292.2.2 The Plane through a Point, Having a Given Normal Di-

rection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.2.3 Equations of a Straight Line in Space . . . . . . . . . . . 32

2.3 Angles and Distances in Space . . . . . . . . . . . . . . . . . . . 342.3.1 Angles: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3.2 Distances in Space . . . . . . . . . . . . . . . . . . . . . . 35

iii

iv PREFACE

2.4 Coordinate Transformations . . . . . . . . . . . . . . . . . . . . . 372.4.1 Translations in Plane and in Space . . . . . . . . . . . . . 372.4.2 Changes of Orthonormal Bases in Plane and in Space . . 38

2.5 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.5.1 De�nition and Examples . . . . . . . . . . . . . . . . . . . 412.5.2 The Standard Form of a Conic . . . . . . . . . . . . . . . 442.5.3 Center, Axes and Asymptotes of a Conic . . . . . . . . . 482.5.4 Alternative De�nitions of Conics. Eccentricity . . . . . . 50

2.6 Quadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.6.1 The Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . 512.6.2 Reduced Canonical Equations of Other Quadrics . . . . . 53

2.7 Generated Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 582.7.1 Cylindrical Surfaces . . . . . . . . . . . . . . . . . . . . . 592.7.2 Conic Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 612.7.3 Conoidal Surfaces with a Directing Plane . . . . . . . . . 632.7.4 Surfaces of Revolution . . . . . . . . . . . . . . . . . . . . 64

3 Di¤erential Geometry 673.1 Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.1.1 Arc Length of a Plane Curve . . . . . . . . . . . . . . . . 683.1.2 Contact Between Two Intersecting Curves . . . . . . . . . 713.1.3 Tangent and Normal Line at a Regular Point . . . . . . . 723.1.4 Osculating Circle; Curvature and Radius of Curvature . . 74

3.2 Space Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773.2.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . 773.2.2 Arc Length of a Space Curve . . . . . . . . . . . . . . . . 783.2.3 The TNB Frame (The Frenet Frame) . . . . . . . . . . . 793.2.4 Curvature and Torsion . . . . . . . . . . . . . . . . . . . . 83

3.3 Di¤erential Geometry of Surfaces . . . . . . . . . . . . . . . . . . 843.3.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . 843.3.2 Curves on a Surface . . . . . . . . . . . . . . . . . . . . . 853.3.3 Tangent Plane and Surface Normal at a Regular Point . . 873.3.4 First Fundamental Form. Applications . . . . . . . . . . . 88

To Students

This text gathers the lecture notes of a 14 weeks course taught by the authorat technical faculties and it is a "user manual" especially designed for thosewho might need in their further work some basic knowledge on Linear Algebraand/or Analytic and Di¤erential 2D- and 3D- Geometry. So, with several simpleexceptions, I did not include any proofs of theorems. Instead of these, I preferredto insert examples and comments or intuitive interpretations.The reader is supposed to have studied at an acceptable level:

� matrices, determinants and linear systems;

� the equation of a line in plane;

� partial derivatives.

For a better understanding of the notions and results, I suggest the followingway of passing through every subsection of the course.

1. read carefully the theoretical results, trying to understand them; basically,all the information in the "theory" is needed in solving the exercises.

2. try to do by yourself the solved examples. If you don�t succeed, repeatstep 1 and eventually, try to recall the necessary high school Mathematicsnotions (don�t hesitate to look them up in books or on the Internet. Asimple Google search could be of real help :-) )

3. Try to solve the exercises inside the subsection.

Feel free to express any comments or questions regarding the contents of thecourse, at: [email protected].

I wish you success!Yours,Nicoleta Brînzei

v

vi TO STUDENTS

Chapter 1

Linear Algebra

1.1 Vector Spaces

1.1.1 De�nition and Examples

Informally speaking, a vector space (or a linear space) is a set of objects (calledvectors) that may be scaled and added according to some rules; these objectscan be numbers, (geometrical) free vectors, matrices, functions etc. The theoryof vector spaces is, in some sense, a "theory of everything", which provides apowerful tool for other branches of Mathematics, such as: Geometry, Analysis,Di¤erential equations.

Let V 6= ? be a non-empty set, and (K;+; �) a commutative �eld. Let usde�ne two operations:

+ : V � V ! V; (u; v)! u+ v;

(internal operation on V ) which will be called vector addition and

�K : K � V ! V; (�; v)! �v;

(external operation) called further, scalar multiplication.

De�nition 1.1 (V;+; �K) is called a vector space over K if:

1. (V;+) is an Abelian group, this is:

(a) V is closed under vector addition: 8 u; v 2 V : u+ v 2 V ;(b) vector addition is associative: for all u; v; w 2 V , we have u + (v +

w) = (u+ v) + w;

(c) vector addition is commutative: 8v; w 2 V : v + w = w + v;(d) vector addition has an identity (a neutral) element: 90 2 V; such that

v + 0 = v, 8v 2 V ;

1

2 CHAPTER 1. LINEAR ALGEBRA

(e) any element v 2 V admits a symmetric (or an additive inverse)(�v) 2 V such that v + (�v) = 0:

2. the scalar multiplication �K has the following properties:

(a) V is closed under scalar multiplication: 8� 2 K; 8v 2 V : �v 2 V ;(b) �K is distributive w.r.t. vector addition: 8� 2 K; 8u; v 2 V : �(u +

v) = �u+ �v;

(c) �K is distributive w.r.t. the addition in K: 8�; � 2 K; 8v 2 V :(�+ �)v = �v + �v;

(d) scalar multiplication is compatible with multiplication in K ("asso-ciativity"): 8�; � 2 K; 8v 2 V : �(�v) = (��)v;

(e) the neutral (identity) element 1 2 K is an identity element for scalarmultiplication: 8v 2 V : 1v = v:

The elements u; v; ::: 2 V are called vectors, while the elements �; �; ::: 2 Kare called scalars.

Rules of computation in vector spaces:

1. Scalar multiplication with the zero vector 0 2 V yields the zero vector:

8� 2 K : �0 = 0:

2. Scalar multiplication by the scalar 0 2 K yields the zero vector:

8v 2 V : 0v = 0:

3. No other scalar multiplication leads to the zero vector: if �v = 0; theneither � = 0; or v = 0:

4. The result of scalar multiplication (�1) is the additive inverse of the vector:

8v 2 V : (�1)v = �v:

If the �eld of scalars K is R; then the vector space is called a real vectorspace; if K = C; then V is called a complex vector space.

Examples of vector spaces

1. Let K = R; and Rn = fx = (x1; x2; :::; xn) j xi 2 R; i = 1; ng; the set ofall ordered n-uples of real numbers, and the operations:�

x = (x1; x2; :::; xn) 2 Rny = (y1; y2; :::; yn) 2 Rn

! x+ y = (x1 + y1; x2 + y2; :::; xn + yn)

1.1. VECTOR SPACES 3

andx = (x1; x2; :::; xn) 2 Rn ! �x = (�x1; �x2; :::; �xn):

Then, (Rn;+; �R) is a vector space, called the real coordinate space Rn:This is perhaps the most important example of vector space; especially,the cases n = 2 and n = 3 are important for Geometry.

2. Let K be a �eld and Mm�n(K) the space of m � n-type matrices withentries aij 2 K. Then, (Mm�n(K);+; �K) is a vector space over K: Inparticular, for K = R; (Mm�n(R);+; �R) is a real vector space.

3. The set K[X] of all polynomials with coe¢ cients in a �eld K is a vectorspace over K:

4. The set F(I) of all real-valued functions de�ned over an interval I � R;

F(I) = ff j f : I ! Rg;

with the usual addition and scalar multiplication of functions,

(f + g)(x) = f(x) + g(x);

(�f)(x) = � � f(x); 8x 2 I; � 2 R;

is a real vector space.

5. ("the last, but not the least"), the set E3 of all free vectors �a in space,with the usual addition (e.g., using the parallelogram rule) and scalarmultiplication is a real vector space.

Exercise 1.2 Let V = (0;1) and K = R: We de�ne the operations:

u; v 2 V ! u� v = u � v� 2 R; v 2 V ! �� v = v�:

Show that (V;�;�) is a vector space over R:

1.1.2 Subspaces in a Vector Space

Let (V;+; �K) be a vector space.

De�nition 1.3 A non-empty subset S of V is called a subspace in V if S;together with the operations + and �K of the space V; restricted to S, is a vectorspace.

In order to check that a subset of a vector space is a subspace, one has thefollowing criterion:

Criterion 1.4 A non-empty set S of a vector space (V;+; �K) is a subspace ifand only if:


1. for any u; v 2 S; the sum u+ v belongs to S and

2. for any scalar � 2 K and any vector u 2 S; we have: �u 2 S:

The above criterion is equivalent to:

Criterion 1.5 A non-empty set S of a vector space (V;+; �K) is a subspace ifand only if:

8�; � 2 V;8u; v 2 S : �u+ �v 2 S:

This is, a subspace is a subset of a vector space which is closed under additionand scalar multiplication, or, equivalently, under formation of linear combina-tions.

Examples:1) Let (V;+; �K) be a vector space. Then f0g and V are subspaces of V:Let us check the above statement:

� For S = f0g; we have: 0 + 0 = 0 2 S and �0 = 0 2 S; 8� 2 K: Hence,S = f0g is a subspace.

� For S = V; there holds: 8 u; v 2 V : u + v 2 V; this is, V is closed underaddition. Also, by the de�nition of a vector space, V is closed under scalarmultiplication: 8� 2 K; 8 u 2 V : �u 2 V; which proves the statement.

2) The solutions of any homogeneous linear system with coe¢ cients in Rform a vector space.Solution: We shall prove that the solutions of a linear system:8>><>>:

a11x1 + a12x2 + ::::+ a1nxn = 0a21x1 + a22x2 + ::::+ a2nxn = 0

:::am1x1 + am2x2 + ::::+ amnxn = 0

;

where aij 2 R; i = 1; :::;m; j = 1; :::; n; form a subspace of the vector space Rn:In order to simplify the writing, let us denote

A =

0BBB@a11 a12 ::: a1na21 a22 ::: a2n...

. . ....

am1 am2 ::: amn

1CCCA ; X =

[email protected]

1CCCA :This is, A is the matrix of the system, and X is the column matrix of theunknowns. Then, our linear system can be written as a unique matrix equation:

AX = 0:

Proof. Let us now prove the statement.

1.1. VECTOR SPACES 5

1. (a) If x = (x1; :::; xn) 2 Rn and y = (y1; :::; yn) 2 Rn are two solutionsof the system, then, with the above notations, we have: AX = 0 andAY = 0: We get immediately:

A(X + Y ) = 0;

which means that x+ y = (x1+ y1; :::; xn+ yn) is again a solution ofthe system.

(b) Let x = (x1; :::; xn) 2 Rn be a solution and � 2 R a scalar. Then,the equality AX = 0 implies

�AX = A(�X) = 0;

which means that �x = (�x1; :::; �xn) is a solution of the system,q.e.d.

The above example can be used as a theorem. For instance, the solutions ofthe system: �

x1 + 2x2 + x3 = 0x1 � 3x3 = 0

(1.1)

form a vector subspace of R3:

Exercise 1.6 Prove that the solutions of the system�x1 + 2x2 + x3 = 1x1 � 3x3 = 2

(1.2)

do not constitute a vector subspace of R3: This is, if in the system (1.1) wechange the free terms, then the solutions of the system do no longer yield avector subspace.

Exercise 1.7 Prove that, in the space of polynomials (C[X];+; �C); the set

Pn = ff 2 C[X] j f = anXn + an�1Xn�1 + ::::+ a1X + a0g

of all polynomials of degree at most n, is a vector subspace.

Exercise 1.8 Let (V;+; �K) be a vector space and U; S � V two subspaces.Prove that:

� the intersection U \ S is a subspace of V ;

� the sum U + S = fu+ s j u 2 U; s 2 Sg is a subspace of V ;

� the union U [ S is generally not a subspace (�nd a counterexample!).


1.2 Basis and Dimension of a Vector Space

1.2.1 Subspace Spanned by a Subset of a Vector Space

Let (V;+; �K) be a vector space and U � V a non-empty subset. Then, the setof all �nite linear combinations of elements in U; namely

[U ] = f�1u1 + ::::�nun j �i 2 K; ui 2 U; i = 1; n; n 2 N�g

is a vector subspace of V; called the subspace spanned by U; or simply, thespan of U: U is called a spanning set (a set of generators) for [U ]:Moreover, it can be proved that [U ] is the smallest subspace of V which

contains U; this is, it is contained in any other subspace containing U:

In particular, if U = fu1; u2; ::::; ung is a �nite set, then its span [U ] is simplythe set of all linear combinations of its elements:

[U ] = f�1u1 + ::::�nun j �i 2 K; i = 1; ng:

Examples:1) Let U = f(1; 2); (3; 4)g � R2: Then,

[U ] = f�(1; 2) + �(3; 4) j �; � 2 Rg = f(�+ 3�; 2�+ 4� j �; � 2 Rg:

2) If U contains only one vector U = fug; then the subspace spanned by Uconsists of all multiples of u :

[U ] = f�u j � 2 Kg:

Exercise 1.9 Show that the following subsets of R3 are subspaces and �nd aspanning subset (a set of generators) for each of them:

1. S1 = f(�; 2�; 3�) j� 2 Rg;

2. S2 = f(�+ �; �� ; 2�) j �; � 2 Rg;

3. S3 = f(x1; x2; x3) j x1 + x2 = 0; x1 � 2x3 = 0g:

1.2.2 Linear Dependence and Independence

Let (V;+; �K) be a vector space and U � V a set of vectors.

De�nition 1.10 A set U of vectors in V is called:

� linearly dependent if one of the vectors of U can be written as a �nitelinear combination of other vectors of U ; for instance

un+1 = �1u1 + ::::+ �nun; �i 2 K; ui 2 U ;

1.2. BASIS AND DIMENSION OF A VECTOR SPACE 7

� linearly independent if none of the vectors of U can be written as alinear combination of other vectors in U:

For instance, the vectors u = (1; 0); v2 = (0; 1); w = (1; 2) in R2 are linearlydependent, because w is a linear combination of u and v:

w = u+ 2v

Still, in practice it would be hard to guess whether and which of the vectorsof a set U could be written as a linear combination of the others. This is whywe shall use the following criterion:

Criterion 1.11 The vectors u1; u2; :::un 2 V are linearly independent if andonly if the equation

�1u1 + �2u2 + ::::+ �nun = 0;

(with the unknowns �1; ::; �n 2 K) has the unique solution �1 = �2 = ::: =�n = 0:

Remark 1.12 An in�nite set U � V is linearly independent if any �nite subsetof U is linearly independent.

Comments of linear dependence/independence: In the above example, thevector w = (1; 2) is a linear combination of u and v; this is, it can be "decom-posed" in terms of u and v: While, u and v are linearly independent, this is,none of them can be expressed (or "decomposed") in terms of the other. Let usimagine a map in which u gives the direction to the North and v; to the East.Then, if we say to someone, "go 3 kilometers to the North and 3 kilometers tothe East", this is, perform a displacement by x = 3u + 3v; we could also sayinstead: "go 2u+v+w". But the latter description is somehow "not necessary",actually, if we ignore altitude, linear combinations of the linearly independentvectors u and v are enough to describe any displacement in the plane. This is,the vector w is redundant.

Exercise 1.13 Which of the following systems of vectors in R2 are linearlyindependent:

1. u = (1; 2);

2. u = (1; 2); v = (3;�1);

3. u = (1; 2); v = (3;�1); w = (4; 1)?

Exercise 1.14 Prove that, in a vector space (V;+; �K) :

� a system consisting of one vector fvg � V is linearly independent ()v 6= 0;

� two vectors u; v 2 V are linearly dependent if and only if one of them is amultiple of the other, this is, there exists an � 2 K such that v = �u:


1.2.3 Basis and Dimension

Let (V;+; �K) be a vector space.

De�nition 1.15 A subset B 6= ? of V is called a basis of V if:

1. B is linearly independent;

2. B spans the entire space V : any v 2 V can be written as a �nite linearcombination of elements of B :

v = �1e1 + :::+ �nen;

where e1; :::; en 2 B; �1; :::; �n 2 K:

Example (the standard basis in Rn): Let, in Rn; B = fe1; :::; eng; where:8>><>>:e1 = (1; 0; 0; :::; 0)e2 = (0; 1; 0; :::; 0)

:::en = (0; 0; 0; :::; 1)

:

Then, B is a basis of Rn; called the standard or the canonical basis.Let us check that B is indeed a basis:

� linear independence: the equality �1e1 + ::::+ �nen = 0 means:

�1(1; 0; :::; 0) + ::::+ �n(0; 0; ::::; 1) = 0;

which is, (�1; ::::; �n) = (0; 0; ::::; 0):

� the "spanning property": Let v = (v1; v2; :::; vn) 2 Rn be arbitrary. Then,the equation

v = �1e1 + :::+ �nen

has the solution �1 = v1; :::; �n = vn; this is, v can indeed be written as alinear combination of e1; ::::; en 2 B:

Properties:

1. Any non-zero vector space admits at least a basis. From now on, whenreferring to vector spaces, we shall only mean non-zero ones.

2. Any linearly independent system of vectors can be completed up to abasis( , can be included in a basis).

3. From any spanning set (system of generators) of a vector space, one canextract a basis.


This is, we can summarize the situation in the following way:

linearly independent set � basis � spanning set. (1.3)

A vector space which admits a basis B consisting of a �nite number of vectorsis called a �nite-dimensional vector space.For �nite-dimensional vector spaces, there exists a "shortcut" for proving

that a certain set is a basis. Namely, it relies on the notion of dimension of avector space.

Theorem 1.16 If V is a �nite-dimensional vector space, then any two bases ofV have the same number of vectors. This number is called the dimension ofthe vector space V :

n = dimV:

Example: The dimension of Rn is n (since the standard basis has n vectors).

From (1.3) we can now infer:

Theorem 1.17 If V is a vector space of dimension n; then any set consistingof precisely n linearly independent vectors of V is a basis of V:

Example: In R2; the vectors u = (1; 2) and v = (2; 1) are linearly indepen-dent (prove it!); since the dimension of R2 is 2, we get that fu; vg is a basis ofR2:

Exercise 1.18 Prove that the vectors u = (1; 0; 0); v = (1; 2; 0) and w =(1; 2; 3) yield a basis of R3:

Another important property of bases is the following:

Theorem 1.19 Let (V;+; �K) be a �nite-dimensional vector space and B =fe1; e2; :::; eng a basis of V: Then any vector x 2 V admits a unique represen-tation:

x = x1e1 + x2e2 + :::+ xnen; (1.4)

with x1; :::; xn 2 K (scalars). The scalars x1; :::; xn are called the coordinatesof x with respect to the basis B (in the basis B).

Proof. Let x 2 V be arbitrary. The "spanning property" in the de�nitionof a basis insures the existence of the scalars x1; :::; xn 2 K such that x =x1e1 + x2e2 + :::+ xnen:Let us prove the uniqueness of this representation: let x01; ::::; x

0n 2 K be

another set of scalars with x = x01e1 + x02e2 + :::+ x

0nen:


By subtracting the above equalities, we are led to:

0 = (x1 � x01)e1 + :::+ (xn � x0n)en:

Since e1; :::; en are linearly independent, we get x1 � x01 = ::: = xn � x0n = 0;q.e.d.Once the basis B is �xed, the column matrix having as entries the coordinates

of some vector x 2 V in the basis B;

X =

[email protected]

1CCCAis called the coordinate vector of x 2 V:

Example: If B = fe1; :::; eng is the standard basis of Rn; we have shownabove that for any x = (x1; x2; :::; xn); we have x = x1e1 + :::: + xnen; whichmeans that, with respect to the standard basis, the coordinates of a vector x 2 Rncoincide with its components. The coordinate vector X is, in this case, preciselythe transpose matrix xt:For instance, the vector v = (1; 3) 2 R2 has, with respect to the standard

basis, the coordinates

X =

�13

�:

1.2.4 Changes of Bases

Let (V;+; �K) be a �nite-dimensional vector space and

B = fe1; ::::; eng; B0 = fe01; :::; e0ng

two bases of V . Then, any vector x 2 V admits a unique decomposition

x = x1e1 + ::::+ xnen

with respect to B; and a unique decomposition

x = x01e01 + ::::+ x

0ne0n

with respect to B0:The problem which arises is obviously the following: which is the relation

between the two sets of coordinates of x; namely, between

X =

[email protected]

1CCCA and X 0 =

[email protected]

1CCCA?


In order to answer this question, let us decompose, one by one, the vectorsof the "new" basis B0 in terms of the basis B: We get:

e01 = s11e1 + s21e2 + :::+ sn1en

e02 = s12e1 + s22e2 + :::+ sn2en

:::

e0n = s1ne1 + s2ne2 + :::+ snnen;

with sij 2 K; i; j = 1; :::; n:The matrix S = (sij)i;j=1;n; having as columns the coordinate vectors of

e01; :::; e0n respectively, in the basis B; is called the matrix of change of bases

from B to B0: We denoteB S! B0:

Then, we get

x =nXj=1

x0je0j =

nXj=1

x0j

nXi=1

sijei

!=

nXi=1

0@ nXj=1

sijx0j

1A ei:On the other side, x =

nPi=1

xiei; since this decomposition is unique, we get:

xi =nPj=1

sijx0j ; or, in terms of matrices, X = SX 0: We have thus proved:

Proposition 1.20 If S is the matrix of change of bases from B to B0; then thecorresponding sets of coordinates of a vector x 2 V are related as follows:

X = SX 0:

The matrix of change S is always invertible, hence we can write also

X 0 = S�1X:

Exercise 1.21 Let B1 = fe1 = (1; 1; 1); e2 = (1; 1; 0); e3 = (1; 2; 3)g and B2 =fe01 = (0; 1; 1); e02 = (1; 0; 0); e03 = (3; 2; 1)g

1. Show that B1 and B2 are bases of R3:

2. Find the expression of the vector x = (2; 2; 1) in the bases B1 and B2:

3. Find the matrix of change B1S! B2 and check, for the vector x above, the

relation X = SX 0:

4. Find the matrices of change from the standard basis B to B1 and B2:Whatdo you notice? Why?


1.3 Linear Transformations


Let (V;+�K) and (W;+; �K) be two vector spaces over the same �eld K:

De�nition 1.22 A function T : V ! W is called a linear transformation(or a morphism of vector spaces) if:

1. for any x; y 2 V : T (x+ y) = T (x) + T (y);

2. for any � 2 K and any x 2 V : T (�x) = �T (x):

The above de�nition is equivalent to:

T (�x+ �y) = �T (x) + �T (y); 8�; � 2 K; 8x; y 2 V: (1.5)

An isomorphism of vector spaces is a one-to-one (=bijective) lineartransformation.Property: If T is a linear transformation, then T (0) = 0:Indeed, we have T (0) = T (0 + 0) = T (0) + T (0): By equating the �rst and

the last term, we get 0 = T (0):Examples:

1. f : R! R; f(x) = ax; where a 2 R is a �xed real number;

2. T : Rm ! Rn; T (x) = A � xt; where A 2Mm�n(R) is a �xed matrix and,for x = (x1; x2; :::xm) 2 Rm; we denoted by xt the transpose matrix of x:

3. D : C1(a; b) ! C0(a; b); D(f) = f 0 (derivation), where C1(a; b) is thevector space of real di¤erentiable functions of class C1 on the interval(a; b) and C0(a; b) is the space of continuous functions on (a; b):

4. I : C0[a; b]! R; I(f) =bRa

f(x)dx (integration over an interval [a; b]).

Exercise 1.23 Prove that the above transformations are linear.

Exercise 1.24 Check which of the following transformations are linear:a) T : R2 ! R3; T (x1; x2) = (x1 + x2; x1 � 3x2; 2x1);b) T : R2 ! R3; T (x1; x2) = (x1 + x2; x1 � 3x2; 2x1 + 1);c) T : R2 ! R3; T (x1; x2) = (x1 � x2; x1 � 3x2; 2x1):

1.3.2 Kernel and Image of a Linear Transformation

De�nition 1.25 Let T : V !W be a linear transformation.

1. The set of all "roots" of T; namely:

kerT = fx 2 V j T (x) = 0g;

is called the kernel of T:

1.3. LINEAR TRANSFORMATIONS 13

2. The set of all values of T; namely

ImT = fT (v) j v 2 V g = fw 2W j 9v 2 V : T (v) = wgis called the image of T:

Exercise 1.26 Prove that kerT is always a subspace of V; this is:

� for any u; v 2 kerT; there holds u+ v 2 kerT ;

� for any � 2 K; and any u 2 kerT; there holds �u 2 kerT:Exercise 1.27 Prove that ImT is a subspace of W:

Exercise 1.28 Prove that: T is injective if and only if kerT = f0g:This means, that for linear transformations, it is not only true that an in-

jective transformation has only one root, but also the converse: if it has onlyone root, then it is injective.

Remark 1.29 T is onto (=Ro: "functie surjectiv¼a") if and only if ImT =W:

Example: Let T : R3 ! R2; T (x1; x2; x3) = (x1 + x2; x3):a) Prove that T is a linear transformation (exercise).b) Find its kernel and its image. Is T injective/onto?Solution: b) We have kerT = fx 2 V j T (x) = 0g: The equation

T (x1; x2; x3) = 0 leads to �x1 + x2 = 0x3 = 0:

This has the solutions x3 = 0; x2 = � 2 R; x1 = ��; this is,kerT = f(��; �; 0) j � 2 Rg:

Since kerT contains more than one element, T is not injective.In order to determine the image of T; we have to take into account that

ImT = fw = (w1; w2) 2 R2 j 9x 2 R3 : T (x) = wg:This is, we have to determine the values of w1; w2 2 R such that the linearsystem �

x1 + x2 = w1x3 = w2

;

in the unknowns x1; x2; x3; is compatible. The matrix of the system is

A =

�1 1 00 0 1

�;

it has rank 2 (=the number of equations), which means that the system iscompatible for any values w1; w2 2 R: This is, ImT = R2; consequently, T isonto.

Exercise 1.30 Find the kernel and the image of the transformation

T : R2 ! R3; T (x1; x2) = (x1 + x2; x1 � 3x2; 2x1);is it injective? Is it onto?


1.3.3 Linear Transformations on Finite-dimensional Vec-tor Spaces

(V;+�K) and (W;+; �K) be two �nite-dimensional vector spaces over the �eldK; with dimV = n; dimW = m; m; n 2 N�: Let

B = fe1; e2; :::; eng; B0 = ff1; :::; fmg

be two bases in V and W; respectively.We denote, for x 2 V; its coordinate vector in the basis B with X; and, for

y = T (x); its coordinate vector w.r.t. B0 with Y :

x = x1e1 + :::+ xnen ! X =

0B@ x1...xn

1CA ;

T (x) = y = y1f1 + :::+ ymfm ! Y =

0B@ y1...yn

1CA :Remark 1.31 If V = Rn and B is the standard basis of Rn; then, forx = (x1; :::; xn); the coordinates in the basis B coincide with the components(x1; :::; xn) :

x = x1e1 + :::+ xnenstandard basis! X =

0B@ x1...xn

1CA :Let us now build a matrix A; having as columns the coordinate vectors (in

the basis B0) of the vectors T (ej) :

T (ej) =mXi=1

aijfi; j = 1; n) A =

0BBB@a11 a12 ::: a1nam1 a22 ::: a2n...

.... . .

...amn a2n ::: amn

1CCCA : (1.6)

The matrix A is called the matrix of T w.r.t. the bases B and B0. Inparticular, if m = n; this is, T : Rn ! Rn; and B = B0; then the matrix A in(1.6) is called simply the matrix of T w.r.t. the basis B.

Proposition 1.32 With the above notations, the values T (x); x 2 V; are com-puted by the rule

Y = AX:

Remark 1.33 In particular, if V = Rn and W = Rm (and, obviously, K = R),then any linear transformation between Rn and Rm can be characterized asabove.

1.3. LINEAR TRANSFORMATIONS 15

Example: Let T : R3 ! R2; T (x1; x2; x3) = (x1 + x2 + x3; x1 � 2x2):a) Prove that T is a linear transformation (homework).b) Find its matrix w.r.t. the standard bases of R3 and R2:c) Calculate T (1; 2; 3) : (i) directly; (ii) by means of the matrix you found

at point b).Solution:

b) We have:

8<: T (e1) = T (1; 0; 0) = (1; 1)T (e2) = T (0; 1; 0) = (1;�2);T (e3) = T (0; 0; 1) = (1; 0):

: Then, A =

�1 1 11 �2 0

�:

(The �rst column represents T (e1); the second T (e2) etc.).c) directly: T (1; 2; 3) = (1 + 2 + 3; 1� 2 � 2) = (6;�3):by means of the matrix:

Y = AX =

�1 1 11 �2 0

�0@ 123

1A =

�6�3

�:

Let T : V ! V (endomorphism) and A be its matrix w.r.t. a basis B: If weperform a change of bases B S!B0 (by means of the matrix of change S), thenthe matrix of T changes by the rule

A0 = S�1AS: (1.7)

Exercise 1.34 Let T : R2 ! R3; T (x1; x2) = (x1 + x2; x1 � 2x2; 3x1):

a) Prove that T is a linear transformation (homework).b) Find its matrix w.r.t. the standard bases of R3 and R2:c) Calculate T (1; 2) : (i) directly; (ii) by means of the matrix you found at

point b).

1.3.4 All Vector Spaces of Dimension n over the SameField Are Isomorphic

Theorem 1.35 Let (V;+; �K) and (W;+; �K) be two �nite dimensional vectorspaces over the same �eld K; having the same dimension n: Then, the two vectorspaces are isomorphic to each other.

Proof. The proof is constructive one: we e¤ectively determine an isomorphismbetween the two spaces. Let B = fe1; ::::; eng be a basis in V and �B = ff1; :::; fngbe a basis in W: Then, any vector in V can be uniquely expressed as

x = x1e1 + :::+ xnen:

We setT (x) = x1f1 + ::::+ xnfn:

As a remark, T actually makes e1 to correspond to f1; e2 to f2 etc. (T (e1) =f1; :::; T (en) = fn).


Exercise 1.36 For T : V !W de�ned above, prove:a) T is a linear transformation;b) T is one-to-one (bijective).

If we set now K = R andW = Rn () dimW = n), then we obtain the mostimportant consequence of the above theorem:

Corollary 1.37 Any real vector space (V;+; �R) of dimension n 2 N� is iso-morphic to Rn:

This is, Rn is the only "model" for real n-dimensional vector spaces, anyother such vector space algebraically "works" in the same way.

1.4 Eigenvectors and Eigenvalues

We take into consideration the case when V =W = Rn:Let T : Rn ! Rn and A be its matrix w.r.t. a basis B: If we perform a

change of bases B S!B0 (by means of the matrix of change S), then the matrixof T changes by the rule

A0 = S�1AS: (1.8)

Problem: In applications (Geometry, Di¤erential equations), it would beuseful to �nd some basis B0 of Rn such that, with respect to it, the matrix of Tshould look like this:

A0 =

0BBB@�1 0 ::: 00 �2 0...

. . ....

0 0 ::: �n

1CCCA :There are two questions which arise:

1. does such a basis exist? In which conditions?

2. If it exists, how can we build it? And how can we �nd the entries �i ofA0?.

If they exist, then A0 is called a diagonal form of T; and B0 is called adiagonal basis.

Let us have a closer look to question 1.This means: does there exist a basis B0 = fe01; :::; e0ng such that T (e01) =

�e01; ::::; T (e0n) = �e

0n?

In order to answer it, let us give some de�nitions:

De�nition 1.38 Let T : Rn ! Rn; having the matrix A w.r.t. a given basis B.

1.4. EIGENVECTORS AND EIGENVALUES 17

1. A real number � 2 R is called an eigenvalue of T (or of A) if there existnonvanishing vectors v 2 Rn (v 6= 0); such that

T (v) = �v: (1.9)

2. If � 2 R is an eigenvalue of T; then any vector v 2 Rn satisfying equation(1.9) is called an eigenvector of T (or of A).

Remark 1.39 For any linear transformation and for any � 2 K; we haveT (0) = �0 = 0; this is why we requested in the de�nition that there shouldexist vectors v with v 6= 0:

This is, our problem can be reformulated: does there exist a basis ofRn; consisting only of eigenvectors? If so, how can we compute eigenval-ues/eigenvectors?

Proposition 1.40 If � is an eigenvalue, then the set

V� = fv 2 Rn j T (v) = vg

of its eigenvectors (including 0) is a subspace of Rn; called the eigenspace of�:

Exercise 1.41 Prove the above statement (u; v 2 V� ) u + v 2 V�; � 2 R;u 2 V� ) �u 2 V�):

We shall start by answering question 2, namely:

Theorem 1.42 Let T : Rn ! Rn be a linear transformation.

1. The eigenvalues of T are the real solutions of the characteristic equa-tion

det(A� �In) = 0: (1.10)

2. If � is an eigenvalue, then the corresponding eigenvectors are given by thesolutions X of the equation

AX = �X: (1.11)

The equation (1.10) is of degree n; so it can have at most n real solutions.

Example: Let T : R3 ! R3; T (x1; x2; x3) = (x2 + x3; x1 + x3; x1 + x2):a) write its matrix in the standard basis of R3;b) determine its eigenvalues, and, for each of them, the corresponding

eigenspace.


a) By computing T (1; 0; 0); T (0; 1; 0); T (0; 0; 1); we get

A =

0@ 0 1 11 0 11 1 0

1A :b) The characteristic equation is det(A� �I3) = 0; this is,��

�� 1 11 �� 11 1 ��

�� = 0 ) 3�� 3 + 2 = 0:

Its solutions are �1 = �2 = �1; �3 = 2: So, we shall have two distincteigenspaces, namely, V�1 and V2:Let us determine V�1 : AX = (�1)X; this is,0@ 0 1 1

1 0 11 1 0

1A0@ x1x2x3

1A =

0@ �x1�x2�x3

1A :This leads to 8<: x2 + x3 = �x1

x1 + x3 = �x2x1 + x2 = �x3

:

The solution is x2 = �; x3 = �; x1 = �� ; �; � 2 R:We are thus able to write

V�1 = f(�� ; �; �) j �; � 2 Rg:

Analogously, we get, for �3 = 2; the equation AX = 2X; which has thesolutions

V2 = f(�; �; �) j � 2 Rg:Diagonal formLet us now answer the �rst question in the beginning of this course. This is,

in what conditions does a basis consisting of eigenvectors exist?Intuitively, we should have "enough" eigenvalues to �ll in the matrix, and

correspondingly, "enough" independent eigenvectors as to form a basis of Rn:This is: we need n real eigenvalues (not necessarily distinct), and n independenteigenvectors.

Proposition 1.43 The eigenvectors corresponding to distinct eigenvalues arelinearly independent.

Proposition 1.44 If � 2 R is an eigenvalue of T and m� is its multiplicity asa root of the characteristic equation (=its algebraic multiplicity), then

dim(V�) � m�:

The number dim(V�) is called the geometric multiplicity of �:

1.4. EIGENVECTORS AND EIGENVALUES 19

The two propositions above mean that, once we have n eigenvalues, the onlyproblem that could arise is when the characteristic equation has multiple roots.In this case, we risk to obtain dim(V�) < m�; which would mean that we cannot�nd n independent eigenvectors.We are now able to state a general criterion:

Theorem 1.45 A linear transformation T admits a diagonal form if and onlyif:

1. all the roots �1; :::; �n of the characteristic equation det(A � �I) = 0 arereal and

2. for every eigenvalue �, its geometric multiplicity equals its algebraic mul-tiplicity:

dim(V�) = m�:

If the above conditions are ful�lled, then the diagonal form of T is

A0 =

0BBB@�1 0 ::: 00 �2 0...

. . ....

0 0 ::: �n

1CCCA ;and the diagonal basis is obtained as the union of the bases in each eigenspaceV�i :

Example: Let T : R3 ! R3; as in the previous example (T (x1; x2; x3) =(x2 + x3; x1 + x3; x1 + x2)).The �rst condition of the criterion is ful�lled, namely: �1 = �2 = �1 2 R;

�3 = 2 2 R:A basis for the eigenspace V�1 = f(�� ; �; �) j �; � 2 Rg is

e1 = (�1; 1; 0); e2 = (�1; 0; 1)) dim(V�1) = 2 = m�1:

In the same way, a basis for V2 = f(�; �; �) j � 2 Rg is

e3 = (1; 1; 1)) dim(V2) = m2:

So, both conditions in the criterion are ful�lled, consequently, T admits a diag-onal form. It is

A0 =

0@ �1 0 00 �1 00 0 2

1A :The diagonal basis is B0 = fe1 = (�1; 1; 0); e2 = (�1; 0; 1); e3 = (1; 1; 1)g:

Remark 1.46 Symmetric matrices A 2Mn(R) always admit a diagonal form.


Exercise 1.47 Check which of the following matrices representing linear trans-formations in the standard bases of Rn admit a diagonal form, and, if so, �ndthe diagonal form/diagonal basis:

a)�4 55 4

�; b)

0@ �1 0 �33 2 3�3 0 1

1A.

1.5 Quadratic forms; Canonical Form

Let V = Rn, B = fe1; ::::; eng a basis of V; and let us denote, for x 2 V;X = (x1; :::; xn)

t its coordinate vector w.r.t. B:

De�nition 1.48 A function h : Rn ! R is called a quadratic form if, w.r.t.

B; h has the form h(x) =nP

i;j=1

aijxixj :

De�nition 1.49 The symmetric matrix A = (aij)i;j=1;n is called the matrix ofh in the basis B:

In order to obtain a symmetric matrix (aij = aji); we take the coe¢ cientsof xixj as follows:a) if i = j; then aij := the coe¢ cient of xixj = x2i ;

b) if i = j; then aij = aji :=1

2� the coe¢ cient of xixj :

Example: Let h : R2 ! R2; h(x1; x2) = 5x21 + 8x1x2 + 5x22: Its matrix inthe standard basis is

A =

�5 44 5

�:

(here, 4 = 8=2).

Proposition 1.50 The values of a quadratic form can be calculated as

h(x) = XtAX:

Proposition 1.51 With respect to changes of bases B S!B0; the matrix of hchanges by the rule

A0 = StAS:

De�nition 1.52 We say that the quadratic form h : Rn ! R is in canonicalform, if in its expression appear only squares: h(x) =

nPi=1

aiix2i (equivalently: if

its matrix is in diagonal form A = diag(aii)).

1.5. QUADRATIC FORMS; CANONICAL FORM 21

Problem (reduction to the canonical form): �nd a basis B0 such that,w.r.t. it, h(x) =

nPi=1

aii (x0i)2; (where xi are the coordinates of x in the basis B0).

There are multiple ways of solving this problem, the only one we indicatehere is themethod of eigenvalues and eigenvectors: we write the matrix ofh in the given basis (usually, the standard one), and we bring it to the diagonalform as in the previous section. The diagonal form of A gives the canonicalform of h : aii = �i; and the basis B0 is none but the diagonal basis of A:

Exercise 1.53 �nd the canonical form for h : R2 ! R2; h(x) = h(x1; x2) =5x21 + 8x1x2 + 5x

22 (given in the standard basis of R2).


Chapter 2

Analytic Geometry

2.1 Free (Spatial) Vectors

2.1.1 The Vector Space of Free Vectors

The notion of vector is one of the most important notions in Physics. Indeed,force, velocity, acceleration, work, momentum, all of them are linked to thisnotion.From now on, by "space", we shall mean the physical space.Let A;B be two points in space.A bound vector

��!AB is a segment AB for which we distinguish two orienta-

tions: from A to B and from B to A: Usually, we denote bound vectors withan arrow. We denote ��!

BA = ��!AB;

the vector��!BA is called the opposite of

��!AB:

For a bound vector��!AB; the point A is called the origin or the tail of the

vector, while B is called the head (tip, destination) of the vector.The zero vector is, by de�nition, the vector

�!0 =

�!AA:

The length (or the norm) ��!AB of the vector ��!AB is, by de�nition, the length

of the segment AB : ��!AB = AB:The direction of the bound vector

��!AB is formed by all the straight lines

which are parallel to the straight line AB:

De�nition 2.1 The free vector AB is the set of all bound vectors which havethe same direction, the same orientation and the same length as the bound vector��!AB:

We usually denote free vectors with a bar: �a:

23

24 CHAPTER 2. ANALYTIC GEOMETRY

This is, if��!AB and

��!CD are two bound vectors with the same direction,

orientation and length, as bound vectors, they are di¤erent, but as free vectors,they are equal to each other:

��!AB 6= ��!CD; AB = CD:

For free vectors, one can de�ne the following operations:

� addition: by the parallelogram rule, or by the triangle rule:

� scalar multiplication: the vector ��a is de�ned by:- the same directionas �a;

- the length j�j �a- the orientation of �a if � is positive, and opposite to �a; if � is negative.:

Theorem 2.2 1. Free vectors on a straight line form a vector space over R:(E1;+; �R):

2. Free vectors in a plane form a vector space over R: (E2;+; �R):

3. Free vectors space form a vector space over R: (E3;+; �R):

Let us take into account the vectors in space (E3). Since E3 is a vector space,it is meaningful to ask ourselves what linear dependence/independence means.A) For one vector �a; linear dependence means that ��a = 0 happens for

some non-zero scalar � 2 R: But, this means that �a itself should be zero.This is:

�a� lin: indep:, �a 6= 0:

B) For two vectors �a;�b; linear dependence means

�a = ��b;

for some � 2 R; this is, �a and �b have the same direction. We say, in this case, thatthey are collinear. Consequently, linear independence means non-collinearity:

�a;�b-lin. indep:, �a;�b� non-collinear.

2.1. FREE (SPATIAL) VECTORS 25

C) For three vectors �a;�b; �c linear dependence means

�c = ��a+ ��b;

for some �; � 2 R; this is, �a and �b are coplanar. Consequently, linear indepen-dence means non-coplanarity:

�a;�b; �c - lin. indep:, �a;�b; �c� non-coplanar.

Now, we are able to establish the dimensionality as vector spaces of E1; E2and E3:

Theorem 2.3 (The fundamental theorem of Analytic Geometry):

1. On a straight line there exists one linearly independent vector. Any twovectors are linearly dependent () dim E1 = 1).

2. In plane, there exist two linearly independent vectors. Any three vectorsare linearly dependent () dim E2 = 2).

3. In space, there exist three linearly independent vectors. Any four vectorsare linearly dependent () dim E3 = 3).

But, as we know from the previous courses, any two �nite dimensional vectorspaces over the same �eld are isomorphic. This is:

E1 ' R1; E2 ' R2; E3 ' R3:

Let us now take into account the space (E3;+; �R) ' (R3+; �R):A Cartesian frame in E3 consists of a point O, called the origin, and an

orthonormal basis B =f�{; �j; �kg; this is, a basis whose vectors have all length1 and are perpendicular("orthogonal") to each other.The isomorphism which links the basis f�{; �j; �kg to the standard basis of R3;

�{$ (1; 0; 0); �j $ (0; 1; 0); k $ (0; 0; 1);

leads to the correspondence

�a = a1�{+ a2�j + a3�k $ (a1; a2; a3) 2 R3:

We denote�a(a1; a2; a3);

and call (a1; a2; a3) the Cartesian coordinates of �a 2 E3:If �a represents the bound vector

�!OA,

�a(a1; a2; a3) = OA;

then we say that the point A has the Cartesian coordinates (a1; a2; a3) :

A(a1; a2; a3)$ OA(a1; a2; a3):

The vector�!OA is called the position vector of the point A:

Now, for any other vector �a which has the same direction, orientation, andlength as OA; we associate the same coordinates (a1; a2; a3):


Proposition 2.4 If A(xA; yA; zA); B(xB ; yB ; zB); then

AB(xB � xA; yB � yA; zB � zA):

Example: If A(2; 3; 4); B(0;�1; 2); then:

OA(2; 3; 4), OA = 2�{+ 3�j + 4�k;

OB(0;�1; 2), OB = ��j + 2�k;AB(xB � xA; yB � yA; zB � zA)) AB(�2;�4; ;�2):

2.1.2 Dot (Scalar) Product of Free Vectors

De�nition 2.5 Let �a;�b 2 E3: The dot (inner, scalar) product �a � �b is thenumber

�a � �b = k�ak � �b � cos�;

where � is the smallest angle between the directions of �a and �b () � 2 (0; 180o)):

� Properties:

1. the dot product is commutative: �a � �b = �b � �a:

2. it is distributive w.r.t. addition: �a � (�b+ �c) = �a � �b+ �a � �c:

3. �a � (��b) = (��a) � b = ��a � �b:

4. if �a and �b are non-zero, then �a � �b = 0 if and only is �a is perpendicular to�b : �a ? �b:

� The expression in coordinates of the scalar product: If �a(a1; a2; a3)and �b(b1; b2; b3); then

�a � �b = a1b1 + a2b2 + a3b3:

� Applications:

1. The length (norm) of the vector �a is

k�ak =p�a � �a =

qa21 + a

22 + a

23:

2. The angle between �a and �b is computed as

cos� =�a � �b

k�ak �b :

3. The unit vector which is collinear to �a is

vers(�a) =�a

k�ak :

2.1. FREE (SPATIAL) VECTORS 27

4. the length of the projection of �a onto �b is given by pr�b�a =k�ak cos�; this is,

pr�b�a =�a � �b �b :

1. Show that the vectors �a(�2; 2; 1) and �b(2; 1; 2) are perpendicular.

2. Let A(1; 2; 0); B(1; 2; 1); C(3; 0; 1): Calculate AB � AC; AB ; AC cos( \AB �AC); prACAB and the unit vector of AC:

2.1.3 The Cross (Vector) Product of Two Vectors

De�nition 2.6 The cross product of �a and �b is a vector, denoted �a��b, char-acterized by:

� direction: it is perpendicular to both �a and �b;

� orientation: given by the right-hand rule;

� length: �a� �b = k�ak �b sin�; where � is the smallest angle between the

directions of �a and �b:

the vector product

Properties:

1. the dot product is anticommutative: �a� �b = ��b� �a:

2. it is distributive w.r.t. addition: �a� (�b+ �c) = �a� b+ �a� �c:

3. �a� (��b) = (��a)� b = ��a� �b:

4. if �a and �b are non-zero, then �a � �b = 0, �a is collinear to �b:


Expression in coordinates:

�a� �b =

��{ �j �ka1 a2 a3b1 b2 b3

�� :Application: The length of �a � �b can be interpreted as the area of the

parallelogram having �a and �b as sides.Consequently, the area of a triangle having �a and �b as sides is half of the

area of the parallelogram, this is:

A� =1

2

�a� �b :Example: Let A(1; 2; 0); B(1; 2; 1); C(3; 0; 1): Calculate the area of the

triangle ABC:Solution: AB(0; 0; 1); AC(2;�2; 1) and

AB �AC =

��{ �j �k0 0 12 �2 1

�� = 2�{+ 2�j:The norm

AB �AC is then AB �AC = q(�2)2 + (�2)2 = 2p2; whichleads to

A�ABC =1

2

AB �AC = p2:Exercise 2.7 Calculate the area of the triangle ABC; where A(1; 0; 0);B(1; 0; 1); C(2; 3; 1) and the height (or altitude) from A:

Exercise 2.8 Find the area of the parallelogram built on the vectors �m = 2�a+

3�b; �n = �a � �b; where k�ak = 1; �b = 2; [(�a;�b) = �

6: (Hint: calculate �m � �n and

take into account that �a� �a = 0; �b� �b = 0; �b� �a = ��a� �b).

2.1.4 Mixed Triple Product

The mixed triple product (also called the box product or scalar tripleproduct) is de�ned as:

(�a;�b; �c) = �a � (�b� �c):Applications:

1. The absolute value of the box product is the volume of the paral-lelepiped built on the three vectors:

Vparall =��(�a;�b; �c)�� :

The volume of the tetrahedron built on �a;�b; �c (as non-coplanar edges)is

Vparall =1

6

��(�a;�b; �c)�� :

2.2. EQUATION OF LINES AND PLANES IN SPACE 29

2. The box product zero if and only if the three vectors are linearly dependent(coplanar).

3. The box product is positive if and only if the system consisting of thethree vectors �a;�b; �c is right-handed.

Exercise 2.9 1. Find � 2 R such that the vectors �a(1; 2; 3); �b(2; 0; 2) and�c(4; 4; �) are coplanar.

2. Determine the volume of the tetrahedron ABCD; where A(1; 2; 3);B(1; 2; 4); C(2; 2; 4); D(2; 3; 4); and the altitude from A of the tetrahedron.

Hint : VABCD =

��(��!AB;�!AC;��!AD)��6

=1

6:

2.2 Equation of Lines and Planes in Space

Basically, in order to uniquely determine a plane in space, one needs either: apoint in the plane and two directions (two vectors) which are parallel to theplane (or contained in it, which is the same, taking into account that we aretalking about free vectors), or a point in the plane and the normal direction ofthe plane.Let fO;�{; �j; �kg denote a Cartesian frame in space.

2.2.1 The Plane Containing a Point and Two Directions

Let (�) be the plane which contains:

� the point M(x0; y0; z0);

� the vectors �v1(l1;m1; n1) and �v2(l2;m2; n2):


Suppose that M; �v1 and �v2 are known. If P (x; y; z) is an arbitrary (moving)point in this plane, then the vectors

MP (x� x0; y � y0; z � z0); �v1(l1;m1; n1); �v2(l2;m2; n2)

are coplanar. This is equivalent to the fact that the mixed product of thesethree vectors vanishes, which is:��

x� x0 y � y0 z � z0l1 m1 n1l2 m2 n2

�� = 0: (2.1)

(the equation of the plane through M(x0; y0; z0); containing �v1 and�v2).Another way of expressing the fact that MP; �v1 and �v2 are coplanar is to

write that they are linearly dependent:

MP = ��v1 + ��v2; �; � 2 R:

In Cartesian coordinates, this is equivalent to:8<: x = x0 + �l1 + �l2y = y0 + �m1 + �m2

z = z0 + �n1 + �n2

(the parametric equations of the plane (�)).

Particular case: the equation of a plane through three non-collinear points Mi(xi; yi; zi); i = 1; 2; 3 :Once we know three non-collinear points in a plane, we know two directions,

for instance,

M1M2(x2 � x1; y2 � y1; z2 � z1); M1M3(x3 � x1; y3 � y1; z3 � z1):

By using the point M1 and these two directions, we can write the equation ofthe plane (M1M2M3) :

(M1M2M3) :

��x� x1 y � y1 z � z1x2 � x1 y2 � y1 z2 � z1x3 � x1 y3 � y1 z3 � z1

�� = 0:Exercise 2.10 Write the equation of the plane:

1. containing M(1;�1; 0) and the vectors �v1(1; 2; 3); �v2(0; 1; 2);

2. through the points M1(1; 0; 1); M2(2; 0; 1); M3(2; 1; 2):

3. containing the Oz axis and the point M(1; 3; 2):


2.2.2 The Plane through a Point, Having a Given NormalDirection

Let M(x0; y0; z0) be a known point in a plane (�) and �N(A;B;C) a normalvector of the plane. Then, for any point P (x; y; z) in the plane, the vector �Nis perpendicular to MP; which is equivalent to the fact that the dot product�N �MP is 0. In coordinates, this yields:

A(x� x0) +B(y � y0) + C(z � z0) = 0: (2.2)

The above equation is called the equation of the plane throughM(x0; y0; z0);with normal direction �N(A;B;C): By computing the brackets,we always get something like this:

Ax+By + Cz +D = 0; (2.3)

which is called the general equation of the plane. This name is motivatedby the fact that, no matter if we start from (2.1) or (2.2), the equation of aplane always looks like above. Conversely, any equation of degree 1 in x; y; zgeometrically means a plane in three dimensional Euclidean space.Example: the coordinate planes xOy; yOz; xOz:The plane xOy contains the point O(0; 0; 0) and has as normal vector

�k(0; 0; 1): Hence, its equation is: (xOy) : 0(x � 0) + 0(y � 0) + 1(z � 0) = 0;which is,

(xOy) : z = 0:

In a similar manner, we get

(yOz) : x = 0; (xOz) : y = 0:

Remark 2.11 The coe¢ cients of x; y and z in the general equation of a planegive exactly the coordinates of the normal vector of the plane. For instance, theplane x+ 3y � z + 5 = 0 has as normal vector �N(1; 3;�1) etc.

Remark 2.12 Since two parallel planes have the same normal direction, theequation of any plane which is parallel to the plane (�) : Ax+By+Cz+D = 0di¤ers from that of (�) only by its free term: Ax+By+Cz +D = �; for some� 2 R:


Exercise 2.13 Write the equation of the plane:

1. through A(1; 2;�1) and having as normal vector �N(2; 3; 4):

2. through A(2; 3; 5) and parallel to the plane (M1M2M3); where M1(1; 0; 0);M2(0; 1; 0); M3(0; 0; 1).

3. through A(x0; y0; z0) and parallel to the plane: a) xOy; b) yOz; c)xOz:

2.2.3 Equations of a Straight Line in Space

Basically, a straight line is uniquely de�ned by the intersection of two non-parallel and distinct planes. This is, it can be described by a linear system asfollows:

(d) :

�A1x+B1y + C1z +D1 = 0A2x+B2y + C2z +D2 = 0

:

Another way of describing a straight line in space is by giving a pointM(x0; y0; z0) on the line and a vector �v(l;m; n) which is parallel to the line(or contained in it).In this case, for any point P (x; y; z); the vector MP (x � x0; y � y0; z � z0)

is collinear to �v(l;m; n); which is equivalent to the fact that the components ofthe two vectors are proportional:

x� x0l

=y � y0m

=z � z0n

(2.4)

(the canonical equations of a line in space).

Remark 2.14 The canonical equations provide useful information about theline: namely, the denominators l;m; n are exactly the components of the direct-ing vector �v of the line, while the quantities subtracted from x; y and z in thenumerators give the coordinates of some point on the line. For instance, theline

x� 11

=y � 32

=z + 1

4

has the directing vector �v(1; 2; 4) and a point on the line is M(1; 3;�1)


If, in (2.4), we denote the common value of the ratios by t we get8<: x = x0 + lty = y0 +mtz = z0 + nt

(2.5)

(the parametric equations of a line).Parametric equations are very useful in Mechanics (in this case, the para-

meter t usually denotes time).

Examples:

1. The equations of the axes Ox; Oy; Oz : For instance, on (Ox) we knowthe point O(0; 0; 0) and the vector �{(1; 0; 0); hence, the canonical equationsare

(Ox) :x� 01

=y � 00

=z � 00

= t

It is admitted to formally write 0 in the denominators! This does notmean that we perform a division by 0, it is just a simpler way of writingthe fact that the numerators and the denominators are proportional. Themeaning of this can be more clearly seen from the parametric equations:

x� 0 = 1 � t = ty � 0 = 0 � t = 0z � 0 = 0 � t = 0:

Actually, 0 in a denominator means that also the corresponding numeratoris 0.

(Ox) can be also given as the intersection of the planes (xOy) and (xOz) :

(Ox) :

�y = 0z = 0

:

The equations of the axes Oy and Oz can be obtained in a similar manner.

2. Let (d) be a line described as the intersection of two planes

(d) :

�x� y = 0x+ 2z = 1:

Let us determine the canonical equations of this line.

First of all, let us notice that the two planes (�1) : x � y = 0 and (�2) :x+2z = 1 indeed determine a line: since their normal vectors �N1(1;�1; 0)and �N2(1; 0; 2) are non-collinear, the planes can neither be parallel, norcoincide, hence their intersection is a line.


The vector of this line is contained both in (�1) and (�2); consequently,it is perpendicular to �N1 and �N2: This means, it is collinear to the crossproduct �N1� �N2: Since we are interested only in its direction, we can takeas directing vector of the line

�v = �N1 � �N2 =

��{ �j �k1 �1 01 0 2

�� = �2�{� 2�j + �k:A point on the line can be obtained from any particular solution of thelinear system which describes (d): This can be performed by giving par-ticular values to one of the unknowns. Let us take, for instance, x = 1:We obtain y = 1; z = 0; this is, M(1; 1; 0):

The canonical equations of the line are, consequently:

x� 1�2 =

y � 1�2 =

z

1:

� Also, a useful notion is that of sheaf of planes. If a straight line (d) isdescribed as the intersection of two planes (P ) and (Q) :

(d) :

�P � A1x+B1y + C1z +D1 = 0Q � A2x+B2y + C2z +D2 = 0

;

then the sheaf of planes through (d) (or determined by (P ) and (Q))is the set of all planes which contain (d): Its equation is

�P + �Q = 0; �; � 2 R:

Equivalently, with � =�

�; this can be written simply as

P + �Q = 0; � 2 R:

(the last form is simpler, still, it requires some attention: there we haveeliminated the case � = 0; this is, the plane Q: This is, in any discussioninvolving the planes of the sheaf, we should take separately the case whenthe plane is Q).

2.3 Angles and Distances in Space

2.3.1 Angles:

1. the angle between two lines (d1) and (d2) is the angle between theirdirecting vectors �v1 and �v2; and its cosine can be obtained by means ofthe dot product �v1 � �v2 :

cos� =�v1 � �v2

k�v1k � k�v2k:

2.3. ANGLES AND DISTANCES IN SPACE 35

2. the angle between a line and a plane is equal to 90o minus the anglebetween the directing vector �v of the line and the normal vector �N of theplane:

cos(90o � �) = sin� = �v � �Nk�vk �

�N :3. the angle between two planes is the same as the angle between theirnormal vectors �N1 and �N2 :

cos� =�N1 � �N2 �N1 � �N2 :

2.3.2 Distances in Space

1. the distance between two points A and B is, obviously, the norm ofthe vector

��!AB :

dist(A;B) = ��!AB =p(xB � xA)2 + (yB � yA)2 + (zB � zA)2:

2. the distance between a point M(x0; y0; z0) and a line (d) :x� x1l

=

y � y1m

=z � z1n

is expressed as the height (altitude) in the parallelogram

built on the vectors MM1 and �v(l;m; n) (where M1(x1; y1; z1) denotes apoint on the line):

dist(M;d) =

MM1 � �v

k�vk :

3. the distance between a point M(x0; y0; z0) and a plane (�) : Ax +By + Cz +D = 0 is given by

dist(M; (�)) =jAx0 +By0 + Cz0 +D0jp

A2 +B2 + C2:

4. the minimum distance between two skew (=non-coplanar) linesin space is expressed as the altitude from M1 in the parallelepiped built


on the vectors M1M2; �v1 and �v2; where Mi and vi denote a point and thedirecting vector of the straight line (di); i = 1; 2 :

dist((d1); (d2)) =

��(M1M2; �v1; �v2)��

kv1 � v2k:

Exercise 2.15 Let A(2; 1; 3); B(1; 2; 3); (d) :x

1=y

2=

z

�2 and (�) : 4x+3z�1 = 0: Find:

1. the distances dist(A;B); dist(A; (d)); dist(A; (�));

2. the angle between the line AB and, (d), respectively, (�):

3. the angle between (d) and (�);

4. the minimum distance between the lines AB and (d):

5. the intersection point between (d) and (�):

Exercise 2.16 Write the equation of a plane:

1. through the point O(0; 0; 0) and which is perpendicular to both (�1) : x +y + z = 3 and (�2) : 3x� z = 2;

2. through the point O(0; 0; 0) and which is perpendicular to (�1) : x+y+z =

3 and parallel to the line (d) :x� 12

=y

3=z + 2

4:

3. which is perpendicular to (�1) : x + y + z = 3 and contains the line

(d) :x� 12

=y

3=z + 2

4:

Exercise 2.17 Find the parameters � and � such that the linex� �1

=y

�=z

1is contained in the plane x+ y + z = 3:

Exercise 2.18 Let M(2; 3; 5) and (d) :x

1=y

1=z

1: Find the projection of M

onto (d) and the re�ection of M in the line (d):

Exercise 2.19 Find the re�ection of the Ox axis in the line (d) :x

1=y

1=z

1(Hint: �nd the re�ections of two conveniently chosen points on Ox in the line(d)).

Exercise 2.20 (*) Find the straight line which is perpendicular to both the lines

Oz and (d) :x� 21

=y + 1

3=z

4and intersects them (the common perpendicular

of the lines Oz and (d)).

2.4. COORDINATE TRANSFORMATIONS 37

2.4 Coordinate Transformations

Let fO;�{; �j; �kg and fO0;�{0; �j0; �k0g denote two Cartesian (orthonormal) frames inspace. Any way of passing from the �rst frame to the latter one can be regardedas the composition of two movements:

1. change the origin O ! O0, and leave the basis unchanged:

fO;�{; �j; �kg ! fO0;�{; �j; �kg:

This is called a translation.

2. leave the origin unchanged, and perform a change of bases:

fO0;�{; �j; �kg ! fO0;�{0; �j0; �k0g:

Let us study separately these two types of transformations.

2.4.1 Translations in Plane and in Space

A. In space:Let us consider the transformation R =fO;�{; �j; �kg ! R0 = fO0;�{; �j; �kg;

where O0 has, with respect to the "old" frame R; the coordinates x0; y0; z0 :

O0(x0; y0; z0):

For a point A in space, let (x; y; z) denote its coordinates with respect to R;and (x0; y0; z0), its coordinates with respect to the "new" frame R0: This is,

�!OA(x; y; z);

��!O0A(x0; y0; z0):

Then, we have��!O0A =

�!OA�

��!OO0; which, in coordinates, leads to:8<: x0 = x� x0y0 = y � y0z0 = z � z0

: (2.6)

The inverse transformation is: 8<: x = x0 + x0y = y0 + y0z = z0 + z0

: (2.7)

The relations (2.6)-(2.7) are the equations of the translation by vector��!OO0(x0; y0; z0) in space.


Example: If A(2; 3; 4) with respect to some frame fO;�{; �j; �kg and we trans-late the frame to the point O0(1; 0; 2); then, in the new frame fO0;�{; �j; �kg; A willhave the coordinates:

x0 = 2� 1 = 1; y0 = 3� 0 = 3; z0 = 4� 2 = 2:

B. In planeIn a similar way, the transformation R =fO;�{; �jg ! R0 = fO0;�{; �jg; where

O0 has, with respect to the "old" frame R; the coordinates x0; y0 :

O0(x0; y0);

yields a transformation of the coordinates of some point A(x; y) as follows:�x0 = x� x0y0 = y � y0

;

�x = x0 + x0y = y0 + y0

Exercise 2.21 Let, in the plane xOy; (�) : x2+ y2� 2x+4y� 4 = 0: Find theequation of (�) after a translation to O0(1; 2):

Exercise 2.22 Show that, in space, performing a translation by an arbitraryvector

��!OO0(a; b; c) always brings the equation of a plane into the equation of a

plane and the equations of a straight line into the equations of a straight line(equivalently: translations preserve collinearity and coplanarity).

2.4.2 Changes of Orthonormal Bases in Plane and inSpace

A. In spaceLet us now preserve the origin O of our Cartesian frame, and perform a

change of orthonormal bases

f�{; �j; �kg ! f�{0; �j0; �k0g:

Let again A be a point whose coordinates are:

� A(x; y; z) with respect to the old frame fO;�{; �j; �kg;

� A(x0; y0; z0) with respect to the new frame fO0;�{0; �j0; �k0g:

As known from Linear Algebra, any change of bases (not necessarily ortho-normal ones) is characterized by the so called matrix of change S = (sij)i;j=1;3,de�ned as follows: 8<: �{0 = s11�{+ s21�j + s31�k

�j0 = s12�{+ s22�j + s32�k�k0 = s13�{+ s23�j + s33�k

:

2.4. COORDINATE TRANSFORMATIONS 39

Then, the columns with the coordinates of A (which are the same as the coor-dinates of

�!OA)

X =

0@ xyz

1A ; X 0 =

0@ x0

y0

z0

1Aare related by

X = SX 0:

Remark 2.23 If both the bases f�{; �j; �kg and f�{0; �j0; �k0g are orthonormal, thenthe matrix of change S has the property S � St = I3; which leads to

detS = �1:

Proposition 2.24 If detS = 1; then the transformation is called a rotation.

Exercise 2.25 1. Show that �{0(2

3;2

3;1

3); �j0(�2

3;1

3;2

3); �k0(

1

3;�23;2

3) is an or-

thonormal basis in space.

2. Find the equation of the plane (�) : x+ y + z = 0 in the basis f�{0; �j0; �k0g:

B. In planeLet fO0;�{0; �j0g be a frame obtained from fO;�{; �jg after a rotation of angle

� 2 [0o; 180o]: Then, we have

�{0 = cos��{+ sin��j�j0 = � sin��{+ cos��j:

This is, the matrix of change of bases is

S =

�cos� � sin�sin� cos�

�; (2.8)

The relation between the "old" coordinates (x; y) and the "new" ones (x0; y0) ofsome point A is given by X = SX 0; which is:�

xy

�=


��x0

y0

�: (2.9)


Example: Find the equation of the line y = x + 5 in the plane xOy; aftera rotation of angle � = 45o:We have �

xy

�=

0B@1p2

� 1p2

1p2

1p2

1CA� x0

y0

�;

which is, 8><>:x =

1p2(x0 � y0)

y =1p2(x0 + y0)

:

By replacing into the equation of the line y = x+ 5; we get

x0 + y0 = x0 � y0 + 5p2 ) y0 =

5p2

2:

Exercise 2.26 Find the equation of the circle x2 + y2 = R2 in the plane xOy;after a rotation of arbitrary angle �:

Also, an important class of changes of bases in plane are re�ections:

� re�ection in the Ox-axis ("vertical �ip"): f�{; �jg ! f�{;��jg; which is,

S =

�1 00 �1

�!�

x = x0

y = �y0 :

� re�ection in the Oy-axis ("horizontal �ip"): f�{; �jg ! f��{; �jg; whichis,

S =

��1 00 1

�!�x = �x0y = y0

:

� re�ection across O : f�{; �jg ! f��{;��jg; which is,

S =

��1 00 �1

�!�x = �x0y = �y0 :

(Re�ection across O can be also regarded as a rotation of angle � = 180o).

Exercise 2.27 Find the equation of the parabola y = x2 + 1 after: a re�ectionin the Ox-axis; a re�ection in the Oy-axis; a re�ection across O; a rotation ofangle � = 90o:

2.5. CONIC SECTIONS 41

2.5 Conic Sections


Conic sections (or, simply, conics) can be de�ned in at least three ways. Asconcerns us, we shall regard them for the beginning as plane algebraic curveswhose equations are of degree 2. This is, a conic is a plane curve whose equationis of the form

a11x2 + 2a12xy + a22y

2 + 2a13x+ 2a23y + a33 = 0;

where aij 2 R; i; j = 1; :::; 3; where at least one of the coe¢ cients a11; a12; a22is nonvanishing.Let us de�ne two quantities related to a conic:

� the determinant � :

� =

��a11 a12 a13a12 a22 a23a13 a23 a33

�� ;obtained as follows: the index 1 corresponds to multiplication by x inthe equation of the conic, the index 2, to multiplication by y; and theindex 3, to multiplication by 1 (this is, the entry a11 is obtained from thecoe¢ cient of x2; the entry a13; from the coe¢ cient of x�1 etc). On the maindiagonal, we put the coe¢ cients of x2; y2 and 1 exactly as they appear inthe equation. All the other entries will be divided by 2 (shared betweenthe "twin" possibilities of obtaining the corresponding term. For instance,the term 2a13 will be shared between x � 1(! a13) and 1 � x(! a31)). Theresult is a symmetric matrix, whose determinant is �:

A conic is called nondegenerate if � 6= 0; and degenerate if its deter-minant � vanishes: � = 0:

� the discriminant � :� =

�� a11 a12a12 a22

�� ;obtained by "cutting o¤" the last line and the last column of �: For �; itssign determines the so-called class of the conic section. Namely, we saythat a conic is

1. "of ellipse type", if � > 0;2. "of hyperbola type", if � < 0;3. "of parabola type", if � = 0:

Example: Let (�) : 3xy + 4y2 � 2x+ 4y + 1 = 0: We have

� =

��0

3

2�1

3

24 2

�1 2 1

�� = �49

46= 0;


hence, the conic is nondegenerate. Its discriminant is

� =

��0

3

23

24

�� =�94< 0;

hence, the conic is of hyperbola type.

Examples of conic sections:

1. the ellipsex2

a2+y2

b2� 1 = 0 :

2. the hyperbolax2

a2� y

2

b2� 1 = 0 :

3. the parabola y2 = 2px; p 6= 0:


4. the empty setx2

a2+y2

b2+ 1 = 0:

5.x2

a2� y2

b2= 0 (the union of two intersecting lines:

x

a+y

b= 0 and

x

a� yb= 0);

6. y2 = 2p (two parallel lines y = �p2p if p > 0; two identical lines

y = 0 if p = 0; and the empty set, if p < 0);

7.x2

a2+y2

b2= 0 (the point O(0; 0)).

Exercise 2.28 Compute � and � for each of the conic sections above and showthat:

� the conics 1)-4) are nondegenerate, while the conics 5)-7) are degenerate;

� the conics 1),4), and 7) are of ellipse type, 2) and 5) are of hyperbola type,while 3) and 6) are of parabola type.

The above examples might seem at �rst sight nothing but a few very simple"naive" examples of conics. Actually, they are the only plane geometric shapesdescribed by second degree equations. And this is why:

Theorem 2.29 For any plane curve whose equation is of degree 2:

a11x2 + 2a12xy + a22y

2 + 2a13x+ 2a23y + a33 = 0;

there exists a transformation of Cartesian frames fO;�{; �jg ! fO0;�{0; �j0g whichtransforms its equation into one of the equations 1)-7). The obtained equation1)-7) is called the standard form (or the reduced canonical form) of theconic.

Moreover, a nondegenerate conic of hyperbola type is actually a hyperbola, anondegenerate conic of parabola type is always a parabola, and a nondegenerateconic of ellipse type, if non-empty, is an ellipse.


2.5.2 The Standard Form of a Conic

Let (�) denote the conic

a11x2 + 2a12xy + a22y

2 + 2a13x+ 2a23y + a33 = 0: (2.10)

Let us notice, for the beginning, that standard forms 1)-7) are particulartypes of equations of conics, in which:

� there is no xy term;

� with the exception of the parabola y2 = 2px; there is no term of degree 1.

This is, we have to "get rid" of the term xy and, as far as possible, of theterms of degree 1. The �rst aim can be attained by a rotation, while, for thesecond one, a translation would su¢ ce.

Step 1: rotation f0;�{; �jg ! fO;�{0; �j0g; (x; y)! (x0; y0):Aim - to get rid of the xy term from the equation (2.10).Obviously, if a12 = 0; this step becomes unnecessary.

In mathematical terms, "getting rid of xy" means reducing the quadraticform

h(x; y) = a11x2 + 2a12xy + a22y

2

to a canonical form. To this aim, let

A =

�a11 a12a12 a22

�be its matrix. Since it is symmetric, it always admits a diagonal form:

A0 =

��1 00 �2

�;

in some basis f�e1; �e2g:

Remark 2.30 The values �1; �2 are the solution of the characteristic equation

det(A� �I2) = 0;

while the vectors �e1; �e2 are to be found as basis vectors in each of the eigenspaces- i.e., particular nonvanishing solutions of the equations:

AX = �iX; i = 1; 2:

It can be proven (by using Linear Algebra tools) that these vectors fe1; e2gare always perpendicular to each other. This is, by setting

�{0 =�e1k�e1k

; �j0 =�e2k�e2k

;


we get an orthonormal basis f�{0; �j0g; in which the equation of the conic lookslike this:

�1(x0)2 + �2(y

0)2 + 2a013x0 + 2a023y

0 + a033 = 0:

Moreover, S can be always regarded as the matrix of a rotation:

S =


�;

where � is given by the coordinates of �{0 : �{0(cos�; sin�):

Remark 2.31 If we index the eigenvalues �1; �2 such that

sign(�1 � �2) = sign(a12);

then the angle � of rotation is between 0o and 90o:

Step 2: translation fO;�{0; �j0g ! fO0;�{0; �j0g; (x0; y0)! (X;Y ):Aim: to get rid of the terms of degree 1 (if possible).So far, the equation of the conic is

(�) : �1(x0)2 + �2(y

0)2 + 2a013x0 + 2a023y

0 + a033 = 0:

We take �1 and �2 as common factors and we group the terms in x0 and,respectively, y0 in order to obtain squares of binomials in the form:

�1 (x0 + x0)

2; �2(y

0 + y0)2:

This is:

(�) : �1

"(x0)2 +

2a013x0

�1+(a013)

2

(�1)2

#+ �2

"(y0)2 +

2a023x0

�2+(a023)

2

(�2)2

#+a033 �

(a013)2

�1� (a

023)

2

�2= 0

:

Let us consider the translation by vector��a

013

�1;�a

023

�2

�:

X = x0 +a013�1; Y = y0 +

a023�2: (2.11)

Also, by denoting the remaining free term by a0033;

a0033 = a033 �

(a013)2

�1� (a

023)

2

�2;

the equation of the conic transforms into

�1X2 + �2Y

2 + a0033 = 0;

which is the standard form of the conic.


Remark 2.32 If one of the eigenvalues �1 or �2 is 0 (one of the squares (x0)2

or (y0)2 is missing), then the corresponding term of degree 1 will not be includedin any square of a binomial (and we just group it with the remaining free term);so in the �nal equation, there will appear a term of degree 1, which means thatthe conic is of parabola type.

Example: Reduce to the standard form and plot the conic: 9x2 � 6xy +y2 + 20x = 0:

We have

� =

��9 �3 10�3 1 010 0 0

�� = �100 6= 0; � =�� 9 �3�3 1

�� = 0;hence, our conic is nondegenerate and of parabola type. As we shall see, it isindeed a parabola.

Step 1 - rotation:

We have A =

�9 �3�3 1

�; Its characteristic equation is�� 9� � �3

�3 1� �

�� = 0; which leads to the eigenvalues � = 10 and � = 0:

Since a12 = �3 < 0; we choose �1 and �2 such that �1 � �2 < 0; this is,

�1 = 0; �2 = 10:

We determine now the eigenvectors corresponding to �1 = 0 :

AX = 0X ;

�9 �3�3 1

��xy

�=

�00

�;

this is,�9x� 3y = 0�3x+ y = 0 : The set of solutions (the eigenspace corresponding to

�1) is

V1 = f(�; 3�) j � 2 Rg:

A basis of this space is �e1(1; 3) (obtained as the particular solution with � = 1).This is,

�{0 =�e1k�e1k

) �{0(1p10;3p10):

But the coordinates (=the components) of �{0 give the �rst column of thematrix of change of bases S: Since S is the matrix of a rotation,

S =


�;


we get cos� =1p10and sin� =

3p10; which means that we can also �ll in the

second column of S; without performing similar computations for �2 :

S =

0B@1p10

� 3p10

3p10

1p10

1CA :The equations of the rotation X = SX 0 are then

x =1p10(x0 � 3y0)

y =1p10(3x0 + y0):

By replacing the above into the equation of the conic, we get

10(y0)2 +20p10(x0 � 3y0) = 0;

after a simpli�cation by 10;

(y0)2 +2p10(x0 � 3y0) = 0:

Step 2: translationSince (x0)2 is missing from the equation, we shall proceed to grouping only

the terms in y0 in order to obtain the square of a binomial, and the remaining

term2p10x0 will be grouped with the remaining free term:

�(y0)2 � 2 � 3p

10y0 +

9

10

�� 9

10+

2p10x0 = 0;

this is, �y0 � 3p

10

�2+

2p10

�x0 � 9

2p10

�= 0:

Let 8><>:X = x0 � 9

2p10

Y = y0 � 3p10

,

8><>:x0 = X +

9

2p10

y0 = Y +3p10

:

denote the translation to O0(x0 =9

2p10; y0 =

3p10): After this translation, the

equation of the conic becomes

Y 2 = � 2p10X;

which is, indeed, the equation of a parabola.


2.5.3 Center, Axes and Asymptotes of a Conic

In some situations, we don�t need the reduced canonical equation of a conic, butonly some data related to it, such as: the coordinates of its center, the slopes ofthe axes, the slopes of its asymptotes. These can be inferred directly from thegeneral equation of a conic, without passing to the standard form.So, let our conic (�) be described by the equation

f(x; y) � a11x2 + 2a12xy + a22y2 + 2a13x+ 2a23y + a33 = 0:

Center of a conic

The coordinates of the center are the solutions (x0; y0) of the system:

1

2f 0x � a11x+ a12y + a13 = 0

1

2f 0y � a12x+ a22y + a23 = 0;

where f 0x and f0y denote the partial derivatives w.r.t. x and y of the function f:

Remark 2.33 For conics of parabola type , the above system does not have anysolutions. Conic sections of parabola type do not have any center.


Axes of a conic

If the conic is an ellipse or a hyperbola, then its axes pass through the center;if it is a parabola, then its axis (together with the tangent at the vertex) passthrough its vertex.The slopes of the axes are the solutions of the second degree equation

a12m2 + (a11 � a22)m� a12 = 0:

The angle of rotation of the axes, namely � = \Ox;Ox0; can be also inferredfrom

tg(2�) =2a12

a11 � a22:

Remark 2.34 The above equation has the discriminant (a11�a22)2+4a212 � 0;this is, it always has real solutions.

Remark 2.35 If a12 = 0; then the angle of rotation � is 0, which means thatthe axes of the conic are parallel to Ox and Oy respectively.

Asymptotes

Also, asymptotes of the conic pass through its center. Their slopes are given bythe equation:

a22m2 + 2a12m+ a11 = 0:

Remark 2.36 The discriminant of the above equation is 4(a212�a11a22) = �4�:Consequently, the only conics which admit real and distinct asymptotes are thoseof hyperbola type.

Exercise 2.37 Let (�) : 3x2 � 6xy + 2x+ 2y = 0:

1. Compute the invariants � and � of the conic and determine its nature(degenerate or nondegenerate) and its type (ellipse, hyperbola or parabola).

2. Find its center, axes and asymptotes (if any).

3. Bring it to the standard form at plot it.


2.5.4 Alternative De�nitions of Conics. Eccentricity

Conic sections are the curves generated by the intersections of a plane with aright circular cone (hence, their name). If the plane is perpendicular to the axisof the cone, one obtains a circle. For a plane that is not perpendicular to theaxis and that intersects only a single nappe of the cone, the curve produced iseither an ellipse or a parabola. The curve produced by a plane intersecting bothnappes is a hyperbola. Degenerate conics can be obtained in a similar manner.

Another de�nition of conic sections is:

A conic is the locus (Ro: "locul geometric") of points P in the plane xOy;with the property that the ratio between the distance to a �xed point F (calleda focus) and the distance to a �xed line d (called the directrix) is a constante, called the eccentricity of the conic.

2.6. QUADRICS 51

� If e = 0; then the conic is a circle;

� if 0 < e < 1; then it is an ellipse;

� if e = 1; it is a parabola, and

� if e > 1; then it is a hyperbola.

Conics have a wide range of applications. For instance, in astronomy: Keplerwas the �rst to notice that planetary orbits were ellipses, and then, Newtonderived the shape of orbits mathematically using calculus. Depending on theenergy of the orbiting body, its trajectory can be an ellipse, a hyperbola or aparabola. Also, conic sections are used in communications (parabolic antennas),which is due to the so-called "optical property" of the parabola: any beam whichis parallel to the axis of the parabola is re�ected (in the parabola, regarded asa "mirror") through its focus.

2.6 Quadrics

A quadric is a set of points in space whose coordinates (x; y; z) are related bya second degree equation:

a11x2+a22y

2+a33z2+2a12xy+2a13xz+2a23yz+2a14x+2a24y+2a34z+a44 = 0:

2.6.1 The Sphere

The sphere is de�ned as the locus of points in space situated at the same distanceR > 0 (called the radius) from a �xed point !(a; b; c); called center.


Let P (x; y; z) denote an arbitrary point of the sphere.The fact that the distance

!P is always R is written, after eliminatingthe square root:

(x� a)2 + (y � b)2 + (z � c)2 = R2 (2.12)

(the equation of the sphere with center !(a; b; c) and radius R). Hence,the sphere is a quadric.

If we perform the computations in the above equation, we obtain a relationof type:

x2 + y2 + z2 +mx+ ny + pz + q = 0; (2.13)

(where m = �2a; n = �2b; p = �2c; q = a2 + b2 + c2 �R2), which is called thegeneral equation of a sphere.This is, any equation of a sphere looks like above. The converse is not true,

this is, not any equation of type (2.13) is the equation of a sphere. This can bechecked by bringing the equation (2.13) to the form (2.12).Examples:

1. The equation x2 + y2 + z2 � 2x+ 4y � 4 = 0 can be written as:

(x2 � 2x+ 1)� 1 + (y2 + 4y + 4)� 4 + z2 � 4 = 0;

this is(x� 1)2 + (y + 2)2 + z2 = 9:

Hence, it represents the sphere of center !(1;�2; 0) and radius R = 3:

2. The equation: x2+ y2+ z2� 2x+4y+6 = 0 becomes, after grouping thesquares

(x� 1)2 + (y + 2)2 + z2 = �1;this is, it cannot represent a sphere, since R2 = �1 has no real solutions.

2.6. QUADRICS 53

Another way of describing the sphere of center !(a; b; c) and radius R is byits parametric equations:8<: x = a+R cos� sin �

y = b+R sin� sin �z = c+R cos �:

:

In particular, for the sphere having as center the origin O(0; 0; 0) and radius R;we get: 8<: x = R cos� sin �

y = R sin� sin �z = R cos �:

The quantities (R;�; �) are called the spherical coordinates of the pointP (x; y; z):

The tangent plane to the sphere

(S) : x2 + y2 + z2 +mx+ ny + pz + q = 0; (2.14)

at one of its points P (x0; y0; z0) 2 (S); has the equation:

xx0 + yy0 + zz0 +mx+ x02

+ ny + y02

+ pz + z02

+ q = 0:

2.6.2 Reduced Canonical Equations of Other Quadrics

The Ellipsoidx2

a2+y2

b2+z2

c2= 1:

I

The numbers a; b; c > 0 are called the semi-axes of the ellipsoid.Intersections with parallel planes to the planes of coordinates are ellipses,

points or the empty set. In order to convince ourselves,let�s take, for instance,


planes which are parallel to (xOy); this is, z = � : we getx2

a2+y2

b2= 1 � �2

c2;

which is: an ellipse, if � 2 (�c; c); a point, if � = �c and the empty set if� 2 R n [�c; c]:If the ellipsoid has two equal semi-axes, then it is called a spheroid; if all

its semi-axes are equal, then it is a sphere.

The Hyperboloid of One Sheetx2

a2+y2

b2� z

2

c2= 1

Its intersections with horizontal planes z = � are ellipses, while the intersec-tions with planes y = const: and x = const: are hyperbolas.The one-sheeted hyperboloid has an interesting property: it is a doubly

ruled surface, this is, such a shape can be built only from rectilinear beams,and this can be performed in two ways (this is why we say it is "doubly" ruled).These beams are called the rulings (rectilinear generatrices) of the hyper-boloid.In order to obtain the equations of its rulings, let us write the equation of

the hyperboloid this way:

x2

a2� z

2

c2= 1� y

2

b2;

and factor the di¤erences of squares:�xa+z

c

��xa� zc

�=�1 +

y

b

��1� y

b

�:

Then,

(G�) :

8<:x

a+z

c= �

�1 +

y

b

��xa� zc

�= 1� y

b

;

2.6. QUADRICS 55

represents a family of straight lines which are entirely situated on the hyper-boloid - hence, they are rectilinear generatrices for this surface.

We can also group the terms with opposite signs:

(G�) :

8<:x

a+z

c= �

�1� y

b

��xa� zc

�= 1 +

y

b

;

which leads to another family of rulings of the one-sheeted hyperboloid.

We should mention several properties of the lines (G�) and (G�) :

� Through each point M(x0; y0; z0) of the hyperboloid there is exactly onegeneratrix (G�) and one generatrix (G�):

� Rulings (G�) obtained for di¤erent values of � do not intersect each other(they are skew lines). The same holds true about rulings (G�):

A notable use of one-sheeted hyperboloids is in cooling towers of powerstations. Such a shape is relatively easy to build and very resistant.

The Hyperboloid of Two Sheetsx2

a2+y2

b2� z

2

c2= �1

Sections with planes parallel to (xOy); this is, with z = � = const:; areellipses, for j�j > c; a point for � = �c and the empty set for j�j < c: Sectionswith planes y = const: and x = const: are hyperbolas.


The elliptic paraboloid z =x2

a2+y2

b2

Sections with planes z = const: are ellipses, while sections with y = const:and x = const: are parabolas.

The Hyperbolic Paraboloid (the "Saddle") z =x2

a2� y

2

b2

Sections with z = � (const.) are hyperbolas, with the exception of the planexOy; where the section consists of two intersecting lines (prove this!). Sectionswith x = const: and y = const: provide parabolas.The hyperbolic paraboloid is also a doubly ruled surface. Its rul-

ings are obtained by factoring the right hand side of the equation: z =�xa+y

b

��xa� yb

�and then grouping terms of degree 1:

(G�) :

8<: ��xa+y

b

�= z

x

a� yb= �

; (G�) :

8<: ��xa� yb

�= z

x

a+y

b= �

:

Properties:

� Through any point of the hyperbolic paraboloid there passes exactly oneruling (generatrix) (G�) and exactly one ruling (G�):

2.6. QUADRICS 57

� Di¤erent rulings belonging to the same family do not intersect each other(they are skew lines).

The Elliptic Conex2

a2+y2

b2� z

2

c2= 0

Sections with horizontal planes z = � 6= 0 are ellipses. For a = b; one gets acircular cone.

Cylinders

A. the elliptic cylinderx2

a2+y2

b2= 1 :

B. the hyperbolic cylinderx2

a2� y

2

b2= 1 :


C. The parabolic cylinder y2 = 2px :

2.7 Generated Surfaces

A generated surface is a surface that can be swept out by moving a givencurve. The di¤erent positions of the moving curve are called the generatricesof the surface.If the moving curve is a line, then the generated surface is called a ruled

surface. In this case, the (rectilinear) generatrices are called the rulings of thesurface.Another important class of generated surfaces is that of revolution sur-

faces, obtained by rotating a given curve around a straight line (in this case,we can regard the surface as generated by circles).

In the following, let us expose some generalities on curves and surfaces in3-dimensional Euclidean space.A surface in E3 is generally de�ned by a single equation in x; y; z:

(S) : F (x; y; z) = 0:

(Remember that a plane was characterized by a single equation!).

2.7. GENERATED SURFACES 59

A curve in 3-dimensional space can be de�ned as the intersection of twosurfaces, this is, it is described by a system of equations:

(C) :

�F (x; y; z) = 0G(x; y; z) = 0

:

(In particular, a straight line is described by two equations in x; y; z).

2.7.1 Cylindrical Surfaces

A cylindrical surface is a surface swept out by a moving line having a �xeddirection, and which intersects a given curve (C); called the directrix of thesurface.In order to deduce the equation of such a surface, let:

� (d) :�P � A1x+B1y + C1z +D1 = 0Q � A2x+B2y + C2z +D2 = 0

be the "initial position" of the

moving lines. Hence, any ruling (rectilinear generatrix) of the surface willbe a line which is parallel to (d):

� (C) :�F (x; y; z) = 0G(x; y; z) = 0

be the directrix curve of the surface.

Let us now deduce the equation of the cylindrical surface.Step 1: We write the equations of all straight lines which have the same

direction as (d) (this is, all lines which are parallel to (d)). Since (d) is given asthe intersection of two planes (P ) and (Q); this can be performed by intersectinga plane which is parallel to (P ) with a plane which is parallel to (Q) :

(G��) :

�P = �Q = �

; �; � 2 R:

Step 2: The rulings of the surface are exactly those lines (G��) whichintersect the curve (C): This is, the system8>><>>:

(G��) :

�P = �Q = �

(�)

(C) :

�F (x; y; z) = 0G(x; y; z) = 0

has to be compatible. Since it is a (generally, nonlinear) system with 3 unknownsx; y; z and 4 equations, it basically requires a condition of compatibility

�(�; �) = 0;

which refers to the parameters � and �: This condition can be obtained byeliminating from the system, one by one, the unknowns x; y; z:


The equation of the surface can now be obtained by simply replacing in thecondition of compatibility, the expressions (�) of � and � in terms of x; y; z :

� = P; � = Q;

this is, the equation of the cylindrical surface is

�(P;Q) = 0:

The reason for the above said is the following: writing the equation of thesurface means to �nd a relation between the coordinates x; y; z of an arbitrarypoint of the surface. Or, points of the surface are precisely the points of thosestraight lines (G��) which intersect (C); this is, of those (G��) for which � and� are related by �(�; �) = 0: But, once � and � stay in this relation, then x; y; zare necessarily related by �(P;Q) = 0:

Example: Write the equation of the cylindrical surface generated by linesparallel to (d) :

x

1=y

2=z

1; and which intersect the curve (C) : y = sinx;

z = 0:First, let us write the equations of (d) in the form P = 0; Q = 0: We get

(d) :

�P � 2x� y = 0Q � x� z = 0 :

Then, any line which is parallel to (d) has the equations

(G��) :

�2x� y = �x� z = � : (�)

The intersection between (G��) and the directrix (C) is given by8>><>>:(G��) :

�2x� y = �x� z = � :

(C) :

�y � sinx = 0z = 0

:

Let us eliminate x; y; z; in order to obtain the condition of compatibility. Byusing the second and the last equation, we get x = �: From the �rst equation,we have now y = 2�� : The third equation then yields

�(�; �) � 2�� sin� = 0:

The coordinates of the points of the surface are related to � and � by (�): Byreplacing into � the expressions (�); we obtain the equation of the surface:

2x� 2z � 2x+ y � sin(x� z) = 0;

this is,(S) : y � 2z � sin(x� z) = 0:


Exercise 2.38 Find the equation of the cylindrical surface generated by straight

lines which are parallel to (d) :�

x+ y = 0x� z � 1 = 0 and intersect the ellipse (C :8<: x2 +

y2

4= 1

z = 0::

2.7.2 Conic Surfaces

A conic surface is swept out by straight lines which pass through a �xed point(called vertex or apex) and intersect a given curve, called the directrix.Let us consider that the vertex V is given as the intersection of three planes,

(P ); (Q) and (R) :

V :

8<: P = 0Q = 0R = 0:

and the directrix curve (C) is again described as the intersection of two surfaces:

(C) :

�F (x; y; z) = 0G(x; y; z) = 0:

Note: If V (x0; y0; z0); then P � x� x0; Q � y � y0; R � z � z0:Step 1: We write the equations of all straight lines passing through V (by

intersecting the sheaf of planes determined by (P ) and (Q) with the sheaf ofplanes determined by (P ) and (R)):

(G��) :

�P = �QP = �R

(�) :


Step 2: The lines (G��) have to intersect (C); this is, the correspondingsystem of equations has to be compatible:8>><>>:

(G��) :

�P = �QP = �R

(C) :

�F (x; y; z) = 0G(x; y; z) = 0:

We get the condition of compatibility: �(�; �) = 0; and, by replacing into thiscondition the expressions (�) of � and �; we get the equation of the surface:

�(P

Q;P

R) = 0:

Example: write the equation of the cone having as vertex V (0; 0; 0) and asdirectrix curve, the parabola: (C) : y2 = 2x; z = 1:The vertex V (0; 0; 0) can be described as the intersection of the planes

P � x = 0

Q � y = 0

R � z = 0:

Then, all lines passing through V are given by

(G��) :

�x = �yx = �z:

(�)

Intersection with (C) leads to the system8>><>>:(G��) :

�x = �yx = �z:

(C) :

�y2 = 2xz = 1

:

By the second and the last equation, we get x = �; from the �rst, y =�

�: Now,

from the third equation, we get

�(�; �) ��

�2� 2� = 0:

But, from (�); we get � = x

y; � =

x

z: Hence, the equation of the surface is

�yz

�2� 2x

z= 0) y2 = 2xz:


Exercise 2.39 Find the equation of the surface generated by straight lines pass-ing through O(0; 0; 0) and which intersect the circle (C) : x2 + y2 = 1; z = 5:

2.7.3 Conoidal Surfaces with a Directing Plane

A conoidal surface is generated by moving straight lines which are parallel toa given plane (P ) and intersect a given line (d) and a given curve (C):Let us suppose

(P ) : P = 0; (d) :

�Q = 0R = 0

; (C) :

�F (x; y; z) = 0G(x; y; z) = 0:

Lines which are parallel to (P ) and intersect (d) are obtained as the intersectionbetween a plane which is parallel to (P ) and one which contains (d) :

(G��) :

�P = �Q = �R

: (�)

Intersection with (C) yields:8>><>>:(G��) :

�P = �Q = �R

(C) :

�F (x; y; z) = 0G(x; y; z) = 0:

;

which produces the condition of compatibility �(�; �) = 0: Then, by (�); we get� = P; � =

Q

R; and the equation of the surface is

�(P;Q

R) = 0:


Example: Find the equation of the surface swept out by lines which areparallel to (P ) = (xOy); and intersect the straight lines Oz and (C) : x+ 2y =3; x = z:

We have (d) = (Oz) :

�Q � x = 0R � y = 0 ; this is, we

get

8>><>>:(G��) :

�z = �x = �y

(C) :

�x+ 2y = 3x = z:

:

From the �rst, the second and the last equation, we have x = �; y =�

�: The

third equation leads to

�+ 2�

�= 3) �(1 +

2

�) = 3

With � = z; � =x

y; the equation of the surface is

z(1 +2y

x) = 3:

Exercise 2.40 Find the equation of the surface generated by lines which inter-sect Oz; the curve (C) : y = 2 + sinx; z = 5 and stay parallel to the plane(P ) : x+ y + z = 0:

2.7.4 Surfaces of Revolution

A surface of revolution is obtained by rotating a plane curve (C) (called thedirectrix) around a straight line, called the axis of rotation.Let (C) : F (x; y; z) = 0; G(x; y; z) = 0 denote the directrix curve, and

(d) :x� x0l

=y � y0m

=z � z0n


be the axis of rotation. Then, the surface can be regarded as swept out bycircles having their centers oh the axis of rotation, and situated in planes whichare perpendicular to the axis. These can be obtained by intersecting of spheresof arbitrary radius � 2 R+; and centers at an arbitrary �xed point on (d) (forinstance, at (x0; y0; z0)), with planes perpendicular to (d): This is, the generatingcircles can be described as:

(G��) :

�(x� x0)2 + (y � y0)2 + (z � z0)2 = �2

lx+my + nz = �: (�)

By intersecting them with (C); the resulting system8>><>>:(G��) :

�(x� x0)2 + (y � y0)2 + (z � z0)2 = �2lx+my + nz = �

:

(C) :

�F (x; y; z) = 0G(x; y; z) = 0

:

(3 unknowns x; y; z, 4 equations) requires a condition of compatibility �(�2; �) =0: By replacing the values of �2 and � from (�); we get the equation of thesurface:

��(x� x0)2 + (y � y0)2 + (z � z0)2; lx+my + nz

�= 0:

Example: Find the equation of the surface obtained by rotating the circle(C) : (x� 4)2 + y2 = 1; z = 0 in the plane xOy; around Oy.We have (d) = (Oy) :

x

0=y

1=z

0; this is, the generating circles are

(G��) :

�x2 + y2 + z2 = �2

y = �:

Intersection with (C) yields8>><>>:x2 + y2 + z2 = �2

y = �(x� 4)2 + y2 = 1

z = 0

:

We get x2 = �2 � �2 ) x = �p�2 � �2. The condition of compatibility of the

above system is:

�(�; �) ��q�2 � �2 � 4

�2+ �2 � 1 = 0:

Since �2 = x2 + y2 + z2; � = y; we get:��px2 + z2 � 4

�2+ y2 � 1 = 0:

The "donut" shape we obtained is called a torus:


Exercise 2.41 Find the equations of the surfaces obtained by rotating theparabola (C) : y = x2; z = 0 around:

1. Oy;

2. Ox:

Chapter 3

Di¤erential Geometry

3.1 Plane Curves

Representations

A plane curve is described by:

1. y = f(x) (explicit representation) or

2. F (x; y) = 0 (implicit representation) or

3.�x = x(t)y = y(t)

; t 2 [a; b] 2 R (parametric representation).

The parametric representation 3) is also met in its vectorial variant

OM � �r(t) = x(t)�{+ y(t)�j:

A point M(x0; y0) of a curve (C) is called regular if the involved functionsf; F; x(t); y(t) are di¤erentiable (and continuous) in M and

1. in implicit representation, the partial derivatives of F do not simultane-ously vanish in M(x0; y0) :

(F 0x(x0; y0))2+�F 0y(x0; y0)

�2> 0;

2. in parametric representation 3), the derivativesdx

dt(t0) and

dy

dt(t0) do not

simultaneously vanish, where t0 is the value of the parameter t correspond-ing to M(x0; y0):

An arc of a curve which consists only of regular points is called a regulararc.In the neighborhood of a regular point one can always pass between the

representations 1),2) and 3).

67

68 CHAPTER 3. DIFFERENTIAL GEOMETRY

If at some point M; the function which describe the curve are di¤erentiableof a certain class p � 1; we say that the curve is of class p at that point.A point of the curve which is not regular is called a singular point. In-

tuitively, singular points M can be points where the functions which appearare not continuous (the curve "breaks" at M), or points where the curve hasan "angle" (which usually corresponds to the situation when the involved func-tions are continuous, but not di¤erentiable, or x0(t) = y0(t) = 0), or "multiplepoints", through which the curve passes more than once).

A curve with singular points

3.1.1 Arc Length of a Plane Curve

Let (C) be given in explicit representation (C) : y = f(x) and_

AB a regular arcof (C):

The length of the arc_

AB is de�ned as the limiting case of the length of a

polygonal line with n sides inscribed in_

AB; and which connects A and B.

�g. 1.2

3.1. PLANE CURVES 69

This polygonal line is chosen such that the length of its greatest side tends to0, when n tends to in�nity. The length Ln of such a polygonal line is computed

by Pitagora�s theorem. If_

AB is a regular arc, then it can be shown that thislimit always exists.

Theorem 3.1 1. If (C) is given in explicit representation

y = f (x) ; x 2 [xA; xB ] � (a; b) ; (3.1)

then the length of the regular arc_

AB is

L _AB

=

xBZxA

q1 + (f 0(x))

2dx: (3.2)

Let now (C) be also given in parametric representation x = x(t); y = y(t):Then, the dependence of y on t is given by

y(t) = f(x(t))) dy

dt=df

dx

dx

dt= f 0(x)

dx

dt:

Usually, derivatives with respect to the parameter t are denoted by dots:

_y =dy

dt; _x =

dx

dt:

With these notations, we get

f 0(x) =_y

_x:

By replacing into the formula of the arc length (3.2), we get

Theorem 3.2 if the arc_

AB of (C) is parametrically represented by�x = x (t)y = y (t)

t 2 [tA; tB ] � (t1; t2) ; (3.3)

then its length is

L _AB

=

tBZtA

q�x2(t) +

�y2(t)dt; (3.4)

where�x =

dx

dt; _y =

dy

dt:

An alternative expression is obtained in di¤erentials. If we take into accountthat

dx = _xdt; dy = _ydt;


then the formula of the arc length can be also written as

L _AB

=

tBZtA

pdx2 + dy2:

The integrand

ds =

q�x2(t) +

�y2(t)dt =

pdx2 + dy2

or, in explicit representation,

ds =p1 + (f 0(x))2dx

is called the element of arc length of the curve (C): With this notation, the

length of the arc_

AB is expressed as:

L _AB

=

Z_AB

ds:

The function s =tRt0

ds (which measures the length of the curve between some

starting point t0 and the current point t) can be used as a parameter on thecurve, called the natural parameter.Intuitively, the element of arc length ds =

pdx2 + dy2 measures the length

of a small (in�nitesimal) arc of the curve, which can be approximated with thelength of a line segment.

the element of arc length

Exercise 3.3 Compute the element of arc length ds =p_x2 + _y2dt for the cy-

cloid:x = a(t� sin t); y = a(1� cos t); t 2 R

and the length of its arc between the points corresponding to t = 0 and t = 2�:


3.1.2 Contact Between Two Intersecting Curves

Given two curves that intersect at some point P; the notion of contact givesinformation about how close to each other the two curves pass in a neighborhoodof the intersection point.Let us suppose that the functions which de�ne the curves are of class n +

1; n � 0:

De�nition 3.4 Let (C) and (C 0) be two curves which intersect at some pointP: We say that (C) and (C 0) have an (n+ 1)-point contact (or, equivalently,a contact of order n), if they have at P (n+ 1) confounded common points.

In order to establish the order of the contact, the algorithm is the following:we write the system of equations which gives the intersection of (C) and (C 0):By eliminating all the unknowns except one, we shall be led to some relation�(x) = 0; �(y) = 0 or �(t) = 0. If the value of the unknown corresponding tothe common point P is a multiple root of order (n+ 1) of �; then at the pointP we have an (n+ 1)-point contact.Let us consider, for instance, the case when the two curves are given in

explicit representation:

(C) : y = f1(x); (C0) : y = f2(x)

and let P (x0; y0) be a common point.The intersection of (C) and (C 0) is given by:�

y = f1(x)y = f2(x)

:

By subtracting the two equations, y is eliminated and we get:

�(x) � f1(x)� f2(x) = 0:

At the point P we have an (n + 1)-point contact, or an n-th order contact, ifand only if x0 is a multiple root of order (n+ 1) of �; this is, i¤:

�(x0) = �0(x0) = �00(x0):::: = �

(n)(x0) = 0;

�(n+1)(x0) 6= 0:

Actually, when we say "n-th order contact", n refers to the last derivativeof � which vanishes at x0; while the expression "n+ 1-point contact" refers tothe number of common points which are confounded at P:

Also, if we have

(C) : F (x; y) = 0; (C 0) : x = x(t); y = y(t);

and the common point P (x0; y0) corresponds to the value t0 of the parameteron (C 0); then, the system which characterizes the intersection between the twocurves is �

F (x; y) = 0x = x(t); y = y(t)

:


We get�(t) � F (x(t); y(t)) = 0:

Then, (n+ 1)-point contact at P means

�(t0) = �0(t0) = �00(t0):::: = �

(n)(t0) = 0;

�(n+1)(t0) 6= 0:

Exercise 3.5 Find the order of the contact at the point P (0; 1); between thecurves (C) : y = ex and:

� (C 0) : y = 1 + x;

� (C 0) : y = 1 + x+ x2

2;

� (C 0) : y = 1 + x+ x2

2+x3

6+ :::+

xn

n!; n � 3:

3.1.3 Tangent and Normal Line at a Regular Point

The tangent line to a curve (C) at a point P 2 (C) is de�ned as the limitposition of a secant line PP 0; when P 0 tends to P: In other words, the tangentline is the straight line which has at least a 2-point contact with the curve at agiven point.In order to write down the equation of a line in plane, one needs a point

M(x0; y0) on the line and its slope m: Then, the equation of the line is

y � y0 = m(x� x0):

This is, once we know a regular point M(x0; y0) of a curve (C); we only needthe slopes of the tangent and of the normal line.

1. If (C) : y = f(x) is given in explicit representation, then, as known,from high school, the slope of the tangent line at M is

mtg = f0(x0):

The normal line is de�ned as the perpendicular atM to the tangent line,hence, its slope is

mN = �1

mtg) mN = �

1

f 0(x0):

2. If (C) : F (x; y) = 0 is given in implicit representation, then (accordingto the Math Analysis), in a neighborhood of the regular point M; it is

de�ned y = f(x); and its derivative at x0 is f 0(x0) = �F 0xF 0yj(x0;y0): Hence,

mtg = �F 0xF 0yj(x0;y0); mN =

F 0yF 0xj(x0;y0):


Since at a regular point, F 0x and F0y do not simultaneously vanish, these two

lines are always well determined (the situation with vanishing denominatorcorresponds to a vertical line, formally, m = �1).

3. In parametric representation, (C) : x = x(t); y = y(t); as shownabove, by passing to the explicit representation y(t) = f(x(t)); we getdy

dt= f 0(x)

dx

dt; this is, f 0(x) =

_y(t)

_x(t): Then, at the point P corresponding

to the value t0 of the parameter, we have

mtg =_y(t0)

_x(t0)) mN = �

_x(t0)

_y(t0):

Proposition 3.6 Two curves (C) and (C 0) have at least a 2-point contact (acontact of order at least 1) at a common point P if and only if they have thesame tangent line at P:

Exercise 3.7 Write the equations of the tangent and of the normal lines of thefollowing curves

1. (C) : y = ex at P (x0 = 0);

2. (C) : x2=3 + y2=3 � 1 = 0 (the astroid), at P ( 1

2p2;1

2p2):

the astroid

3. the ellipse (C) : x = 2 cos t; y = sin t; at A(t0 = 0) and at B(0; 1):

Exercise 3.8 Prove Proposition (3.6) for the curves (C) : y = f1(x) and (C 0) :y = f2(x); at the common point P (x0; y0):


3.1.4 Osculating Circle; Curvature and Radius of Curva-ture

Let(C) : x = x(t); y = y(t)

be a parametrized curve of class at least 2, and P (x0; y0) 2 (C) a regular point,corresponding to some value t0 of the parameter.We are looking for a circle that has contact of maximal order with (C) at

the point P (which best approximates the curve on a neighborhood of P ). Inorder to uniquely de�ne a circle, on needs three non-collinear points. So, wemight expect that (C) should have at P (at least) a 3-point contact with a givencircle (�):

De�nition 3.9 The osculating circle (or the circle of curvature) of thecurve (C) at the point P 2 (C) (if it exists) is the circle which has at least a3-point contact with (C) at the point P:

osculating circles of a plane curve

Note: The term "osculating" comes from the Latin "osculate", which means"to kiss"; intuitively, the osculating circle "kisses" the curve (C) at the contactpoint.In order to deduce the equation of a circle, one needs the coordinates of its

center !(a; b) and its radius Rc; once we know these, its equation is:

(�) : (x� a)2 + (y � b)2 �R2c = 0:

The conditions of 3-point contact (at least) between (C) and (�) at P (x0; y0)$t0 are obtained from

�(t) � (x� a)2 + (y � b)2 �R2c )) �(t0) = �

0(t0) = �00(t0) = 0:


By performing all computations, one can prove that the coordinates a; b of thecenter and the radius Rc are given by

a = x0 �_y( _x2 + _y2)

_x��y � ��

x _yjt0 ; b = y0 +

_x( _x2 + _y2)

_x��y � ��

x _yjt0

Rc =( _x2 + _y2)3=2�� _x��y � ��

x _y�� jt0 :

where the notation jt0 means that all the involved derivatives are calculated att = t0:

Remark 3.10 At the points P where the denominator _x��y � ��

x _y vanishes, thecurve has no osculating circle. Such points are called in�ection points of thecurve.

Intuitively, the bigger the radius Rc; the less the curve is "bent" at P andconversely: the smaller Rc; the more is (C) bent at P: This "bending" of a curveis seized in the mathematical notion of curvature.

De�nition 3.11 The number

K =_x��y � ��

x _y

( _x2 + _y2)3=2jt0

is called the curvature of (C) at the point P (t = t0); and the number

R =1

K=( _x2 + _y2)3=2

_x��y � ��

x _yjt0

is called the radius of curvature of (C) at P:

Remark 3.12 The radius of curvature R is actually the radius of the osculatingcircle, taken with a plus or a minus sign:

jRj = Rc:

This sign tells "on which side" of the curve the osculating circle is located.A change of the sign of R (equivalently, of the curvature K) means that theosculating circle passes on the other side of the curve. The points at which thecurvature vanishes are precisely the in�ection points of the curve.

Remark 3.13 The centers of all osculating circles of the curve (C) describe acurve called the evolute of (C):


Remark 3.14 If two curves have at a common point at least a 3-point contact,then at that point, they have the same osculating circle and the same curvature.

Example: Find the osculating circle and the curvature of the parabola(C) : y = x2 at the point A(0; 0):The osculating circle (�) : (x� a)2+(y� b)2�R2c = 0 has at least a 3-point

contact with (C) at A:By intersecting (C) with (�), we get

�(x) � (x� a)2 + (x2 � b)2 �R2c ;

the conditions of 3-point contact at A(x = 0) are �(0) = �0(0) = �00(0) = 0;which is:

�(0) � a2 + b2 �R2c = 0�0(x)jx=0 � 2(x� a) + 2(x2 � b) � 2x jx=0 = 0�00(x)jx=0 � 2 + 12x2 � 4bjx=0 = 0;

this is, 8<: a2 + b2 �R2c = 0�2a = 02� 4b = 0;

which leads to a = 0; b =1

2; Rc =

1

2: Obviously, then, the curvature must be 2

or -2.In order to compute the curvature, we need a parametric representation of

the curve. The simplest choice is by setting t as one of the coordinates x or y;for instance:

x = t

y = t2:

The point A(x = 0; y = 0) is obtained for t0 = 0: Then, _x = 1; �x = 0;_y = 2tjt0=0 = 0; �y = 2; and we get, at the point A

_x��y � ��

x _y = 2; _x2 + _y2 = 1:

Then, the curvature is

K =_x��y � ��

x _y

( _x2 + _y2)3=2=2

1= 2;

and the radius of curvature is R =1

K=1

2:

Exercise 3.15 Show that the radius of curvature R =1

Kof the circle (C) :

x = r cos t; y = r sin t; t 2 [0; 2�] at an arbitrary point A(t = t0) coincides withits radius r: Equivalently: the curvature of a circle is a constant, namely, the

inverse of its radius, K =1

r:

3.2. SPACE CURVES 77

Exercise 3.16 Show that the curvature of the straight line (d) : y = mx+ n atany of its points is 0 (all points of a line are in�ection points).

Exercise 3.17 Determine the osculating circle, the curvature and the radius ofcurvature of the ellipse

x = 2 cos t; y = sin t

at the point A(0; 1):

Exercise 3.18 For a curve (C) given in explicit representation y = f(x); showthat:

1. in�ection points A(x = x0) are characterized by f 00(x0) = 0:

2. the curvature K at an arbitrary point P (x; y) is given by

K =f 00(x)

1 + (f 0(x))2 :

3. There holds the equivalence: f is convex at P (this is, f 00(x) � 0) if andonly if the curvature of (C) at P is positive; f is concave at P if and onlyif its graph (C) has negative curvature at P:

(Hint: choose t = x as a parameter).

3.2 Space Curves

3.2.1 Representations

A space curve is described by:

1.�F (x; y; z) = 0G(x; y; z) = 0

(as the intersection of two surfaces) or

2.

8<: x = x(t)y = y(t)z = z(t)

; t 2 [a; b] 2 R (parametric representation), which is also

met in vectorial variant:

OM � �r(t) = x(t)�{+ y(t)�j + z(t)�k:;

where OM is the position vector of the arbitrary point M of the curve.

A point M(x0; y0; z0) of a curve (C) is called regular if the involved func-tions F;G; x(t); y(t); z(t) are di¤erentiable (and continuous) in M and


1. in implicit representation, the Jacobian of F and G does not vanish atM(x0; y0; z0) : �� F 0x F 0y F 0z

G0x G0y G0z

�� 6= 0 at (x0; y0; z0);

2. in parametric representation 3), the derivativesdx

dt(t0),

dy

dt(t0) and

dz

dt(t0)

do not simultaneously vanish, where t0 is the value of the parameter tcorresponding to M(x0; y0; z0):

An arc of a curve which consists only of regular points is called a regulararc. If at a point M; the functions which describe the curve are di¤erentiableof a certain class p � 1; we say that the curve is of class p at that point.In the neighborhood of a regular point one can always pass between the

representations 1) and 2).

3.2.2 Arc Length of a Space Curve

Let (C) be given in parametric representation

�r(t) = x(t)�{+ y(t)�j + z(t)�k

and_

AB denote a regular arc of (C); corresponding to t 2 [tA; tB ]:The length of the arc

_

AB is de�ned in the same way as for plane curves, thisis, as the limiting case of the length of a polygonal line with n sides inscribed

in_

AB, chosen such that the length of its greatest side tends to 0, when n tends

to in�nity. If_

AB is a regular arc, then it can be shown that its arc length iswell de�ned (=it uniquely exists).

Theorem 3.19 if the arc_

AB of (C) is parametrically represented by8<: x = x (t)y = y (t)z = z(t)

t 2 [tA; tB ] ; (3.5)

then its length is

L _AB

=

tBZtA

p_x2(t) + _y2(t) + _z2(t)dt; (3.6)

where the dots denote derivatives w.r.t. t :

_x =dx

dt; _y =

dy

dt; _z =

dz

dt:


In di¤erentials, we have dx = _xdt; dy = _ydt; dz = _zdt; hence, the formulaof the arc length can be also written as

L _AB

=

tBZtA

pdx2 + dy2 + dz2:

The integrand

ds =p_x2(t) + _y2(t) + _z2(t)dt =

pdx2 + dy2 + dz2

is called the element of arc length of the curve (C): With this notation, the

length of the arc_

AB is expressed as:

L _AB

=

Z_AB

ds:

The function s =tRt0

ds (which measures the length of the curve between some

starting point t0 and the current point t) can be used as a parameter on thecurve, called the natural parameter.The element of arc length ds =

pdx2 + dy2 + dz2 measures the length of

a small (in�nitesimal) arc of the curve, which, in theory, is approximated withthe length of a line segment.

3.2.3 The TNB Frame (The Frenet Frame)

The TNB-frame (also known as the Frenet Frame), was discovered by JeanFrederic Frenet (1816-1900). It is an orthonormal frame, moving along the curve.The initials "TNB" come from the vectors which constitute the frame, namely:the unit tangent vector �� ; the unit normal vector (or, more rigorously, theunit principal normal vector) ��; and the unit binormal vector ��:Let us describe in the following, the faces and the edges of this frame. To

this aim, let us recall some basic things about lines and planes in space:- a line can be uniquely de�ned by a point M(x0; y0; z0) and a direction

�v(l;m; n); this way:

(line) :x� x0l

=y � y0m

=z � z0n

:

- the plane which passes through the point M(x0; y0; z0) and has as normaldirection �N = �v(l;m; n); has the equation:

(plane) : l(x� x0) +m(y � y0) + n(z � z0) = 0:

Let (C) be a curve of class at least 2, given in parametric representation:

�r(t) = x(t)�{+ y(t)�j + z(t)�k;

and let M(x0; y0; z0) be a regular point of the curve, corresponding to the valuet0 of the parameter.


� The tangent line (tg) and the normal plane (�N ) :The tangent line atM(t = t0) has as directing vector, the tangent vector(velocity vector, derivative vector), having as Cartesian coordinates,the derivatives at t0 of the "trajectory" coordinates x; y; z :

��r(t0) = _x(t0)�{+ _y(t0)�j + _z(t0)�k: (3.7)

Shortly, if we omit the argument t0, we can write:��r( _x; _y; _z): Since M is

a regular point, the tangent vector at M exists and does not vanish. Theequations of the tangent line are:

(tg) :x� x0_x(t0)

=y � y0_y(t0)

=z � z0_z(t0)

The normal plane (�N ) is de�ned as the plane through M; which isperpendicular to the tangent line. Hence, its equation is:

(�N ) : _x(t0)(x� x0) + _y(t0) (y � y0) + _z(t0) (z � z0) = 0:

� The binormal line (b) and the osculating plane (�Osc):Let us suppose that at the point M; there exists the second derivative

��r(t0) = �x(t0)�{+ �y(t0)�j + �z(t0)�k;

(the acceleration vector) and the cross product��r(t0)�

��r(t0) does not

vanish, this is, the velocity vector��r(t0) and the acceleration vector

��r(t0)

are nonvanishing and non-collinear. Then, the two vectors, together withthe point M; de�ne a plane, called the osculating plane of (C) at M:The vector

�b(t0) =��r(t0)�

��r(t0); (3.8)

which is actually, the normal vector of the osculating plane, is called thebinormal vector of (C) at M: Let us suppose that, after computingthe cross product, we get for �b(t0) the coordinates �b(l;m; n): Then, thebinormal line has the equations:

(b) :x� x0l

=y � y0m

=z � z0n

:

The osculating plane is de�ned by:

(�Osc) : l(x� x0) +m(y � y0) + n(z � z0) = 0:

� The (principal) normal line (np) and the rectifying plane (�R) :The normal line (or the principal normal line) of the curve is de�nedas the intersection between the normal plane (�N ) and the osculating plane


(�Osc). This is, it is perpendicular to both the tangent vector and to thebinormal one. Hence, the principal normal vector �np can be obtained asthe cross product of the two vectors:

�np = �b(t0)��r(t0); (3.9)

which is,

�np =� ��r(t0)�

��r(t0)

��

��r(t0): (3.10)

This choice of the order �b(t0) ��r(t0) (instead of

��r(t0) � �b(t0)) insures

that the vectors��r, �np and �b constitute a right-handed system at M (the

orientation of �b can be obtained by the right hand rule from those of��r

and �np).

Consequently, if, by performing computations, we are led to

�np(A;B;C);

then, the equations of the principal normal line are

(np) :x� x0A

=y � y0B

=z � z0C

;

while the rectifying plane (which is de�ned as the plane perpendicularto np at M) is given by:

(�R) : A(x� x0) +B(y � y0) + C(z � z0) = 0:

the TNB (Frenet) frame


By (3.7), (3.8) and (3.9)-(3.10), we get

Theorem 3.20 The unit vectors of the TNB frame at the point M(t = t0) are:

� the unit tangent vector �� =��r(t0) ��r(t0) ;

� the unit (principal) normal vector �� =

� ��r(t0)�

��r(t0)

��

��r(t0) � ��r(t0)� ��

�r(t0)��

��r(t0)

;� the unit binormal vector �� =

��r(t0)�

��r(t0) ��r(t0)� ��r(t0)

Example: Find the unit vectors of the TNB (Frenet) frame of the curve

(C) : �r(t) = et�{+ e�t�j + t2�k at the point M(1; 1; 0):The point M has the Cartesian coordinates x = 1; y = 1; z = 0; hence

x0 = 1 = et0

y0 = 1 = e�t0

z0 = 0 = t20:

All three relations above yield for M the value t0 = 0:Further, let us calculate the derivatives of the position vector �r at M :

��r(t0) = (et�{� e�t�j + 2t�k)jt0=0 = �{� �j��r(t0) = (et�{+ e�t�j + 2�k)jt0=0 = �{+ �j + 2�k:

This is,��r(1;�1; 0);

��r(1; 1; 2): The unit tangent vector is �� =

��r(t0) ��r(t0) )

��(1p2;� 1p

2; 0):

In order to calculate the unit binormal vector, we need

�b =��r(t0)�

��r(t0) =

��{ �j �k1 �1 01 1 2

�� = �2�{� 2�j + 2�k:Then, �� =

�b �b ) ��(� 1p3;� 1p

3;1p3):

The principal normal vector is �np = �b��r; this is,

�np =

��{ �j �k�2 �2 21 �1 0

�� = 2�{+ 2�j + 4�k:Its unit vector is �� =

�npk�npk

) ��(1p6;1p6;2p6):


3.2.4 Curvature and Torsion

Let now (C) : �r(t) = x(t)�{ + y(t)�j + z(t)�k be a curve of class at least 3, and

M(t = t0) a regular point with��r �

��r 6= 0 (this is, such that the TNB frame

exists at M).The curvature of (C) at the point M(t = t0) is a number which measures

the "deviation" of the curve from its tangent line in a neighborhood of M: It isgiven by:

K(t0) =

��r � ��r ��r 3 jt0 :

A point at which the curvature vanishes (this is,��r �

��r = 0) is called an

in�ection point of the curve. At in�ection points, there is no TNB frame.(!) Do not confound this formula with that of the curvature of

a plane curve! The curvature of a space curve, as de�ned above, is alwaysnonnegative. On the contrary, for plane curves, the curvature can have a minussign.

Proposition 3.21 For straight lines, the curvature is 0 at each point.(a straightline consists only of in�ection points). Conversely, if the curvature of a curveis 0 at each point, then the curve is a straight line.

This is easy to see, if we take into account that a straight line is describedparametrically by:

x = x0 + lt; y = y0 +mt; z = z0 + nt;

where x0; y0; z0; l;m; n are constants. Then, we get��r(l;m; n);

��r = �0; which

leads to K = 0:

The torsion � of the space curve (C) at a regular point M(t = t0) is anumber which indicates how much the curves deviates from a plane (namely,from the osculating plane at M) in a neighborhood of M: It is given by theformula:

�(t0) =

� ��r ;

��r ;

��r�

��r � ��r 2 jt0 :

Proposition 3.22 For plane curves, the torsion is identically 0. Conversely,if the torsion identically vanishes, then the curve is contained in a plane.

A space curve which is contained in no plane is called a skew curve.The sign of the torsion provides information about the side of the osculating

plane in which the curve "bends" around M: Namely:

� if � > 0 at M; then, in a neighborhood of M; the curve "bends" on theside indicated by the binormal vector �b;


� if � < 0 at M; then, in a neighborhood of M; the curve "bends" on theopposite side of �b;

� if � changes the sign at M; then the curve "pierces" the osculating planeat M:

Exercise 3.23 Calculate the curvature and the torsion of the curve (C) : �r(t) =et�{+ e�t�j + t2�k at the point M(1; 1; 0):

Exercise 3.24 Find the curvature and the torsion of the curve (C) : �r(t) =a cos t�{+ a sin t�j + bt�k (where a; b 2 R are constants) at an arbitrary point.

Exercise 3.25 Show that for plane curves (C) : �r(t) = x(t)�{+y(t)�j+0�k (in the

plane xOy), the formula of the curvature of a space curve Kspace =

��r � ��r ��r 3 jt0

gives the absolute value of the curvature of (C) computed by the "plane" formula

Kplane =_x�y � �x _y

j _x2 + _y2j3=2:

3.3 Di¤erential Geometry of Surfaces

3.3.1 Representations

A surface (in a general sense) is a set (�) of points in Euclidean space whoseposition vectors obey an equation of type:

(�) : �r(u; v) = x(u; v)�{+ y(u; v)�j + z(u; v)�k; (u; v) 2 A�B � R2;

this is, �r = �r(u; v); or, equivalently,

(�) :

8<: x = x(u; v);y = y(u; v);z = z(u; v):

(3.11)

This is called the parametric representation of (�); while u and v are calledthe parameters (or the curvilinear coordinates) of the representation.We shall denote the partial derivatives of �r(u; v) by indices, as follows:

�r1 =@�r

@u; �r2 =

@�r

@u; �r11 =

@2�r

@u2; �r12 =

@2�r

@u@vetc.

A point of M the surface is called regular, if, at M; �r1 and �r2 exist andtheir cross product is nonzero:

�r1 � �r2 6= 0

(this is, the two vectors are nonvanishing and non-collinear). A point of thesurface which is not regular is called a singular point.

3.3. DIFFERENTIAL GEOMETRY OF SURFACES 85

� Other description of a surface is

(�) : z = z(x; y); (3.12)

which is called the explicit Cartesian equation. At regular points, z has tobe (continuous and) di¤erentiable.

� Finally, a surface or a fragment of surface can be represented also as:

F (x; y; z) = 0; (3.13)

which is called the implicit Cartesian equation. At regular points, Fhas to be di¤erentiable and to obey the condition:

(F 0x)2 + (F 0x)

2 + (F 0x)2 > 0; (3.14)

which is, at least one of the partial derivatives F 0x; F0y; F

0z has to be non-

vanishing.

If the functions which describe the surface are di¤erentiable of some classp � 1 at some point, we say that the surface is of class p at that point.Example: For the elliptic paraboloid z = x2 + y2 (! explicit representa-

tion), we have

� the implicit representation: x2 + y2 � z = 0:

� a parametric representation is x = u cos v; y = u sin v; z = u2; or, invectorial form,

~r(u; v) = u cos v~i+ u sin v~j + u2~k:

3.3.2 Curves on a Surface

Let (S) be a surface of class p � 1 in the representation

�r = �r(u; v); (3.15)

and P 2 (S) a regular point. A curve (C) on (S) is de�ned by establishing alink between the parameters, for instance:

u = u(t); v = v(t); t 2 A � R; (3.16)

or, as well,v = v(u) or u = u(v) or h(u; v) = 0: (3.17)

A peculiar importance have the curves u = u0 (constant) and v = v0 (con-stant), which are called coordinate curves; they compose a grid, such thatthrough every point P (u = u0; v = v0) of (S) there passes exactly a curve ofeach family, namely: (�u) : v = v0; (C2) : (�v) : u = u0:


coordinate curves (�u); (�v)

This is, P represents the intersection point of the coordinate curves v =v0 and u = u0; the situation is similar to that one in the plane xOy; whereM(x0; y0) lies at the intersection of the straight lines x = x0 and y = y0; there,(x0,y0) were called rectilinear (or Cartesian) coordinates. This similarity is whatgives us the right to call u and v; curvilinear coordinates on (S):Let now P denote a point of an arbitrary curve (C) � (S); represented as

in (3.16). This is, along (C); the position vector is �r = �r(u(t); v(t)): Then, thetangent vector at P to (C) is

��r =

d�r

dt=@�r

@u

du

dt+@�r

@v

dv

dt= �r1 _u+ �r2 _v; (3.18)

where we used the notations �r1 =@�r

@u; �r2 =

@�r

@v: This relation shows that actu-

ally, the tangent vector��r is coplanar with �r1 and �r2:

Example: For the coordinate curves, we have: (�u) : (u = t; v = const:))_u = 1; _v = 0 and (�v) : (u = const:; v = t) ) _u = 0; _v = 1: This is, the

tangent vectors to the coordinate curves v = const are �r1 =@�r

@u; similarly, the

tangent vectors to the coordinate curves u = const are �r2 =@�r

@v:

If P 2 (S) is a regular point, then, at P; the two vectors �r1 and �r2 arenonvanishing and non-collinear )since �r1 � �r2 6= 0). Hence, together with thepoint P , they uniquely de�ne a plane P: Moreover, by (3.18), we conclude thatall tangent lines to curves through P belong to this plane.

Exercise 3.26 Describe the coordinate curves of the surface: ~r(u; v) = cosu~i+sinu ~j + v~k: Also, �nd the implicit Cartesian equation of the surface and recog-nize it.


3.3.3 Tangent Plane and Surface Normal at a RegularPoint

De�nition 3.27 The tangent plane at a regular point P (x0; y0; z0) 2 (S) isthe plane which contains the tangent lines at P of all curves on (S); passingthrough P:

the tangent plane of a surface

Let us suppose that P corresponds to u = u0; v = v0: As shown above, thetangent plane is actually the plane which passes through P and contains thedirections �r1 and �r2: This is, its equation is:��

x� x0 y � y0 z � z0@x

@u(u0; v0)

@y

@u(u0; v0)

@z

@u(u0; v0)

@x

@v(u0; v0)

@y

@v(u0; v0)

@z

@v(u0; v0)

�� = 0: (3.19)

The normal line at a regular point P 2 (S) to (S) is the perpendicularthrough P to the tangent plane of (S) at P: Since the vectors �r1; �r2 belong tothe tangent plane at P to (S), it follows that the normal vector of the surfaceat P is �!

N = (�r1 � �r2) j(u0;v0): (3.20)

The unit surface normal at the point P is:

�!n =�!N �!N = �r1 � �r2

k�r1 � �r2k: (3.21)

Example: Let (S) be the surface x = u cos v; y = u sin v; z = hv (helicoid).Determine the tangent plane and the normal line at the point A(1; 0; 0):


The point A corresponds to the values u0 = 1; v0 = 0 of the parameters.We have: ~r1 = cos v~i + sin v~jjA = ~i; ~r2 = �u sin v~i + u cos v~j + h~kjA = ~j + h~k:Then, the tangent plane is described by the equation:��

x� 1 y z1 0 00 1 h

�� = 0;or �hy + z = 0: The surface normal vector at A is the normal vector of this

plane, hence ~N(0;�h; 1); the unit surface normal is ~n�0;

�hp1 + h2

;1p1 + h2

�;

and the equations of the normal line are:

x� 10

=y

�h =z

1:

� � If (S) is given by in implicit form (3.13):

F (x; y; z) = 0;

then the equation of the tangent plane to the surface (S) :F (x; y; z) = 0 at a regular point P (x0; y0; z0) 2 (S) is:

(x�x0)F 0x(x0; y0; z0)+(y�y0)F 0y(x0; y0; z0)+(z�z0)F 0z(x0; y0; z0) = 0:(3.22)

The surface normal�!N; which is also denoted by gradF (and called

the gradient of F ), is

�!N � gradF (F 0x; F 0y; F 0z)j(x0;y0;z0) (3.23)

Exercise 3.28 Find the tangent planes and the corresponding normal lines ofthe surfaces

1. (S) : ~r(u; v) = u cos v~i+ u sin v~j + u~k; (circular cone) at A(u = 1; v = 0);

2. (S) : x2 + y2 + z2 �R2 = 0 (the sphere with centre O and radius R) at apoint A(x0; y0; z0) 2 (S);

3. (S) : z = x2 � y2 (the "saddle"), at A(1; 1; 0):

3.3.4 First Fundamental Form. Applications

Let(S) : ~r = ~r(u; v)

be a surface of class p � 1; (C) : u = u(t); v = v(t); a curve on (S) and a regular

arcaAB of the curve (C); corresponding to the values t 2 [a; b] : A(t = a);

B(t = b): We intend to calculate the length of the arcaAB � (S): Taking into


account thataAB is an arc of a space curve, its arc length is de�ned by the

integralbRa

��r(t) dt.We have

��r = ~r1 _u+ ~r2 _v, hence, we obtain ��r 2 = �

�r ��r = (�r1 � �r1) _u2 + 2(�r1 � �r2) _u _u2 + (�r2 � �r2) _v2:

By denoting

E = �r1 � �r1 = k�r1k2 ; F = �r1 � �r2; G = �r2 � �r2 = k�r2k2 ; (3.24)

we get: ��r 2 = E _u2 + 2F _u _v +G _v2; (3.25)

We get the formula for the arc length of a curve on (S):

L aAB

=

bZa

pE _u2 + 2F _u _v +G _v2dt: (3.26)

Moreover, from (3.25), we can see that the arc length element of the curve(C) is

ds2 = Edu2 + 2Fdudv +Gdv2: (3.27)

De�nition 3.29 The di¤erential quadratic form ds2 in (3.27) is called the�rst fundamental form of the surface (S):

Other applications of the �rst fundamental form are to the calculus of anglesbetween curves on the surface and areas of regions of (�):

� The angle � between two intersecting curves (C) and (C�) on (S);is, by de�nition, is the angle between the tangent lines to (C) and (C�)at their common point.

To this aim, let us consider the curves (C) and (C�) given on (S) by

(C) : u = u(t); v = v(t)

(C�) : u� = u�(t�); v� = v�(t�):

If P 2 (C1) \ (C2) is a common point, then, there holds

Proposition 3.30 The angle � between two curves on (S) is given by:

cos� =E _u _u� + F ( _u _v� + _u� _v) +G _v _v�p

E _u2 + 2F _u _v +G _v2pE _u�2 + 2F _u� _v� +G _v�2

: (3.28)


Example: If (C) are (C�) the coordinate curves u = const: and v = const:;

then _u = 0; _v� = 0; which means that�~r = ~r2 _v and

�~r�= ~r1 _u

�: It follows thatthe angle between the coordinate curves is

cos� =~r1 � ~r2k~r1k k~r2k

=FpEG

; (3.29)

� The area A(�) of a regular fragment (�) of the surface (S) is

A(�) =

Z ZU

pEG� F 2dudv; (3.30)

where U � R2 is the domain of the values (u; v) corresponding to thefragment (�): The expression

d� =pEG� F 2dudv (3.31)

is called the area element of the surface (S); and represents the areaof a small (in�nitesimal) "curvilinear parallelogram" of the surface (�);determined by two pairs of neighboring coordinate curves.

Example: For the circular cone ~r(u; v) = u cos v~i+u sin v~j+u~k; v 2 [0; 2�);u 2 R; determine:a) its �rst fundamental form;b) the length of the curve u = 1 between the points corresponding to v0 = 0

and v1 = �;c) the angle of the curves u = 1 and u+ v = 1;d) the area of the curvilinear parallelogram built on the curves: u = 1; u = 2;

v = 0; v =�

2:

Bibliography

[1] M. Anastasiei: Geometrie: Curbe si suprafete, Ed. Cermi, Iasi, 2003.

[2] Gh. Atanasiu, E. Stoica, N. Brînzei: Curbe si suprafete, Ed. Matrix Rom,Bucuresti, 2005.

[3] Gh. Atanasiu, Gh. Munteanu, M. P¼aun: Curs de Algebr¼a liniar¼a, Geometrieanalitic¼a, Geometrie diferential¼a si Ecuatii diferentiale (partea a II-a), Univ.�Transilvania�, Brasov, 1993.

[4] Gh. Atanasiu, Gh. Munteanu, M. Postolache : Algebr¼a liniar¼a, Geometrieanalitic¼a, Geometrie diferential¼a si Ecuatii diferentiale - culegere de prob-leme, Ed. All Educational, Bucuresti, Editia I, 1994, Editia a II-a, 1998,Editia a III-a, Ed. Fair Partners, Bucuresti, 2002.

[5] Gh. Atanasiu, E. Stoica: Algebr¼a liniar¼a, Geometrie analitic¼a, Ed. Fair Part-ners, Bucuresti, 2003.

[6] C. Udriste, V. Balan: Analytic and Di¤erential Geometry, Geometry BalkanPress, Bucuresti, 1999.

[7] C. Udriste, I. Boca: Linear Algebra, Geometry Balkan Press, Bucharest,1999.

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