\chapter{Analytic Geometry}   % correction 2001-08-02
\label{vect-chap}
   
\Citation{In Geometry (which is the only science that it hath please God hitherto to bestow on mankind) men 
  begin at settling the significations of their words; which \dots they call Definitions.\\ 
  \auteur{Thomas}{Hobbes} {1588}{1679}{English philosopher}{Leviathan\,\footnotemark}{1651}} 
   
\footnotetext{Hobbes \pron{h£obz} --- leviathan \pron{l£i£hv£p£e£fn}} 
   
   
For sake of convenience, we will write $\EuScript{L} \parallel \EuScript{M}$ to denote that the lines $\EuScript{L}$ and 
$\EuScript{M}$ are parallel.\index{$\parallel$} 
   
\section{Parallelograms}   
   
\begin{Def}   
  Let $A$, $B$, $C$ and $D$ be four points in the plane. $ABCD$ is a \nuvo{parallelogram} \pron{£bp£ar£a£hlel£egr£am}
  iff its diagonals bisect \pron{b£p£hsekt} each other.
\end{Def}   
   
For sake of convenience, we will write $ABCD \ \# $ as an abbreviation of the sentence `$ABCD$ is a 
parallelogram'.\index{$\#$} 
   
Parallelograms have many properties of which we state but a few. The first two properties given below involve the four 
sides of the quadrilateral $ABCD$. The third one involves only two of them. 
   
\begin{prop}   
  $ABCD$ is a parallelogram iff its opposite sides are parallel, \idest\  iff $(AB) \parallel (CD)$ and $(AD) \parallel 
  (BC)$. 
\end{prop}   
   
\begin{prop}   
  $ABCD$ is a parallelogram iff its opposite sides have the same length, \idest\  iff $AB = CD$ and $AD = BC$. 
\end{prop}   
   
\begin{prop}   
  $ABCD$ is a parallelogram iff two opposite sides are parallel and have equal lengths, \idest\  iff \parex\  $(AB) 
  \parallel (CD)$ and $AB = CD$. 
\end{prop}   
   
\section{Vectors}   
   
\subsection{Equality}   
   
Let $A$ and $B$ be two points. The vector $\vecteur{AB}$ is represented by an arrow from $A$ to $B$.  $A$ is the 
\nuvo{beginning point} and $B$ the \nuvo{end point}.\index{point!beginning}\index{point!end} If $\vecteur{AB} = 
\vecteur{CD}$ we can decide to denote that vector with a single sign as \parex\ \vect{u}.  So $\vecteur{AB} = 
\vecteur{CD} = \vect{u}$.  In such a case the \emph{ordered pairs} \couple{A}{B} and \couple{C}{D} are said to represent 
the vector \vect{u}.\index{ordered pair} 
   
{\small In English texts one can find \textbf{u} or $\underline{u}$ instead of \vect{u} to denote the vector 
  `\emph{u}'.} 
   
\begin{Def}   
  The \nuvo{magnitude} \pron{£hmagn£itj£Ud} (or \nuvo{length} or \nuvo{modulus} \pron{£hm£odj£ul£es}) of the vector
  $\vecteur{AB}$ is the length of $[AB]$ \idest\ $AB$. The magnitude of a vector $\vect{u}$ is denoted by
  \norme{\vect{u}}. % prononciations ajoutées 2003-06-15
\end{Def}   
   
\begin{Def}   
  Let $A$, $B$, $C$ and $D$ be four points in the plane. The vectors \vecteur{AB} and \vecteur{CD} are equal iff $ABDC$ 
  is a parallelogram. $$\dps \vecteur{AB} = \vecteur{CD} \Iff ABDC \ \# $$ 
\end{Def}   
   
\begin{rem}   
  If $\vecteur{AB} = \vecteur{CD}$ we can say that those vectors have the same \nuvo{direction} 
  ---~$(AB)\parallel(CD)$~--- the same \nuvo{sense} ---~$B$ and $D$ are on the same side of $(AC)$~--- and the same 
  length. 
\end{rem}   
   
\begin{prop}   
  $\vecteur{AB} = \vecteur{CD}$ iff $D$ is the image under $t_{\vecteur{AB}}$ of $C$. 
\end{prop}   
   
\begin{prop}   
  Let \vect{u} be any vector and let $A$ be any point in the plane. There exists one and only one point $B$ such that 
  $\vect{u} = \vecteur{AB}$. 
\end{prop}   
   
So one can always represent any given vector with a given beginning point. Obviously one can also represent any given 
vector with a given end point. 
   
\subsection{Sum}   
   
Let \vect{u} and \vect{v} be two vectors. There are two ways of defining the sum of the vectors \vect{u} and \vect{v} 
depending on how they are represented. 
   
\begin{prop}[Triangle law]   \label{prop-trianglelaw}
  If $A$, $B$ and $C$ are points such that $\vect{u} = \vecteur{AB}$ and $\vect{v} = \vecteur{BC}$ then 
  $$\vect{u} + \vect{v} = \vecteur{AB} + \vecteur{BC} = \vecteur{AC}.$$ 
\end{prop}   
   
\begin{prop}[Parallelogram law]   
  If $A$, $B$ and $C$ are points such that $\vect{u} = \vecteur{AB}$ and $\vect{v} = \vecteur{AC}$.  Let $D$ be the point 
  such that $CABD$ is a parallelogram. Then $$\vect{u} + \vect{v} = \vecteur{AB} + \vecteur{AC} = \vecteur{AD}.$$ 
\end{prop}   
   

\begin{fgre}
  \IG{vect01}
\end{fgre}
   
\medskip 
   
In fact if $CABD$ is a parallelogram, $$\vecteur{AB} + \vecteur{AC} = \vecteur{AB} + \vecteur{BD}$$ 
and we can conclude 
using the triangle law\footnote{which is known in French as `la relation de Chasles'.}. 
   
\begin{Def}   
  The \emph{null vector} \vect{0} is equal to \vecteur{AA} for any point $A$. For any vector \vect{u}: 
  $$\vect{u} + \vect{0} = \vect{0} + \vect{u} = \vect{u}.$$ 
\end{Def}\index{$\vect{0}$}\index{null!vector}   
   
\begin{Def}   
  \vect{v} is said to be the \nuvo{inverse} of \vect{u} iff $$\vect{u} + \vect{v} = \vect{v} + \vect{u} = \vect{0}.$$ 
\end{Def}   
   
We denote the (additive) inverse of \vect{u} by $-\vect{u}$.

\bigskip

\begin{rem}[a]
  In French one must use the word `oppos{\'e}'.
\end{rem}

\vspace{\stretch{1}}
   
If $A$ and $B$ are such that $\dps \vect{u} = \vecteur{AB}$ then $\dps -\vect{u} = \vecteur{BA}$ for 
$$ 
\vecteur{AB}+\vecteur{BA}=\vecteur{AA}=\vect{0} \text{ and }\vecteur{BA}+\vecteur{AB}=\vecteur{BB}=\vect{0}. 
$$ 

\vspace{\stretch{1}}   

\begin{prop}[Properties of the sum]   
  Let \vect{u}, \vect{v} and \vect{w} be vectors. 
\begin{itemize}   
\item $\parent{\vect{u} + \vect{v}}+\vect{w} = \vect{u} + \parent{\vect{v}+\vect{w}}$; 
\item $\vect{u} + \vect{v} = \vect{v} + \vect{u}$; 
\end{itemize}   
\end{prop}   

\vspace{\stretch{1}}
   
\begin{fgre}
  \IG{vect02}
\end{fgre}

\vspace{\stretch{1}}
   
\subsection{Multiplication by a scalar}   
   
\begin{rem}   
  When dealing with vectors, any real number is called a \nuvo{scalar} \pron{£hsk£ql£e£Z}. 
\end{rem}   
   
{\small If $\vecteur{AB} = \vect{u}$, we know how to represent $\vect{u} + \vect{u}$, $\vect{u} + \vect{u} + \vect{u}$ 
  or even $\vect{u} + \vect{u} + \vect{u} + \vect{u}$ and it would be agreable to be allowed to write $2\vect{u}$, 
  $3\vect{u}$ and $4\vect{u}$ instead of those rather boring long sums. That will be the case at the end of this section 
  the aim of which is to define the multiplication of a vector by a scalar (or number).} 

\newpage
   
\begin{Def}   
  Let \vect{u} be a vector and $k$ be a number.  We define the product of \vect{u} by $k$, denoted by $k \vect{u}$, as 
  follows: 
\begin{itemize}   
\item If $\vect{u} = \vect{0}$ or $k=0$ then $k\vect{u} = \vect{0}$. 
\item If $\vect{u} \neq \vect{0}$ and $k>0$ then $k\vect{u}$ has the same direction and sense as \vect{u} and 
  $\norme{k\vect{u}} = k\times\norme{\vect{u}}$. 
\item If $\vect{u} \neq \vect{0}$ and $k<0$ then $k\vect{u}$ has the same direction and sense as $-\vect{u}$ and 
  $\norme{k\vect{u}} = -k\times\norme{\vect{u}}$. 
\end{itemize}\end{Def}\index{$k\vect{u}$}   
   

\begin{fgre}
  \IG{vect03}
\end{fgre}
   
\begin{Def}   
  Vectors \vect{u} and \vect{v} are said to be \nuvo{parallel} iff there exists a scalar $k$ such that $\vect{u} = k 
  \vect{v}$ or $\vect{v} = k \vect{u}$. 
\end{Def}   
   
When necessary we will write $\vect{u} \parallel \vect{v}$ to denote that vectors \vect{u} and \vect{v} are 
parallel. 

\begin{rem}[a]
  In French one must use the word `colinéaire' in such a case.
\end{rem}
   
For each vector \vect{u} the following equality holds: $\vect{0} = 0\vect{u}$. So \vect{0} is parallel to any vector. 
   
\begin{prop}   
  Let $A$, $B$, $C$ and $D$ be four distinct points. 
\begin{itemize}   
\item \vecteur{AB} and \vecteur{CD} are parallel iff $(AB)$ and $(CD)$ are parallel; 
\item \vecteur{AB} and \vecteur{AC} are parallel iff $C$ is on $(AB)$. 
\end{itemize}   
\end{prop}   

The last proposition is very useful to prove that three points are collinear.

\newpage
   
\begin{prop}[Properties of the multiplication by a scalar]   
  Let \vect{u} and \vect{v} be vectors and $a$ and $b$ be scalars. 
\begin{itemize}   
\item $a\parent{\vect{u} + \vect{v}} = a\vect{u} + a\vect{v}$; 
\item $\parent{a+b}\vect{u} = a\vect{u} + b\vect{u}$; 
\item $a\parent{b\vect{u}} = \parent{a\times b}\vect{u}$; 
\item $0\vect{u} = \vect{0}$; 
\item $(-1)\vect{u} = -\vect{u}$. 
\end{itemize}   
\end{prop}   
   
\subsection{Basis}   
   
\begin{Def}   
  The ordered pair of vectors \couple{\vect{u}}{\vect{v}} is said to be a \nuvo{basis}\,\footnotemark{} \pron{£hb£qs£is} iff 
  \vect{u} and \vect{v} are not parallel. 
\end{Def}\footnotetext{\emph{plur.} bases \pron{£hb£qs£Iz}}   
   
\textbf{Question}: Let $A$ and $C$ be two points, \Li and \Lip be two intersecting lines.  Determine $B$ and $D$ such that 
$ABCD$ is a parallelogram, $(AB)$ is parallel to \Li and $(AD)$ is parallel to \Lip. 
   
\textbf{Answer}: \textsl{$B$ is the common point of the line parallel to \Li passing through $A$ and the line parallel 
  to \Lip passing through $C$. Similarly, $D$ is the common point of the line parallel to \Lip passing through $A$ and 
  the line parallel to \Li passing through $C$.} 
   
\smallskip 
   
If \couple{\vect{u}}{\vect{v}} is a basis, there are three non-collinear points $O$, $U$ and $V$ such that $\vect{u} = 
\vecteur{OU}$ and $\vect{v} = \vecteur{OV}$. In that case, $(OU)$ and $(OV)$ intersect. Let \vect{w} be a vector. There 
are two points $A$ and $C$ such that $\vect{w} = \vecteur{AC}$. Applying the preceding result, we can construct $ABCD$, 
parallelogram such that $(AB) \parallel (OU)$ and $(AD) \parallel (OV)$. Then $\vecteur{AB} \parallel \vecteur{OU}$ and 
there exists one (and only one) number $x$ such that $\vecteur{AB} = x\,\vecteur{OU}$. By the same reasoning we obtain a 
number $y$ such that $\vecteur{AD} = y\,\vecteur{OV}$. Then using the parallelogram law we prove that $\vect{w} = 
x\vect{u} + y\vect{v}$. 
   
That result is important and we restate it in the following proposition. 

\newpage   

\begin{prop}   
  If \couple{\vect{u}}{\vect{v}} is a basis, for any \vect{w}, there exist two numbers $x$ and $y$ such that $\vect{w} = x 
  \vect{u} + y \vect{v}$. 
   
  Moreover if $\vect{w} = x \vect{u} + y \vect{v} = a \vect{u} + b \vect{v}$ then $x=a$ and $y=b$. 
\end{prop}   
   
\begin{Def}   
  If \couple{\vect{u}}{\vect{v}} is a basis, the \nuvo{coordinates} of a vector \vect{w} are the numbers $x$ and $y$ 
  such that $\vect{w} = x \vect{u} + y \vect{v}$. 
   
  We write $\vect{w} : \couple{x}{y}$. 
\end{Def}   
   
Coordinates are sometimes written in a column like this $\dps \begin{pmatrix}x\\y\end{pmatrix}$. 
   
   
Let \base{} be a basis, let \vect{u} and \vect{v} be two vectors the coordinates of which are \couple{u}{u'} and 
\couple{v}{v'} respectively, and let $a$ be a number. 
\begin{align*}   
    \vect{u}+\vect{v} &= \parent{u\vecti + u'\vectj} + \parent{v\vecti + v'\vectj} \\ 
    &= \parent{u+v}\vecti + \parent{u'+v'}\vectj, 
    \text{and}& \\ 
    a\vect{u} &= a\parent{u\vecti + u'\vectj} = au\vecti + au'\vectj. 
\end{align*} 
   
Therefore we can state the proposition: 
   
\begin{prop}   
  Let $\vect{u} : \couple{u}{u'}$ and $\vect{v} : \couple{v}{v'}$. 
\begin{itemize}    
\item $\vect{u}=\vect{0}$ iff $u =0$ and $u'=0$; 
\item $\vect{u}=\vect{v}$ iff $u =v$ and $u'=v$; 
\item $\vect{u}+\vect{v} : \couple{u+v}{u'+v'}$; 
\item for each number $k$, $k\vect{u} : \couple{ku}{ku'}$. 
\end{itemize}    
\end{prop}   
   
\begin{Def}   
  Let \base{} be a basis and $O$ be a point in the plane. The ordered triple \pron{tr£ipl} \repere is said to be a
  Cartesian\footnotemark  \emph{coordinate system}.\index{coordinate!system}\index{system!coordinate}
   
  Let $M$ be a point in the plane. The coordinates of $M$ ---~relatively to the coordinate system \repere~--- are 
  \couple{x}{y} iff the coordinates of \vecteur{OM} are \couple{x}{y} in the basis \base. 
\end{Def}  \footnotetext{Cartesian \pron{k£A£ht£Iz£I£en} of or relating to René \textsc{Descartes} \Detantatant{1596}{1650} or his work in
  philosophy, science, and mathematics.}
   
\begin{prop}   
  If $A$ and $B$ have coordinates \couple{a}{a'} and \couple{b}{b'} respectively then \vecteur{AB} has coordinates 
  \couple{b-a}{b'-a'}. 
\end{prop}   
   
\begin{prop}[Parallelism criterion]  \label{paralcrit} 
  Let {\rmfamily\upshape\base} be a basis, \vect{u} and \vect{v} two vectors the respective coordinates of which are 
  \couple{u}{u'} and \couple{v}{v'}. 
   
  \vect{u} and \vect{v} are parallel iff $uv'-u'v = 0$. 
\end{prop}   
\begin{rem}   
  Criterion \pron{kr£p£ht£Qr£Qn} \emph{plur.} criteria \pron{kr£p£ht£Qr£e}: a principle or standard that a thing is 
  judged by. 
\end{rem}   

\input{fini/colinear}

\subsection{Midpoint}   
   
One knows that the midpoint \pron{£hm£idp£rnt} of $[AB]$ is the point $I$ such that $I$ is on $(AB)$ and is equidistant from $A$ and 
$B$.\index{midpoint} 
   
\begin{prop}   
  $I$ is the midpoint of $[AB]$ iff one of the following equalities holds: 
\begin{itemize}   
\item $\dps\vecteur{AI} = \vecteur{IB}$; 
\item $\dps\vecteur{AI} = \dfrac{1}{2}\vecteur{AB}$ or $\dps\vecteur{IB} = \dfrac{1}{2}\vecteur{AB}$; 
\item $\dps\vecteur{AB} = 2\vecteur{AI}$ or $\dps\vecteur{AB} = 2\vecteur{IB}$; 
\item $\dps\vecteur{IA} + \vecteur{IB}= \vect{0}$. 
\end{itemize}\end{prop}\index{midpoint!vectors}   
   
\begin{prop}   
  If $A$ and $B$ have coordinates \couple{a}{a'} and \couple{b}{b'} respectively then the midpoint of $[AB]$ has 
  coordinates \couple{\dfrac{a+b}{2}}{\dfrac{a'+b'}{2}}. 
\end{prop}\index{midpoint!coordinates}   
   
\begin{prop}   
  If $I$ is the midpoint of $[AB]$ then for any point $M$ \quad $\dps\vecteur{MA} + \vecteur{MB} = 2\vecteur{MI}$. 
\end{prop}   
   
That is easily proved for 
\begin{align*}   
  \vecteur{MA} + \vecteur{MB} &= \vecteur{MI} + \vecteur{IA} +\vecteur{MI} + \vecteur{IB}\\ 
  &=2\,\vecteur{MI} + \vecteur{IA} + \vecteur{IB}\\ 
  &= 2\,\vecteur{MI} + \vect{0} = 2\,\vecteur{MI}\quad \textsc{qed}. 
\end{align*}   
   
\begin{prop}   
  If $I$ and $J$ are the midpoints of $[AB]$ and $[AC]$ respectively then $\vecteur{IJ} = \dfrac{1}{2}\vecteur{BC}$. 
\end{prop}   
   
\section{Equations of lines}   
   
Let \repere be a coordinate system in the plane. 
   
\begin{Def}   
  Let \vect{u} be a vector and \Li be a line. 
   
  \vect{u} is a \emph{direction vector} of \Li iff there exist two points $A$ and $B$ on \Li such that $\vect{u} = 
  \vecteur{AB}$.\index{direction!vector}\index{vector!direction} 
\end{Def}   

\begin{prop}
  If \vect{u} is a direction vector of \Li then, for any real $k$ different from $0$, $k\,\vect{u}$ is also a direction
  vector of \Li.
\end{prop}

Let $A$ and $B$ be two points. Two cases occur: or $x_A$ equals $x_B$ or not. 

\medskip
 
In the first case, \vecteur{AB} has coordinates \couple{0}{y_B - y_A}. Let $M$ be a point with coordinates 
\couple{x}{y}. $M$ is on $(AB)$ if and only if \vecteur{AM} and \vecteur{AB} are parallel \idest\  
$(y_B - y_A)(x-x_A) - 0\times(y-y_A) = 0$ which is equivalent to $x=x_A$. 

In that case, line $(AB)$ is parallel to the $y$-axis.

\medskip

In the second case, \vecteur{AB} has coordinates \couple{x_B-x_A}{y_B - y_A} and $x_B-x_A \neq 0 $. Let $M$ be a point
with coordinates \couple{x}{y}. $M$ is on $(AB)$ if and only if \vecteur{AM} and \vecteur{AB} are parallel \idest\\ 
$(y_B - y_A)\times(x-x_A) - (x_B-x_A)\times(y-y_A) = 0$\\ 
which is equivalent to\\ 
$(y_B - y_A)x - (x_B-x_A)y + \parent{(x_B-x_A)y_A -(y_B - y_A)x_A}=0 $\\
and as that equation characterises the fact that a point is on $(AB)$ or not, it is named `equation of $(AB)$'.%
\index{equation}\index{equation!line}\index{line!equation}

That equation is equivalent to\\
$ (x_B-x_A)y = (y_B - y_A)x  + \parent{(x_B-x_A)y_A -(y_B - y_A)x_A}=0 $, and
as $x_B-x_A \neq 0 $, one can write
$$
y = \dfrac{y_B - y_A}{x_B-x_A}x + \dfrac{(x_B-x_A)y_A -(y_B - y_A)x_A}{x_B-x_A}
$$
which is also an equation of $(AB)$.

In such an equation the coefficient of $x$ is called the \nuvo{slope} \pron{sl£wp} or \nuvo{gradient}
\pron{£hgr£qd£i£ent} of $(AB)$. The number $\dfrac{(x_B-x_A)y_A -(y_B - y_A)x_A}{x_B-x_A}$ is called the
\nuvo{intercept}. It is the ordinate of the point at which $(AB)$ cuts the $y$-axis. The last equation is said to have
the \nuvo{slope-intercept form}.

We can sum that up with:

\begin{prop}   
  Let $A : \couple{x_{A}}{y_{A}}$ and $B : \couple{x_{B}}{y_{B}}$ be two points in the plane. 
\begin{itemize}   
\item If $x_A = x_B$ then $(AB)$ has equation $x = x_A$. 
\item If $x_A \neq x_B$ then the slope of $(AB)$ is $m = \dfrac{y_B - y_A}{x_B - x_A}$ 
\end{itemize}   
\end{prop}  

If a line \Li is parallel to the $y$-axis then all its points have the same abscissa. That is why such a line has
equation $x=k$ for a certain real number $k$.

If contrariwise \Li is not parallel to the $y$-axis, then there exist two points which have different abscissae 
and  ---~it is the second case stated above~--- 
it is the possible to find two real numbers $m$ and $p$ such that \Li has equation $y=mx+p$. 
The points $A$: \couple{0}{p} and $B$: \couple{1}{m+p} are on \Li and so the vector \vect{u}: \couple{1}{m} is a
direction vector of \Li. 

We can now state:

\begin{prop}[Equation of a line]   
  \textbf{1} \quad Each line parallel to the $y$-axis has an equation of the form $x = a$ ($a \in \reel$) where 
  $a$ is the abscissa of every point on the line.\\[3pt] 
  Each line intersecting the $y$-axis has an equation of the form $y = mx+p$ with $m$ and $p$ two real numbers. 
   
\newpage
   
  \textbf{2} \quad The set of the points $M : \couple{x}{y}$ such that: 
\begin{itemize}   
\item $x=k$ with $k$ a real, is a line parallel to the $y$-axis. 
\item $y=mx+p$ is a line which passes through the point with coordinates \couple{0}{p} and has \vect{d}: \couple{1}{m} 
  as direction vector. 
\end{itemize}   
\end{prop}   
   
\begin{fgre}
  \IG{vect04}
\end{fgre}
   
Let $a$, $b$ and $c$ be real numbers such that $a$ and $b$ are not simultaneously null. Consider the equation \graf{e}: 
$ax+by+c=0$. First consider that $b=0$. Then $a \neq 0$ and so \graf{e} is equivalent to $x = -\dfrac{c}{a}$ so it is an
equation of a line parallel to the $y$-axis. Then \couple{0}{1} is a direction vector of that line and so is
\couple{0}{a} for $a$ is not null.

If $b \neq 0$ then \graf{e} is equivalent to $y = -\dfrac{a}{b}x-\dfrac{c}{b}$. So \graf{e} is an equation of line with
direction vector \couple{1}{-\dfrac{a}{b}}. We can multiply that vector by $-b$ to obtain an other direction vector, for
$b$ is not null, and we can state the following proposition.

\newpage

\begin{prop}   
  Let $a$, $b$ and $c$ be three reals such that $a$ and $b$ are not simultaneously null.  The set of the points $M: 
  \couple{x}{y}$ such that $ax+by+c=0$ is a line with direction vector \vect{V}: \couple{-b}{a}. 
\end{prop}   
   
If two lines are parallel they have the same direction vectors. So if \couple{1}{p} is a direction vector of the first
then it is also a direction vector of the second. If a line does not intersect the $y$-axis it has a slope $p$ and
direction vector \couple{1}{p}. Therefore:
   
\begin{prop}   
  Let \Li and \Lip be two lines intersecting the $y$-axis. \Li and \Lip are parallel iff they have the same slope. 
\end{prop}   
   
\section{Orthonormal bases and coordinate systems}   
   
\begin{Def}   
  The vectors \vect{u} and \vect{v} are \nuvo{orthogonal} \pron{£O£h£f£og£enl} iff one of the following conditions 
  holds: 
\begin{enumerate}   
\item \vect{u} or \vect{v} is equal to \vect{0}; 
\item any line with direction vector \vect{u} is perpendicular to any line with direction vector \vect{v}. 
\end{enumerate}   
One can write $\vect{u}\perp\vect{v}$ to denote the fact that \vect{u} and \vect{v} are orthogonal. 
\end{Def}   
   
\begin{Def}   
\begin{itemize}   
\item The basis \base is said to be \nuvo{orthogonal} iff $\vecti \perp \vectj$. 
\item The coordinate system \repere is said to be orthogonal iff \base is an orthogonal basis. 
\item The basis \base is said to be \nuvo{orthonornal} \pron{£O£f£e£hn£Om(£e)l} iff\\ 
  $\vecti \perp \vectj$ and $\norme{\vecti}=\norme{\vectj}=1$. 
\item The coordinate system \repere is said to be orthonormal iff \base is an orthonormal basis. 
\end{itemize}   
\end{Def}   
   
\newpage

\begin{prop}   
  Let \repere be an orthonormal coordinate system. 
\begin{itemize}   
\item Let $A : \couple{x_{A}}{y_{A}}$ and $B : \couple{x_{B}}{y_{B}}$ be two points in the plane. Then 
  $$ 
  AB = \norme{\vecteur{AB}}=\sqrt{\parent{x_B - x_A}^{2}+\parent{y_B - y_A}^{2}}.
  $$ 
\item Let \vect{u}: \couple{x}{y}. Then $\dps \norme{\vect{u}}=\sqrt{x^2 + y^2}$. 
\end{itemize}   
\end{prop}   

\vspace*{\stretch{1}}   

$$ \includegraphics[keepaspectratio=true, totalheight=8cm]{figures/Vignette04}$$

\vspace*{\stretch{1}}

   
\endinput 
   
% corrections du 17/01/2001   
   
%%% Local Variables:    
%%% mode: latex   
%%% TeX-master: "../2dcours"   
%%% End:    

% correction pour v1.1 2001-08-02
% correction pour v1.2 2002-03-11
% correction pour v1.3 2003-06-15
% correction pour v1.4 2003-07-04
