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\begin{document}

\title{Episode 17 -- Cross sections and coordinates}
\subtitle{European section -- Season 2}
\date{}

\begin{frame}
  \titlepage
\end{frame}


\begin{frame}{Cartesian equations and planes}

\begin{definition}
The set of points $M(x,y,z)$ in the 3D space such that $$ax+by+cz=d$$ where $a$, $b$, $c$ and $d$ are real numbers is a
\emph{plane}.
\end{definition}\pause

\begin{definition}
The previous equality, true only for the points on the plane, is a \emph{cartesian equation} of the plane.
\end{definition}\pause

\begin{theorem}
A plane has infinitely many different cartesian equations.
\end{theorem}

\end{frame}

\begin{frame}{How to find a cartesian equation}

\begin{block}{An example}
Find a cartesian equation of the plane $\mathscr{P}$  passing through the points of coordinates $P(1,0,0)$,
$Q(\frac12,1,0)$ and $R(0,0,1)$.
\end{block}\pause

We know that a cartesian equation has the form $$ax+by+cz=d.$$\pause
As the points $P$, $Q$, $R$ are on the plane, we deduce that $a$, $b$, $c$ and $d$ are solutions to the system \pause
$$\left\{
\begin{array}{rcl}
a\times 1+b\times 0+c\times 0 &=& d \\ \pause
a\times \frac12+b\times 1+c\times 0 &=& d \\ \pause
a\times 0+b\times 0+c\times 1 &=& d 
\end{array}
\right.$$\pause
From the first equation we get $a=d$,\pause\ and from the last $c=d$.\pause\ Then, the second equation is equivalent
to\pause\ $\frac12d+b=d$, or \pause\ $b=\frac12d$.\pause

\end{frame}

\begin{frame}{How to find a cartesian equation}

As there are infininitely many cartesian equation, we can choose for $d$ any number (except $0$ in this case). Let's
choose\pause\ $d=2$.\pause\ 

\bigskip

Then we deduce $a=2$, $c=2$ and $b=1$, so a cartesian equation of the plane is \pause
$$2x+y+2z=2.$$

\end{frame}



\begin{frame}{Parametric equations and lines}

\begin{definition}
The set of points $M(x,y,z)$ in the 3D space such that $$\left\{
\begin{array}{rcl}
x &=& x_A+at \\ 
y &=& y_A+bt \\
z &=& z_A+ct
\end{array}
\right.$$ where $x_A$, $y_A$, $z_A$, $a$, $b$ and $c$ are real numbers, is a
\emph{line} passing through point $A$ and with directing vector $\vec{u}(a,b,c)$. 
\end{definition}\pause

\begin{definition}
The previous system, true only for the points on the line, is a set of \emph{parametric equations} of the line.
\end{definition}\pause

\begin{theorem}
A line has infinitely many different sets of parametric equations.
\end{theorem}

\end{frame}

\begin{frame}{How to find parametric equations}

\begin{block}{An example}
Find a set of parametric equations of the line $\mathscr{L}$ passing through the point of coordinates $T(1,1,0)$ and
with directing vector $\vec{u}(0,0,1)$.
\end{block}\pause

According to the definition, a set of parametric equations of this line is \pause
$$\left\{
\begin{array}{rcl}
x &=& 1+0\times t \\ \pause
y &=& 1+0\times t \\ \pause
z &=& 0+1\times t\pause
\end{array}
\right.
\textrm{ or }
\left\{
\begin{array}{rcl}
x &=& 1 \\ \pause
y &=& 1 \\ \pause
z &=& t
\end{array}
\right.$$ 

\end{frame}

\begin{frame}{Intersections of lines and planes}

\begin{block}{Method}
To find the coordinates of the intersection of a plane with a line (if it exists), just solve the system made of a
cartesian equation of the plane and a set of parametric equations of the line.\pause\ 
\begin{itemize}
\item If the line intersects the plane in one just point, the solving will give one value for $t$, that can be used in
the parametric equations to find the coordinates of the point.\pause
\item If the line is parallel to the plane, then it will be impossible to solve the system.\pause
\item If the line is included in the plane, then there will be an infinite number of solutions.
\end{itemize}

\end{block}

\end{frame}

\begin{frame}{Intersections of lines and planes} 

\begin{block}{An example}
Find the coordinates of the intersection of the plane $\mathscr{P}$ with the line $\mathscr{L}$, if it exists.
\end{block}\pause

To do so, we have to solve the system :
$$
\left\{
\begin{array}{rcl}
2x+y+2z &=& 2 \\ \pause
x &=& 1 \\ 
y &=& 1 \\
z &=& t
\end{array}
\right.$$ 

We can replace $x$, $y$ and $z$ by their expressions as a function of $t$ in the first equation. That gives : \pause
\begin{eqnarray*}
2\times 1+1+2t &=& 2 \\ \pause
3+2t &=& 2 \\ \pause
2t &=& -1 \\ \pause
t &=& -\frac12
\end{eqnarray*} \pause

\end{frame}

\begin{frame}{Intersections of lines and planes} 

Then, we use this value of $t$ and the parametric equations to find the coordinates of the point :

$$\left\{
\begin{array}{rcl}
x &=& 1 \\ \pause
y &=& 1 \\ \pause
z &=& -\frac12
\end{array}
\right.$$ 

So the plane and the line intersection in the point of coordinates $$\left(1,1,-\frac12\right).$$

\end{frame}


\end{document}
