From 319c48adeb1d36ac6a28b856e39ee427fdf6398e Mon Sep 17 00:00:00 2001
From: "David A. Madore" <david+git@madore.org>
Date: Thu, 6 Apr 2023 19:58:58 +0200
Subject: =?UTF-8?q?Write=20a=20(far=20too=20long)=20exercise=20on=20Pl?=
 =?UTF-8?q?=C3=BCcker=20coordinates.?=
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---
 controle-20230412.tex | 423 ++++++++++++++++++++++++++++++++++++++++++++++----
 1 file changed, 395 insertions(+), 28 deletions(-)

diff --git a/controle-20230412.tex b/controle-20230412.tex
index c7dec35..c6100bb 100644
--- a/controle-20230412.tex
+++ b/controle-20230412.tex
@@ -1,6 +1,7 @@
 %% This is a LaTeX document.  Hey, Emacs, -*- latex -*- , get it?
 \documentclass[12pt,a4paper]{article}
-\usepackage[francais]{babel}
+\usepackage[a4paper,margin=2.5cm]{geometry}
+\usepackage[english]{babel}
 \usepackage[utf8]{inputenc}
 \usepackage[T1]{fontenc}
 %\usepackage{ucs}
@@ -24,11 +25,11 @@
 %\externaldocument{notes-accq205}[notes-accq205.pdf]
 %
 \theoremstyle{definition}
-\newtheorem{comcnt}{Tout}
+\newtheorem{comcnt}{Whatever}
 \newcommand\thingy{%
 \refstepcounter{comcnt}\smallskip\noindent\textbf{\thecomcnt.} }
-\newcommand\exercice{%
-\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercice~\thecomcnt.}\par\nobreak}
+\newcommand\exercise{%
+\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercise~\thecomcnt.}\par\nobreak}
 \renewcommand{\qedsymbol}{\smiley}
 %
 \newcommand{\id}{\operatorname{id}}
@@ -37,6 +38,7 @@
 \newcommand{\val}{\operatorname{val}}
 %
 \DeclareUnicodeCharacter{00A0}{~}
+\DeclareUnicodeCharacter{A76B}{z}
 %
 \DeclareMathSymbol{\tiret}{\mathord}{operators}{"7C}
 \DeclareMathSymbol{\traitdunion}{\mathord}{operators}{"2D}
@@ -52,9 +54,9 @@
 \newcommand{\spaceout}{\hskip1emplus2emminus.5em}
 \newif\ifcorrige
 \corrigetrue
-\newenvironment{corrige}%
+\newenvironment{answer}%
 {\ifcorrige\relax\else\setbox0=\vbox\bgroup\fi%
-\smallbreak\noindent{\underbar{\textit{Corrigé.}}\quad}}
+\smallbreak\noindent{\underbar{\textit{Answer.}}\quad}}
 {{\hbox{}\nobreak\hfill\checkmark}%
 \ifcorrige\par\smallbreak\else\egroup\par\fi}
 %
@@ -62,12 +64,12 @@
 %
 \begin{document}
 \ifcorrige
-\title{ACCQ205\\Contrôle de connaissances — Corrigé\\{\normalsize Courbes algébriques}}
+\title{ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}}
 \else
-\title{ACCQ205\\Contrôle de connaissances\\{\normalsize Courbes algébriques}}
+\title{ACCQ205\\Final exam\\{\normalsize Algebraic curves}}
 \fi
 \author{}
-\date{12 avril 2023}
+\date{2023-04-12}
 \maketitle
 
 \pretolerance=8000
@@ -75,38 +77,31 @@
 
 \vskip1truein\relax
 
-\noindent\textbf{Consignes.}
+\noindent\textbf{Instructions.}
 
-Les exercices sont totalement indépendants.  Ils pourront être traités
-dans un ordre quelconque, mais on demande de faire apparaître de façon
-très visible dans les copies où commence chaque exercice.
+The different exercises below are completely independent.  They can be
+answered in any order, but candidates are asked to label very clearly
+on their papers where each exercise starts.
 
-La longueur du sujet ne doit pas effrayer : l'énoncé du dernier
-exercice est long parce que beaucoup de rappels ont été faits et que
-la rédaction des questions cherche à donner tous les éléments
-nécessaires pour passer d'une question aux suivantes.
-
-La difficulté des questions étant variée, il vaut mieux ne pas rester
-bloqué trop longtemps.
+\medbreak
 
-Si on ne sait pas répondre rigoureusement, une réponse informelle peut
-valoir une partie des points.
+Answers can be written in English or French.
 
 \medbreak
 
-L'usage de tous les documents (notes de cours manuscrites ou
-imprimées, feuilles d'exercices, livres) est autorisé.
+Use of written documents of any kind (such as handwritten or printed
+notes, exercise sheets or books) is authorized.
 
-L'usage des appareils électroniques est interdit.
+Use of electronic devices of any kind is prohibited.
 
 \medbreak
 
-Durée : 2h
+Duration: 2 hours
 
 \ifcorrige
-Ce corrigé comporte \textcolor{red}{XXX} pages (page de garde incluse).
+This answer key has \textcolor{red}{XXX} pages (cover page included).
 \else
-Cet énoncé comporte \textcolor{red}{XXX} pages (page de garde incluse).
+This exam has \textcolor{red}{XXX} pages (cover page included).
 \fi
 
 \vfill
@@ -123,7 +118,379 @@ Git: \input{vcline.tex}
 %
 %
 
+\exercise
+
+\textit{The goal of this exercise is to study a representation of
+  lines in $\mathbb{P}^3$.}
+
+We fix a field $k$.  Recall that \emph{points} in $\mathbb{P}^3(k)$
+are given by quadruplets $(x_0{:}x_1{:}x_2{:}x_3)$ of “homogeneous
+coordinates” in $k$, not all zero, defined up to a common
+multiplicative constant, and that \emph{planes} in $\mathbb{P}^3(k)$
+are of the form $\{(x_0{:}x_1{:}x_2{:}x_3) \in \mathbb{P}^3(k) : u_0
+x_0 + \cdots + u_3 x_3 = 0\}$ (for some $u_0,\ldots,u_3$, not all
+zero, defined up to a common multiplicative constant) which we can
+denote as $[u_0{:}u_1{:}u_2{:}u_3]$ (a point of the
+“dual” $\mathbb{P}^3$).  Our goal is to find a representation for
+lines.
+
+It may be convenient, if so desired, to call $\langle w\rangle$ the
+point in projective space $\mathbb{P}^{m-1}(k)$ defined by a vector
+$w\neq 0$ in $k^m$ (i.e., if $w = (w_0,\ldots,w_m)$ then $\langle w
+\rangle = (w_0{:}\cdots{:}w_m)$), that is, the class of $w$ under
+collinearity.
+
+\textbf{(1)} Given $x := (x_0,\ldots,x_3) \in k^4$ and $y :=
+(y_0,\ldots,y_3) \in k^4$, let us call $x\wedge y := (w_{0,1},
+w_{0,2}, w_{0,3}, w_{1,2}, w_{1,3}, w_{2,3}) \in k^6$ where $w_{i,j}
+:= x_i y_j - x_j y_i$.  What is $(\lambda x)\wedge(\mu y)$ in relation
+to $x\wedge y$?  Under what necessary and sufficient condition do we
+have $x\wedge y = 0$?  What is $x\wedge(\lambda x+\mu y)$ in relation
+to $x\wedge y$?
+
+\begin{answer}
+The $w_{i,j}$ are bilinear in $x,y$ (they are $2\times 2$
+determinants) so $(\lambda x)\wedge(\mu y) = \lambda\mu(x\wedge y)$.
+Vanishing of $w_{i,j}$ means $(x_i,x_j)$ is proportional to
+$(y_i,y_j)$ so vanishing of all the $w_{i,j}$ means precisely that
+$x$ or $y$ is zero or that $x$ and $y$ are collinear.  Again by
+bilinearity, we have $x\wedge(\lambda x+\mu y) = \lambda(x\wedge x) +
+\mu(x\wedge y)$ which is just $\mu(x\wedge y)$ since $x\wedge x$
+is $0$.
+\end{answer}
+
+\textbf{(2)} Show that if $V \subseteq k^4$ is a $2$-dimensional
+vector subspace, then the set of $x\wedge y$ for $x,y\in V$ is a
+$1$-dimensional subspace of $k^6$.
+
+\begin{answer}
+Consider $u,v$ a basis of $V$: then $u\wedge v$ is nonzero, and any
+element of $V$ can be written $\lambda u + \mu v$, and then $(\lambda
+u + \mu v) \wedge (\lambda' u + \mu' v) = (\lambda \mu' - \lambda'
+\mu)(u \wedge v)$ by (1) (or by composition of determinants).  In
+other words, any $x\wedge y$ with $x,y\in V$ is collinear to $u\wedge
+v$, and the latter is nonzero, and since $\lambda \mu' - \lambda' \mu$
+can obviously take every value in $k$, we see that $\{x\wedge y :
+x,y\in V\}$ is the line spanned by $u\wedge v$.
+\end{answer}
+
+\textbf{(3)} Deduce from (2) that if $L \subseteq \mathbb{P}^3(k)$ is
+a line, then $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0
+w_{1,2}{:}w_{1,3}{:}w_{2,3})$, where $w_{i,j} := x_i y_j - x_j y_i$ as
+above, and $(x_0{:}x_1{:}x_2{:}x_3)$ and $(y_0{:}y_1{:}y_2{:}y_3)$ are
+two distinct points on $L$, is a well-defined point in
+$\mathbb{P}^5(k)$, not depending on the chosen points on $L$ nor on
+the homogeneous coordinates representing them.
+
+\begin{answer}
+Calling $\langle w\rangle$ the point in projective space
+$\mathbb{P}^{m-1}(k)$ defined by a vector $w\neq 0$ in $k^m$, if $L$
+is a line in $\mathbb{P}^3(k)$ we can see it as $\{\langle v\rangle :
+v\in V\}$ for a $2$-dimensional vector subspace $V \subseteq k^4$, and
+we have seen that $\langle x\wedge y\rangle \in \mathbb{P}^4(k)$
+exists when $x$ and $y$ are not collinear (so that $x\wedge y \neq
+0$), i.e., when $\langle x\rangle \neq \langle y\rangle$, and does not
+depend on the $x,y \in V$.
+\end{answer}
+
+The $w_{i,j}$ in question are known as the \textbf{Plücker
+  coordinates} of $L$.
+
+\textbf{(4)} Show that any point $(z_0{:}z_1{:}z_2{:}z_3)$ on the
+line $L$ as above satisfies
+\[
+w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0 = 0
+\tag{$*$}
+\]
+— and analogously by replacing $0,1,2$ by three distinct coordinates
+in $\{0,1,2,3\}$.
+
+\begin{answer}
+Expanding the $3\times 3$ determinant
+\[
+\left|
+\begin{matrix}
+x_0&y_0&z_0\\
+x_1&y_1&z_1\\
+x_2&y_2&z_2\\
+\end{matrix}
+\right|
+\]
+gives $w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0$.  If $\langle
+z\rangle$ is in the line through $\langle x\rangle$ and $\langle
+y\rangle$, meaning the three vectors $x,y,z$ are linearly dependent,
+then this determinant is zero.  The same holds, of course, for any
+other choice of coordinates instead of $0,1,2$.
+\end{answer}
+
+\textbf{(5)} Deduce from (4) that the $w_{i,j}$ satisfy the following
+relation:
+\[
+w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + w_{0,3} w_{1,2} = 0
+\tag{$\dagger$}
+\]
+
+\begin{answer}
+By (4), we have $w_{0,1} x_2 - w_{0,2} x_1 + w_{1,2} x_0 = 0$ and
+$w_{0,1} y_2 - w_{0,2} y_1 + w_{1,2} y_0 = 0$.  Adding $y_3$ times the
+first and $-x_3$ times the second gives the stated
+relation ($\dagger$).
+\end{answer}
+
+The projective algebraic variety defined by ($\dagger$) in
+$\mathbb{P}^5$ is known as the \textbf{Plücker quadric}.  In other
+words, we have shown how to associate to any line $L$ in
+$\mathbb{P}^3(k)$ a $k$-point on the Plücker quadric.  We now consider
+the converse.
+
+\textbf{(6)} Assuming $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0
+w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$
+satisfies ($\dagger$) (viꝫ. belongs to the Plücker quadric), and
+assuming also that $w_{0,3} \neq 0$, show that the two points
+$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and
+$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ are meaningful and distinct, and
+that the line joining them has the Plücker coordinates
+$(w_{0,1}:\cdots:w_{2,3})$ that were given.  (\emph{Hint:}
+\underline{first} compute $(w_{0,3},w_{1,3},w_{2,3},0) \wedge
+(0,w_{0,1},w_{0,2},w_{0,3})$ and use the result, with the Plücker
+relation and the fact that $w_{0,3} \neq 0$ to conclude.)
+
+\begin{answer}
+We straightforwardly compute $(w_{0,3},w_{1,3},w_{2,3},0) \wedge
+(0,w_{0,1},w_{0,2},w_{0,3}) = (w_{0,3} w_{0,1}, w_{0,3} w_{0,2},
+w_{0,3}^2, \penalty0 w_{1,3} w_{0,2} - w_{2,3} w_{0,1}, w_{1,3}
+w_{0,3}, w_{2,3} w_{0,3})$.  By Plücker's relation ($\dagger$),
+$w_{1,3} w_{0,2} - w_{2,3} w_{0,1} = w_{1,2} w_{0,3}$, so we get
+$w_{0,3}$ times $(w_{0,1}, w_{0,2}, w_{0,3}, \penalty0 w_{1,2},
+w_{1,3}, w_{2,3})$.  Assuming $w_{0,3} \neq 0$, this is a nonzero
+vector, which implies (by question (1)) that
+$(w_{0,3},w_{1,3},w_{2,3},0)$ and $(0,w_{0,1},w_{0,2},w_{0,3})$ are
+nonzero and non-collinear, so that the two points
+$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and
+$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ are meaningful and distinct, and
+by the computation we have just done, the Plücker coordinates of the
+line joining them is the set of coordinates $(w_{0,1}:\cdots:w_{2,3})$
+that were given.
+\end{answer}
+
+\textbf{(7)} Deduce from (6) that any $(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:}
+\penalty0 w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$
+satisfying ($\dagger$) (viꝫ. belonging to the Plücker quadric) is the
+set of Plücker coordinates of a (clearly unique) line in
+$\mathbb{P}^3(k)$.  (\emph{Hint:} what needs to be proved is that the
+assumption $w_{0,3} \neq 0$ in (6) is harmless: explain how it can be
+arranged by a judicious permutation of coordinates.)
+
+\begin{answer}
+We have seen in (6) that when $w_{0,3} \neq 0$ then
+$(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:} \penalty0
+w_{1,2}{:}w_{1,3}{:}w_{2,3})$ are the Plücker coordinates of a line
+in $\mathbb{P}^3(k)$.  But all $w_{i,j}$ cannot be zero (as they are
+given in $\mathbb{P}^5$), and we can always permute coordinates in
+such a way that any $w_{i,j} \neq 0$, which is sure to exist, becomes
+$w_{0,3}$, and the formula $w_{0,1} w_{2,3} - w_{0,2} w_{1,3} +
+w_{0,3} w_{1,2} = 0$ is invariant under any permutation of coordinates
+(for this is is enough to check a cyclic permutation and a
+transposition; keep in mind that $w_{j,i} = -w_{i,j}$ when rewriting
+so as $i<j$), so we have confirmed the result in all cases.
+\end{answer}
+
+At this point, we have established a bijection between the set of
+lines $L$ in $\mathbb{P}^3(k)$ and the set of $k$-points in the
+Plücker quadric defined by ($\dagger$) in $\mathbb{P}^5$; we know how
+to compute Plücker coordinates from two distinct points lying on $L$
+(by definition).  We now wish to compute Plücker coordinates for a
+line defined as the the intersection of two plines.
+
+\textbf{(8)} Rephrase (4) to deduce that, if $L$ is a line with
+Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$, then the planes
+$[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ and $[0 : w_{2,3} : {-w_{1,3}}
+  : w_{1,2}]$ contain $L$.  Now consider these as points in the
+dual $\mathbb{P}^3$: show that the Plücker coordinates of the line
+$L^*$ joining the two points in question are: $[w_{2,3} : {-w_{1,3}} :
+  w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$, provided $w_{1,2} \neq
+0$.
+
+\begin{answer}
+The relation ($*$) of (4), namely $w_{1,2} z_0 - w_{0,2} z_1 + w_{0,1}
+z_2 = 0$, means precisely that $(z_0{:}z_1{:}z_2{:}z_3)$ is on the
+plane $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$.  Shifting coordinates
+cyclically, it is also on the plane $[0 : w_{2,3} : {-w_{1,3}} :
+  w_{1,2}]$.  Computing the Plücker coordinates as defined in
+questions (1) to (3) for the line through these two (dual) points
+gives $[w_{1,2} w_{2,3} : {-w_{1,2} w_{1,3}} : w_{1,2}^2 : {w_{0,2}
+    w_{1,3} - w_{0,1} w_{2,3}} : {-w_{0,2} w_{1,2}} : w_{0,1}
+  w_{1,2}]$ (provided not all are zero).  By Plücker's
+relation ($\dagger$), $w_{0,2} w_{1,3} - w_{0,1} w_{2,3} = w_{0,3}
+w_{1,2}$, and now we can divide all coordinates by $w_{1,2}$ if it is
+nonzero, giving $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
+  {-w_{0,2}} : w_{0,1}]$ (we already know that these six coordinates
+cannot vanish simultaneously, so $w_{1,2} \neq 0$ is the only
+condition we need).
+\end{answer}
+
+\textbf{(9)} Deduce from (8) that if $L$ is a line in
+$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1} : w_{0,2} :
+w_{0,3} : \penalty0 w_{1,2} : w_{1,3} : w_{2,3})$, and if $L^*$
+denotes the “dual” line in the dual $\mathbb{P}^3$, that is, the line
+consisting of all points corresponding to planes containing $L$, then
+$L^*$ has (dual) Plücker coordinates $[w_{2,3} : {-w_{1,3}} : w_{1,2}
+  : w_{0,3} : {-w_{0,2}} : w_{0,1}]$.  (\emph{Hint:} the only thing
+that needs to be proved is that the assumption $w_{1,2} \neq 0$ in (8)
+is harmless: explain how it can be arranged by a judicious permutation
+of coordinates.)
+
+\begin{answer}
+We have seen in (8) that when $w_{1,2} \neq 0$ then the line $L^*$
+dual to $L$ is given by $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
+  {-w_{0,2}} : w_{0,1}]$.  But all $w_{i,j}$ cannot be zero
+(question (3)), and we can always permute coordinates in such a way
+that any $w_{i,j} \neq 0$, which is sure to exist, becomes $w_{1,2}$,
+and the formula $(w_{0,1} : w_{0,2} : w_{0,3} : \penalty0 w_{1,2} :
+w_{1,3} : w_{2,3}) \mapsto [w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
+  {-w_{0,2}} : w_{0,1}]$ is covariant under any permutation of
+coordinates (for this is is enough to check a cyclic permutation and a
+transposition), so we have confirmed the result in all cases.
+\end{answer}
+
+\textbf{(10)} If $[u_0{:}u_1{:}u_2{:}u_3]$ and
+$[v_0{:}v_1{:}v_2{:}v_3]$ are distinct planes in $\mathbb{P}^3(k)$,
+how can we compute the Plücker coordinates of their line of
+intersection?  (\emph{Hint:} first describe the Plücker coordinates of
+the line $L^*$ joining the corresponding points in the “dual”
+$\mathbb{P}^3$, and then apply the result of (9), together with
+projective duality.)
+
+\begin{answer}
+Le line $L^*$ joining the points $[u_0{:}u_1{:}u_2{:}u_3]$ and
+$[v_0{:}v_1{:}v_2{:}v_3]$ of the dual $\mathbb{P}^3$ is given by the
+Plücker coordinates $w_{i,j} = u_i v_j - u_j v_i$.  We then obtain the
+Plücker coordinates of $L$ as $(w_{2,3} : {-w_{1,3}} : w_{1,2} :
+w_{0,3} : {-w_{0,2}} : w_{0,1})$ (the formula given in (9) for
+computing the Plücker coordinates of $L^*$ from those of $L$: it is
+involutive and projective duality ensures that it also computes the
+Plücker coordinates of $L$ from those of $L^*$).
+\end{answer}
+
+\textbf{(11)} Explain how, by expanding a $4\times 4$ determinant
+expressing the fact that four points $(x_0{:}x_1{:}x_2{:}x_3)$,
+$(y_0{:}y_1{:}y_2{:}y_3)$, $(z_0{:}z_1{:}z_2{:}z_3)$ and
+$(p_0{:}p_1{:}p_2{:}p_3)$ in $\mathbb{P}^3$ are coplanar, we can
+obtain a formula for the plane through a line $L$ defined by its
+Plücker coordinates and a point $(z_0{:}z_1{:}z_2{:}z_3)$ not situated
+on $L$.  (It is not required to go through the full computations.)
+
+\begin{answer}
+Consider the $4\times 4$ determinant
+\[
+\left|
+\begin{matrix}
+x_0&y_0&z_0&p_0\\
+x_1&y_1&z_1&p_1\\
+x_2&y_2&z_2&p_2\\
+x_3&y_3&z_3&p_3\\
+\end{matrix}
+\right|
+\]
+whose vanishing expresses the fact that $x,y,z,p$ are in a common
+hyperplane in $k^4$, i.e., that the points $\langle x\rangle$,
+$\langle y\rangle$, $\langle z\rangle$ and $\langle p\rangle$ are
+coplanar.  Expanding it with respect to the final column $p$ gives a
+linear condition for $p$ to be in the plane $P$ spanned by $\langle
+x\rangle$, $\langle y\rangle$, $\langle z\rangle$, whose coefficients
+are $3\times 3$ determinants, which are the coordinates of the
+plane $P$ in the dual $\mathbb{P}^3$.  Expanding those $3\times 3$
+determinants with respect to the final column $z$ writes them in terms
+of the Plücker coordinates of the line $L$ through $\langle x\rangle$,
+and $\langle y\rangle$, and the point $\langle z\rangle$.
+
+To be precise (although this was not asked), we get:
+\[
+\begin{aligned}
+\relax [\;
+& - w_{1,2} z_3 + w_{1,3} z_2 - w_{2,3} z_1 \\
+:\;
+& w_{2,3} z_0 + w_{0,2} z_3 - w_{0,3} z_2 \\
+:\;
+& w_{0,3} z_1 - w_{1,3} z_0 - w_{0,1} z_3 \\
+:\;
+& w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0
+\;]
+\end{aligned}
+\]
+(the last coordinate being precisely given by ($*$)).
+\end{answer}
+
+For your additional information: if $L$ and $L'$ are two lines in
+$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$
+and $(w'_{0,1}:\cdots:w'_{2,3})$ respectively, then $L$ and $L'$ meet
+in a common point (or equivalently, belong to a common plane)
+iff\footnote{This relation (the “polarization” of the quadratic
+  relation ($\dagger$)) can be interpreted by saying that the line
+  in $\mathbb{P}^5$ joining the two points $(w_{0,1}:\cdots:w_{2,3})$
+  and $(w'_{0,1}:\cdots:w'_{2,3})$ on the Plücker quadric is entirely
+  contained in said quadric.}
+\[
+w_{0,1} w'_{2,3} - w_{0,2} w'_{1,3} + w_{0,3} w'_{1,2}
++ w_{2,3} w'_{0,1} - w_{1,3} w'_{0,2} + w_{1,2} w'_{0,3} = 0
+\tag{$\ddagger$}
+\]
+(it is not required to prove this).
+
+\textbf{(12)} Briefly summarize all of the above, emphasizing how we
+have obtained formulæ allowing algorithmic computation of all possible
+geometric constructions between points, lines and planes
+in $\mathbb{P}^3$.
+
+\begin{answer}
+For projective subspaces in $\mathbb{P}^3$ we can represent:
+\begin{itemize}
+\item \textbf{points} by their homogeneous coordinates
+  $(x_0{:}x_1{:}x_2{:}x_3)$ (which are arbitrary not all zero, defined
+  up to a common multiplicative constant),
+\item \textbf{lines} by their Plücker coordinates
+  $(w_{0,1} : w_{0,2} : w_{0,3}  :  \penalty0
+  w_{1,2} : w_{1,3} : w_{2,3})$ (which are not all zero, and subject
+  to the sole condition ($*$) of belonging to the Plücker quadric,
+  defined up to a common multiplicative constant), or equivalently by
+  their dual Plücker coordinates which are the same up to a
+  permutation and some changes of sign, $[w_{2,3} : {-w_{1,3}} :
+    w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$,
+\item \textbf{planes} by their dual coordinates
+  $[u_0{:}u_1{:}u_2{:}u_3]$ (which are arbitrary not all zero, defined
+  up to a common multiplicative constant).
+\end{itemize}
 
+We can then compute:
+\begin{itemize}
+\item whether a point lies on a line by checking the relation ($*$)
+  and all its permutations of coordinates (cyclic permutations
+  suffice),
+\item whether a point lies on a plane by the relation $u_0 x_0 +
+  \cdots + u_3 x_3 = 0$,
+\item whether a line lies in a plane by checking whether the dual
+  point of the plane lies on the dual line (this gives $w_{2,3} u_2 +
+  w_{1,3} u_1 + w_{0,3} u_0 = 0$ and cyclic permutations),
+\item the line joining two distinct points by computing the Plücker
+  coordinates as $2\times 2$ determinants as defined in question (1),
+\item the plane though a line and a point not lying on it by the
+  formula found as explained in question (11),
+\item the intersection line of two distinct planes by computing dual
+  Plücker coordinates as explained in question (10),
+\item the point of intersection of a line and a plane by taking the
+  dual of the formula for the plane through a line and a point,
+\item a sample point on a line as $(w_{0,3} : w_{1,3} : w_{2,3} : 0)$
+  or some coordinate permutation thereof (as shown in question (6)),
+\item a sample plane through a line dually to the previous point,
+  namely $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ (as shown in
+  question (8)),
+\item whether two lines meet, or equivalently, belong to a common
+  plane, by using the criterion stated before (12), in which case
+  their point of intersection can be computed by intersecting one of
+  the lines with a sample plane through the other (as explained in the
+  previous items), and their common plane can be computed dually.
+\end{itemize}
+\end{answer}
 
 
 
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