Montrer que 4 points sont coplanaires

Calculons maintenant les vecteurs : $\overrightarrow{AB} ,\overrightarrow{AC}$ et $\overrightarrow{AD}$.
$\overrightarrow{AB} \left(\begin{array}{c} {1} \\ {2} \\ {-4} \end{array}\right)$ , $\overrightarrow{AC} \left(\begin{array}{c} {4} \\ {-3} \\ {17} \end{array}\right)$ et $\overrightarrow{AD} \left(\begin{array}{c} {7} \\ {5} \\ {-1} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {1} \\ {2} \\ {-4} \end{array}\right)=a\left(\begin{array}{c} {4} \\ {-3} \\ {17} \end{array}\right)+b\left(\begin{array}{c} {7} \\ {5} \\ {-1} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {1} \\ {2} \\ {-4} \end{array}\right)=\left(\begin{array}{c} {4a} \\ {-3a} \\ {17a} \end{array}\right)+\left(\begin{array}{c} {7b} \\ {5b} \\ {-b} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left\{\begin{array}{ccccc} {4a} & {+} & {7b} & {=} & {1} \\ {-3a} & {+} & {5b} & {=} & {2} \\ {17a} & {-} & {b} & {=} & {-4} \end{array}\right.$ $\;\;$ Il nous faut donc résoudre ce système linéaire .
$\left\{\begin{array}{ccccc} {4a} & {+} & {7b} & {=} & {1} \\ {-3a} & {+} & {5b} & {=} & {2} \\ {} & {-} & {b} & {=} & {-4-17a} \end{array}\right.$
$\left\{\begin{array}{ccccc} {4a} & {+} & {7b} & {=} & {1} \\ {-3a} & {+} & {5b} & {=} & {2} \\ {} & {} & {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccccc} {4a} & {+} & {7\times \left(4+17a\right)} & {=} & {1} \\ {-3a} & {+} & {5\times \left(4+17a\right)} & {=} & {2} \\ {} & {} & {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccccc} {4a} & {+} & {28+119a} & {=} & {1} \\ {-3a} & {+} & {20+85a} & {=} & {2} \\ {} & {} & {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccc} {123a} & {=} & {1-28} \\ {82a} & {=} & {2-20} \\ {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccc} {123a} & {=} & {-27} \\ {82a} & {=} & {-18} \\ {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccc} {a} & {=} & {\frac{-27}{123} } \\ {a} & {=} & {\frac{-18}{82} } \\ {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccc} {a} & {=} & {-\frac{9}{41} } \\ {a} & {=} & {-\frac{9}{41} } \\ {b} & {=} & {4+17a} \end{array}\right.$
$\left\{\begin{array}{ccc} {a} & {=} & {-\frac{9}{41} } \\ {b} & {=} & {4+17\times \left(-\frac{9}{41} \right)} \end{array}\right.$
$\left\{\begin{array}{ccc} {a} & {=} & {-\frac{9}{41} } \\ {b} & {=} & {\frac{11}{41} } \end{array}\right.$
Il en résulte que $\overrightarrow{AB} =-\frac{9}{41}\overrightarrow{AC} +\frac{11}{41}\overrightarrow{AD}$.
Les points $A,B,C$ et $D$ sont donc coplanaires.

Calculons maintenant les vecteurs : $\overrightarrow{AB} ,\overrightarrow{AC}$ et $\overrightarrow{AD}$.
$\overrightarrow{AB} \left(\begin{array}{c} {1} \\ {2} \\ {-3} \end{array}\right)$ , $\overrightarrow{AC} \left(\begin{array}{c} {2} \\ {-1} \\ {-1} \end{array}\right)$ et $\overrightarrow{AD} \left(\begin{array}{c} {-1} \\ {1} \\ {0} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {1} \\ {2} \\ {-3} \end{array}\right)=a\left(\begin{array}{c} {2} \\ {-1} \\ {-1} \end{array}\right)+b\left(\begin{array}{c} {-1} \\ {1} \\ {0} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {1} \\ {2} \\ {-3} \end{array}\right)=\left(\begin{array}{c} {2a} \\ {-a} \\ {-a} \end{array}\right)+\left(\begin{array}{c} {-b} \\ {b} \\ {0} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left\{\begin{array}{ccccc} {2a} & {-} & {b} & {=} & {1} \\ {-a} & {+} & {b} & {=} & {2} \\ {-a} & {} & {} & {=} & {-3} \end{array}\right. \Leftrightarrow \left\{\begin{array}{ccccc} {2a} & {-} & {b} & {=} & {1} \\ {-a} & {+} & {b} & {=} & {2} \\ {a} & {} & {} & {=} & {3} \end{array}\right.$
Comme $a=3$, on substitue $a$ dans les deux premières lignes et l'on doit obtenir les deux mêmes valeurs pour $b$ .
Cela donne :
$\left\{\begin{array}{ccccc} {2a} & {-} & {b} & {=} & {1} \\ {-a} & {+} & {b} & {=} & {2} \\ {a} & {} & {} & {=} & {3} \end{array}\right.$
D'où : $\left\{\begin{array}{ccc} {b} & {=} & {5} \\ {b} & {=} & {5} \\ {a} & {=} & {3} \end{array}\right.$
Il en résulte que $\overrightarrow{AB} =3\overrightarrow{AC} +5\overrightarrow{AD}$.
Les points $A,B,C$ et $D$ sont donc coplanaires.

Calculons maintenant les vecteurs : $\overrightarrow{AB} ,\overrightarrow{AC}$ et $\overrightarrow{AD}$.
$\overrightarrow{AB} \left(\begin{array}{c} {2} \\ {1} \\ {1} \end{array}\right)$ , $\overrightarrow{AC} \left(\begin{array}{c} {-3} \\ {1} \\ {0} \end{array}\right)$ et $\overrightarrow{AD} \left(\begin{array}{c} {7} \\ {1} \\ {2} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {2} \\ {1} \\ {1} \end{array}\right)=a\left(\begin{array}{c} {-3} \\ {1} \\ {0} \end{array}\right)+b\left(\begin{array}{c} {7} \\ {1} \\ {2} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {2} \\ {1} \\ {1} \end{array}\right)=\left(\begin{array}{c} {-3a} \\ {a} \\ {0} \end{array}\right)+\left(\begin{array}{c} {7b} \\ {b} \\ {2b} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left\{\begin{array}{ccccc} {-3a} & {+} & {7b} & {=} & {2} \\ {a} & {+} & {b} & {=} & {1} \\ {} & {} & {2b} & {=} & {1} \end{array}\right. \Leftrightarrow \left\{\begin{array}{ccccc} {-3a} & {+} & {7b} & {=} & {2} \\ {a} & {+} & {b} & {=} & {1} \\ {} & {} & {b} & {=} & {\frac{1}{2}} \end{array}\right.$
Comme $b=\frac{1}{2}$, on substitue $b$ dans les deux premières lignes et l'on doit obtenir les deux mêmes valeurs pour $a$ .
Cela donne :
$\left\{\begin{array}{ccccc} {-3a} & {+} & {7\times \frac{1}{2} } & {=} & {2} \\ {a} & {+} & {\frac{1}{2} } & {=} & {1} \\ {} & {} & {b} & {=} & {\frac{1}{2} } \end{array}\right.$
$\left\{\begin{array}{ccccc} {-3a} & {} & {} & {=} & {2-\frac{7}{2} } \\ {a} & {} & {} & {=} & {1-\frac{1}{2} } \\ {} & {} & {b} & {=} & {\frac{1}{2} } \end{array}\right.$
$\left\{\begin{array}{ccccc} {-3a} & {} & {} & {=} & {-\frac{3}{2} } \\ {a} & {} & {} & {=} & {\frac{1}{2} } \\ {} & {} & {b} & {=} & {\frac{1}{2} } \end{array}\right.$
$\left\{\begin{array}{ccccc} {a} & {} & {} & {=} & {-\frac{3}{2} \div \left(-3\right)} \\ {a} & {} & {} & {=} & {\frac{1}{2} } \\ {} & {} & {b} & {=} & {\frac{1}{2} } \end{array}\right.$
$\left\{\begin{array}{ccccc} {a} & {} & {} & {=} & {\frac{1}{2} } \\ {a} & {} & {} & {=} & {\frac{1}{2} } \\ {} & {} & {b} & {=} & {\frac{1}{2} } \end{array}\right.$
Il en résulte que $\overrightarrow{AB} =\frac{1}{2}\overrightarrow{AC} +\frac{1}{2}\overrightarrow{AD}$.
Les points $A,B,C$ et $D$ sont donc coplanaires.

Calculons maintenant les vecteurs : $\overrightarrow{AB} ,\overrightarrow{AC}$ et $\overrightarrow{AD}$.
$\overrightarrow{AB} \left(\begin{array}{c} {-2} \\ {-1} \\ {1} \end{array}\right)$ , $\overrightarrow{AC} \left(\begin{array}{c} {0} \\ {-2} \\ {-1} \end{array}\right)$ et $\overrightarrow{AD} \left(\begin{array}{c} {2} \\ {-3} \\ {-3} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {-2} \\ {-1} \\ {1} \end{array}\right)=a\left(\begin{array}{c} {0} \\ {-2} \\ {-1} \end{array}\right)+b\left(\begin{array}{c} {2} \\ {-3} \\ {-3} \end{array}\right)$
$\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD} \Leftrightarrow \left(\begin{array}{c} {-2} \\ {-1} \\ {1} \end{array}\right)=\left(\begin{array}{c} {0} \\ {-2a} \\ {-a} \end{array}\right)+\left(\begin{array}{c} {2b} \\ {-3b} \\ {-3b} \end{array}\right)$
On obtient le système suivant :
$\left\{\begin{array}{ccccc} {} & {} & {2b} & {=} & {-2} \\ {-2a} & {-} & {3b} & {=} & {-1} \\ {-a} & {-} & {3b} & {=} & {1} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {\frac{-2}{2} } \\ {-2a} & {-} & {3b} & {=} & {-1} \\ {-a} & {-} & {3b} & {=} & {1} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {-1} \\ {-2a} & {-} & {3\times \left(-1\right)} & {=} & {-1} \\ {-a} & {-} & {3\times \left(-1\right)} & {=} & {1} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {-1} \\ {-2a} & {+} & {3} & {=} & {-1} \\ {-a} & {+} & {3} & {=} & {1} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {-1} \\ {-2a} & {} & {} & {=} & {-1-3} \\ {-a} & {} & {} & {=} & {1-3} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {-1} \\ {-2a} & {} & {} & {=} & {-4} \\ {-a} & {} & {} & {=} & {-2} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {-1} \\ {a} & {} & {} & {=} & {\frac{-4}{-2} } \\ {a} & {} & {} & {=} & {2} \end{array}\right.$
$\left\{\begin{array}{ccccc} {} & {} & {b} & {=} & {-1} \\ {a} & {} & {} & {=} & {2} \\ {a} & {} & {} & {=} & {2} \end{array}\right.$
Il en résulte que $\overrightarrow{AB} =2\overrightarrow{AC} -\overrightarrow{AD}$.
Les points $A,B,C$ et $D$ sont donc coplanaires.