Géométrie vectorielle, droites et plans de l'espace

Montrer que 4 points sont coplanaires - Exercice 1

8 min
25
On donne les points A(1;4;−8)A\left(1;4;-8\right) , B(2;6;−12)B\left(2;6;-12\right) , C(5;1;9)C\left(5;1;9\right) et D(8;9;−9)D\left(8;9;-9\right).
Question 1

Les quatre points sont-ils coplanaires ?

Correction
Les points A,B,CA,B,C et DD sont coplanaires s'il existe deux réels aa et bb tels que : AB→=aAC→+bAD→\overrightarrow{AB} =a\overrightarrow{AC} +b\overrightarrow{AD}
Calculons maintenant les vecteurs : AB→,AC→\overrightarrow{AB} ,\overrightarrow{AC} et AD→\overrightarrow{AD} .
AB→(12−4)\overrightarrow{AB} \left(\begin{array}{c} {1} \\ {2} \\ {-4} \end{array}\right) , AC→(4−317)\overrightarrow{AC} \left(\begin{array}{c} {4} \\ {-3} \\ {17} \end{array}\right) et AD→(75−1)\overrightarrow{AD} \left(\begin{array}{c} {7} \\ {5} \\ {-1} \end{array}\right)
AB→=aAC→+bAD→⇔(12−4)=a(4−317)+b(75−1)\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)
AB→=aAC→+bAD→⇔(12−4)=(4a−3a17a)+(7b5b−b)\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)
AB→=aAC→+bAD→⇔{4a+7b=1−3a+5b=217a−b=−4\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 .
{4a+7b=1−3a+5b=2−b=−4−17a\left\{\begin{array}{ccccc} {4a} & {+} & {7b} & {=} & {1} \\ {-3a} & {+} & {5b} & {=} & {2} \\ {} & {-} & {b} & {=} & {-4-17a} \end{array}\right.
{4a+7b=1−3a+5b=2b=4+17a\left\{\begin{array}{ccccc} {4a} & {+} & {7b} & {=} & {1} \\ {-3a} & {+} & {5b} & {=} & {2} \\ {} & {} & {b} & {=} & {4+17a} \end{array}\right.
{4a+7×(4+17a)=1−3a+5×(4+17a)=2b=4+17a\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.
{4a+28+119a=1−3a+20+85a=2b=4+17a\left\{\begin{array}{ccccc} {4a} & {+} & {28+119a} & {=} & {1} \\ {-3a} & {+} & {20+85a} & {=} & {2} \\ {} & {} & {b} & {=} & {4+17a} \end{array}\right.
{123a=1−2882a=2−20b=4+17a\left\{\begin{array}{ccc} {123a} & {=} & {1-28} \\ {82a} & {=} & {2-20} \\ {b} & {=} & {4+17a} \end{array}\right.
{123a=−2782a=−18b=4+17a\left\{\begin{array}{ccc} {123a} & {=} & {-27} \\ {82a} & {=} & {-18} \\ {b} & {=} & {4+17a} \end{array}\right.
{a=−27123a=−1882b=4+17a\left\{\begin{array}{ccc} {a} & {=} & {\frac{-27}{123} } \\ {a} & {=} & {\frac{-18}{82} } \\ {b} & {=} & {4+17a} \end{array}\right.
{a=−941a=−941b=4+17a\left\{\begin{array}{ccc} {a} & {=} & {-\frac{9}{41} } \\ {a} & {=} & {-\frac{9}{41} } \\ {b} & {=} & {4+17a} \end{array}\right.
{a=−941b=4+17×(−941)\left\{\begin{array}{ccc} {a} & {=} & {-\frac{9}{41} } \\ {b} & {=} & {4+17\times \left(-\frac{9}{41} \right)} \end{array}\right.
{a=−941b=1141\left\{\begin{array}{ccc} {a} & {=} & {-\frac{9}{41} } \\ {b} & {=} & {\frac{11}{41} } \end{array}\right.
Il en résulte que AB→=−941AC→+1141AD→\overrightarrow{AB} =-\frac{9}{41}\overrightarrow{AC} +\frac{11}{41}\overrightarrow{AD} .
Les points A,B,CA,B,C et DD sont donc coplanaires.