13.11 1) Vérifions que les points A, B,C etD sont non coplanaires.
−−−−→
AB =
−1 2 0
−−−−→
AC =
−3 2
−1
−−−−→
AD =
−2
−1
−3
−1 −3 −2 2 2 −1 0 −1 −3
= 6 + 4 + 0−0−(−1)−18 =−76= 0
Par conséquent, les pointsA,B,C etD forment bien un tétraèdre.
2) aire ABC = 1
2k−AB−−−→×−AC−−−→k= 1 2
−1 2 0
×
−3 2
−1
= 1 2
−2
−1 4
=
= 1 2
p(−2)2+ (−1)2+ 42 = 1 2
√21 =
√21 2
3) aire ABD = 1
2k−AB−−−→×−BD−−−→k= 1 2
−1 2 0
×
−1
−3
−3
= 1 2
−6
−3 5
=
= 1 2
p(−6)2+ (−3)2+ 52 = 1 2
√70 =
√70 2
4) aire ACD = 1
2k−AC−−−→×−CD−−−→k= 1 2
−3 2
−1
×
1
−3
−2
= 1 2
−7
−7 7
=
= 1 2
−7
1 1
−1
= 1 2|7|
1 1
−1
= 1 2·7p
12+ 12+ (−1)2 =
= 7 2
√3 = 7√ 3 2
5) aire BCD = 1
2k−BC−−−→×−BD−−−→k= 1 2
−2 0
−1
×
−1
−3
−3
= 1 2
−3
−5 6
=
= 1 2
p(−3)2+ (−5)2+ 62 = 1 2
√70 =
√70 2 6) L’aire totale du tétraèdre ABCD vaut :
√21 2 +
√70
2 + 7√ 3 2 +
√70 2 =
√21 2 +√
70 + 7√ 3 2 .
Géométrie : produit vectoriel Corrigé 13.11
