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Retour au chapitre
Plan, produit scalaire, orthogonalité et distance dans l'espace
Exercices types :
1
1
1
ère
partie - Exercice 3
5 min
10
On considère deux vecteurs de l'espace
u
→
\overrightarrow{u}
u
et
v
→
\overrightarrow{v}
v
tels que :
∥
u
→
∥
=
4
\left\| \overrightarrow{u} \right\| =4
∥
∥
u
∥
∥
=
4
;
∥
v
→
∥
=
5
\left\| \overrightarrow{v} \right\| =5
∥
∥
v
∥
∥
=
5
et
∥
u
→
+
v
→
∥
=
8
\left\| \overrightarrow{u} +\overrightarrow{v} \right\| =8
∥
∥
u
+
v
∥
∥
=
8
Question 1
Calculer
u
→
⋅
v
→
\overrightarrow{u} \cdot \overrightarrow{v}
u
⋅
v
Correction
Soient deux vecteurs
u
→
\overrightarrow{u}
u
et
v
→
\overrightarrow{v}
v
alors :
∥
u
→
+
v
→
∥
2
=
∥
u
→
∥
2
+
2
u
→
⋅
v
→
+
∥
v
→
∥
2
\left\| \overrightarrow{u} +\overrightarrow{v} \right\|^{2} =\left\| \overrightarrow{u} \right\| ^{2} +2\overrightarrow{u} \cdot \overrightarrow{v} +\left\| \overrightarrow{v} \right\| ^{2}
∥
∥
u
+
v
∥
∥
2
=
∥
∥
u
∥
∥
2
+
2
u
⋅
v
+
∥
∥
v
∥
∥
2
∥
u
→
+
v
→
∥
2
=
∥
u
→
∥
2
+
2
u
→
⋅
v
→
+
∥
v
→
∥
2
\left\| \overrightarrow{u} +\overrightarrow{v} \right\|^{2} =\left\| \overrightarrow{u} \right\| ^{2} +2\overrightarrow{u} \cdot \overrightarrow{v} +\left\| \overrightarrow{v} \right\| ^{2}
∥
∥
u
+
v
∥
∥
2
=
∥
∥
u
∥
∥
2
+
2
u
⋅
v
+
∥
∥
v
∥
∥
2
équivaut successivement à :
8
2
=
4
2
+
2
u
→
⋅
v
→
+
5
2
8^{2} =4^{2} +2\overrightarrow{u} \cdot \overrightarrow{v} +5^{2}
8
2
=
4
2
+
2
u
⋅
v
+
5
2
64
=
16
+
2
u
→
⋅
v
→
+
25
64=16+2\overrightarrow{u} \cdot \overrightarrow{v} +25
64
=
16
+
2
u
⋅
v
+
25
64
−
16
−
25
=
2
u
→
⋅
v
→
64-16-25=2\overrightarrow{u} \cdot \overrightarrow{v}
64
−
16
−
25
=
2
u
⋅
v
23
=
2
u
→
⋅
v
→
23=2\overrightarrow{u} \cdot \overrightarrow{v}
23
=
2
u
⋅
v
Il vient alors que :
u
→
⋅
v
→
=
23
2
\overrightarrow{u} \cdot \overrightarrow{v}=\frac{23}{2}
u
⋅
v
=
2
23