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Produit scalaire
Propriétés de calculs du produit scalaire : symétrie, bilinéarité - Exercice 4
5 min
15
Soient
u
→
\overrightarrow{u}
u
et
v
→
\overrightarrow{v}
v
deux vecteurs tels que :
∥
u
→
∥
=
6
\left\| \overrightarrow{u} \right\| =6
∥
∥
u
∥
∥
=
6
;
∥
v
→
∥
=
5
\left\| \overrightarrow{v} \right\| =5
∥
∥
v
∥
∥
=
5
et
u
→
⋅
v
→
=
−
3
\overrightarrow{u} \cdot \overrightarrow{v} =-3
u
⋅
v
=
−
3
Question 1
Calculer
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)
(
u
−
v
)
⋅
(
3
u
+
2
v
)
Correction
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
3
u
→
⋅
u
→
+
2
u
→
⋅
v
→
−
3
v
→
⋅
u
→
−
2
v
→
⋅
v
→
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=3\overrightarrow{u} \cdot \overrightarrow{u} +2\overrightarrow{u} \cdot \overrightarrow{v} -3\overrightarrow{v} \cdot \overrightarrow{u} -2\overrightarrow{v} \cdot \overrightarrow{v}
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
3
u
⋅
u
+
2
u
⋅
v
−
3
v
⋅
u
−
2
v
⋅
v
Soient deux vecteurs
u
→
\overrightarrow{u}
u
et
v
→
\overrightarrow{v}
v
. Le produit scalaire est symétrique alors :
u
→
⋅
v
→
=
v
→
⋅
u
→
\overrightarrow{u} \cdot \overrightarrow{v} =\overrightarrow{v} \cdot \overrightarrow{u}
u
⋅
v
=
v
⋅
u
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
3
u
→
⋅
u
→
+
2
u
→
⋅
v
→
−
3
u
→
⋅
v
→
−
2
v
→
⋅
v
→
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=3\overrightarrow{u} \cdot \overrightarrow{u} +2\overrightarrow{u} \cdot \overrightarrow{v} -3\overrightarrow{u} \cdot \overrightarrow{v} -2\overrightarrow{v} \cdot \overrightarrow{v}
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
3
u
⋅
u
+
2
u
⋅
v
−
3
u
⋅
v
−
2
v
⋅
v
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
3
u
→
⋅
u
→
−
u
→
⋅
v
→
−
2
v
→
⋅
v
→
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=3\overrightarrow{u} \cdot \overrightarrow{u} -\overrightarrow{u} \cdot \overrightarrow{v} -2\overrightarrow{v} \cdot \overrightarrow{v}
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
3
u
⋅
u
−
u
⋅
v
−
2
v
⋅
v
Soient un vecteur
u
→
\overrightarrow{u}
u
alors :
u
→
⋅
u
→
=
∥
u
→
∥
2
\overrightarrow{u} \cdot \overrightarrow{u} =\left\| \overrightarrow{u} \right\| ^{2}
u
⋅
u
=
∥
∥
u
∥
∥
2
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
3
∥
u
→
∥
2
−
u
→
⋅
v
→
−
2
∥
v
→
∥
2
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=3\left\| \overrightarrow{u} \right\| ^{2} -\overrightarrow{u} \cdot \overrightarrow{v} -2\left\| \overrightarrow{v} \right\| ^{2}
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
3
∥
∥
u
∥
∥
2
−
u
⋅
v
−
2
∥
∥
v
∥
∥
2
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
3
×
6
2
−
(
−
3
)
−
2
×
5
2
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=3\times 6^{2} -\left(-3\right)-2\times 5^{2}
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
3
×
6
2
−
(
−
3
)
−
2
×
5
2
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
3
×
36
+
3
−
2
×
25
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=3\times 36+3-2\times 25
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
3
×
36
+
3
−
2
×
25
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
108
+
3
−
50
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=108+3-50
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
108
+
3
−
50
Ainsi :
(
u
→
−
v
→
)
⋅
(
3
u
→
+
2
v
→
)
=
61
\left(\overrightarrow{u} -\overrightarrow{v} \right)\cdot \left(3\overrightarrow{u} +2\overrightarrow{v} \right)=61
(
u
−
v
)
⋅
(
3
u
+
2
v
)
=
61