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Plan, produit scalaire, orthogonalité et distance dans l'espace
Propriétés de calculs du produit scalaire : symétrie, bilinéarité - Exercice 3
5 min
15
Soient
∥
u
→
∥
\left\| \overrightarrow{u} \right\|
∥
∥
u
∥
∥
et
∥
v
→
∥
\left\| \overrightarrow{v} \right\|
∥
∥
v
∥
∥
tels que :
∥
u
→
∥
=
5
4
\left\| \overrightarrow{u} \right\| =\frac{5}{4}
∥
∥
u
∥
∥
=
4
5
;
∥
v
→
∥
=
2
\left\| \overrightarrow{v} \right\| =2
∥
∥
v
∥
∥
=
2
et
u
→
⋅
v
→
=
3
5
\overrightarrow{u} \cdot \overrightarrow{v} =\frac{3}{5}
u
⋅
v
=
5
3
Question 1
Calculer
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
Correction
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
8
u
→
⋅
u
→
−
6
u
→
⋅
v
→
+
4
v
→
⋅
u
→
−
3
v
→
⋅
v
→
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=8\overrightarrow{u} \cdot \overrightarrow{u} -6\overrightarrow{u} \cdot \overrightarrow{v} +4\overrightarrow{v} \cdot \overrightarrow{u} -3\overrightarrow{v} \cdot \overrightarrow{v}
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
8
u
⋅
u
−
6
u
⋅
v
+
4
v
⋅
u
−
3
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
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
8
u
→
⋅
u
→
−
6
u
→
⋅
v
→
+
4
u
→
⋅
v
→
−
3
v
→
⋅
v
→
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=8\overrightarrow{u} \cdot \overrightarrow{u} -6\overrightarrow{u} \cdot \overrightarrow{v} +4\overrightarrow{u} \cdot \overrightarrow{v} -3\overrightarrow{v} \cdot \overrightarrow{v}
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
8
u
⋅
u
−
6
u
⋅
v
+
4
u
⋅
v
−
3
v
⋅
v
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
8
u
→
⋅
u
→
−
2
u
→
⋅
v
→
−
3
v
→
⋅
v
→
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=8\overrightarrow{u} \cdot \overrightarrow{u} -2\overrightarrow{u} \cdot \overrightarrow{v} -3\overrightarrow{v} \cdot \overrightarrow{v}
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
8
u
⋅
u
−
2
u
⋅
v
−
3
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
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
8
∥
u
→
∥
2
−
2
u
→
⋅
v
→
−
3
∥
v
→
∥
2
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=8\left\| \overrightarrow{u} \right\| ^{2} -2\overrightarrow{u} \cdot \overrightarrow{v} -3\left\| \overrightarrow{v} \right\| ^{2}
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
8
∥
∥
u
∥
∥
2
−
2
u
⋅
v
−
3
∥
∥
v
∥
∥
2
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
8
×
(
5
4
)
2
−
2
×
3
5
−
3
×
2
2
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=8\times \left(\frac{5}{4} \right)^{2} -2\times \frac{3}{5} -3\times 2^{2}
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
8
×
(
4
5
)
2
−
2
×
5
3
−
3
×
2
2
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
8
×
25
16
−
2
×
3
5
−
3
×
4
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=8\times \frac{25}{16} -2\times \frac{3}{5} -3\times 4
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
8
×
16
25
−
2
×
5
3
−
3
×
4
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
25
2
−
6
5
−
12
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=\frac{25}{2} -\frac{6}{5} -12
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
2
25
−
5
6
−
12
Ainsi :
(
2
u
→
+
v
→
)
⋅
(
4
u
→
−
3
v
→
)
=
−
7
10
\left(2\overrightarrow{u} +\overrightarrow{v} \right)\cdot \left(4\overrightarrow{u} -3\overrightarrow{v} \right)=-\frac{7}{10}
(
2
u
+
v
)
⋅
(
4
u
−
3
v
)
=
−
10
7