1. Définition avec les normes et l’angle
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Exercice 1 — ⭐
∥ u ⃗ ∥ = 5 \|\vec u\| = 5 ∥ u ∥ = 5 ∥ v ⃗ ∥ = 4 \|\vec v\| = 4 ∥ v ∥ = 4 u ⃗ ⋅ v ⃗ \vec u \cdot \vec v u ⋅ v ( u ⃗ , v ⃗ ) (\vec u, \vec v) ( u , v ) π 3 \dfrac\pi3 3 π 3 π 4 \dfrac{3\pi}{4} 4 3 π π 2 \dfrac\pi2 2 π A B C ABC A B C 6 6 6 A B → ⋅ A C → \overrightarrow{AB} \cdot \overrightarrow{AC} A B ⋅ A C B A → ⋅ B C → \overrightarrow{BA} \cdot \overrightarrow{BC} B A ⋅ B C A B → ⋅ C A → \overrightarrow{AB} \cdot \overrightarrow{CA} A B ⋅ C A
Voir le corrigé
a) 5 × 4 × 1 2 = 10 5 \times 4 \times \dfrac12 = 10 5 × 4 × 2 1 = 10 20 × ( − 2 2 ) = − 10 2 20 \times \left(-\dfrac{\sqrt2}{2}\right) = -10\sqrt2 20 × ( − 2 2 ) = − 10 2 20 × 0 = 0 20 \times 0 = 0 20 × 0 = 0
A B → ⋅ A C → = 6 × 6 × cos 60 ° = 18 \overrightarrow{AB} \cdot \overrightarrow{AC} = 6 \times 6 \times \cos 60° = 18 A B ⋅ A C = 6 × 6 × cos 60° = 18 B A → ⋅ B C → = 18 \overrightarrow{BA} \cdot \overrightarrow{BC} = 18 B A ⋅ B C = 18 C A → = − A C → \overrightarrow{CA} = -\overrightarrow{AC} C A = − A C A B → ⋅ C A → = − A B → ⋅ A C → = − 18 \overrightarrow{AB} \cdot \overrightarrow{CA} = -\overrightarrow{AB} \cdot \overrightarrow{AC} = -18 A B ⋅ C A = − A B ⋅ A C = − 18
2. Avec la projection orthogonale
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Exercice 2 — ⭐⭐
A B C D ABCD A B C D 3 3 3 I I I [ B C ] [BC] [ B C ]
A B → ⋅ A I → \overrightarrow{AB} \cdot \overrightarrow{AI} A B ⋅ A I A D → ⋅ A I → \overrightarrow{AD} \cdot \overrightarrow{AI} A D ⋅ A I A B → ⋅ C I → \overrightarrow{AB} \cdot \overrightarrow{CI} A B ⋅ C I
Voir le corrigé
Le projeté de I I I ( A B ) (AB) ( A B ) B B B A B → ⋅ A I → = A B × A B = 9 \overrightarrow{AB} \cdot \overrightarrow{AI} = AB \times AB = 9 A B ⋅ A I = A B × A B = 9
Le projeté de I I I ( A D ) (AD) ( A D ) K K K [ A D ] [AD] [ A D ] A K = B I = 1,5 AK = BI = 1{,}5 A K = B I = 1 , 5 A K → \overrightarrow{AK} A K A D → \overrightarrow{AD} A D A D → ⋅ A I → = 3 × 1,5 = 4,5 \overrightarrow{AD} \cdot \overrightarrow{AI} = 3 \times 1{,}5 = 4{,}5 A D ⋅ A I = 3 × 1 , 5 = 4 , 5
C I → \overrightarrow{CI} C I ( B C ) (BC) ( B C ) ( A B ) (AB) ( A B ) A B → ⋅ C I → = 0 \overrightarrow{AB} \cdot \overrightarrow{CI} = 0 A B ⋅ C I = 0
3. Orthogonalité
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Exercice 3 — ⭐⭐⭐
A B C D ABCD A B C D 3 3 3 I I I [ B C ] [BC] [ B C ] J J J [ C D ] [CD] [ C D ]
Justifie que A I → = A B → + 1 2 A D → \overrightarrow{AI} = \overrightarrow{AB} + \dfrac12\overrightarrow{AD} A I = A B + 2 1 A D B J → = A D → − 1 2 A B → \overrightarrow{BJ} = \overrightarrow{AD} - \dfrac12\overrightarrow{AB} B J = A D − 2 1 A B
Calcule A I → ⋅ B J → \overrightarrow{AI} \cdot \overrightarrow{BJ} A I ⋅ B J ( A I ) (AI) ( A I ) ( B J ) (BJ) ( B J )
Voir le corrigé
A I → = A B → + B I → = A B → + 1 2 B C → = A B → + 1 2 A D → \overrightarrow{AI} = \overrightarrow{AB} + \overrightarrow{BI} = \overrightarrow{AB} + \dfrac12\overrightarrow{BC} = \overrightarrow{AB} + \dfrac12\overrightarrow{AD} A I = A B + B I = A B + 2 1 B C = A B + 2 1 A D B C → = A D → \overrightarrow{BC} = \overrightarrow{AD} B C = A D B J → = B C → + C J → = A D → + 1 2 C D → = A D → − 1 2 A B → \overrightarrow{BJ} = \overrightarrow{BC} + \overrightarrow{CJ} = \overrightarrow{AD} + \dfrac12\overrightarrow{CD} = \overrightarrow{AD} - \dfrac12\overrightarrow{AB} B J = B C + C J = A D + 2 1 C D = A D − 2 1 A B C D → = − A B → \overrightarrow{CD} = -\overrightarrow{AB} C D = − A B Avec A B → ⋅ A D → = 0 \overrightarrow{AB} \cdot \overrightarrow{AD} = 0 A B ⋅ A D = 0 A B 2 = A D 2 = 9 AB^2 = AD^2 = 9 A B 2 = A D 2 = 9 A I → ⋅ B J → = A B → ⋅ A D → − 1 2 A B 2 + 1 2 A D 2 − 1 4 A D → ⋅ A B → = 0 − 4,5 + 4,5 − 0 = 0 \overrightarrow{AI} \cdot \overrightarrow{BJ} = \overrightarrow{AB} \cdot \overrightarrow{AD} - \dfrac12 AB^2 + \dfrac12 AD^2 - \dfrac14 \overrightarrow{AD} \cdot \overrightarrow{AB} = 0 - 4{,}5 + 4{,}5 - 0 = 0 A I ⋅ B J = A B ⋅ A D − 2 1 A B 2 + 2 1 A D 2 − 4 1 A D ⋅ A B = 0 − 4 , 5 + 4 , 5 − 0 = 0 ( A I ) (AI) ( A I ) ( B J ) (BJ) ( B J ) perpendiculaires .
4. Propriétés de calcul
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Exercice 4 — ⭐⭐⭐
On sait que ∥ u ⃗ ∥ = 4 \|\vec u\| = 4 ∥ u ∥ = 4 ∥ v ⃗ ∥ = 3 \|\vec v\| = 3 ∥ v ∥ = 3 ∥ u ⃗ − v ⃗ ∥ = 13 \|\vec u - \vec v\| = \sqrt{13} ∥ u − v ∥ = 13
Calcule u ⃗ ⋅ v ⃗ \vec u \cdot \vec v u ⋅ v
Calcule ∥ u ⃗ + v ⃗ ∥ \|\vec u + \vec v\| ∥ u + v ∥
Déduis-en une mesure de l’angle ( u ⃗ , v ⃗ ) (\vec u, \vec v) ( u , v )
Voir le corrigé
u ⃗ ⋅ v ⃗ = 1 2 ( ∥ u ⃗ ∥ 2 + ∥ v ⃗ ∥ 2 − ∥ u ⃗ − v ⃗ ∥ 2 ) = 1 2 ( 16 + 9 − 13 ) = 6 \vec u \cdot \vec v = \dfrac12\left(\|\vec u\|^2 + \|\vec v\|^2 - \|\vec u - \vec v\|^2\right) = \dfrac12(16 + 9 - 13) = 6 u ⋅ v = 2 1 ( ∥ u ∥ 2 + ∥ v ∥ 2 − ∥ u − v ∥ 2 ) = 2 1 ( 16 + 9 − 13 ) = 6 ∥ u ⃗ + v ⃗ ∥ 2 = 16 + 2 × 6 + 9 = 37 \|\vec u + \vec v\|^2 = 16 + 2 \times 6 + 9 = 37 ∥ u + v ∥ 2 = 16 + 2 × 6 + 9 = 37 ∥ u ⃗ + v ⃗ ∥ = 37 \|\vec u + \vec v\| = \sqrt{37} ∥ u + v ∥ = 37 cos ( u ⃗ , v ⃗ ) = u ⃗ ⋅ v ⃗ ∥ u ⃗ ∥ ∥ v ⃗ ∥ = 6 12 = 1 2 \cos(\vec u, \vec v) = \dfrac{\vec u \cdot \vec v}{\|\vec u\|\|\vec v\|} = \dfrac{6}{12} = \dfrac12 cos ( u , v ) = ∥ u ∥∥ v ∥ u ⋅ v = 12 6 = 2 1 60 ° 60° 60° π 3 \dfrac\pi3 3 π
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Exercice 5 — ⭐⭐
Dans le triangle A B C ABC A B C A B = 6 AB = 6 A B = 6 A C = 4 AC = 4 A C = 4 B A C ^ = 120 ° \widehat{BAC} = 120° B A C = 120° B C BC B C
Un triangle a pour côtés 5 5 5 6 6 6 7 7 7
Voir le corrigé
B C 2 = 36 + 16 − 2 × 6 × 4 × cos 120 ° = 52 − 48 × ( − 1 2 ) = 52 + 24 = 76 BC^2 = 36 + 16 - 2 \times 6 \times 4 \times \cos 120° = 52 - 48 \times \left(-\dfrac12\right) = 52 + 24 = 76 B C 2 = 36 + 16 − 2 × 6 × 4 × cos 120° = 52 − 48 × ( − 2 1 ) = 52 + 24 = 76 B C = 76 = 2 19 BC = \sqrt{76} = 2\sqrt{19} B C = 76 = 2 19 Le plus grand angle est opposé au plus grand côté (7 7 7 cos α = 5 2 + 6 2 − 7 2 2 × 5 × 6 = 12 60 = 1 5 > 0 \cos\alpha = \dfrac{5^2 + 6^2 - 7^2}{2 \times 5 \times 6} = \dfrac{12}{60} = \dfrac15 > 0 cos α = 2 × 5 × 6 5 2 + 6 2 − 7 2 = 60 12 = 5 1 > 0 aigu (donc le triangle a
ses trois angles aigus).
Exercice 6 — ⭐⭐⭐
A B C D ABCD A B C D A B = 5 AB = 5 A B = 5 A D = 3 AD = 3 A D = 3 B A D ^ = 60 ° \widehat{BAD} = 60° B A D = 60° A C AC A C B D BD B D
Voir le corrigé A B → ⋅ A D → = 5 × 3 × cos 60 ° = 7,5 \overrightarrow{AB} \cdot \overrightarrow{AD} = 5 \times 3 \times \cos 60° = 7{,}5 A B ⋅ A D = 5 × 3 × cos 60° = 7 , 5
A C → = A B → + A D → \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD} A C = A B + A D A C 2 = A B 2 + 2 A B → ⋅ A D → + A D 2 = 25 + 15 + 9 = 49 AC^2 = AB^2 + 2\,\overrightarrow{AB} \cdot \overrightarrow{AD} + AD^2 = 25 + 15 + 9 = 49 A C 2 = A B 2 + 2 A B ⋅ A D + A D 2 = 25 + 15 + 9 = 49 A C = 7 AC = 7 A C = 7 B D → = A D → − A B → \overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} B D = A D − A B B D 2 = A D 2 − 2 A B → ⋅ A D → + A B 2 = 9 − 15 + 25 = 19 BD^2 = AD^2 - 2\,\overrightarrow{AB} \cdot \overrightarrow{AD} + AB^2 = 9 - 15 + 25 = 19 B D 2 = A D 2 − 2 A B ⋅ A D + A B 2 = 9 − 15 + 25 = 19 B D = 19 BD = \sqrt{19} B D = 19
(On pouvait aussi appliquer Al-Kashi dans les triangles A B D ABD A B D 60 ° 60° 60° A B C ABC A B C 120 ° 120° 120°
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Exercice 7 — ⭐⭐⭐⭐
A A A B B B A B = 6 AB = 6 A B = 6 I I I [ A B ] [AB] [ A B ]
Montre que, pour tout point M M M M A → ⋅ M B → = M I 2 − 9 \overrightarrow{MA} \cdot \overrightarrow{MB} = MI^2 - 9 M A ⋅ M B = M I 2 − 9
Détermine l’ensemble des points M M M M A → ⋅ M B → = 0 \overrightarrow{MA} \cdot \overrightarrow{MB} = 0 M A ⋅ M B = 0 M A → ⋅ M B → = 16 \overrightarrow{MA} \cdot \overrightarrow{MB} = 16 M A ⋅ M B = 16 M A → ⋅ M B → = − 9 \overrightarrow{MA} \cdot \overrightarrow{MB} = -9 M A ⋅ M B = − 9 M A → ⋅ M B → = − 10 \overrightarrow{MA} \cdot \overrightarrow{MB} = -10 M A ⋅ M B = − 10
Voir le corrigé
M A → ⋅ M B → = ( M I → + I A → ) ⋅ ( M I → − I A → ) = M I 2 − I A 2 = M I 2 − 3 2 = M I 2 − 9 \overrightarrow{MA} \cdot \overrightarrow{MB} = (\overrightarrow{MI} + \overrightarrow{IA}) \cdot (\overrightarrow{MI} - \overrightarrow{IA}) = MI^2 - IA^2 = MI^2 - 3^2 = MI^2 - 9 M A ⋅ M B = ( M I + I A ) ⋅ ( M I − I A ) = M I 2 − I A 2 = M I 2 − 3 2 = M I 2 − 9 a) M I 2 = 9 MI^2 = 9 M I 2 = 9 M I = 3 MI = 3 M I = 3 I I I 3 3 3 [ A B ] [AB] [ A B ] M I 2 = 25 MI^2 = 25 M I 2 = 25 M I = 5 MI = 5 M I = 5 I I I 5 5 5 M I 2 = 0 MI^2 = 0 M I 2 = 0 I I I M I 2 = − 1 MI^2 = -1 M I 2 = − 1 vide .