3 types d’exercices, trois exemples corrigés pour chacun. Tout se fait sans calculatrice. Pour une série sans fin, avec ta maîtrise de la notion, passe par l’exerciseur.
Dans un repère orthonormé, u ⃗ ⋅ v ⃗ = x x ′ + y y ′ \vec{u} \cdot \vec{v} = xx' + yy' u ⋅ v = x x ′ + y y ′
1. Pour quelle valeur de m les vecteurs u et v sont-ils orthogonaux ?
u ⃗ ( − 2 ; m ) v ⃗ ( − 4 ; − 4 ) \vec{u}\,(-2 \,;\, m) \qquad \vec{v}\,(-4 \,;\, -4) u ( − 2 ; m ) v ( − 4 ; − 4 ) Corrigé u ⃗ ⋅ v ⃗ = − 2 × ( − 4 ) + m × ( − 4 ) = 8 − 4 m \vec{u} \cdot \vec{v} = -2 \times (-4) + m \times (-4) = 8 - 4m u ⋅ v = − 2 × ( − 4 ) + m × ( − 4 ) = 8 − 4 m u ⃗ ⋅ v ⃗ = 0 ⟺ m = 2 \vec{u} \cdot \vec{v} = 0 \iff m = 2 u ⋅ v = 0 ⟺ m = 2 Réponse m = 2 2 2
2. Calcule le produit scalaire (repère orthonormé).
u ⃗ ( − 3 ; 4 ) v ⃗ ( 1 ; 6 ) \vec{u}\,(-3 \,;\, 4) \qquad \vec{v}\,(1 \,;\, 6) u ( − 3 ; 4 ) v ( 1 ; 6 ) Corrigé u ⃗ ⋅ v ⃗ = x x ′ + y y ′ \vec{u} \cdot \vec{v} = xx' + yy' u ⋅ v = x x ′ + y y ′ u ⃗ ⋅ v ⃗ = − 3 × 1 + 4 × 6 = 21 \vec{u} \cdot \vec{v} = -3 \times 1 + 4 \times 6 = 21 u ⋅ v = − 3 × 1 + 4 × 6 = 21 Réponse u · v = 21 21 21
3. Calcule le produit scalaire (repère orthonormé).
u ⃗ ( − 5 ; 5 ) v ⃗ ( 6 ; 4 ) \vec{u}\,(-5 \,;\, 5) \qquad \vec{v}\,(6 \,;\, 4) u ( − 5 ; 5 ) v ( 6 ; 4 ) Corrigé u ⃗ ⋅ v ⃗ = x x ′ + y y ′ \vec{u} \cdot \vec{v} = xx' + yy' u ⋅ v = x x ′ + y y ′ u ⃗ ⋅ v ⃗ = − 5 × 6 + 5 × 4 = − 10 \vec{u} \cdot \vec{v} = -5 \times 6 + 5 \times 4 = -10 u ⋅ v = − 5 × 6 + 5 × 4 = − 10 Réponse u · v = − 10 -10 − 10
u ⃗ ⋅ v ⃗ = ∥ u ⃗ ∥ × ∥ v ⃗ ∥ × cos ( u ⃗ , v ⃗ ) \vec{u} \cdot \vec{v} = \|\vec{u}\| \times \|\vec{v}\| \times \cos(\vec{u}, \vec{v}) u ⋅ v = ∥ u ∥ × ∥ v ∥ × cos ( u , v ) cos π 3 = 1 2 \cos\frac{\pi}{3} = \frac{1}{2} cos 3 π = 2 1 cos π = − 1 \cos\pi = -1 cos π = − 1
4. Calcule le produit scalaire u · v.
∥ u ⃗ ∥ = 7 ∥ v ⃗ ∥ = 4 ( u ⃗ , v ⃗ ) = π 3 \|\vec{u}\| = 7 \qquad \|\vec{v}\| = 4 \qquad (\vec{u}, \vec{v}) = \dfrac{\pi}{3} ∥ u ∥ = 7 ∥ v ∥ = 4 ( u , v ) = 3 π Corrigé u ⃗ ⋅ v ⃗ = ∥ u ⃗ ∥ × ∥ v ⃗ ∥ × cos ( u ⃗ , v ⃗ ) \vec{u} \cdot \vec{v} = \|\vec{u}\| \times \|\vec{v}\| \times \cos(\vec{u}, \vec{v}) u ⋅ v = ∥ u ∥ × ∥ v ∥ × cos ( u , v ) u ⃗ ⋅ v ⃗ = 7 × 4 × 1 2 = 14 \vec{u} \cdot \vec{v} = 7 \times 4 \times \dfrac{1}{2} = 14 u ⋅ v = 7 × 4 × 2 1 = 14 Réponse u · v = 14 14 14
5. Calcule le produit scalaire u · v.
∥ u ⃗ ∥ = 6 ∥ v ⃗ ∥ = 8 ( u ⃗ , v ⃗ ) = 0 \|\vec{u}\| = 6 \qquad \|\vec{v}\| = 8 \qquad (\vec{u}, \vec{v}) = 0 ∥ u ∥ = 6 ∥ v ∥ = 8 ( u , v ) = 0 Corrigé u ⃗ ⋅ v ⃗ = ∥ u ⃗ ∥ × ∥ v ⃗ ∥ × cos ( u ⃗ , v ⃗ ) \vec{u} \cdot \vec{v} = \|\vec{u}\| \times \|\vec{v}\| \times \cos(\vec{u}, \vec{v}) u ⋅ v = ∥ u ∥ × ∥ v ∥ × cos ( u , v ) u ⃗ ⋅ v ⃗ = 6 × 8 × 1 = 48 \vec{u} \cdot \vec{v} = 6 \times 8 \times 1 = 48 u ⋅ v = 6 × 8 × 1 = 48 Réponse u · v = 48 48 48
6. Calcule le produit scalaire u · v.
∥ u ⃗ ∥ = 4 ∥ v ⃗ ∥ = 5 ( u ⃗ , v ⃗ ) = π 2 \|\vec{u}\| = 4 \qquad \|\vec{v}\| = 5 \qquad (\vec{u}, \vec{v}) = \dfrac{\pi}{2} ∥ u ∥ = 4 ∥ v ∥ = 5 ( u , v ) = 2 π Corrigé u ⃗ ⋅ v ⃗ = ∥ u ⃗ ∥ × ∥ v ⃗ ∥ × cos ( u ⃗ , v ⃗ ) \vec{u} \cdot \vec{v} = \|\vec{u}\| \times \|\vec{v}\| \times \cos(\vec{u}, \vec{v}) u ⋅ v = ∥ u ∥ × ∥ v ∥ × cos ( u , v ) u ⃗ ⋅ v ⃗ = 4 × 5 × 0 = 0 \vec{u} \cdot \vec{v} = 4 \times 5 \times 0 = 0 u ⋅ v = 4 × 5 × 0 = 0 Réponse u · v = 0 0 0
B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C ^ BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos\widehat{BAC} B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C cos 60 ° = 1 2 \cos 60° = \frac{1}{2} cos 60° = 2 1 cos 120 ° = − 1 2 \cos 120° = -\frac{1}{2} cos 120° = − 2 1
7. Calcule la longueur BC avec la formule d’Al-Kashi.
A B = 10 A C = 6 B A C ^ = 120 ° AB = 10 \qquad AC = 6 \qquad \widehat{BAC} = 120° A B = 10 A C = 6 B A C = 120° Corrigé B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C ^ BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos \widehat{BAC} B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C B C 2 = 10 2 + 6 2 − 2 × 10 × 6 × ( − 1 2 ) = 196 BC^2 = 10^2 + 6^2 - 2 \times 10 \times 6 \times \left(-\dfrac{1}{2}\right) = 196 B C 2 = 1 0 2 + 6 2 − 2 × 10 × 6 × ( − 2 1 ) = 196 B C = 196 = 14 BC = \sqrt{196} = 14 B C = 196 = 14 Réponse BC = 14 14 14
8. Calcule la longueur BC avec la formule d’Al-Kashi.
A B = 8 A C = 7 B A C ^ = 120 ° AB = 8 \qquad AC = 7 \qquad \widehat{BAC} = 120° A B = 8 A C = 7 B A C = 120° Corrigé B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C ^ BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos \widehat{BAC} B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C B C 2 = 8 2 + 7 2 − 2 × 8 × 7 × ( − 1 2 ) = 169 BC^2 = 8^2 + 7^2 - 2 \times 8 \times 7 \times \left(-\dfrac{1}{2}\right) = 169 B C 2 = 8 2 + 7 2 − 2 × 8 × 7 × ( − 2 1 ) = 169 B C = 169 = 13 BC = \sqrt{169} = 13 B C = 169 = 13 Réponse BC = 13 13 13
9. Calcule la longueur BC avec la formule d’Al-Kashi.
A B = 3 A C = 5 B A C ^ = 120 ° AB = 3 \qquad AC = 5 \qquad \widehat{BAC} = 120° A B = 3 A C = 5 B A C = 120° Corrigé B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C ^ BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos \widehat{BAC} B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos B A C B C 2 = 3 2 + 5 2 − 2 × 3 × 5 × ( − 1 2 ) = 49 BC^2 = 3^2 + 5^2 - 2 \times 3 \times 5 \times \left(-\dfrac{1}{2}\right) = 49 B C 2 = 3 2 + 5 2 − 2 × 3 × 5 × ( − 2 1 ) = 49 B C = 49 = 7 BC = \sqrt{49} = 7 B C = 49 = 7 Réponse BC = 7 7 7