D.S de MATHÉMATIQUES

1)
• $\textcolor{red}{\vec{AB} \cdot \vec{AC} =} (\vec{AH}+\vec{HB}) \cdot \vec{AC} = \vec{AH} \cdot \vec{AC} + \vec{HB} \cdot \vec{AC} = \vec{AH} \cdot \vec{AC} = -2 \times 1 = \textcolor{red}{-2}$.

• A(1;2); B(-2;1); C(5/2;-1/2). $\vec{AB}(-3;-1), \vec{AC}(3/2;-5/2)$.
  $\textcolor{red}{\vec{AB} \cdot \vec{AC} =} -3 \times \frac{3}{2} + (-1) \times (-\frac{5}{2}) = \textcolor{red}{-2}$.

• $\textcolor{red}{\vec{AB} \cdot \vec{AC} =} \vec{AH} \cdot \vec{AC} = (3+2) \times 3 = \textcolor{red}{15}$.

2) a) $\textcolor{red}{(3\vec{u}-\vec{v}) \cdot \vec{v} =} 3\vec{u}\cdot\vec{v} - \vec{v}\cdot\vec{v} = 3\vec{u}\cdot\vec{v} - \|\vec{v}\|^2 = 3\times 4 - 5^2 = \textcolor{red}{-13}$.

b) $\textcolor{red}{\|\vec{u}-\vec{v}\|^2 =} \|\vec{u}\|^2-2\vec{u}\cdot\vec{v}+\|\vec{v}\|^2 = 1^2-2\times 4+5^2=\textcolor{red}{18}$.
    $\textcolor{red}{\|\vec{u}+\vec{v}\|^2 =} \|\vec{u}\|^2+2\vec{u}\cdot\vec{v}+\|\vec{v}\|^2 = 1^2+2\times 4+5^2=\textcolor{red}{6\sqrt{34}}$.
    $\textcolor{red}{\|\vec{u}-\vec{v}\| \times \|\vec{u}+\vec{v}\| =} \sqrt{18} \times \sqrt{34} = \textcolor{red}{6\sqrt{17}}$.

1) a) $\textcolor{red}{\vec{AC} \cdot \vec{AB} =} \vec{AJ} \cdot \vec{AB} = 2 \times 4 = \textcolor{red}{8}$.

b) $\textcolor{red}{\vec{CK} \cdot \vec{BC} =} -\vec{CK} \cdot \vec{CB} = -\|\vec{CK}\| \times \|\vec{CB}\| \times \cos(\widehat{KCB}) = -\frac{1}{2}CA \times CB \times \cos(\frac{\pi}{3}) = -\frac{1}{2}\times 4 \times 4 \times \frac{1}{2} = \textcolor{red}{-4}$.

c) $\textcolor{red}{\vec{BA} \cdot \vec{DC} =} -BA \times DC = -4 \times 2 = \textcolor{red}{-8}$.

d) $\vec{AC} \cdot \vec{AD} = \vec{AD} \cdot \vec{AD} = AD^2$ et $AC^2=AD^2+DC^2 \implies AD^2=AC^2-DC^2=4^2-2^2=12$. Donc $\textcolor{red}{\vec{AC} \cdot \vec{AD} = 12}$.

e) $\textcolor{red}{\vec{DK} \cdot \vec{CB} =} (\vec{DC}+\vec{CK})\cdot\vec{CB} = \vec{DC}\cdot\vec{CB}+\vec{CK}\cdot\vec{CB} = -\vec{CD}\cdot\vec{CB}-\vec{CK}\cdot\vec{BC} = -CD \times CB \times \cos(\frac{\pi}{6}+\frac{\pi}{2})+4 = -2\times 4 \times \cos(\frac{2\pi}{3})+4 = -8 \times (-\frac{1}{2})+4 = \textcolor{red}{8}$.

f) $\textcolor{red}{\vec{AB} \cdot \vec{CK} =} \vec{AB} \cdot \vec{KA} = -\vec{AB} \cdot \vec{AK} = -AB \times AK \times \cos(\widehat{BAK}) = -4 \times 2 \times \cos(\frac{\pi}{3}) = -4 \times 2 \times \frac{1}{2} = \textcolor{red}{-4}$.

2) $\vec{AC} \cdot \vec{AD} = \vec{AC} \cdot \vec{AH} = AC \times AH = 4AH$ et $\vec{AC} \cdot \vec{AD} = 12$ d'après 1) d). D'où $4AH=12 \implies \textcolor{red}{AH=3}$.

a) $\textcolor{red}{\vec{AH} \cdot \vec{FE} =} -AH \times FE = -5 \times 5 = \textcolor{red}{-25}$.

b) $\textcolor{red}{\vec{AE} \cdot \vec{BG} =} \vec{AB} \cdot \vec{BG} = 10 \times 5 = \textcolor{red}{50}$.

c) $\textcolor{red}{\vec{AH} \cdot \vec{DG} =} \vec{AH} \cdot \vec{AG} = AH \times AG = 5 \times 15 = \textcolor{red}{75}$.

d) $\textcolor{red}{\vec{DB} \cdot \vec{FG} =} \vec{DB} \cdot \vec{EB} = \vec{CB} \cdot \vec{EB} = CB \times EB = 10 \times 5 = \textcolor{red}{50}$.

e) $\textcolor{red}{\vec{IB} \cdot \vec{HF} =} (\vec{IH}+\vec{HB})\cdot(\vec{HG}+\vec{GF}) = \vec{IH}\cdot\vec{HG}+\vec{IH}\cdot\vec{GF}+\vec{HB}\cdot\vec{HG}+\vec{HB}\cdot\vec{GF} = 0+HB \times HG - IH \times GF+0 = 5 \times 10 - 10 \times 5 = \textcolor{red}{0}$.

f) $\textcolor{red}{\vec{BI} \cdot \vec{AC} =} (\vec{BC}+\vec{CI})\cdot(\vec{AB}+\vec{BC}) = \vec{BC}\cdot\vec{AB}+\vec{CI}\cdot\vec{AB}+\vec{BC}\cdot\vec{BC}+\vec{CI}\cdot\vec{BC} = 0-CI \times AB + BC^2+0 = -5 \times 10 + 10^2 = \textcolor{red}{50}$.