5.
\(\overrightarrow{BA}.\overrightarrow{BC}=BA.BC.cos\widehat{CBA}=a.a.cos60^0=\dfrac{a^2}{2}\)
\(\overrightarrow{AB}.\overrightarrow{BC}=-\overrightarrow{BA}.\overrightarrow{BC}=-\dfrac{a^2}{2}\)
6.
\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(3;-4\right)\\\overrightarrow{BC}=\left(4;3\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}AB=\sqrt{3^2+\left(-4\right)^2}=5\\BC=\sqrt{4^2+3^2}=5\\\overrightarrow{AB}.\overrightarrow{BC}=3.4-4.3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}AB=BC\\AB\perp BC\end{matrix}\right.\)
\(\Rightarrow\) Tam giác ABC vuông cân tại B
\(AB+AC+BC=5+5+5\sqrt{2}=5\left(2+\sqrt{2}\right)\)
b. Gọi \(M\left(x;y\right)\Rightarrow\overrightarrow{AM}=\left(x+1;y-8\right)\)
\(\overrightarrow{AC}=\left(7;-1\right)\Rightarrow2\overrightarrow{AB}+3\overrightarrow{AC}=\left(27;-11\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=27\\y-8=-11\end{matrix}\right.\) \(\Rightarrow M\left(26;-3\right)\)
c. Do tam giác ABC vuông tại B nên trực tâm H trùng B
\(\Rightarrow H\left(2;4\right)\)
