\(\overrightarrow{u}=i+3j=\left(1;3\right)\)

\(\overrightarrow{v}=2\cdot j-2i=\left(-2;2\right)\)

\(\overrightarrow{u}\cdot\overrightarrow{v}=1\cdot\left(-2\right)+3\cdot2=-2+6=4\)

\(\overrightarrow{a}=2\overrightarrow{i}-4\overrightarrow{j}\Rightarrow\overrightarrow{a}=\left(2;-4\right)\)

\(\overrightarrow{b}=-5\overrightarrow{i}+3\overrightarrow{j}\Rightarrow\overrightarrow{b}=\left(-5;3\right)\)

\(\Rightarrow\overrightarrow{u}=2\overrightarrow{a}-\overrightarrow{b}=2\left(2;-4\right)-\left(-5;3\right)=\left(9;-11\right)\)

\(\overrightarrow{u}\) ⊥\(\overrightarrow{v}\)

=>\(\overrightarrow{u}\cdot\overrightarrow{v}=0\)

=>\(\left(\overrightarrow{a}-3\cdot\overrightarrow{b}\right)\left(2\cdot\overrightarrow{a}+\overrightarrow{b}\right)=0\)

=>\(2\cdot\left(\overrightarrow{a}\right)^2+\overrightarrow{a}\cdot\overrightarrow{b}-6\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot\left(\overrightarrow{b}\right)^2=0\)

=>\(2\cdot2-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot3=0\)

=>\(-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}=5\)

=>\(\overrightarrow{a}\cdot\overrightarrow{b}=-1\)

a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)

a) ta có vector AA'+vectorBB'+vectorCC'=1/2(vectorAB+vectorAC+vectorBA+vectorBC+vectorCA+vectorCB)=vector 0

t/c trung tuyến

\(\overrightarrow{u}+2\overrightarrow{v}-3\overrightarrow{w}+\overrightarrow{x}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{x}=3\overrightarrow{w}-\overrightarrow{u}-2\overrightarrow{v}=3\left(-5;7\right)-\left(2;-5\right)-2\left(3;4\right)=\left(-23;18\right)\)

\(\overrightarrow{v}=3\overrightarrow{i}-m\overrightarrow{j}\Rightarrow\overrightarrow{v}=\left(3;-m\right)\)

Để \(\overrightarrow{u};\overrightarrow{v}\) cùng phương:

\(\Leftrightarrow\frac{3}{-2}=\frac{-m}{1}\Rightarrow m=\frac{3}{2}\)