cho vecto u=i+3j và vecto v=(2;-1) . tính vecto u.v
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\(\overrightarrow{u}=i+3j=\left(1;3\right)\)
\(\overrightarrow{v}=2\cdot j-2i=\left(-2;2\right)\)
\(\overrightarrow{u}\cdot\overrightarrow{v}=1\cdot\left(-2\right)+3\cdot2=-2+6=4\)
a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)
\(\overrightarrow{a}=2\overrightarrow{i}-4\overrightarrow{j}\Rightarrow\overrightarrow{a}=\left(2;-4\right)\)
\(\overrightarrow{b}=-5\overrightarrow{i}+3\overrightarrow{j}\Rightarrow\overrightarrow{b}=\left(-5;3\right)\)
\(\Rightarrow\overrightarrow{u}=2\overrightarrow{a}-\overrightarrow{b}=2\left(2;-4\right)-\left(-5;3\right)=\left(9;-11\right)\)
\(\overrightarrow{u}+2\overrightarrow{v}-3\overrightarrow{w}+\overrightarrow{x}=\overrightarrow{0}\)
\(\Rightarrow\overrightarrow{x}=3\overrightarrow{w}-\overrightarrow{u}-2\overrightarrow{v}=3\left(-5;7\right)-\left(2;-5\right)-2\left(3;4\right)=\left(-23;18\right)\)
\(\overrightarrow{u}\) ⊥\(\overrightarrow{v}\)
=>\(\overrightarrow{u}\cdot\overrightarrow{v}=0\)
=>\(\left(\overrightarrow{a}-3\cdot\overrightarrow{b}\right)\left(2\cdot\overrightarrow{a}+\overrightarrow{b}\right)=0\)
=>\(2\cdot\left(\overrightarrow{a}\right)^2+\overrightarrow{a}\cdot\overrightarrow{b}-6\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot\left(\overrightarrow{b}\right)^2=0\)
=>\(2\cdot2-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot3=0\)
=>\(-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}=5\)
=>\(\overrightarrow{a}\cdot\overrightarrow{b}=-1\)


Ta có:
\(\overrightarrow{u}=\overrightarrow{i}+3\overrightarrow{j}=\overrightarrow{u}=\left(1;3\right)\\ \Rightarrow\overrightarrow{u}.\overrightarrow{v}=\left(1;3\right).\left(2;-1\right)=1.2+3.\left(-1\right)=-1\)