Trong không gian với hệ tọa độ Oxyz cho tam giác ABC biết A(2; 1...

Ta có \)">\overrightarrow{AB}=(1;-1;2)\Rightarrow \overrightarrow{{{i}_{AB}}}=\dfrac{1}{\left| \overrightarrow{AB} \right|}.\overrightarrow{AB}=\left( \dfrac{1}{\sqrt{6}};-\dfrac{1}{\sqrt{6}};\dfrac{2}{\sqrt{6}} \right)\overrightarrow{{{i}_{AB}}}=\overrightarrow{AE}\overrightarrow{AC}=\left( 2;2;-4 \right)\Rightarrow \overrightarrow{{{i}_{AC}}}=\dfrac{1}{\left| \overrightarrow{AC} \right|}\overrightarrow{AC}=\left( \dfrac{1}{\sqrt{6}};\dfrac{1}{\sqrt{6}};-\dfrac{2}{\sqrt{6}} \right)\overrightarrow{{{i}_{AC}}}=\overrightarrow{AF}\overrightarrow{AM}=\overrightarrow{AE}+\overrightarrow{AF}=\left( \dfrac{2}{\sqrt{6}};0;0 \right)=\dfrac{2}{\sqrt{6}}(1;0;0)\Rightarrow \overrightarrow{AM}\overrightarrow{{{u}_{1}}}=(1;0;0)\left\{ \begin{aligned}
& x=2+t \\
& y=1 \\
& z=0 \\
\end{aligned} \right.,(t\in \mathbb{R})$