Trong không gian với hệ tọa độ $Oxyz$, cho hình vuông $ABCD$ biết...

Ta có : \)">\overrightarrow{AB}=\left( 0;0;-4 \right)=-4\left( 0;0;1 \right)AB\overrightarrow{k}=\left( 0;0;1 \right)\left( ABCD \right)\left[ \overrightarrow{OA};\overrightarrow{OB} \right]=\left( 0;4;0 \right)=4\left( 0;1;0 \right)\overleftarrow{j}=\left( 0;1;0 \right)\left( ABCD \right)\left\{ \begin{aligned}
& AD\bot AB \\
& AD\subset \left( ABCD \right) \\
\end{aligned} \right. \left\{ \begin{aligned}
& \overrightarrow{AD}\bot \overrightarrow{k} \\
& \overrightarrow{AD}\bot \overrightarrow{j} \\
\end{aligned} \right.AD\left[ \overrightarrow{j};\overrightarrow{k} \right]=\left( 1;0;0 \right)AD\left\{ \begin{aligned}
& x=1+t \\
& y=0 \\
& z=1 \\
\end{aligned} \right. D\left( 1+t;0;1 \right)AD=AB\Leftrightarrow \sqrt{{{t}^{2}}+{{0}^{2}}+{{\left( 1-1 \right)}^{2}}}=4\Leftrightarrow \left[ \begin{aligned}
& t=4 \\
& t=-4 \\
\end{aligned} \right.DD\left( -3;0;1 \right)IABCDBDI=\left( -1;0;-1 \right)dABCD\overrightarrow{j}=\left( 0;1;0 \right)dd:\left\{ \begin{aligned}
& x=-1 \\
& y=t \\
& z=-1 \\
\end{aligned} \right.$.