Giải câu 3 bài: Hệ tọa độ trong không gian.
Ta có:
$\overrightarrow{AB}=(1;1;1)$
$\overrightarrow{AD}=(0;-1;0)$
Mà $\overrightarrow{AD}=\overrightarrow{BC}$
<=> $\left\{\begin{matrix}x_{A}-2=0 & & \\ y_{C}-1=-1 & & \\ z_{C}-2=0 & & \end{matrix}\right.<=>\left\{\begin{matrix}x_{C}=2 & &\\y_{C}=0 & & \\ z_{C}=2 & & \end{matrix}\right.$
=> $C(2;0;2)$
=> $\overrightarrow{CC'}=(2;5;-7)=\overrightarrow{AA'}=\overrightarrow{BB'}=\overrightarrow{DD'}$
Ta có: $\left\{\begin{matrix}x_{A}-1=2 & & \\ y_{A}-0=5 & & \\ z_{A}-1=-7 & & \end{matrix}\right.<=>\left\{\begin{matrix}x_{A}=3 & &\\y_{A}=5 & & \\ z_{A}=-6 & & \end{matrix}\right.$
=> $A(3;5;-6)$
Ta có: $\left\{\begin{matrix}x_{B}-2=2 & & \\ y_{B}-1=5 & & \\ z_{B}-2=-7 & & \end{matrix}\right.<=>\left\{\begin{matrix}x_{B}=4 & &\\y_{B}=6 & & \\ z_{B}=-5 & & \end{matrix}\right.$
=> $B(4;6;-5)$
Ta có: $\left\{\begin{matrix}x_{D}-1=2 & & \\ y_{D}+1=5 & & \\ z_{D}-1=-7 & & \end{matrix}\right.<=>\left\{\begin{matrix}x_{D}=3 & &\\y_{D}=4 & & \\ z_{D}=-6 & & \end{matrix}\right.$
=> $D(3;4;-6)$