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: $\left\{\begin{matrix}x_{D}-1=2 &  & \\ y_{D}+1=5 &  & \\ z_{D}-1=-7 &  & \end{matrix}\right.<=>\left\{\begin{matrix}x_{D}=3 &  &\\y_{D}=4 &  & \\ z_{D}=-6 &  & \end{matrix}\right.$

=> $D(3;4;-6)$