Cho hình hộp ABCD.A'B'C'D' biết $A=\left(1;0;1\right), B=\left(2;1;2\right), D=\left(1;-1;1\right), C\text{'}=\left(4;5;-5\right)$. Tính tọa độ các đỉnh

còn lại của hình hộp.

Giải

Ta có: $$\begin{gathered} \overrightarrow{AB}=\left(1;1;1\right) \\ \overrightarrow{AD}=\left(0;-1;0\right) \\ \overrightarrow{BC}=\overrightarrow{AD}\iff \left\{\begin{array}{l}{x}_C-2=0\\ y_C-1=-1\\ z_C-2=0\end{array}\iff \left\{\begin{array}{l}{x}_C=2\\ y_C=0\\ z_C=2\end{array}\right.\right. \end{gathered}$$

Vậy $C=(2;0;2)$

Suy ra $\overrightarrow{CC\text{'}}=(2;5;-7)$

Từ $\overrightarrow{AA\text{'}}=\overrightarrow{BB\text{'}}=\overrightarrow{DD\text{'}}=\overrightarrow{CC\text{'}}=\left(2;5;-7\right)$

Suy ra $$\begin{gathered} \left\{\begin{array}{l}{x}_{A\text{'}}-1=2\\ y_{A\text{'}}-0=5\\ z_{A\text{'}}-1=-7\end{array}\right. \\ \iff \left\{\begin{array}{l}{x}_{A\text{'}}=3\\ y_{A\text{'}}=5\\ z_{A\text{'}}=-6\end{array}\right. \end{gathered}$$

Vậy $A\text{'}=(3;5;-6)$.

Tương tự $B\text{'}=(4;6;-5), D\text{'}=(3;4;-6)$