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Câu hỏi: Trong không gian tọa độ Oxyz, cho hình hộp $ABCD.{A}'{B}'{C}'{D}'$ với các điểm $A\left( -1;1;2 \right)$, $B\left( -3;2;1 \right)$, $D\left( 0;-1;2 \right)$ và ${A}'\left( 2;1;2 \right)$. Tìm tọa độ đỉnh ${C}'$.
A. ${C}'\left( 1;0;1 \right)$
B. ${C}'\left( -3;1;3 \right)$
C. ${C}'\left( 0;1;0 \right)$
D. ${C}'\left( -1;3;1 \right)$
Gọi $C\left( a;b;c \right)$, ta có $\overrightarrow{DC}=\left( a;b+1;c-2 \right); \overrightarrow{AB}=\left( -2;1;-1 \right)$
Ta có $\overrightarrow{DC}=\overrightarrow{AB}\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b+1=1 \\
& c-2=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=0 \\
& c=1 \\
\end{aligned} \right.\to C\left( -2;0;1 \right)$.
Gọi ${C}'\left( m;n;p \right), \overrightarrow{C{C}'}=\left( m+2;n;p-1 \right); \overrightarrow{A{A}'}=\left( 3;0;0 \right)$
Ta có $\overrightarrow{C{C}'}=\overrightarrow{A{A}'}\Leftrightarrow \left\{ \begin{aligned}
& m+2=3 \\
& n=0 \\
& p-1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m=1 \\
& n=0 \\
& p=1 \\
\end{aligned} \right. $. Vậy $ {C}'\left( 1;0;1 \right)$.
A. ${C}'\left( 1;0;1 \right)$
B. ${C}'\left( -3;1;3 \right)$
C. ${C}'\left( 0;1;0 \right)$
D. ${C}'\left( -1;3;1 \right)$
Gọi $C\left( a;b;c \right)$, ta có $\overrightarrow{DC}=\left( a;b+1;c-2 \right); \overrightarrow{AB}=\left( -2;1;-1 \right)$
Ta có $\overrightarrow{DC}=\overrightarrow{AB}\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b+1=1 \\
& c-2=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=0 \\
& c=1 \\
\end{aligned} \right.\to C\left( -2;0;1 \right)$.
Gọi ${C}'\left( m;n;p \right), \overrightarrow{C{C}'}=\left( m+2;n;p-1 \right); \overrightarrow{A{A}'}=\left( 3;0;0 \right)$
Ta có $\overrightarrow{C{C}'}=\overrightarrow{A{A}'}\Leftrightarrow \left\{ \begin{aligned}
& m+2=3 \\
& n=0 \\
& p-1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m=1 \\
& n=0 \\
& p=1 \\
\end{aligned} \right. $. Vậy $ {C}'\left( 1;0;1 \right)$.
Đáp án A.