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Câu hỏi: Cho ba điểm $A\left( 1;2;-1 \right), B\left( 2;-1;3 \right), C\left( -3;5;1 \right).$ Tìm tọa độ điểm $D$ sao cho $ABCD$ là hình bình hành.
A. $D=\left( -2 ;2 ;5 \right).$
B. $D=\left( -4 ;8 ;-5 \right).$
C. $D=\left( -2 ;8 ;-3 \right).$
D. $D=\left( -4 ;8 ;-3 \right).$
A. $D=\left( -2 ;2 ;5 \right).$
B. $D=\left( -4 ;8 ;-5 \right).$
C. $D=\left( -2 ;8 ;-3 \right).$
D. $D=\left( -4 ;8 ;-3 \right).$
Tứ giác ${ABCD}$ là hình bình hành
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{A}}+{{x}_{C}}={{x}_{B}}+{{x}_{D}} \\
& {{y}_{A}}+{{y}_{C}}={{y}_{B}}+{{y}_{D}} \\
& {{z}_{A}}+{{z}_{C}}={{z}_{B}}+{{z}_{D}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{D}}={{x}_{A}}+{{x}_{C}}-{{x}_{B}}=1-3-2=-4 \\
& {{y}_{D}}={{y}_{A}}+{{y}_{C}}-{{y}_{B}}=2+5+1=8 \\
& {{z}_{D}}={{z}_{A}}+{{z}_{C}}-{{z}_{B}}=-1+1-3=-3 \\
\end{aligned} \right.\Rightarrow D\left( -4,8,-3 \right).$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{A}}+{{x}_{C}}={{x}_{B}}+{{x}_{D}} \\
& {{y}_{A}}+{{y}_{C}}={{y}_{B}}+{{y}_{D}} \\
& {{z}_{A}}+{{z}_{C}}={{z}_{B}}+{{z}_{D}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{D}}={{x}_{A}}+{{x}_{C}}-{{x}_{B}}=1-3-2=-4 \\
& {{y}_{D}}={{y}_{A}}+{{y}_{C}}-{{y}_{B}}=2+5+1=8 \\
& {{z}_{D}}={{z}_{A}}+{{z}_{C}}-{{z}_{B}}=-1+1-3=-3 \\
\end{aligned} \right.\Rightarrow D\left( -4,8,-3 \right).$
Đáp án D.