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Câu hỏi: Trong không gian với hệ tọa độ $Oxyz$ cho $A\left( -1; 2; 3 \right)$, $B\left( 1; 0; 2 \right).$ Tìm tọa độ điểm $M$ thỏa mãn $\overrightarrow{AB}=2.\overrightarrow{MA}$ ?
A. $M\left( -2;3;\dfrac{7}{2} \right)$.
B. $M\left( -2;3;7 \right)$.
C. $M\left( -4;6;7 \right)$.
D. $M\left( -2;-3;\dfrac{7}{2} \right)$.
A. $M\left( -2;3;\dfrac{7}{2} \right)$.
B. $M\left( -2;3;7 \right)$.
C. $M\left( -4;6;7 \right)$.
D. $M\left( -2;-3;\dfrac{7}{2} \right)$.
Ta có: $\overrightarrow{AB}=2.\overrightarrow{MA}\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{B}}-{{x}_{A}}=2\left( {{x}_{A}}-{{x}_{M}} \right) \\
& {{y}_{B}}-{{y}_{A}}=2\left( {{y}_{A}}-{{y}_{M}} \right) \\
& {{z}_{B}}-{{z}_{A}}=2\left( {{z}_{A}}-{{z}_{M}} \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{M}}=\dfrac{3{{x}_{A}}-{{x}_{B}}}{2} \\
& {{y}_{M}}=\dfrac{3{{y}_{A}}-{{y}_{B}}}{2} \\
& {{z}_{M}}=\dfrac{3{{z}_{A}}-{{z}_{B}}}{2} \\
\end{aligned} \right.\Leftrightarrow M\left( -2; 3; \dfrac{7}{2} \right)$.
& {{x}_{B}}-{{x}_{A}}=2\left( {{x}_{A}}-{{x}_{M}} \right) \\
& {{y}_{B}}-{{y}_{A}}=2\left( {{y}_{A}}-{{y}_{M}} \right) \\
& {{z}_{B}}-{{z}_{A}}=2\left( {{z}_{A}}-{{z}_{M}} \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{M}}=\dfrac{3{{x}_{A}}-{{x}_{B}}}{2} \\
& {{y}_{M}}=\dfrac{3{{y}_{A}}-{{y}_{B}}}{2} \\
& {{z}_{M}}=\dfrac{3{{z}_{A}}-{{z}_{B}}}{2} \\
\end{aligned} \right.\Leftrightarrow M\left( -2; 3; \dfrac{7}{2} \right)$.
Đáp án A.