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Câu hỏi: Trong không gian $Oxyz$, cho tứ diện $ABCD$ với $A\left( 1;-4;2 \right),B\left( 2;1;-3 \right),C\left( 3;0;-2 \right)$ và $D\left( 2;-5;-1 \right)$. Điểm $G$ thỏa mãn $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}$ có tọa độ là:
A. $G\left( 2;-1;-1 \right)$.
B. $G\left( 2;-2;-1 \right)$.
C. $G\left( 0;-1;-1 \right)$.
D. $G\left( 6;-3;-3 \right)$.
A. $G\left( 2;-1;-1 \right)$.
B. $G\left( 2;-2;-1 \right)$.
C. $G\left( 0;-1;-1 \right)$.
D. $G\left( 6;-3;-3 \right)$.
Tọa độ điểm G thỏa mãn:
$\left\{ \begin{aligned}
& {{x}_{G}}=\dfrac{{{x}_{A}}+{{x}_{B}}+{{x}_{C}}+{{x}_{D}}}{4}=\dfrac{1+2+3+2}{4}=2 \\
& {{y}_{G}}=\dfrac{{{y}_{A}}+{{y}_{B}}+{{y}_{C}}+{{y}_{D}}}{4}=\dfrac{-4+1+0+\left( -5 \right)}{4}=-2 \\
& {{z}_{G}}=\dfrac{{{z}_{A}}+{{z}_{B}}+{{z}_{C}}+{{z}_{D}}}{4}=\dfrac{2+\left( -3 \right)+\left( -2 \right)+\left( -1 \right)}{4}=-1 \\
\end{aligned} \right.\Rightarrow G\left( 2;-2;-1 \right)$
$\left\{ \begin{aligned}
& {{x}_{G}}=\dfrac{{{x}_{A}}+{{x}_{B}}+{{x}_{C}}+{{x}_{D}}}{4}=\dfrac{1+2+3+2}{4}=2 \\
& {{y}_{G}}=\dfrac{{{y}_{A}}+{{y}_{B}}+{{y}_{C}}+{{y}_{D}}}{4}=\dfrac{-4+1+0+\left( -5 \right)}{4}=-2 \\
& {{z}_{G}}=\dfrac{{{z}_{A}}+{{z}_{B}}+{{z}_{C}}+{{z}_{D}}}{4}=\dfrac{2+\left( -3 \right)+\left( -2 \right)+\left( -1 \right)}{4}=-1 \\
\end{aligned} \right.\Rightarrow G\left( 2;-2;-1 \right)$
Đáp án B.