Google
Câu hỏi: Trong không gian với hệ tọa độ $Oxyz$, cho tam giác $ABC$ với $A\left( -2;4;1 \right)$, $B\left( 1;1;-6 \right)$, $C\left( 0;-2;3 \right)$. Tìm tọa độ trọng tâm $G$ của tam giác $ABC$.
A. $G\left( -\dfrac{1}{2};\dfrac{5}{2};-\dfrac{5}{2} \right)$.
B. $G\left( -1;3;-2 \right)$.
C. $G\left( \dfrac{1}{3};-1;\dfrac{2}{3} \right)$.
D. $G\left( -\dfrac{1}{3};1;-\dfrac{2}{3} \right)$.
A. $G\left( -\dfrac{1}{2};\dfrac{5}{2};-\dfrac{5}{2} \right)$.
B. $G\left( -1;3;-2 \right)$.
C. $G\left( \dfrac{1}{3};-1;\dfrac{2}{3} \right)$.
D. $G\left( -\dfrac{1}{3};1;-\dfrac{2}{3} \right)$.
Ta có: $\left\{ \begin{aligned}
& {{x}_{G}}=\dfrac{{{x}_{A}}+{{x}_{B}}+{{x}_{C}}}{3}=\dfrac{-2+1+0}{3}=-\dfrac{1}{3} \\
& {{y}_{G}}=\dfrac{{{y}_{A}}+{{y}_{B}}+{{y}_{C}}}{3}=\dfrac{4+1-2}{3}=1 \\
& {{z}_{G}}=\dfrac{{{z}_{A}}+{{z}_{B}}+{{z}_{C}}}{3}=\dfrac{1-6+3}{3}=-\dfrac{2}{3} \\
\end{aligned} \right. $ nên $ G\left( -\dfrac{1}{3};1;-\dfrac{2}{3} \right)$.
& {{x}_{G}}=\dfrac{{{x}_{A}}+{{x}_{B}}+{{x}_{C}}}{3}=\dfrac{-2+1+0}{3}=-\dfrac{1}{3} \\
& {{y}_{G}}=\dfrac{{{y}_{A}}+{{y}_{B}}+{{y}_{C}}}{3}=\dfrac{4+1-2}{3}=1 \\
& {{z}_{G}}=\dfrac{{{z}_{A}}+{{z}_{B}}+{{z}_{C}}}{3}=\dfrac{1-6+3}{3}=-\dfrac{2}{3} \\
\end{aligned} \right. $ nên $ G\left( -\dfrac{1}{3};1;-\dfrac{2}{3} \right)$.
Đáp án D.