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Câu hỏi: Biết A(1,1,0); B(2,0,3); C(3,2,-3), tọa độ trọng tâm G của $\Delta ABC$ là
A. $G\left( 2,1,-1 \right)$
B. $G\left( 2,1,0 \right)$
C. $G\left( 2,0,-1 \right)$
D. $G\left( -2,1,0 \right)$
A. $G\left( 2,1,-1 \right)$
B. $G\left( 2,1,0 \right)$
C. $G\left( 2,0,-1 \right)$
D. $G\left( -2,1,0 \right)$
Ta có $G\left( \dfrac{{{x}_{A}}+{{x}_{B}}+{{x}_{C}}}{3};\dfrac{{{y}_{A}}+{{y}_{B}}+{{y}_{C}}}{3};\dfrac{{{z}_{A}}+{{z}_{B}}+{{z}_{C}}}{3} \right)\Rightarrow G\left( 2,1,0 \right)$
Đáp án B.