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