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Câu hỏi: Trong không gian $Oxyz$, cho $A\left( 1;1;-3 \right),B\left( 3;-1;1 \right)$. Gọi $G$ là trọng tâm tam giác $OAB$, $\overrightarrow{OG}$ có độ dài bằng:
A. $\dfrac{2\sqrt{5}}{3}$.
B. $\dfrac{2\sqrt{5}}{5}$.
C. $\dfrac{3\sqrt{5}}{3}$.
D. $\dfrac{3\sqrt{5}}{2}$.
Ta có $G=\left( \dfrac{4}{3};0;-\dfrac{2}{3} \right)$ $\Rightarrow \overrightarrow{OG}=\left( \dfrac{4}{3};0;-\dfrac{2}{3} \right)$
$\Rightarrow \left| \overrightarrow{OG} \right|=\sqrt{{{\left( \dfrac{4}{3} \right)}^{2}}+{{0}^{2}}+{{\left( -\dfrac{2}{3} \right)}^{2}}}=\dfrac{2\sqrt{5}}{3}$
A. $\dfrac{2\sqrt{5}}{3}$.
B. $\dfrac{2\sqrt{5}}{5}$.
C. $\dfrac{3\sqrt{5}}{3}$.
D. $\dfrac{3\sqrt{5}}{2}$.
Ta có $G=\left( \dfrac{4}{3};0;-\dfrac{2}{3} \right)$ $\Rightarrow \overrightarrow{OG}=\left( \dfrac{4}{3};0;-\dfrac{2}{3} \right)$
$\Rightarrow \left| \overrightarrow{OG} \right|=\sqrt{{{\left( \dfrac{4}{3} \right)}^{2}}+{{0}^{2}}+{{\left( -\dfrac{2}{3} \right)}^{2}}}=\dfrac{2\sqrt{5}}{3}$
Đáp án A.