16.
a.
Theo tính chất trọng tâm ta có: \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)\)
\(=3\overrightarrow{MG}+\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)=3\overrightarrow{MG}+\overrightarrow{0}\)
\(=3\overrightarrow{MG}\) (đpcm)
b.
Câu b đề yêu cầu biểu diễn vecto \(\overrightarrow{AN}\) và vecto nào vậy nhỉ?
\(BN=2NC\Rightarrow BN=\dfrac{2}{3}BC\)
\(\Rightarrow\overrightarrow{BN}=\dfrac{2}{3}\overrightarrow{BC}\)
Từ đó: \(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
c.
Do AH là trung tuyến \(\Rightarrow\overrightarrow{AH}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\Rightarrow\overrightarrow{AH}+\overrightarrow{AC}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\)
Đặt \(T=\left|\overrightarrow{AH}+\overrightarrow{AC}\right|=\left|\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\right|\)
\(\Rightarrow T^2=\dfrac{1}{4}AB^2+\dfrac{9}{4}AC^2+\dfrac{3}{2}\overrightarrow{AB}.\overrightarrow{AC}\)
\(=\dfrac{1}{4}a^2+\dfrac{9}{4}a^2+\dfrac{3}{2}.a.a.cos60^0=\dfrac{13a^2}{4}\)
\(\Rightarrow T=\dfrac{a\sqrt{13}}{2}\)
