a.
Áp dụng định lý cossin:
\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=a\sqrt{7}\)
Áp dụng công thức trung tuyến:
\(AM=\sqrt{\dfrac{2\left(AB^2+AC^2\right)-BC^2}{4}}=\dfrac{a\sqrt{19}}{2}\)
Áp dụng định lý hàm sin:
\(R=\dfrac{BC}{2sinA}=\dfrac{a\sqrt{21}}{3}\)
b.
Do AM là trung tuyến \(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
\(5\overrightarrow{NA}+7\overrightarrow{NC}=0\Leftrightarrow5\overrightarrow{NA}+7\left(\overrightarrow{NA}+\overrightarrow{AC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow12\overrightarrow{NA}+7\overrightarrow{AC}=0\Leftrightarrow\overrightarrow{AN}=\dfrac{7}{12}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=-\overrightarrow{AB}+\dfrac{7}{12}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{BN}=\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)\left(-\overrightarrow{AB}+\dfrac{7}{12}\overrightarrow{AC}\right)\)
\(=-\dfrac{1}{2}AB^2+\dfrac{7}{24}AC^2-\dfrac{5}{24}\overrightarrow{AB}.\overrightarrow{AC}\)
\(=-\dfrac{1}{2}.\left(2a\right)^2+\dfrac{7}{24}.\left(3a\right)^2-\dfrac{5}{24}.2a.3a.coss60^0\)
\(=0\)
\(\Rightarrow AM\perp BN\) (ddpcm)
