Lời giải:
Xử lý điều kiện đề bài:
\(\overrightarrow{NC}+2\overrightarrow{NB}=\overrightarrow{0}\Rightarrow \overrightarrow{BN}+\overrightarrow{NC}+2\overrightarrow{NB}=\overrightarrow{BN}\)
\(\Leftrightarrow \overrightarrow{BC}=\overrightarrow{BN}-2\overrightarrow{NB}=3\overrightarrow{BN}\)
\(\overrightarrow{IN}=2\overrightarrow{AI}\Rightarrow 3\overrightarrow{AI}=\overrightarrow{AI}+\overrightarrow{IN}=\overrightarrow{AN}\)
Áp dụng kết quả trên ta có:
a)
\(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{3}(\overrightarrow{BA}+\overrightarrow{AC})\)
\(=\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AI}=\frac{1}{3}\overrightarrow{AN}=\frac{1}{3}(\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC})=\frac{2}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}\)
\(\overrightarrow{BI}=\overrightarrow{AI}-\overrightarrow{AB}=\frac{2}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}-\overrightarrow{AB}=\frac{-7}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}\)
\(\overrightarrow{CI}=\overrightarrow{AI}-\overrightarrow{AC}=\frac{2}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}-\overrightarrow{AC}=\frac{2}{9}\overrightarrow{AB}-\frac{8}{9}\overrightarrow{AC}\)
b)
Với $K\in AC$, tồn tại $m\in\mathbb{R}$ sao cho \(\overrightarrow{AK}=m\overrightarrow{AC}\)
\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=-\overrightarrow{AB}+m\overrightarrow{AC}\)
\(\overrightarrow{BI}=\frac{-7}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}\)
Để $B,K,I$ thẳng hàng: \(\frac{-1}{\frac{-7}{9}}=\frac{m}{\frac{1}{9}}\Rightarrow m=\frac{1}{7}\)
