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Dựa theo đề bài ta có hình vẽ: A B C M N I
Ta có: MA = 2MB; BN = 5CN => BN = 5/6 BC
Có \(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=-\overrightarrow{BA}+\frac{5}{6}\overrightarrow{BC}\)
Áp dụng định lí menelaus cho tam giác ABN
\(\frac{MA}{MB}.\frac{CB}{CN}.\frac{IN}{IA}=1\)=> \(\frac{2}{1}.\frac{6}{1}.\frac{IN}{IA}=1\Rightarrow IA=12IN\)=> \(\overrightarrow{AI}=\frac{12}{13}\overrightarrow{AN}\)
Ta có: \(\overrightarrow{BI}=\overrightarrow{BA}+\overrightarrow{AI}=\overrightarrow{BA}+\frac{12}{13}\overrightarrow{AN}=\overrightarrow{BA}+\frac{12}{13}\left(-\overrightarrow{BA}+\frac{5}{6}\overrightarrow{BC}\right)\)rút gọn tính tiếp nhé
\(AM=\frac{1}{2}MB\Rightarrow\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}\)
\(AN=3NC\Rightarrow\overrightarrow{AN}=\frac{3}{4}\overrightarrow{AC}\)
\(\overrightarrow{AK}=\frac{1}{2}\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AN}=\frac{1}{2}.\frac{1}{3}\overrightarrow{AB}+\frac{1}{2}.\frac{3}{4}\overrightarrow{AC}=\frac{1}{6}\overrightarrow{AB}+\frac{3}{8}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{6}\\n=\frac{3}{8}\end{matrix}\right.\) \(\Rightarrow mn=\frac{1}{16}\)
a/ Có AM= 3MB\(\Rightarrow\overrightarrow{AM}=3\overrightarrow{MB}\)
Theo quy tắc 3 điểm=> \(\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{AM}\)
và \(\overrightarrow{CM}=\overrightarrow{CB}-\overrightarrow{MB}\)
Cộng vế vs vế=> \(2\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{AM}+\overrightarrow{CB}-\overrightarrow{MB}\)
\(\Leftrightarrow2\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{MB}+\overrightarrow{CB}\)
\(\Leftrightarrow2\overrightarrow{CM}=\overrightarrow{CA}+\frac{2}{3}\overrightarrow{AB}+\overrightarrow{CB}\)
\(\Leftrightarrow6\overrightarrow{CM}=3\overrightarrow{CA}+2\overrightarrow{\:AB}+3\overrightarrow{CB}\)
\(\Leftrightarrow6\overrightarrow{CM}=\overrightarrow{CA}+5\overrightarrow{CB}\) ( vì \(2\overrightarrow{CA}+2\overrightarrow{AB}=2\overrightarrow{CB}\) )
b/ Làm tương tự câu a
c/ Theo quy tắc trung điểm có:
\(2\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{CB}\)
\(2\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}=2\overrightarrow{AN}+\overrightarrow{BA}+2\overrightarrow{CM}+\overrightarrow{AC}\)
\(=2\overrightarrow{AN}+\overrightarrow{BC}+2\overrightarrow{CM}\)
Có \(\overrightarrow{BC}=2\overrightarrow{BN}=2\left(\overrightarrow{BA}+\overrightarrow{AN}\right)\)
=>\(\overrightarrow{AB}=2\overrightarrow{AN}+2\overrightarrow{BA}+2\overrightarrow{AN}+2\overrightarrow{CM}\)
\(\Leftrightarrow3\overrightarrow{AB}=4\overrightarrow{AN}+2\overrightarrow{CM}\Leftrightarrow\overrightarrow{AB}=\frac{4}{3}\overrightarrow{AN}+\frac{2}{3}\overrightarrow{CM}\)