\(\overrightarrow{MA}=-\frac{3}{4}\overrightarrow{BM};\overrightarrow{AN}=-3\overrightarrow{CN};\overrightarrow{CP}=\frac{1}{4}\overrightarrow{PB}\)
a/ \(VT=\overrightarrow{AM}+\overrightarrow{MC}+\overrightarrow{CB}=\overrightarrow{AC}+\overrightarrow{CB}=\overrightarrow{AB}=\overrightarrow{AN}+\overrightarrow{NB}=VP\)
b/ \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=-\frac{3}{4}\overrightarrow{BM}-3\overrightarrow{CN}=\frac{3}{4}\overrightarrow{MB}+3\overrightarrow{NC}\)
\(VP=\frac{15}{4}\overrightarrow{AB}+\frac{3}{4}\overrightarrow{BC}=\frac{15}{4}\overrightarrow{AM}+\frac{15}{4}\overrightarrow{MB}+\frac{3}{4}\overrightarrow{BN}+\frac{3}{4}\overrightarrow{NC}\)
\(=\frac{15}{4}.\frac{3}{4}\overrightarrow{BM}+\frac{15}{4}\overrightarrow{MB}+\frac{3}{4}\overrightarrow{BA}+\frac{3}{4}\overrightarrow{AN}+\frac{3}{4}\overrightarrow{NC}\)
Có \(\overrightarrow{MA}=-\frac{3}{4}\overrightarrow{BM}\Rightarrow\overrightarrow{MB}+\overrightarrow{BA}=-\frac{3}{4}\overrightarrow{BM}\Leftrightarrow\overrightarrow{MB}=4\overrightarrow{AB}\)
\(\Rightarrow VP=\frac{15}{16}\overrightarrow{MB}+\frac{3}{4}.\frac{1}{4}\overrightarrow{BM}+\frac{3}{4}.\left(-3\right)\overrightarrow{CN}+\frac{3}{4}\overrightarrow{NC}\)
\(=\frac{3}{4}\overrightarrow{MB}+3\overrightarrow{NC}=VT\)
c/ Có \(\overrightarrow{MN}=\frac{15}{4}\overrightarrow{AB}+\frac{3}{4}\overrightarrow{BC}\) (1)
\(\overrightarrow{CP}=\frac{1}{4}\overrightarrow{PB}\Leftrightarrow\overrightarrow{CM}+\overrightarrow{MP}=\frac{1}{4}\overrightarrow{PM}+\frac{1}{4}\overrightarrow{MB}\)
\(\Leftrightarrow\frac{3}{4}\overrightarrow{MP}=\frac{1}{4}\overrightarrow{MB}+\overrightarrow{MC}\)
Có \(\overrightarrow{MB}=4\overrightarrow{AB};\overrightarrow{MC}+\overrightarrow{CA}=-\frac{3}{4}\overrightarrow{BC}-\frac{3}{4}\overrightarrow{CM}\Leftrightarrow\frac{1}{4}\overrightarrow{MC}=\overrightarrow{AC}+\frac{3}{4}\overrightarrow{CB}\Leftrightarrow\overrightarrow{MC}=4\overrightarrow{AC}+3\overrightarrow{CB}\)
\(\Rightarrow\overrightarrow{MC}=4\overrightarrow{AC}+3\overrightarrow{CB}=4\overrightarrow{AB}+4\overrightarrow{BC}+3\overrightarrow{CB}=4\overrightarrow{AB}+\overrightarrow{CB}\)
\(\Rightarrow\frac{3}{4}\overrightarrow{MP}=\frac{1}{4}.4\overrightarrow{AB}+4\overrightarrow{AB}+\overrightarrow{BC}\Leftrightarrow\overrightarrow{MP}=\frac{20}{3}\overrightarrow{AB}+\frac{4}{3}\overrightarrow{BC}\) (2)
Có \(\overrightarrow{MN}=\frac{9}{16}\overrightarrow{MP}\Rightarrow\) thẳng hàng
