Cho tam giác ABC, M, N, P được xác định bởi véctơ vecto MA= -3/4 vecto BM , vecto AN

Trả lời

Ta có: $\overrightarrow{MA}=-\frac{3}{4}\overrightarrow{BM}=\frac{3}{4}\overrightarrow{MB};\overrightarrow{AN}=-3\overrightarrow{CN}=3\overrightarrow{NC};\overrightarrow{CP}=\frac{1}{4}\overrightarrow{PB}$

Ta lại có: $\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=\frac{3}{4}\overrightarrow{MB}+3\overrightarrow{NC}=\frac{3}{4}\overrightarrow{MP}+\frac{3}{4}\overrightarrow{PB}+3\overrightarrow{NP}+3\overrightarrow{PC}$

_{$\iff \overrightarrow{MN}=\frac{3}{4}\overrightarrow{MP}+3\overrightarrow{CP}+3\overrightarrow{NP}-3\overrightarrow{CP}=\frac{3}{4}\overrightarrow{MP}+3\overrightarrow{NP}$}

_{$\iff \overrightarrow{MP}+\overrightarrow{PN}=\frac{3}{4}\overrightarrow{MP}+3\overrightarrow{NP}$}

_{$\iff \frac{1}{4}\overrightarrow{MP}=4\overrightarrow{NP}\iff \overrightarrow{MP}=16\overrightarrow{NP}$}

Do đó M, N, P thằng hàng.