Question

Đề là chứng minh M N và P thẳng hàng nha cái chỗ khó nhìn là thoả mãn Giúp mik với

Answer 1

3.

\(\overrightarrow{IA}+3\overrightarrow{IC}=\overrightarrow{0}\Rightarrow\overrightarrow{IB}+\overrightarrow{BA}+3\overrightarrow{IB}+3\overrightarrow{BC}=\overrightarrow{0}\)

\(\Rightarrow4\overrightarrow{IB}+\overrightarrow{BA}+3\overrightarrow{BC}=\overrightarrow{0}\Rightarrow4\overrightarrow{IB}=\overrightarrow{AB}+3\overrightarrow{CB}\) (1)

\(\overrightarrow{JA}+2\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\Rightarrow\overrightarrow{JB}+\overrightarrow{BA}+2\overrightarrow{JB}+3\overrightarrow{JB}+3\overrightarrow{BC}=\overrightarrow{0}\)

\(\Rightarrow6\overrightarrow{JB}+\overrightarrow{BA}+3\overrightarrow{BC}=0\Rightarrow6\overrightarrow{JB}=\overrightarrow{AB}+3\overrightarrow{CB}\) (2)

(1);(2) \(\Rightarrow4\overrightarrow{IB}=6\overrightarrow{JB}\Rightarrow\overrightarrow{IB}\) và \(\overrightarrow{JB}\) cùng phương

Hay I; J; B thẳng hàng

Answer 2

4.

\(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{0}\Rightarrow\overrightarrow{PA}+\overrightarrow{PA}+\overrightarrow{AB}=0\Rightarrow\overrightarrow{PA}=-\dfrac{1}{2}\overrightarrow{AB}\)

\(\overrightarrow{NA}=3\overrightarrow{CN}\Rightarrow\overrightarrow{NA}=3\overrightarrow{CA}+3\overrightarrow{AN}\Rightarrow4\overrightarrow{AN}=3\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{AN}=\dfrac{3}{4}\overrightarrow{AC}\)

\(\overrightarrow{MB}=3\overrightarrow{MC}\Rightarrow\overrightarrow{MB}=3\overrightarrow{MB}+3\overrightarrow{BC}\)

\(\Rightarrow2\overrightarrow{BM}=3\overrightarrow{BC}\Rightarrow\overrightarrow{BM}=\dfrac{3}{2}\overrightarrow{BC}=\dfrac{3}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\dfrac{3}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{PN}=\overrightarrow{PA}+\overrightarrow{AN}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\) (1)

\(\overrightarrow{PM}=\overrightarrow{PB}+\overrightarrow{BM}=\dfrac{1}{2}\overrightarrow{AB}-\dfrac{3}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}=-\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}=2\left(-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)\) (2)

(1);(2) \(\Rightarrow\overrightarrow{PM}=2\overrightarrow{PN}\Rightarrow\) P, M, N thẳng hàng