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Lời giải:
a)
\(\bullet \overrightarrow{IM}=\frac{1}{2}\overrightarrow{BM}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AM})=\frac{1}{2}(\overrightarrow{BA}+\frac{1}{2}\overrightarrow{AC})\)
\(=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}\)
\(\bullet \overrightarrow{AI}=\overrightarrow{AM}+\overrightarrow{MI}=\frac{1}{2}\overrightarrow{AC}-\overrightarrow{IM}=\frac{1}{2}\overrightarrow{AC}-(-\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC})\)
\(=\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}\)
b)
Để \(\overline{A,I,K}\) thì tồn tại \(m\in\mathbb{R}|\overrightarrow{AI}=m\overrightarrow{AK}\)
\(\Leftrightarrow \overrightarrow{AI}=m(\overrightarrow{AB}+\overrightarrow{BK})\)
\(\Leftrightarrow \overrightarrow{AI}=m(\overrightarrow{AB}+x\overrightarrow{BC})\)
\(\Leftrightarrow \overrightarrow{AI}=m\overrightarrow{AB}+mx(\overrightarrow{BA}+\overrightarrow{AC})\)
\(\Leftrightarrow \frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}=(m-mx)\overrightarrow{AB}+mx\overrightarrow{AC}\)
\(\Rightarrow m-mx=\frac{1}{2}; mx=\frac{1}{4}\Rightarrow m=\frac{3}{4}; x=\frac{1}{3}\)
b) giả sử ta có A, I, K thẳng hàng=> ta có tỉ lệ \(\dfrac{AI}{AK}\)(1)
AK= AB+ BK
AK= AB+ xBC
AK= AB+ xBA+ x AC
AK= (1-x) AB+ xAC(2)
mà từ câu a) ta đã tìm được AI= 1/2AB+ 1/4AC(3)
từ (1), (2) và (3)=> \(\dfrac{1}{2-2x}=\dfrac{1}{4x}\)=> x=1/3
Bài 1 và Bài 2 tương tự nhau nên mk sẽ chỉ CM bài 1 thôi nha
Có \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\overrightarrow{AB}+\overrightarrow{CD}=0\)
\(\Rightarrow\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}=0\)
\(\Leftrightarrow\overrightarrow{AD}+\overrightarrow{CB}=0\Leftrightarrow\overrightarrow{AD}=\overrightarrow{BC}\)
Bài 3:
Xét \(\Delta AIP\) theo quy tắc trung điểm có:
\(\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}}{2}\)
Làm tương tự vs các tam giác còn lại
\(\Rightarrow\overrightarrow{IB}=\frac{\overrightarrow{IN}+\overrightarrow{IC}}{2}\)
\(\Rightarrow\overrightarrow{IA}=\frac{\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
Cộng vế vs vế
\(\Rightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}+\overrightarrow{IN}+\overrightarrow{IC}+\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
\(\Leftrightarrow2\overrightarrow{IA}+2\overrightarrow{IB}+2\overrightarrow{IC}=\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\)
\(\Leftrightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\left(đpcm\right)\)
\(\overrightarrow{BM}=\overrightarrow{BC}-2\overrightarrow{AB}\Leftrightarrow\overrightarrow{BI}+\overrightarrow{IM}=\overrightarrow{BC}-2\left(\overrightarrow{AC}+\overrightarrow{CB}\right)\)
\(\Leftrightarrow\frac{1}{2}\overrightarrow{BC}+\overrightarrow{IM}=\overrightarrow{BC}-2\overrightarrow{AC}+2\overrightarrow{BC}\Rightarrow\overrightarrow{IM}=\frac{5}{2}\overrightarrow{BC}-2\overrightarrow{AC}\)
\(\overrightarrow{CI}+\overrightarrow{IN}=x\overrightarrow{AC}-\overrightarrow{BC}\Rightarrow-\frac{1}{2}\overrightarrow{BC}+\overrightarrow{IN}=x\overrightarrow{AC}-\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{IN}=-\frac{1}{2}\overrightarrow{BC}+x\overrightarrow{AC}=-\frac{1}{5}\left(\frac{5}{2}\overrightarrow{BC}-5x.\overrightarrow{AC}\right)\)
Để MN qua I hay I;M;N thẳng hàng \(\Leftrightarrow5x=2\Rightarrow x=\frac{2}{5}\)
\(\overrightarrow{AD}=2\overrightarrow{DB}\Rightarrow\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\) ; \(\overrightarrow{CE}=3\overrightarrow{EA}\Rightarrow\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
Lại có M là trung điểm DE
\(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AE}\right)=\dfrac{1}{2}\left(\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}\)
I là trung điểm BC \(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{MI}=\overrightarrow{MA}+\overrightarrow{AI}=\overrightarrow{AI}-\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{8}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
A B C D I M
a)
\(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=\dfrac{1}{2}\left(\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\).
b)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+x\overrightarrow{BC}\)\(=\overrightarrow{AB}+x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\left(1-x\right)\overrightarrow{AB}+x\overrightarrow{AC}\).
c) A, M, I thẳng hàng khi và chỉ khi hai véc tơ \(\overrightarrow{AM};\overrightarrow{AI}\) cùng phương
hay \(\dfrac{1-x}{\dfrac{1}{2}}=\dfrac{x}{\dfrac{3}{8}}\Leftrightarrow\dfrac{3}{8}\left(1-x\right)=\dfrac{1}{2}x\)
\(\Leftrightarrow\dfrac{7}{8}x=\dfrac{3}{8}\)\(\Leftrightarrow x=\dfrac{3}{7}\).