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a) \(\overrightarrow{BI}=\overrightarrow{BC}+\overrightarrow{CI}=\overrightarrow{BC}+\dfrac{1}{4}\overrightarrow{CA}=\overrightarrow{BA}+\overrightarrow{AC}+\dfrac{1}{4}\overrightarrow{CA}\)
\(=\overrightarrow{BA}+\overrightarrow{AC}-\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AC}+\overrightarrow{BA}=\dfrac{3}{4}\overrightarrow{AC}-\overrightarrow{AB}\).
b) Có \(\overrightarrow{BJ}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{2}{3}\overrightarrow{AB}=\dfrac{3}{2}\left(\dfrac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)=\dfrac{3}{2}\overrightarrow{BI}\).
Vì vậy 3 điểm B, I, J thẳng hàng.
c)
Trên cạnh AC lấy điểm K sao cho \(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AC}\).
Tại điểm K dựng điểm T sao cho \(\overrightarrow{KT}=-\dfrac{3}{2}\overrightarrow{AB}=\dfrac{3}{2}\overrightarrow{BA}\).
\(\overrightarrow{BJ}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}=\overrightarrow{AK}+\overrightarrow{KT}=\overrightarrow{AT}\).
Dựng điểm T sao cho \(\overrightarrow{BJ}=\overrightarrow{AT}\).
a/ \(\overrightarrow{AC}=3\overrightarrow{AM};\overrightarrow{BN}=\frac{1}{2}\overrightarrow{BC}\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}=\frac{1}{3}\overrightarrow{CA}+\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}\)
\(=\frac{1}{3}\overrightarrow{CB}+\frac{1}{3}\overrightarrow{CD}+\overrightarrow{DC}+\frac{1}{2}\overrightarrow{BC}=\frac{2}{3}\overrightarrow{DC}+\frac{1}{6}\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{AC}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)
Hmm, MN làm sao vuông góc vs BC đc. Nó chỉ vuông góc khi M là TĐ của AC thôi, bởi N là trung điểm của BC rồi mà, hại não :((
2/\(\overrightarrow{BK}=\frac{4}{13}\overrightarrow{BA}\Rightarrow\overrightarrow{BC}+\overrightarrow{CK}=\frac{4}{13}\overrightarrow{BC}+\frac{4}{13}\overrightarrow{CA}\)
\(\Leftrightarrow\overrightarrow{CK}=\frac{9}{13}\overrightarrow{CB}+\frac{4}{13}\overrightarrow{CA}\)
\(\overrightarrow{GB}+\overrightarrow{GM}+\overrightarrow{GN}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{GC}+\overrightarrow{CB}+\overrightarrow{GC}+\overrightarrow{CM}+\overrightarrow{GC}+\overrightarrow{CN}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GC}+\overrightarrow{CB}+\overrightarrow{CN}+\overrightarrow{NM}+\overrightarrow{CN}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GC}+\overrightarrow{CB}+2\overrightarrow{CN}+\frac{1}{2}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{CA}=\overrightarrow{0}\)
Ta có : \(\overrightarrow{CN}=\frac{1}{2}\overrightarrow{CB}\Rightarrow3\overrightarrow{GC}+\overrightarrow{CB}+\overrightarrow{CB}+\frac{1}{2}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{CA}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{CG}=\frac{2}{3}\overrightarrow{CB}+\frac{1}{6}\overrightarrow{BA}+\frac{1}{18}\overrightarrow{CA}\)
\(\Leftrightarrow\overrightarrow{CG}=\frac{2}{3}\overrightarrow{CB}+\frac{1}{6}\overrightarrow{BC}+\frac{1}{6}\overrightarrow{CA}+\frac{1}{18}\overrightarrow{CA}\)
\(=\frac{1}{2}\overrightarrow{CB}+\frac{2}{9}\overrightarrow{CA}\)
Có \(\overrightarrow{CK}=\frac{18}{13}\overrightarrow{CG}\Rightarrow\) C,G,K thẳng hàng
\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)
\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)
\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)
\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)
\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)
\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)
\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\) \(trung\) \(điểm\) \(BC)\)
\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)
Đok đề cứ thấy sai sai... Sao cho J lại thoả mãn \(\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AC}-\frac{2}{3}\overrightarrow{AB}\) :))