Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\overrightarrow{AM}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\overrightarrow{AC}\)
b: \(=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
\(=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
c: \(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)=\overrightarrow{0}\)
MN là đường trung bình của tam giác ABC
\(\Rightarrow\overrightarrow{MN}=\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
Từ giả thiết:
\(\overrightarrow{KM}=-2\overrightarrow{KN}=-2\left(\overrightarrow{KM}+\overrightarrow{MN}\right)\)
\(\Rightarrow3\overrightarrow{KM}=2\overrightarrow{NM}\Rightarrow\overrightarrow{KM}=\dfrac{2}{3}\overrightarrow{NM}\)
\(\Rightarrow\overrightarrow{MK}=\dfrac{2}{3}\overrightarrow{MN}=\dfrac{2}{3}\left(-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
M là trung điểm AB \(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}\)
Do đó:
\(\overrightarrow{AK}=\overrightarrow{AM}+\overrightarrow{MK}=\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)
\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)
\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)
\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)
\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)
\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)
\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\) \(trung\) \(điểm\) \(BC)\)
\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)