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A B C D I K
a)
- \(\overrightarrow{BI}=\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\) (t/c trung điểm)
\(=\frac{1}{2}\left(\overrightarrow{BA}+\frac{1}{2}\overrightarrow{BC}\right)\)
\(=\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\)
- \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)
\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AC}\)
\(=\overrightarrow{BA}+\frac{1}{3}\left(\overrightarrow{BC}-\overrightarrow{BA}\right)\)
\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}-\frac{1}{3}\overrightarrow{BA}\)
\(=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}\)
b) Ta có: \(\overrightarrow{BK}=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}=\frac{4}{3}\left(\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\right)=\frac{4}{3}\overrightarrow{BI}\)
=> B,K,I thẳng hàng
c) \(27\overrightarrow{MA}-8\overrightarrow{MB}=2015\overrightarrow{MC}\)
\(\Leftrightarrow27\left(\overrightarrow{MC}+\overrightarrow{CA}\right)-8\left(\overrightarrow{MC}+\overrightarrow{CB}\right)=2015\overrightarrow{MC}\)
\(\Leftrightarrow27\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{MC}-8\overrightarrow{CB}-2015\overrightarrow{MC}=\overrightarrow{0}\)
\(\Leftrightarrow-1996\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{CB}=\overrightarrow{0}\)
\(\Leftrightarrow1996\overrightarrow{CM}=8\overrightarrow{CB}-27\overrightarrow{CA}\)
\(\Leftrightarrow\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)
Vậy: Dựng điểm M sao cho \(\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)
Câu 1:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
M là trung điểm BC \(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{IM}=2\overrightarrow{AI}\Rightarrow\overrightarrow{IA}+\overrightarrow{AM}=2\overrightarrow{AI}\)
\(\Rightarrow-\overrightarrow{AI}+\overrightarrow{AM}=2\overrightarrow{AI}\)
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{3}\overrightarrow{AM}=\dfrac{1}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{BI}=\overrightarrow{BA}+\overrightarrow{AI}=-\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}=-\dfrac{5}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}\)
Đặt \(\overrightarrow{AK}=x.\overrightarrow{AC}\)
\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=-\overrightarrow{AB}+x.\overrightarrow{AC}\)
Do B, I, K thẳng hàng \(\Rightarrow\overrightarrow{BK}\) và \(BI\) cùng phương
\(\Rightarrow\dfrac{-1}{\left(-\dfrac{5}{6}\right)}=\dfrac{x}{\left(\dfrac{1}{6}\right)}\Rightarrow x=\dfrac{1}{5}\)
\(\Rightarrow\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AC}=\dfrac{1}{5}\left(\overrightarrow{AK}+\overrightarrow{KC}\right)=\dfrac{1}{5}\overrightarrow{AK}+\dfrac{1}{5}\overrightarrow{KC}\)
\(\Rightarrow\dfrac{4}{5}\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{KC}\)
\(\Rightarrow4.\overrightarrow{AK}=1.\overrightarrow{KC}\Rightarrow4.\overrightarrow{KA}=1.\overrightarrow{CK}\)
\(\Rightarrow\left\{{}\begin{matrix}n=4\\m=1\end{matrix}\right.\)
Các kí hiệu em ghi như IM=2AI và nKA=mCK nó là đoạn thẳng hay có vecto?
Câu 1:
Gọi E là trung điểm của KC
=>AK=KE=EC
Xét ΔBKC có CM/CB=CE/CK
nên ME//BK
Xét ΔAME có AI/AM=AK/AE
nên IK//ME
=>IK//BK
=>B,I,K thẳng hàng
moi nguoi oi giup minh voi