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Ta có: \(\overrightarrow{MB}=3\overrightarrow{MC}\Rightarrow\overrightarrow{MB}=3\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)
\(\Rightarrow\overrightarrow{MB}=3\overrightarrow{MB}+3\overrightarrow{BC}\)
\(\Rightarrow-\overrightarrow{MB}=3\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{BM}=\dfrac{2}{3}\overrightarrow{BC}\). Mà \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}\) nên \(\overrightarrow{BM}=\dfrac{2}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
Theo quy tắc 3 điểm, ta có
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\Rightarrow\overrightarrow{AM}=\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}\)
\(\Rightarrow\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\) hay \(\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{u}+\dfrac{3}{2}\overrightarrow{v}\)
= 3
=> 
)
= 
- 
nên 
= 
+ 
a.
\(\overrightarrow{AM}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CM}+\overrightarrow{BM}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{BM}\)
b.
\(\overrightarrow{AE}=3\overrightarrow{EM}=3\overrightarrow{EA}+3\overrightarrow{AM}\Rightarrow4\overrightarrow{AE}=3\overrightarrow{AM}\Rightarrow\overrightarrow{AE}=\dfrac{3}{4}\overrightarrow{AM}\)
\(\Rightarrow\overrightarrow{AE}=\dfrac{3}{4}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{3}{8}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}=-\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}=-\dfrac{5}{8}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=-\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}=\dfrac{8}{5}\overrightarrow{BE}\)
\(\Rightarrow\) B, E, K thẳng hàng
\(\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{CM}=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CB}=\overrightarrow{AC}+\dfrac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{AB}\right)\)
\(=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AB}\).
a/ \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{EF}+\overrightarrow{FE}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\)
b/ Theo tính chất trung tuyến:
\(\left\{{}\begin{matrix}\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AK}\\\overrightarrow{BA}+\overrightarrow{BC}=2\overrightarrow{BM}\end{matrix}\right.\) \(\Rightarrow\overrightarrow{AC}+\overrightarrow{BC}=2\overrightarrow{AK}+2\overrightarrow{BM}\)
\(\overrightarrow{AC}=\overrightarrow{AK}+\overrightarrow{KC}=\overrightarrow{AK}+\frac{1}{2}\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{BC}=\overrightarrow{AK}+2\overrightarrow{BM}-\frac{1}{2}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\)
\(\Rightarrow\overrightarrow{AC}=\overrightarrow{AK}+\frac{1}{2}\left(\frac{3}{2}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\right)=...\)
\(\overrightarrow{AB}=\overrightarrow{AC}-\overrightarrow{BC}=...\)
Theo tính chất trung điểm
\(\overrightarrow{AE}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{AC}\)\(=\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\)\(=\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{AB}=\overrightarrow{u}+\dfrac{1}{2}\overrightarrow{v}\).
a/ \(\overrightarrow{DA}-\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{BD}=\overrightarrow{BA}\)
\(\overrightarrow{OD}-\overrightarrow{OC}=\overrightarrow{OD}+\overrightarrow{CO}=\overrightarrow{CD}\)
Mà \(\overrightarrow{BA}=\overrightarrow{CD}\) (t/c hình bình hành) \(\Rightarrow\) đpcm
b/ Theo tính chất trung tuyến:
\(\left\{{}\begin{matrix}\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AK}\\\overrightarrow{BA}+\overrightarrow{BC}=2\overrightarrow{BM}\end{matrix}\right.\) \(\Rightarrow\overrightarrow{AC}+\overrightarrow{BC}=2\overrightarrow{AK}+2\overrightarrow{BM}\)
\(\Rightarrow\overrightarrow{AC}+\overrightarrow{BA}+\overrightarrow{AC}=2\overrightarrow{AK}+2\overrightarrow{BM}\)
\(\Rightarrow2\overrightarrow{AC}-\overrightarrow{AB}=2\overrightarrow{AK}+2\overrightarrow{BM}\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AK}\\2\overrightarrow{AC}-\overrightarrow{AB}=2\overrightarrow{AK}+2\overrightarrow{BM}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AC}=\frac{4}{3}\overrightarrow{AK}+\frac{2}{3}\overrightarrow{BM}\\\overrightarrow{AB}=\frac{2}{3}\overrightarrow{AK}-\frac{2}{3}\overrightarrow{BM}\end{matrix}\right.\)
Gọi G là giao điểm của AK, BM thì G là trọng tâm của tam giác.
Ta có
= 
=> 
=
Theo quy tắc 3 điểm đối với tổng vec tơ:
AK là trung tuyến thuộc cạnh BC nên
Từ đây ta có
= 
+
=> 
= -
- 
.
BM là trung tuyến thuộc đỉnh B nên
=>
= 
+
.