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Gọi tọa độ điểm \(M\) là \(M\left(x;y\right).\)
\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;-y\right);\overrightarrow{MC}=\left(2-x;-5-y\right).\)
Ta có: \(\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\overrightarrow{0}.\)
\(\left\{{}\begin{matrix}1-x+4-x-3\left(2-x\right)=0.\\3-y-y-3\left(-5-y\right)=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2x+5-6+3x=0.\\3-2y+15+3y=0.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0.\\y+18=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1.\\y=-18.\end{matrix}\right.\) \(\Rightarrow M\left(1;-18\right).\)
câu 1: \(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}=4\overrightarrow{AG}\) Ta có vế trái
\(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}=\overrightarrow{AE}+\overrightarrow{EB}+\overrightarrow{AG}+\overrightarrow{GC}+\overrightarrow{AG}+\overrightarrow{GD}\\ =2\overrightarrow{AE}+2\overrightarrow{AG}+\overrightarrow{GC}+\overrightarrow{GD}\\ =2\overrightarrow{AG}+2\overrightarrow{GE}+2\overrightarrow{AG}+\overrightarrow{GC}+\overrightarrow{GD}\\ =4\overrightarrow{AG}+2\overrightarrow{GE}+\overrightarrow{GC}+\overrightarrow{GD}\\ =4\overrightarrow{AG}+2\overrightarrow{GE}+\overrightarrow{GF}+\overrightarrow{FC}+\overrightarrow{GF}+\overrightarrow{FD}\\ =4\overrightarrow{AG}+2\left(\overrightarrow{GF}+\overrightarrow{GE}\right)+\overrightarrow{FC}+\overrightarrow{FD}\\ =4\overrightarrow{AG}\left(đpcm\right)\)
a: \(\left|\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AC}\right|=2\cdot AC=2\cdot5=10\)
b: \(\left|\overrightarrow{AM}+\overrightarrow{AN}\right|=\left|\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}+\dfrac{\overrightarrow{AD}+\overrightarrow{AC}}{2}\right|\)
\(=\left|\dfrac{3\cdot\overrightarrow{AC}}{2}\right|=\dfrac{3}{2}AC=\dfrac{3}{2}\cdot5=\dfrac{15}{2}=7.5\)
\(\overrightarrow{AH}=\frac{2}{3}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}\Leftrightarrow2\overrightarrow{AC}-\overrightarrow{AB}=3\overrightarrow{AH}\)
Gọi I là trung điểm AC
Ta có : \(BG=GH=2GI\Rightarrow GI=IH\)
Tứ giác \(AGCH\)có 2 đường chéo cắt nhau tại trung điểm mỗi đường là hình bình hành
\(\Rightarrow AH=GC\)
\(2\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{BC}\)
\(=\overrightarrow{AH}+\overrightarrow{HC}+\overrightarrow{BH}+\overrightarrow{HC}=\overrightarrow{AH}+2\overrightarrow{GH}+2\overrightarrow{HC}\)
\(=\overrightarrow{AH}+2\overrightarrow{GH}+2\left(\overrightarrow{HG}+\overrightarrow{GC}\right)=\overrightarrow{AH}+2\overrightarrow{GC}=\overrightarrow{AH}+2\overrightarrow{AH}=3\overrightarrow{AH}\)
A B C H G I
Gọi \(M\left(a;b\right)\)
\(\Rightarrow\overrightarrow{MB}=\left(2-a;3-b\right)\Rightarrow2\overrightarrow{MB}=\left(4-2a;6-2b\right)\)
\(\overrightarrow{MC}=\left(-1-a;-2-b\right)\Rightarrow3\overrightarrow{MC}=\left(-3-3a;-6-3b\right)\)
\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(1-5a;-5b\right)=\overrightarrow{0}\)
\(\Rightarrow\left\{{}\begin{matrix}1-5a=0\\-5b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{5}\\b=0\end{matrix}\right.\) \(\Rightarrow M\left(\frac{1}{5};0\right)\)
a: vecto AB-vecto AD
=vecto DA+vecto AB
=vecto DB
-vecto CD-veco BC
=vecto CB-vecto CD
=vecto DC+vecto CB=vecto DB
=>vecto AB+vecto CD=vecto AD-vecto BC
b: \(\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\)
\(\overrightarrow{CD}-\overrightarrow{BD}=\overrightarrow{CD}+\overrightarrow{DB}=\overrightarrow{CB}\)
Do đó: \(\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CD}-\overrightarrow{BD}\)
=>\(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AC}-\overrightarrow{BD}\)
c: \(\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{DA}+\overrightarrow{AB}=\overrightarrow{DB}\)
\(\overrightarrow{CB}-\overrightarrow{CD}=\overrightarrow{DC}+\overrightarrow{CB}=\overrightarrow{DB}\)
Do đó: \(\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{CB}-\overrightarrow{CD}\)
=>\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}\)