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\(\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|\overrightarrow{AB}-\overrightarrow{AC}\right|\)
\(\Leftrightarrow\left|\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right|=\left|\overrightarrow{AB}+\overrightarrow{CA}\right|\)
\(\Leftrightarrow\left|3\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right|=\left|\overrightarrow{CB}\right|\)
\(\Leftrightarrow\left|3\overrightarrow{MG}\right|=\left|\overrightarrow{CB}\right|\)
\(\Leftrightarrow MG=\dfrac{1}{3}BC\)
Tập hợp M là đường tròn tâm G bán kính \(R=\dfrac{BC}{3}\)
Do G là trọng tâm ABC \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
Do I là trung điểm BC \(\Rightarrow\overrightarrow{MB}+\overrightarrow{MC}=2\overrightarrow{MI}\)
\(2\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=3\left|\overrightarrow{MB}+\overrightarrow{MC}\right|\)
\(\Leftrightarrow2\left|\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right|=3.\left|2\overrightarrow{MI}\right|\)
\(\Leftrightarrow2.\left|3\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right|=6\left|\overrightarrow{MI}\right|\)
\(\Leftrightarrow6\left|\overrightarrow{MG}\right|=6\left|\overrightarrow{MI}\right|\)
\(\Leftrightarrow MG=MI\)
Tập hợp M là đường trung trực của đoạn thẳng IG
d, Lấy P, Q sao cho \(4\overrightarrow{PA}-\overrightarrow{PB}+\overrightarrow{PC}=\overrightarrow{0};2\overrightarrow{QA}-\overrightarrow{QB}-\overrightarrow{QC}=\overrightarrow{0}\)
Ta có \(\left|4\overrightarrow{MA}-\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|4\text{ }\overrightarrow{MP}+4\overrightarrow{PA}-\overrightarrow{PB}+\overrightarrow{PC}\right|=\left|4\overrightarrow{MP}\right|=4MP\)
\(\left|2\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC}\right|=\text{ }\left|2\overrightarrow{QA}-\overrightarrow{QB}-\overrightarrow{QC}\right|=0\)
\(\Rightarrow4MP=0\Rightarrow M\equiv P\)
Gọi G là trọng tâm tam giác, I là trung điểm BC, N là trung điểm của AC
a, Ta có \(\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|3\overrightarrow{MG}\right|=3MG\)
\(\frac{3}{2}\left|\overrightarrow{MB}+\overrightarrow{MC}\right|=\frac{3}{2}\left|2\overrightarrow{MI}\right|=3MI\)
\(\Rightarrow MG=MI\Rightarrow M\) thuộc đường trung trực của BC
b, \(\left|\overrightarrow{MA}+\overrightarrow{MC}\right|=\left|2\overrightarrow{MN}\right|=2MN\)
\(\left|\overrightarrow{MA}-\overrightarrow{MB}\right|=\left|\overrightarrow{BA}\right|=BA\)
\(\Rightarrow2MN=BA\Rightarrow M\in\left(N;\frac{BA}{2}\right)\)
\(\text{a) }\left|2\overrightarrow{MA}+3\overrightarrow{MB}\right|=\left|3\overrightarrow{MB}-2\overrightarrow{MC}\right|\\ \Rightarrow\left(2\overrightarrow{MA}+3\overrightarrow{MB}\right)^2=\left(3\overrightarrow{MB}-2\overrightarrow{MC}\right)^2\\ \Rightarrow\left(2\overrightarrow{MA}+3\overrightarrow{MB}\right)^2-\left(3\overrightarrow{MB}-2\overrightarrow{MC}\right)^2=0\\ \Rightarrow\left(2\overrightarrow{MA}+3\overrightarrow{MB}-3\overrightarrow{MB}+2\overrightarrow{MC}\right)\left(2\overrightarrow{MA}+3\overrightarrow{MB}+3\overrightarrow{MB}-2\overrightarrow{MC}\right)=0\\ \Rightarrow\left(2\overrightarrow{MA}+2\overrightarrow{MC}\right)\left[2\left(\overrightarrow{MA}-\overrightarrow{MC}\right)+6\overrightarrow{MB}\right]=0\\ \Rightarrow\left(\overrightarrow{MA}+\overrightarrow{MC}\right)\left(\overrightarrow{CA}+3\overrightarrow{MB}\right)=0\\ \Rightarrow\left[{}\begin{matrix}\overrightarrow{MA}+\overrightarrow{MC}=0\\\overrightarrow{CA}+3\overrightarrow{MB}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\overrightarrow{MA}=-\overrightarrow{MC}\\\overrightarrow{CA}=-3\overrightarrow{MB}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}M;A;C\text{ thẳng hàng };M\text{ nằm giữa }A;C\\MA=MC\end{matrix}\right.\\\left\{{}\begin{matrix}CA//MB\\CA=3MB\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}M\text{ là trung điểm }AC\\CA//MB;CA=3MB\end{matrix}\right.\)
Vậy......
\(b\text{) }\left|4\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|2\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC}\right|\\ \Rightarrow\left(4\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)^2=\left(2\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC}\right)^2\\ \Rightarrow\left(4\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)^2-\left(2\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC}\right)^2=0\\ \Rightarrow\left(4\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}-2\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)\left(4\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+2\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC}\right)=0\\ \Rightarrow\left(2\overrightarrow{MA}+2\overrightarrow{MB}+2\overrightarrow{MC}\right)\cdot6\overrightarrow{MA}=0\\ \Rightarrow\overrightarrow{MA}\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)=0\\ \Rightarrow\left[{}\begin{matrix}\overrightarrow{MA}=0\\\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}M\equiv A\\M\text{ là trọng tâm }\Delta ABC\end{matrix}\right.\)Vậy...........
Qua A dựng đường thẳng d song song BC, trên d lấy điểm I sao cho \(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{BC}\)
\(\Rightarrow3\overrightarrow{IA}=2\overrightarrow{BC}\Rightarrow3\overrightarrow{IA}+2\overrightarrow{CB}=\overrightarrow{0}\)
Ta có:
\(\left|3\overrightarrow{MA}+2\overrightarrow{MB}-2\overrightarrow{MC}\right|=\left|\overrightarrow{MB}-\overrightarrow{MC}\right|\)
\(\Leftrightarrow\left|3\overrightarrow{MA}+2\left(\overrightarrow{MB}+\overrightarrow{CM}\right)\right|=\left|\overrightarrow{MB}+\overrightarrow{CM}\right|\)
\(\Leftrightarrow\left|3\overrightarrow{MA}+2\overrightarrow{CB}\right|=\left|\overrightarrow{CB}\right|\)
\(\Leftrightarrow\left|3\overrightarrow{MI}+3\overrightarrow{IA}+2\overrightarrow{CB}\right|=\left|\overrightarrow{CB}\right|\)
\(\Leftrightarrow\left|3\overrightarrow{MI}\right|=\left|\overrightarrow{CB}\right|\)
\(\Leftrightarrow MI=\dfrac{1}{3}BC\)
Tập hợp M là đường tròn tâm I bán kính \(\dfrac{BC}{3}\)