Question

[Thanh Trúc]

a, \(\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|\)=\(\frac{3}{2}\)\(\left|\overrightarrow{MB}+\overrightarrow{MC}\right|\)

b, \(\left|\overrightarrow{MA}+\overrightarrow{BC}\right|\)=\(\left|\overrightarrow{MA}-\overrightarrow{MB}\right|\)

Answer 1

[Trần Quốc Lộc]

\(\text{a) }\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\frac{3}{2}\left|\overrightarrow{MB}+\overrightarrow{MC}\right|\\ \Rightarrow\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)^2=\left(\frac{3}{2}\overrightarrow{MB}+\frac{3}{2}\overrightarrow{MC}\right)^2\\ \Rightarrow\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)^2-\left(\frac{3}{2}\overrightarrow{MB}+\frac{3}{2}\overrightarrow{MC}\right)^2=0\\ \Rightarrow\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\frac{3}{2}\overrightarrow{MB}+\frac{3}{2}\overrightarrow{MC}\right)\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}-\frac{3}{2}\overrightarrow{MB}-\frac{3}{2}\overrightarrow{MC}\right)=0\\ \Rightarrow\left[\overrightarrow{MA}+\frac{5}{2}\left(\overrightarrow{MB}+\overrightarrow{MC}\right)\right]\left[\overrightarrow{MA}-\frac{1}{2}\left(\overrightarrow{MB}+\overrightarrow{MC}\right)\right]=0\)

Gọi I là trung điểm BC

\(\Rightarrow\left(\overrightarrow{MA}+5\overrightarrow{MI}\right)\left(\overrightarrow{MA}-\overrightarrow{MI}\right)=0\\ \Rightarrow\left[{}\begin{matrix}\overrightarrow{MA}+5\overrightarrow{MI}=0\\\overrightarrow{MA}-\overrightarrow{MI}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\overrightarrow{MA}=-5\overrightarrow{MI}\\\overrightarrow{IA}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}M;A;I\text{ thẳng hàng },M\text{ nằm giữa }AI\text{ và }MA=5MI\\I\equiv A\end{matrix}\right.\)

Vậy với A là trung điểm BC thì M tùy ý.

Với A không là trung điểm BC thì \(M;A;I\text{ thẳng hàng },M\text{ nằm giữa }AI\text{ và }MA=5MI\)

Answer 2

[Trần Quốc Lộc]

\(\text{b) }\left|\overrightarrow{MA}+\overrightarrow{BC}\right|=\left|\overrightarrow{MA}-\overrightarrow{MB}\right|\\ \Leftrightarrow\left(\overrightarrow{MA}+\overrightarrow{BC}\right)^2-\left(\overrightarrow{MA}-\overrightarrow{MB}\right)^2=0\\ \Leftrightarrow\left(\overrightarrow{MA}+\overrightarrow{BC}+\overrightarrow{MA}-\overrightarrow{MB}\right)\left(\overrightarrow{MA}+\overrightarrow{BC}-\overrightarrow{MA}+\overrightarrow{MB}\right)=0\\ \Leftrightarrow\left(\overrightarrow{MA}+\overrightarrow{BC}+\overrightarrow{MA}-\overrightarrow{MB}\right)\left(\overrightarrow{BC}+\overrightarrow{MB}\right)=0\\ \Leftrightarrow\left(\overrightarrow{MA}+\overrightarrow{BC}+\overrightarrow{BA}\right)\left(\overrightarrow{BC}+\overrightarrow{MB}\right)=0\)

Gọi D là trung điểm AC

\(\Leftrightarrow\left(\overrightarrow{MA}+2\overrightarrow{BD}\right)\left(\overrightarrow{BC}+\overrightarrow{MB}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\overrightarrow{MA}+2\overrightarrow{BD}=0\\\overrightarrow{BC}+\overrightarrow{MB}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\overrightarrow{MA}=-2\overrightarrow{BD}\\\overrightarrow{BC}-\overrightarrow{BM}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\overrightarrow{MA}=-2\overrightarrow{BD}\\\overrightarrow{MC}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}MA//BD;MA=2BD\\M\equiv C\end{matrix}\right.\)

Vậy......