Cho ba điểm $A(1;0), B(0;3), C(-3;-5)$. Tìm điểm $M$ thuộc trục $Ox$ mà $T=|2 \overrightarrow{MA}-3 \overrightarrow{MB}+2 \overrightarrow{MC}|$ bé nhất.
Gọi $I(x;y)$ là điểm thỏa mãn: $2 \overrightarrow{IA}-3 \overrightarrow{IB}+2 \overrightarrow{IC}=\overrightarrow{0}$
$\Leftrightarrow \left\{ \begin{array}{l} 2(1-x)-3(-x)+2(-3-x)=0\\ 2(-y)-3(3-y)+2(-5-y)=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x=-4\\ y=-19  \end{array} \right.$
Ta có $T= |2 \overrightarrow{MA}-3 \overrightarrow{MB}+2 \overrightarrow{MC}|$
$= |2(\overrightarrow{MI}+\overrightarrow{IA})-3 (\overrightarrow{MI}+\overrightarrow{IB})+2(\overrightarrow{MI}+\overrightarrow{IC})|$
$=|\overrightarrow{MI}+2 \overrightarrow{IA}-3 \overrightarrow{IB}+2 \overrightarrow{IC}|=|\overrightarrow{MI}|=MI$
Vì $I$ cố định và $M$ thuộc trục $Ox $ nên $T$ bé nhất khi $M$ là hình chiếu $I$ lên $Ox$. Vậy $M(-4;0)$.          

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