Trong mặt phẳng tọa độ Oxy cho A(1;0);B(2;-1);C(3;2) và đường thẳng $\Delta :x-2y-1=0$ .Tìm $M \in \Delta $ sao cho $\left| {\overrightarrow{MA}}-2\overrightarrow{MB}+3 \overrightarrow{MC}\right|$ đạt GTNN
$M \in \Delta \Rightarrow M(2a+1;a)$
Ta có $\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}=(4-4a;-2a+8)$
$\left| {\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}} \right|=\sqrt{(4-4a)^{2}+(-2a+8)^{2}}=\sqrt{20a^{2}-64a+80}$
$\left| {\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}} \right|$$_{min}\Leftrightarrow (20a^2-64a+80)_{min}$
Mà $(20a^2-64a+80)_{min}=\frac{144}{5}$ khi $a=\frac{8}{5}\Rightarrow M(\frac{21}{5};\frac{8}{5})$
Vậy $M(\frac{21}{5};\frac{8}{5})$

Gọi I là điểm sao cho $\underset{IA}{\rightarrow} -2\underset{IB}{\rightarrow} +3 \underset{IC}{\rightarrow}=0  $ thì $ I(3,4) $
Ta có $ \left| { \underset{MA}{\rightarrow} -2\underset{MB}{\rightarrow} +3\underset{MC}{\rightarrow} } \right| =\left| {2\underset{MI}{\rightarrow}} \right| $ $=$ $2MI$
Để MI nhỏ nhất thì M là hình chiếu của I trên $\Delta$  suy ra $ M( \frac{21}{5}, \frac{8}{5}) $ 

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