Question

Cho tam giác ABC biết A(2;5), B(-1;8),C(4;-3). Tìm tọa độ điểm M ∈ Ox sao cho:

a)\(\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|\) đạt GTNN.

b) /\(\left|2\overrightarrow{MA}-\overrightarrow{MB}+3\overrightarrow{MC}\right|\) đạt giá trị nhỏ nhất.

Answer 1

Gọi \(M\left(x;0\right)\Rightarrow\overrightarrow{MA}\left(2-x;5\right)\) ; \(\overrightarrow{MB}=\left(-1-x;8\right)\); \(\overrightarrow{MC}=\left(4-x;-3\right)\)

a/ \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(5-3x;10\right)\)

\(\Rightarrow T=\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\sqrt{\left(5-3x\right)^2+10^2}\ge10\)

\(T_{min}=10\) khi \(5-3x=0\Rightarrow x=\frac{5}{3}\Rightarrow M\left(\frac{5}{3};0\right)\)

b/ \(2\overrightarrow{MA}-\overrightarrow{MB}+3\overrightarrow{MC}=\left(17-4x;-7\right)\)

\(\Rightarrow A=\left|2\overrightarrow{MA}-\overrightarrow{MB}+3\overrightarrow{MC}\right|=\sqrt{\left(17-4x\right)^2+\left(-7\right)^2}\ge7\)

\(A_{min}=7\) khi \(17-4x=0\Rightarrow x=\frac{17}{4}\Rightarrow M\left(\frac{17}{4};0\right)\)