Question

Cho tam giác ABC vuông cân tại A, cạnh AB = 2, M là điểm tùy ý, độ dài \(\left|3\overrightarrow{MA}-5\overrightarrow{MB}+2\overrightarrow{MC}\right|=...\)

Answer 1

Đặt \(\overrightarrow{u}=3\overrightarrow{MA}-5\overrightarrow{MB}+2\overrightarrow{MC}=3\left(\overrightarrow{MA}-\overrightarrow{MB}\right)+2\left(\overrightarrow{MC}-\overrightarrow{MB}\right)\)

\(=3\left(\overrightarrow{BM}+\overrightarrow{MA}\right)+2\left(\overrightarrow{BM}+\overrightarrow{MC}\right)=3\overrightarrow{BA}+2\overrightarrow{BC}\)

\(=3\overrightarrow{BA}+2\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=5\overrightarrow{BA}+2\overrightarrow{AC}\)

\(\Rightarrow\left|\overrightarrow{u}\right|^2=25AB^2+4AC^2+20\overrightarrow{BA}.\overrightarrow{AC}=29AB^2\)

\(\Rightarrow\left|\overrightarrow{u}\right|=AB\sqrt{29}=2\sqrt{29}\)