Cho tam giác ABC. Biết AB = 8; AC = 9; BC = 11. M là trung điểm của BC, N là điểm trên đoạn AC sao cho \(AN=x\left(0< x< 9\right)\). Hãy khai triển \(\overrightarrow{MN}\) qua \(\overrightarrow{c}=\overrightarrow{AC},\overrightarrow{b}=\overrightarrow{AB}\).
\(\overrightarrow{MN}=\left(\dfrac{1}{2}-\dfrac{x}{9}\right)\overrightarrow{c}+\dfrac{1}{2}\overrightarrow{b}\) \(\overrightarrow{MN}=\left(-\dfrac{x}{9}+\dfrac{1}{2}\right)\overrightarrow{c}-\dfrac{1}{2}\overrightarrow{b}\) \(\overrightarrow{MN}=\left(\dfrac{x}{9}+\dfrac{1}{2}\right)\overrightarrow{c}-\dfrac{1}{2}\overrightarrow{b}\) \(\overrightarrow{MN}=\left(\dfrac{x}{9}-\dfrac{1}{2}\right)\overrightarrow{c}-\dfrac{1}{2}\overrightarrow{b}\) Hướng dẫn giải:
\(\overrightarrow{MN}=\overrightarrow{MC}+\overrightarrow{CN}\)
\(\overrightarrow{MC}=\frac{1}{2}\overrightarrow{BC}=\frac{1}{2}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
\(\frac{\overrightarrow{CN}}{\overrightarrow{AC}}=-\frac{\left(9-x\right)}{9}=\frac{x}{9}-1\)
\(\Rightarrow\overrightarrow{CN}=\left(\frac{x}{9}-1\right)\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{MN}=\frac{1}{2}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)+\left(\frac{x}{9}-1\right)\overrightarrow{AC}\)
\(=\left(\frac{x}{9}-\frac{1}{2}\right)\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\)
Vậy \(\overrightarrow{MN}=\left(\dfrac{x}{9}-\dfrac{1}{2}\right)\overrightarrow{c}-\dfrac{1}{2}\overrightarrow{b}\)
