Đặt \(\overrightarrow{A'K}=x.\overrightarrow{A'I}=x\left(\dfrac{1}{2}\overrightarrow{A'B'}+\dfrac{1}{2}\overrightarrow{A'C'}\right)=\dfrac{x}{2}\left(\overrightarrow{a}+\overrightarrow{b}\right)\)
\(\overrightarrow{B'K}=\overrightarrow{B'A'}+\overrightarrow{A'K}=-\overrightarrow{a}+\dfrac{x}{2}\overrightarrow{a}+\dfrac{x}{2}\overrightarrow{b}=\left(\dfrac{x}{2}-1\right)\overrightarrow{a}+\dfrac{x}{2}\overrightarrow{b}\)
\(B'D'=\overrightarrow{B'A'}+\overrightarrow{A'C'}+\overrightarrow{C'D'}=-\overrightarrow{a}+\overrightarrow{c}-\overrightarrow{a}=-2\overrightarrow{a}+\overrightarrow{c}\)
Do B', K, D' thẳng hàng
\(\Rightarrow\dfrac{\dfrac{x}{2}-1}{-2}=\dfrac{\dfrac{x}{2}}{1}\Leftrightarrow x=\dfrac{2}{3}\) \(\Rightarrow\overrightarrow{A'K}=\dfrac{1}{3}\overrightarrow{a}+\dfrac{1}{3}\overrightarrow{b}\)
\(\Rightarrow\overrightarrow{C'K}=\overrightarrow{C'A'}+\overrightarrow{A'K}=-b+\dfrac{1}{3}\overrightarrow{a}+\dfrac{1}{3}\overrightarrow{b}=\dfrac{1}{3}\overrightarrow{a}-\dfrac{2}{3}\overrightarrow{b}\)
\(\Rightarrow\overrightarrow{DK}=\overrightarrow{DC}+\overrightarrow{CC'}+\overrightarrow{C'K}=\overrightarrow{a}+\overrightarrow{c}+\dfrac{1}{3}\overrightarrow{a}-\dfrac{2}{3}\overrightarrow{b}\)
\(=\dfrac{4}{3}\overrightarrow{a}-\dfrac{2}{3}b+\overrightarrow{c}=\dfrac{1}{3}\left(4\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}\right)\)
Đáp án A
