Đặt \(\overrightarrow{C'N}=x.\overrightarrow{C'D'}\)
\(\overrightarrow{AM}=\overrightarrow{AD}+\overrightarrow{DM}=\overrightarrow{AD}+\dfrac{1}{3}\overrightarrow{DC}=\overrightarrow{A'D'}+\dfrac{1}{3}\overrightarrow{D'C'}\)
\(\overrightarrow{B'N}=\overrightarrow{B'C'}+\overrightarrow{C'N}=\overrightarrow{A'D'}+x.\overrightarrow{C'D'}\)
\(AM\perp B'N\Rightarrow\overrightarrow{AM}.\overrightarrow{B'N}=0\)
\(\Rightarrow\left(\overrightarrow{A'D'}-\dfrac{1}{3}\overrightarrow{C'D'}\right)\left(\overrightarrow{A'D'}+x.\overrightarrow{C'D'}\right)=0\)
\(\Leftrightarrow A'D'^2-\dfrac{1}{3}x.C'D'^2=0\) (do \(A'D'\perp C'D'\Rightarrow\overrightarrow{A'D'}.\overrightarrow{C'D'}=0\))
\(\Rightarrow4-\dfrac{4}{3}x=0\Rightarrow x=3\)
Vậy N là điểm trên C'D' thỏa mãn \(\overrightarrow{C'N}=3\overrightarrow{C'D'}\)
