Cho hình hộp ABCD.A'B'C'D' có \(\overrightarrow{AB}=\overrightarrow{a}\) \(\overrightarrow{AC}=\overrightarrow{b}\) \(\overrightarrow{AA'}=\overrightarrow{c}\) . Gọi I là trung
điểm của \(B'C'\) , K là giao điểm của A 'I và B'D'. Phân tích \(\overrightarrow{DK}\) theo \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\)
\(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\Rightarrow\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}\)
Theo Talet: \(\dfrac{A'K}{IK}=\dfrac{B'I}{A'D'}=\dfrac{1}{2}\Rightarrow A'K=\dfrac{2}{3}A'I\)
\(\Rightarrow\overrightarrow{A'K}=\dfrac{2}{3}\overrightarrow{A'I}=\dfrac{2}{3}\left(\overrightarrow{A'B'}+\overrightarrow{B'I}\right)=\dfrac{2}{3}\left(\overrightarrow{A'B'}+\dfrac{1}{2}\overrightarrow{B'C'}\right)\)
\(=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{a}+\dfrac{1}{3}\left(\overrightarrow{b}-\overrightarrow{a}\right)=\dfrac{1}{3}\overrightarrow{a}+\dfrac{1}{3}\overrightarrow{b}\)
\(\Rightarrow\overrightarrow{DK}=\overrightarrow{DD'}+\overrightarrow{D'A'}+\overrightarrow{A'K}=\overrightarrow{AA'}-\overrightarrow{BC}+\overrightarrow{A'K}\)
\(=\overrightarrow{c}-\left(\overrightarrow{b}-\overrightarrow{a}\right)+\dfrac{1}{3}\overrightarrow{a}+\dfrac{1}{3}\overrightarrow{b}\)
\(=\dfrac{4}{3}\overrightarrow{a}-\dfrac{2}{3}\overrightarrow{b}+\overrightarrow{c}\)
