\(\overrightarrow {AK} \)
Giải chi tiết:

Vì \(A'D'\parallel B'I\) nên áp dụng định lí Ta-lét ta có: \(\dfrac{{A'K}}{{KI}} = \dfrac{{A'D'}}{{B'I}} = 2\).
\( \Rightarrow A'K = 2KI \Rightarrow A'K = \dfrac{2}{3}A'I\).
Mà \(\overrightarrow {A'K} ,\,\,\overrightarrow {A'I} \) là 2 vectơ cùng hướng nên \(\overrightarrow {A'K} = \dfrac{2}{3}\overrightarrow {A'I} \).
\(\begin{array}{l}\overrightarrow {AK} = \overrightarrow {AA'} + \overrightarrow {A'K} \\\,\,\,\,\,\,\,\,\, = \overrightarrow c + \dfrac{2}{3}\overrightarrow {A'I} \\\,\,\,\,\,\,\,\,\, = \overrightarrow c + \dfrac{2}{3}\left( {\overrightarrow {A'B'} + \overrightarrow {B'I} } \right)\\\,\,\,\,\,\,\,\,\, = \overrightarrow c + \dfrac{2}{3}\left( {\overrightarrow a + \dfrac{1}{2}\overrightarrow {B'C'} } \right)\\\,\,\,\,\,\,\,\, = \overrightarrow c + \dfrac{2}{3}\overrightarrow a + \dfrac{1}{3}\left( {\overrightarrow {A'C'} - \overrightarrow {A'B'} } \right)\\\,\,\,\,\,\,\,\, = \overrightarrow c + \dfrac{2}{3}\overrightarrow a + \dfrac{1}{3}\left( {\overrightarrow b - \overrightarrow a } \right)\\\,\,\,\,\,\,\,\, = \dfrac{1}{3}\overrightarrow a + \dfrac{1}{3}\overrightarrow b + \overrightarrow c \end{array}\)
Chọn B.