Đáp án:
$A.\,\overrightarrow{DK}=\dfrac13\left(4\overrightarrow{a}- 2\overrightarrow{b} + 3\overrightarrow{c}\right)$
Giải thích các bước giải:
Ta có:
$A'D'//B'C'$
$\to A'D'//B'I$
$\to \dfrac{D'K}{KB'}=\dfrac{A'D'}{BI}= 2$ (Theo định lý $Thales$)
$\to \dfrac{D'K}{D'B'}=\dfrac23$
$\to D'K=\dfrac23D'B'$
$\to \overrightarrow{D'K}=\dfrac23\overrightarrow{D'B'}$
$\to \overrightarrow{D'K}=\dfrac23\left(\overrightarrow{A'B'} - \overrightarrow{A'D'}\right)$
$\to \overrightarrow{D'K}=\dfrac23\left(\overrightarrow{A'B'}- \overrightarrow{A'C'} - \overrightarrow{C'D'}\right)$
$\to \overrightarrow{D'K}=\dfrac23\left(2\overrightarrow{A'B'}- \overrightarrow{A'C'}\right)$
Khi đó:
$\quad \overrightarrow{DK} = \overrightarrow{DD'} + \overrightarrow{D'K}$
$\to \overrightarrow{DK} = \overrightarrow{AA'} + \dfrac23\left(2\overrightarrow{A'B'}- \overrightarrow{A'C'}\right)$
$\to \overrightarrow{DK} = \dfrac43\overrightarrow{AB} - \dfrac23\overrightarrow{AC} + \overrightarrow{AA'}$
$\to \overrightarrow{DK} = \dfrac43\overrightarrow{a} - \dfrac23\overrightarrow{b}+ \overrightarrow{c}$
$\to \overrightarrow{DK}=\dfrac13\left(4\overrightarrow{a}- 2\overrightarrow{b} + 3\overrightarrow{c}\right)$
