Đáp án:
$\vec v=\left({\dfrac{16}{13};-\dfrac{24}{13}}\right)$
Lời giải:
$\begin{array}{l} d\text{ đi qua }A\left( {0;1} \right),VTPT\,\overrightarrow n = \left( {2; - 3} \right) \Rightarrow VTCP\,\overrightarrow u = \left( {3;2} \right)\\ \text{Gọi }\overrightarrow v = \left( {a;b} \right)\\ \overrightarrow v \bot d \Rightarrow \overrightarrow v .\overrightarrow u = 0 \Leftrightarrow 3a + 2b = 0\\ \text{Gọi }A' = {T_{\overrightarrow v }}\left( A \right) \Rightarrow A'\left( {a;b + 1} \right)\\ A' \in d' \Rightarrow 2a - 3\left( {b + 1} \right) - 5 = 0 \Leftrightarrow 2a - 3b - 8 = 0\\ \text{Ta có hệ }\left\{ \begin{array}{l} 3a + 2b = 0\\ 2a - 3b = 8 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = \dfrac{{16}}{{13}}\\ b = - \dfrac{{24}}{{13}} \end{array} \right. \end{array}$
