Góc giữa hai đường thẳng \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 + t\end{array} \right.\) và \(\left\{ \begin{array}{l}x = 1 - 3t\\y = - 2 + t\end{array} \right.\) là:

* Hướng dẫn giải

\(\left( {{d_1}} \right):\,\,\left\{ \begin{array}{l}x = 1 + 2t\\y =  - 2 + t\end{array} \right. \Rightarrow {\vec u_{{d_1}}} = \left( {2;\,\,1} \right)\)

\(\left( {{d_2}} \right):\,\,\left\{ \begin{array}{l}x = 1 - 3t\\y =  - 2 + t\end{array} \right. \Rightarrow {\vec u_{{d_2}}} = \left( { - 3;\,\,1} \right)\)

\({\rm{cos}}\left( {{d_1},{d_2}} \right) = {\rm{cos}}\left( {{{\vec u}_{{d_1}}},{{\vec u}_{{d_2}}}} \right)\)\( = \dfrac{{\left| {{{\vec u}_{{d_1}}}.{{\vec u}_{{d_2}}}} \right|}}{{\left| {{{\vec u}_{{d_1}}}} \right|.\left| {{{\vec u}_{{d_2}}}} \right|}}\)\( = \dfrac{{\left| {2.\left( { - 3} \right) + 1.1} \right|}}{{\sqrt {{2^2} + {1^2}} .\sqrt {{{\left( { - 3} \right)}^2} + {1^2}} }}\)\( = \dfrac{5}{{\sqrt 5 .\sqrt {10} }} = \dfrac{1}{{\sqrt 2 }}\)

\( \Rightarrow \left( {{d_1},\,\,{d_2}} \right) = {45^0}\)

Vậy góc giữa hai đường thẳng trên bằng \({45^0}\).

Chọn A.