Xét vị trí tương đối của hai đường thẳng sau \({d_1}:\left\{ \begin{array}{l} x = - 1 + t\\ y = - 2 - 2t \end{array} \right.

Ta có:

\(\left. \begin{array}{l} {d_1}:\left\{ \begin{array}{l} x = - 1 + t\\ y = - 2 - 2t \end{array} \right. \to {{\vec u}_1} = \left( {1; - 2} \right)\\ {d_2}:\left\{ \begin{array}{l} x = 2 - 2t'\\ y = - 8 + 4t' \end{array} \right. \to B\left( {2; - 8} \right) \in {d_2},\,\,{{\vec u}_2} = \left( { - 2;4} \right) \end{array} \right\} \to \left\{ \begin{array}{l} \frac{1}{{ - 2}} = \frac{{ - 2}}{4}\\ B \in {d_1} \leftrightarrow t = 3 \end{array} \\ \right. \to {d_1} \equiv {d_2}.\)