Cho hai đường thẳng \({d_1}:3x + 4y + 12 = 0\) và ...

A

- Hướng dẫn giải

Ta có:

\(\begin{array}{l} \left\{ \begin{array}{l} {d_1}:3x + 4y + 12 = 0 \to {{\vec n}_1} = \left( {3;4} \right)\\ {d_2}:\left\{ \begin{array}{l} x = 2 + at\\ y = 1 - 2t \end{array} \right. \to {{\vec n}_2} = \left( {2;a} \right) \end{array} \right.\\ \Rightarrow \frac{1}{{\sqrt 2 }} = \cos {45^ \circ } = \cos \varphi = \frac{{\left| {6 + 4a} \right|}}{{\sqrt {25} .\sqrt {{a^2} + 4} }} \end{array}\)

\(\Leftrightarrow 25\left( {{a^2} + 4} \right) = 8\left( {4{a^2} + 12a + 9} \right)\\ \Leftrightarrow 7{a^2} + 96a - 28 = 0 \Leftrightarrow \left[ \begin{array}{l} a = - 14\\ a = \frac{2}{7} \end{array} \right..\)