Đường thẳng \(\Delta\)

C

- Hướng dẫn giải

\(\left\{ \begin{array}{l}{d_1}:2x + y - 3 = 0\\{d_2}:x - 2y + 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right. \Rightarrow {d_1} \cap {d_2} = A\left( {1;1} \right) \in \Delta .\)

Ta có \({d_3}:y - 1 = 0 \to {\vec n_3} = \left( {0;1} \right),\) gọi \({\vec n_\Delta } = \left( {a;b} \right),\,\,\varphi = \left( {\Delta ;{d_3}} \right)\).

Khi đó

\(\begin{array}{l}\frac{1}{{\sqrt 2 }} = \cos \varphi = \frac{{\left| b \right|}}{{\sqrt {{a^2} + {b^2}} .\sqrt {0 + 1} }}\\ \Leftrightarrow {a^2} + {b^2} = 2{b^2}\\ \Leftrightarrow \left[ \begin{array}{l}a = b \to a = b = 1 \to \Delta :x + y - 2 = 0\\a = - b \to a = 1,\,b = - 1 \to \Delta :x - y = 0\end{array} \right..\end{array}\)