\(d_1\) : \(x-y+1=0\Rightarrow\overrightarrow{n_{d1}}=\left(1;-1\right)\)
Gọi vtecto pháp tuyến của d là \(\overrightarrow{n_d}=\left(a;b\right)\)
\(cos60^0=\frac{\left|a-b\right|}{\sqrt{1^2+1^2}\sqrt{a^2+b^2}}=\frac{1}{2}\)
\(\Leftrightarrow2\left(a-b\right)^2=a^2+b^2\Leftrightarrow a^2-4ab+b^2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\left(2+\sqrt{3}\right)b\\a=\left(2-\sqrt{3}\right)b\end{matrix}\right.\)
Chọn \(a=1\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(1;2+\sqrt{3}\right)\\\left(a;b\right)=\left(1;2-\sqrt{3}\right)\end{matrix}\right.\)
Phương trình d: \(\left[{}\begin{matrix}x+\left(2+\sqrt{3}\right)y=0\\x+\left(2-\sqrt{3}\right)y=0\end{matrix}\right.\)
