\(\overrightarrow{n_{d1}}=\left(1;2\right)\) ; \(\overrightarrow{n_{d2}}=\left(3;m\right)\)
Ta có: cos(d1;d2) = \(\left|cos(\overrightarrow{n_{d1};}\overrightarrow{n_{d2}})\right|\) = \(\frac{\sqrt{2}}{2}\)
=> \(\frac{3+2m}{\sqrt{\left(3+m^2\right)5}}\) = \(\frac{\sqrt{2}}{2}\) ⇔ 2(3 + 2m) = \(\sqrt{10\left(3+m^2\right)}\)
=> ĐK: 3 + 2m > 0 ⇔ m > \(\frac{-3}{2}\)