Đặt \(\dfrac{x-3}{2}=\dfrac{y+5}{4}=\dfrac{z-7}{2}=c\)
=>\(\left\{{}\begin{matrix}x-3=2c\\y+5=4c\\z-7=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3+2c\\y=-5+4c\\z=7+2c\end{matrix}\right.\)
=>(d) có vecto chỉ phương là \(\overrightarrow{a}=\left(2;4;2\right)\)
Đặt \(\dfrac{x-1}{3}=\dfrac{y+7}{3}=\dfrac{z}{6}=k\)
=>\(\left\{{}\begin{matrix}x-1=3k\\y+7=3k\\z=6k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+3k\\y=-7+3k\\z=6k\end{matrix}\right.\)
=>(d') có vecto chỉ phương là \(\overrightarrow{b}=\left(3;3;6\right)\)
Gọi góc giữa (d) và (d') là \(\alpha\)
\(cos\alpha=\dfrac{\left|2\cdot3+4\cdot3+2\cdot6\right|}{\sqrt{2^2+4^2+2^2}\cdot\sqrt{3^2+3^2+6^2}}=\dfrac{30}{2\sqrt{6}\cdot3\sqrt{6}}=\dfrac{5}{6}\)
=>\(\alpha\simeq33^033'\)
