(d): Đặt \(\dfrac{x-1}{1}=\dfrac{y+2}{2}=\dfrac{z-1}{1}=d\)
=>\(\left\{{}\begin{matrix}x-1=d\\y+2=2d\\z-1=d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+d\\y=-2+2d\\z=1+d\end{matrix}\right.\)
=>(d) có vecto chỉ phương là \(\overrightarrow{a}=\left(1;2;1\right)\)
(d'): Đặt \(\dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z-1}{1}=k\)
=>\(\left\{{}\begin{matrix}x-1=k\\y+2=k\\z-1=k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+k\\y=-2+k\\z=1+k\end{matrix}\right.\)
=>(d') có vecto chỉ phương là \(\overrightarrow{b}=\left(1;1;1\right)\)
Gọi góc giữa (d) và (d') là \(\alpha\)
\(cos\alpha=\dfrac{\left|1\cdot1+2\cdot1+1\cdot1\right|}{\sqrt{1^2+2^2+1^2}\cdot\sqrt{1^2+1^2+1^2}}=\dfrac{4}{\sqrt{6}\cdot\sqrt{3}}=\dfrac{4}{\sqrt{18}}=\sqrt{\dfrac{16}{18}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}\)
=>\(\alpha\simeq19^028'\)
