Trong không gian Oxyz, cho mặt phẳng (P):4y−z+3=0 và hai đường

Gọi $M=d\cap {\text{Δ}}_1\Rightarrow M\left(1+t_1;-2+4t_1;2+3t_1\right)$

$N=d\cap {\text{Δ}}_2\Rightarrow N\left(-4+5t_2;-7+9t_2;t_2\right)$

$\Rightarrow \overrightarrow{MN}=\left(5t_2-t_1-5;9t_2-4t_1-5;t_2-3t_1-2\right)$

Vì $d\perp \left(P\right):4y-z+3=0$ có 1 VTPT là $\overrightarrow{n}\left(0;4;-1\right)$ nên $\overrightarrow{MN}$ và $\overrightarrow{n}$  là 2 vectơ cùng phương.

$\begin{array}{l}\Rightarrow \overrightarrow{MN}=k\overrightarrow{n}\left(k\ne 0\right)\iff \left\{\begin{array}{c}5t_2-t_1-5=0\\ 9t_2-4t_1-5=4k\\ t_2-3t_1-2=-k\end{array}\right.\iff \left\{\begin{array}{c}{t}_1=5t_2-5\\ 9t_2-4t_1-5=4k\\ 4t_2-12t_1-8=-4k\end{array}\right.\\ \iff \left\{\begin{array}{c}{t}_1=5t_2-5\\ 13t_2-16t_1-13=0\\ t_2-3t_1-2=-k\end{array}\right.\iff \left\{\begin{array}{c}{t}_1=5t_2-5\\ 13t_2-16(5t_2-5)-13=0\\ t_2-3t_1-2=-k\end{array}\right.\\ \iff \left\{\begin{array}{c}{t}_1=5t_2-5\\ -67t_2+67=0\\ t_2-3t_1-2=-k\end{array}\right.\iff \left\{\begin{array}{c}{t}_2=1\\ t_1=0\\ k=1\end{array}\right.\end{array}$

$\Rightarrow M\left(1;-2;2\right),N\left(1;2;1\right)\Rightarrow \overrightarrow{MN}=\left(0;4;-1\right)$

Vậy phương trình đường thẳng d đi qua M và có 1 VTCP$\overrightarrow{MN}\left(0;4;-1\right)$ là:

$\left\{\begin{array}{c}x=1\\ y=-2+4t\\ z=2-t\end{array}\right.$

Đáp án cần chọn là: A