Cho hai đường thẳng chéo nhau

D

- Hướng dẫn giải

Đáp án D

Gọi $\left\{\begin{array}{l}M\left(2+t;0;0\right)\in d_1\\ N\left(1+s;1+s;1\right)\in d_2\end{array}\right.$ sao cho $\overrightarrow{MN}.\overrightarrow{v_1}=\overrightarrow{MN}.\overrightarrow{v_2}=0$ ($\overrightarrow{v_1},\overrightarrow{v_2}$ là các VTCP của $d_1,d_2$)

$\overrightarrow{MN}=\left(s-t-1;s+1;1\right)$

$$\begin{gathered} \left\{\begin{array}{l}\overrightarrow{MN}.\overrightarrow{v_1}=s-t-1=0\\ \overrightarrow{MN}.\overrightarrow{v_2}=s-t-1+s+1=0\end{array}\right. \\ \iff \left\{\begin{array}{l}s-t=1\\ 2s-t=0\end{array}\right.\Rightarrow \left\{\begin{array}{l}s=-1\\ t=-2\end{array}\right. \end{gathered}$$.

Vậy$M\left(0;0;0\right),N\left(0;0;1\right),\left|MN\right|=1$.