Cho hai đường thẳng $d$ và $d'$ chéo nhau có phương trình:
$d:\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z}{-1}  ;    d':\left\{ \begin{array}{l} x=-2t\\ y=-5+3t\\z=4 \end{array} \right.  (t\in R)$
Tính khoảng cách giữa $d$ và $d'$
$d$ đi qua $A(1;2;0)$ và có vecto chỉ  phương $\overrightarrow{u}_d $
$d'$ đi qua $B(0;-5;4)$ và có vecto chỉ phương $\overrightarrow{u}_d=(-2;3;0) $
Vecto pháp tuyến của $mp(\alpha )$ chứa $d$ và song song với $d'$ là:
$\overrightarrow n  = \left( {\left| \begin{array}{l}
- 2\\
3
\end{array} \right.\,\,\,\,\,\,\left. \begin{array}{l}
1\\
0
\end{array} \right|;\left| \begin{array}{l}
- 1\\
0
\end{array} \right.\,\,\,\,\,\,\left. \begin{array}{l}
    2\\
- 2
\end{array} \right|;\left| \begin{array}{l}
2\\
  2
\end{array} \right.\,\,\,\,\,\,\left. \begin{array}{l}
- 2\\
    3
\end{array} \right|} \right) = (3;2;2)$
Phương trình mặt phẳng $(\alpha )$:
$3(x-1)+2(y-2)+2(z-0)=0\Rightarrow  3x+2y+2z-7=0$
Khoảng cách từ điểm $B(0;-5;4)\in d'$ đến $(\alpha )$
$d(B,(\alpha )) = \frac{{|3.0 + 2.( - 5) + 2.4 - 7|}}{{\sqrt {9 + 4 + 4} }} = \frac{{9\sqrt {17} }}{{17}}$
Vậy: $d(d,d' ) =\frac{{9\sqrt {17} }}{{17}}$

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