Khoảng cách giữa hai đường thẳng \({d_1}:\left\{ \begin{array}{l}x = 2 + 2t\\y =  - 1 + t\\z = 1\end{array} \right.,{d_2}:\left\{ \begin{array}{l}x = 1\\y = 1 + t\\z = 3 - t\end{array} \right.\) là:

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] = \overrightarrow 0 \)

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] = \left[ {\overrightarrow u ,\overrightarrow {MM'} } \right]\)          

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] = \left[ {\overrightarrow u ,\overrightarrow {MM'} } \right] = \overrightarrow 0 \)           

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] \ne \left[ {\overrightarrow u ,\overrightarrow {MM'} } \right]\)

Song song.

Trùng nhau.

Cắt nhau.

Chéo nhau.

\(\left\{ \begin{array}{l}\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] \ne \overrightarrow 0 \\\left[ {\overrightarrow u ,\overrightarrow {u'} } \right]\overrightarrow {MM'}  = 0\end{array} \right.\)    

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] \ne \overrightarrow 0 \)         

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right]\overrightarrow {MM'}  = 0\)            

\(\left[ {\overrightarrow u ,\overrightarrow {u'} } \right] = \overrightarrow 0 \)

\(d//d'\)

\(d \equiv d'\)

\(d\) cắt \(d'\)

\(d\) chéo \(d'\)

Song song.

Trùng nhau.

Cắt nhau.

Chéo nhau.

\({d_1}:\left\{ \begin{array}{l}x = 3t\\y = 1 + t\\z = 5t\end{array} \right.\).

\({d_2}:\left\{ \begin{array}{l}x = 2\\y = 2 + t\\z = 1 + t\end{array} \right.\).

\({d_3}:\dfrac{{x - 2}}{3} = \dfrac{y}{2} = \dfrac{{z - 1}}{{ - 5}}\).

\({d_4}:\dfrac{{x + 2}}{2} = \dfrac{y}{{ - 1}} = \dfrac{{z + 1}}{2}\).

\(d\left( {A,d'} \right) = \dfrac{{\left| {\left[ {\overrightarrow {AM'} ,\overrightarrow {u'} } \right]} \right|}}{{\left| {\overrightarrow {u'} } \right|}}\)                 

\(d\left( {A,d'} \right) = \dfrac{{\left| {\left[ {\overrightarrow {AM'} ,\overrightarrow {u'} } \right]} \right|}}{{\overrightarrow {u'} }}\)

\(d\left( {A,d'} \right) = \dfrac{{\left[ {\overrightarrow {AM'} ,\overrightarrow {u'} } \right]}}{{\overrightarrow {u'} }}\)           

\(d\left( {A,d'} \right) = \dfrac{{\left| {\overrightarrow {AM'} .\overrightarrow {u'} } \right|}}{{\left| {\overrightarrow {u'} } \right|}}\)