Skew lines

Lines that are neither parallel nor intersect, are called skew.

Example

$\text{g: } \vec{x} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + r \cdot \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix}$

$\text{h: } \vec{x} = \begin{pmatrix} 6 \\ 4 \\ 8 \end{pmatrix} + s \cdot \begin{pmatrix} 0 \\ 0 \\ 8 \end{pmatrix}$

1. Direction vectors not collinear?

   First, examine if the direction vectors are multiples of each other (=collinear).

   $\vec{a}=t\cdot\vec{b}$

   $\begin{pmatrix} 0 \\ 4 \\ 0\end{pmatrix}=t\cdot\begin{pmatrix} 0 \\ 0 \\ 8 \end{pmatrix}$

   Determine $t$ for every row

   1. $0=t\cdot0$
   2. $4=t\cdot0$
   3. $0=t\cdot8$
   
   
   1. $0=0$
   2. $4=0$
   3. $t=0$

   A contradiction results in the second row. Thus the direction vectors are not collinear.

2. Equate $g$ and $h$

   By equating the linear equations we try to find a point of intersection. If a contradiction occurs the lines are skew.

   $\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + r \cdot \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix}$ $= \begin{pmatrix} 6 \\ 4 \\ 8 \end{pmatrix} + s \cdot \begin{pmatrix} 0 \\ 0 \\ 8 \end{pmatrix}$

   Now we set up an equation system and solve it.

   1. $2+r\cdot0=6+s\cdot0$
   2. $4+r\cdot4=4+s\cdot0$
   3. $6+r\cdot0=8+s\cdot8$
   
   
   1. $2=6$
   2. $4+4r=4$
   3. $6=8+8s$

   A contradiction results from I. (false statement)
   => The lines are skew