Math Relative position of lines Skew lines

Skew lines

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

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Requirements

1. Direction vectors are not collinear
2. A false statement results when we equate the linear equations

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