Intersecting lines
Two lines can intersect and then have a point of intersection.
Requirements
- Direction vectors are not collinear
- When equating the linear equations, we receive a distinct result
Calculating point of intersection
To calculate the point of intersection, we follow the scheme above. The result from step 2 for $r$ or $s$ is then inserted into the corresponding linear equation.
Example
$\text{g: } \vec{x} = \begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + r \cdot \begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}$
$\text{h: } \vec{x} = \begin{pmatrix} 4 \\ 4 \\ 6 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$
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Direction vectors not collinear?
First, we examine if the direction vectors are multiples of each other (=collinear).$\vec{a}=t\cdot\vec{b}$
$\begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}=t\cdot\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$
Calculate $t$ for every row
- $9=t\cdot1$
- $8=t\cdot1$
- $7=t\cdot2$
- $t=9$
- $t=8$
- $t=3.5$
$t$ does not have the same value everywhere. Thus the direction vectors are not collinear. Therefore the lines are skew or have a point of intersection.
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Equate $g$ and $h$
Now equate the linear equations and calculate $r$ and $s$.$\begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + r \cdot \begin{pmatrix} 9 \\ 8 \\ 7\end{pmatrix}$ $= \begin{pmatrix} 4 \\ 4 \\ 6 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$
Now we set up an equation system and solve it.
- $10+9r=4+s$
- $9+8r=4+s$
- $7+7r=6+2s$
Equation method: Equate I. and II.
$4+s=4+s$
$10+9r=9+8r\quad|-8r$
$10+r=9\quad|-10$
$r=-1$Compute $s$ with e.g. I. (for that insert $r=-1$)
$10+9\cdot(-1)=4+s$
$1=4+s\quad|-4$
$s=-3$For validation insert $r=-1$ and $s=-3$ into III.
$7+7\cdot(-1)=6+2\cdot(-3)$
$0=0$$r=-1$ and $s=-3$ result from all equations
=> The lines intersect. -
Determining point of intersection
To determine the point of intersection we insert $r=-1$ into $g$.$\begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + (-1) \cdot \begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}$ $=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$
=> The lines intersect in point $S(1|1|0)$.