Math Relative position of lines Determining point of intersection

# Intersecting lines

Two lines can intersect and then have a point of intersection.

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### Requirements

1. Direction vectors are not collinear
2. 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

$\text{g: } \vec{x} = \begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + r \cdot \begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}$

$\text{h: } \vec{x} = \begin{pmatrix} 4 \\ 4 \\ 6 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

1. #### 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

1. $9=t\cdot1$
2. $8=t\cdot1$
3. $7=t\cdot2$

1. $t=9$
2. $t=8$
3. $t=3.5$

$t$ 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.

2. #### 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.

1. $10+9r=4+s$
2. $9+8r=4+s$
3. $7+7r=6+2s$

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.

3. #### 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 $S(1|1|0)$.