# Checking if the following lines are coplanar

1. Sep 20, 2016

### Kernul

1. The problem statement, all variables and given/known data
Set an orthonormal reference system in the euclidean space $E^3$ and consider the following lines:
$$r: \begin{cases} x = 3 \tau \\ y = 1 + \tau \\ z = 3 - 2 \tau \end{cases}s: \begin{cases} x + z - 4 = 0 \\ x - 3y + z + 2 = 0 \end{cases}t: \begin{cases} 4x - 2y + 5z - 3 = 0 \\ 2y + z + 1 = 0 . \end{cases}$$
Find the mutual position of each possible couple of lines.

2. Relevant equations

3. The attempt at a solution
I've calculated the mutual positions of the three couple of lines($r$ and $s$, $s$ and $t$, and $r$ and $t$) but I'm not sure about it so here it's how I proceeded:

Before starting, I transformed $r$ from the parametric form into the Cartesian form, so:

$$r: \begin{cases} x - 3y + 3 = 0 \\ 2y + z - 5 = 0 . \end{cases}$$

Now, to know if the lines are parallel, orthogonal, etc, I first find out about the directional vector of each line by finding the determinant of the matrices made of the components of the lines:
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 0 \\ 0 & -2 & 1 \end{vmatrix} = (-3, -1, -2) \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 0 & 1 \\ 1 & -3 & 1 \end{vmatrix} = (3, 0, -3) \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & -2 & 5 \\ 0 & 2 & 1 \end{vmatrix} = (-12, -4, 8)$

Then I find a point for each line:
$P_r (0, 1, 3), P_s (2, 2, 2), P_t (-1, -1, 1)$

After that I do the determinant with the first row as the difference between the two points of the two lines, and on the second and third raw I put the directional vectors of the two lines.
For $r$ and $s$:
$\begin{vmatrix} 0-2 & 1-2 & 3-2 \\ -3 & -1 & -2 \\ 3 & 0 & -3 \end{vmatrix} = 12$
Non-coplanar.

$\begin{vmatrix} 2-(-1) & 2-(-1) & 2-1 \\ 3 & 0 & -3 \\ -12 & -4 & 8 \\ \end{vmatrix} = -10$
Non-coplanar.

$\begin{vmatrix} 0-(-1) & 1-(-1) & 3-1 \\ -3 & -1 & -2 \\ -12 & -4 & 8 \\ \end{vmatrix} = 80$
Non-coplanar.

The thing is that all three of them are askew (non-coplanar). Could someone check it out and tell me if they are really all non-coplanar or maybe I did it wrong?

2. Sep 20, 2016

### Ssnow

Hi, finding the directional vector of the first line $r$, in the determinant the third line is $0 \ \ 2 \ \ 1$ and not $0 \ \ -2 \ \ 1$, this can change something later ... the procedure seems reasonable ...

Ssnow

3. Sep 20, 2016

### Kernul

Wow... that sign changed everything...
$r$ and $s$ are intersecting lines.
$s$ and $t$ are still non-coplanar. (because the directional vector wasn't needed here)
$r$ and $t$ are parallel.
Thank you very much for noticing that mistake.