Checking if the following lines are coplanar

[Kernul]

Homework Statement

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.

Homework Equations

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?

 

[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

 

[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.