I came across this silly exercise but it's giving me a headache, because I cannot figure whether or not I'm not doing it right.
I have two lines of cartesian equations given
$$ r : \left\{ \begin{array}{l} x = 2z - 1 \\ y = z + 1 \end{array}\right. \;\;\; s : \left\{ \begin{array}{l} z = 2 \\ x + y - 1 = 0 \end{array}\right. $$
And I need to find the line passing through origin $O = (0,0,0)$ and intersecting both $r$ and $s$.
I don't think the exercise has solution.
My attempt: I wrote both lines in parametric form (I don't think this was necessary but it made the equations a bit easier to read)
$$ r : \left\{ \begin{array}{l} x = 2t - 1 \\ y = t + 1 \\ z = t \end{array}\right. \;\;\; s : \left\{ \begin{array}{l} x = -s + 1 \\ y = s \\ z = 2 \end{array}\right. $$
Then I imposed the collinearity among the origin, generic point for $r$ and generic point for $s$ which is equivalent to impose
$$ \text{rank} \begin{pmatrix} 2t - 1 & 1 - s \\ t + 1 & s \\ t & 2 \end{pmatrix} = 1 $$
which gave me the system (non linear but easy)
$$ \left\{ \begin{array}{l} 2t + 2 - ts = 0 \\ 3t - 2 + ts = 0 \end{array} \right. $$
Subtracting the second equation from the first gives me $t = 0$, but substituting this value in any of the two equations gives me a contradiction $1 = 0$, which makes me say there's no such line.
Am I missing anything?
I agree. I worked it slightly differently.
$$x_1 -2z_1 = -1$$ $$y_1 - z_1 = 1$$
$$x_2 + y_2 = 1$$ $$z_2 = 2$$
And some scaling of a solution to the first set equals the 2nd set so that we get a line through the origin. $(x_1,y_1,z_1) = c (x_2,y_2,z_2)$.
So substituting:
$$z_1/c = 2$$ $$x_1/c + y_1/c = 1$$ $$z_1 - 2c = 0$$ $$x_1+y_1 -c =0$$
Using the first 2 and last 2 equations I got the solution $(x_1,y_1,z_1,c) = (-1,1,0,0)$
Since $z_1 = c = 0$ there is no scaling that makes a solution.