I have the equation for the first line in $\Bbb{R}^4$ $$q : [1,3,1,1] + t(-1,2,-1,1)$$ where $t$ is a parameter. Then I have my plane given by these two equations $$\rho : x_1 + x_2 - x_3 -3x_4 = -2 \quad\text{and}\quad 2x_1 + x_2 -2x_4 = 5$$ I have to construct a line $p$ such that it intersects both the plane and the line $q$ and I know that it has the form of $p: A + s(1,1,2,0)$, where $A$ is some point.
Now I would know how to solve this if I was given the point instead of the vector but in this situation I have no idea.
Does anyone have any tips on how to proceed? Thank you for your help in advance.
If we take any line with direction vector which is NOT parallel to the plane, then it will intersect the plane at some point. Try to visualize this, if you are having a hard time then visualize it in 3 dimensions instead. Now we simply need to ensure that this line will intersect q. To ensure this we can simply take any point on q and incorporate it into our line. The easiest point to take would be [1,3,1,1]. Now we just need to add to this any direction vector, not parallel to the plane, with a parameter.
In short, you already know how to solve this question if you were given a point instead of a line. So just do that! Take any point on q and solve as you usually would. Be sure to double check that it works