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I've been trying to find out how to compute the corresponding point on the line $x=1$, starting from point $(95,30)$ knowing the normalized direction vector $(-0.9734,0.2290)$.

My idea is to compute at which point the "projection" of my normalized direction vector intersects on line $x=1$, but I fail every time since I think I ignore too much.

I think this is super basic trigonometry but if someone could point me out where can I read about it, it would be perfect!

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It sounds like what you want to do is make a line from $(95,30)$ in the desired direction and find where it intersects the line $x=1$. This point of intersection if found by setting the line equations equal, plug in $x=1$ and solve for $y$.

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    so should I treat the direction vector as a line? I want to travel along the direction vector (starting from (95,30)) and suddenly find myself on a point which belongs to line x=1.. think this site is sort of what I need but I can't seem to really relate it.. http://wwwf.imperial.ac.uk/metric/metric_public/vectors/vector_coordinate_geometry/vector_equation_of_line.html – 2017-02-01
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    You should use the direction vector to make a line, a line that is through the point $(95,30)$, and in the direction of said vector. – 2017-02-01