Find the point Q that is symmetric to point $P(-1,-2,1)$ in relation to line $l: \frac{x}{-2}=\frac{y-3}{4}=\frac{z-4}{1}$ as well as projection P' of point P onto line l.
1) I guess we start off with a parallel vector with a line (ie, $\langle -2,4,1 \rangle$) and then rotate it such way that it points to point P. If we call the line's vector $\langle-2,4,1 \rangle =C$, then we need to find some point that lies on the line X such that vector XP is orthogonal to vector c. That way we'd have a vector pointing from the line to the P.
2) Then we take that vector, basically scale it by $-1$ and say - "there should be point Q"?
3) Point P' would in this case be exactly point X?
Is this thought-process valid what-so-ever?
Thanks! :)