I really don't know how to search for this specific question.
So, I'll try my best to explain my issue.
I have the point P1 (pink) and the normal vector M (white) of its line, Given an arbitrary point P2 (blue) and an arbitrary normal vector N (yellow).... I have to find the Green point, which intersects the white line
This would be a case where I have to find the point where two lines intersects. But I don't have two points to represent each line. I only have one of the points and their normals.
Thanks.
If $P_{i=1,2}=P(x_i,y_i,z_i)$ and $N_{i=1,2}=\langle a_i,b_i,c_i\rangle$ then you know the equations of the lines
You can get the parametric forms of the lines:
$$L_i:\ \ \begin{matrix} x = x_i + a_i t\\y = y_i + b_i t\\z = z_i + c_i t\end{matrix}$$
Now suppose there exists a point $Q=P(p,q,r)$ for which $L_1(Q)=L_2(Q)$. Then there exist $s,t$ such that
$$\begin{matrix}p = x_1+a_1t=x_2+a_2s\\q = y_1+b_1t=y_2+b_2s\\r = z_1+c_1t=z_2+c_2s \end{matrix}$$
which gives us a system of three equations in two unknowns:
$$\begin{matrix}a_1t-a_2s = x_2-x_1\\b_1t-b_2s = y_2-y_1\\c_1t-c_2s = z_2-z_1 \end{matrix}$$
The only way for this system to not have a unique solution is if all three of these linear equations are multiples of eachother. In this case the normal vectors (and hence lines) would be parallel. If the system is inconsistent, there is no intersection.
Example 1 (do not intersect): Given $N_1=\langle 1,1,0\rangle, N_2=\langle1,-1,1\rangle$ and $P_1=(1,-2,3), P_2=(-1,1,1)$ we get the system
$$\begin{matrix}t-s = -2\\t+s = 3\\-s = -2 \end{matrix}$$
This system is inconsistent, and so the lines do not intersect.
Example 2 (do intersect): $N_1=\langle -2,2,3\rangle, N_2=\langle 5,3,2\rangle$ and $P_1=(0,1,0), P_2=(-9,2,4)$
$$\begin{matrix}-2t-5s = -9\\2t-3s = 1\\3t-2s = 4 \end{matrix}$$
Adding equations (1) and (2) you can see that $s=1$ and plugging this into any of the three you will get $t=2$ (test consistency by trying all 3). So the point of intersection is $(-4,5,6)$