Finding the plane where the line lies on

Question

Having a line and a normal orthogonal to it, how do I find the plane in which the normal will come out from and the line will lie on?

Answer 1

In the three-dimensional affine space $\mathbb{A^3}_{\mathbb{R}}$ (I'm assuming $\mathbb{K}=\mathbb{R}$) a line is defined, in its cartesian form, by the intersection of two planes.

Let $l$:\begin{cases} ax+by+cz+d=0 \\ a'x+b'y+c'z+d'=0 \end{cases} and $\textbf{u}$ a vector orthogonal to the line.

We can consider the pencil of planes with support the line $l$ such that $$\Lambda_{\mu,\eta}: \mu (ax+by+cz+d) + \eta(a'x+b'y+c'z+d')=0$$
that can be written as $x(\mu a + \eta a')+y(\mu b + \eta b')+z(\mu c + \eta c')+\mu d+\eta d'=0$, where $$\textbf{v}=(\mu a + \eta a', \mu b + \eta b',\mu c + \eta c')$$ is the parametrization of its orthogonal vector.

So, in order to find the plane we only have to solve the system $\textbf{v}= \rho \textbf{u}$ (the two vectors have to be proportional).