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?

  • 1
    $\begingroup$ $\mathbf{n}\cdot(\mathbf{x}-\mathbf{a})=0$ is normal plane equation where $\mathbf{n}$ is the normal, $\mathbf{x}=(x,y,z)^T$ and $\mathbf{a}$ is a point on the plane (you can take any point on the line for $\mathbf{a}$) $\endgroup$ Jul 2, 2020 at 12:40
  • $\begingroup$ So,let $l$ be a line with a vector $u$ perpendicular to the line, you have to find a plane $\pi$ such that its orthogonal vector is $u$ and $l \subset \pi$, right? $\endgroup$
    – Vajra
    Jul 2, 2020 at 12:41
  • $\begingroup$ @AlexeyBurdinwhat would T be? @psidaga exactly. $\endgroup$ Jul 2, 2020 at 12:43
  • $\begingroup$ $T$ means transpose: $(x,y,z)^T=\pmatrix {x\\y\\z}$ $\endgroup$ Jul 2, 2020 at 12:44
  • $\begingroup$ Ok, perfect. Are you in 3-dimensional space or in a general n-dimensional affine space? $\endgroup$
    – Vajra
    Jul 2, 2020 at 12:45

1 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).

  • $\begingroup$ That solves my problem. But I was looking into the documentation of my programming language/library and found out that a plane can be generated by using 3 points that are on it as long as they are not forming a line. Probably very dummy question: is it possible to get multiple perpendiculars of the unit normal by just swapping x, y and z in different combinations? $\endgroup$ Jul 2, 2020 at 13:22
  • $\begingroup$ Sorry, I forgot the terms d and d' (they were not really important for our problem). Now it's correct at all I think. $\endgroup$
    – Vajra
    Jul 2, 2020 at 13:24
  • $\begingroup$ I'm not understanding if you are looking for the equation of the plane for those three points or for another thing... $\endgroup$
    – Vajra
    Jul 2, 2020 at 13:26
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    $\begingroup$ Never mind. I got the problem solved with your main answer already. Thanks. $\endgroup$ Jul 2, 2020 at 13:37
  • $\begingroup$ What is $\rho$ ? $\endgroup$
    – user253751
    Jul 3, 2020 at 13:51

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