Suppose you're given three planes $\pi_1, \pi_2, \pi_3$, meeting at a single point $O$. These planes divide the $3D$ space into $8$ parts. The task is to find all possible lines that make the same angle with all three planes, meaning that points on the line are equidistant from all three planes.

My attempt:

Since points on the line are equidistant from $\pi_1$ and $\pi_2$, then it lies on their angle bisecting plane. If $\beta_1$ is the bisecting plane of $\pi_1$ and $\pi_2$ and $\beta_2$ is the bisecting plane of $\pi_1$ and $\pi_3$, then the line is the intersection of $\beta_1$ and $\beta_2$. Since there are two possible bisecting planes $\beta_1$ and two possible bisecting planes $\beta_2$, the total number of possible lines is $4$.

  • $\begingroup$ How do you define the distance between a line and a plane? $\endgroup$
    – user1551
    Aug 16, 2023 at 8:39
  • $\begingroup$ What I meant, was that points on the line are equidistant from the three planes. I'll clarify this in the question. Thanks for your comment. $\endgroup$
    – of course
    Aug 16, 2023 at 9:05

2 Answers 2


Let us assume WLOG that the equations of the planes are under the normalized form :

$$(P_k) \ : \ a_kx+b_ky+c_kz=0 \ \text{with} \ a_k^2+b_k^2+c_k^2=1$$

with independant normal vectors $(a_k,b_k,c_k)$.

(indeed one can change in particular the common point to be the origin of coordinates).

In this case, the distance between a point $M(x,y,z)$ and plane $(P_k)$ is given by :


and the oriented distance by


Therefore, the locus of points that are at equal distance from the 3 planesare defined by the system of equations :

$$\delta(M,(P_1))=\delta(M,(P_2))=\delta(M,(P_3)) \tag{1}$$

(we don't know really know that it is a union of lines even if we have a strong intuition that such is the case).

(1) is equivalent to :

$$\begin{cases}\Delta(M,(P_1)))&=&\color{red}{u}\Delta(M,(P_2))\\ \Delta(M,(P_2)))&=&\color{red}{v}\Delta(M,(P_3))\\ \Delta(M,(P_3)))&=&\color{red}{w}\Delta(M,(P_1))\end{cases}\tag{2}$$

where $\color{red}{(u,v,w)}$ can be $\color{red}{(-,-,-)}$, $\color{red}{(-,-,+)}$, ... to $\color{red}{(+,+,+)}$.

Only some of the combinations will give rise to solutions (in this regard, see the important Edit below).

Let us consider first the example of the 3 planes with equations

$$\begin{cases}2x+2y+z&=&0\\ x-2y-2z&=&0\\ -2x+2y-z&=&0;\end{cases}$$

(Notice that we have taken coefficients such that the common norm is simple : $\sqrt{2^2+2^2+1^2}=3$)

Among the 8 possible sign combinations, 4 of them will give rise to straight lines :

Here is one :

$$\color{red}{(+,+,+)}\begin{cases}\Delta(M,(P_1)))&=&\color{red}{+}\Delta(M,(P_2))\\ \Delta(M,(P_2)))&=&\color{red}{+}\Delta(M,(P_3))\\ \Delta(M,(P_3)))&=&\color{red}{+}\Delta(M,(P_1))\end{cases} \iff \pmatrix{x\\y\\z}=\pmatrix{2r\\-2r\\r}$$

Here are the 3 others :

$$\color{red}{(+,-,-)} \pmatrix{x\\y\\z}=\pmatrix{-3r\\0\\r}$$

$$\color{red}{(-,+,-)} \pmatrix{x\\y\\z}=\pmatrix{-r/3\\0\\r}$$

$$\color{red}{(-,-,+)} \pmatrix{x\\y\\z}=\pmatrix{0\\r\\0}$$

enter image description here

Fig. 1 : The 3 great circles are the intersection of the 3 planes with the unit sphere : they delimitate 8 spherical triangles paired 2 by 2 (each triangle paired with its "antipodal" triangle). In black and red, the 4 "bissecting" lines ; for example the red one intersects the unit sphere in two points which are the equivalent of the incenters of the corresponding spherical triangle.

Here is the SAGE program that I have used :

x,y,z,u,v,w=var('x y z u v w')
u=1;v=1;w=1;#to be changed

(I am much indebted to @Oscar Lanzi who has pointed a big error of mine).

Important Edit :

The fact that there are exactly 4 lines is rather intuitive geometricaly but deserves an algebraic proof.

In fact (2) is equivalent to this matrix equation describing a certain kernel :

$$\underbrace{\pmatrix{1&-u&0\\ 0&1&-v\\-w&0&1}}_{M_{u,v,w} \ \text{with rank} \ r}\underbrace{\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3}}_P \pmatrix{x\\y\\z}=\pmatrix{0\\0\\0};$$

According to the values $\pm1$ of $u,v,w$, the value of rank $r$ will be $2$ or $3$ :

$$\begin{array}{|r|r|r|r|c|r|} \hline u&v&w&r&d=3-r\\ \hline -1&-1&-1&3&0&\text{point}\\ -1&-1&1&2&1&\text{line}\\ -1&1&-1&2&1&\text{line}\\ -1&1&1&3&0&\text{point}\\ 1&-1&-1&2&1&\text{line}\\ 1&-1&1&3&0&\text{point}\\ 1&1&-1&3&0&\text{point}\\ 1&1&1&2&1&\text{line}\\ \hline \end{array}$$

  • $\begingroup$ @Oscar Lanzi I was unsatisfied with my answer. I have finally found a satisfying algebraic explanation for the different cases. Satisfying because, with this methodology, one can tackle the general case of an "equidistant set of points" for $k$ hyperplanes in a $n$-dimensional space... $\endgroup$
    – Jean Marie
    Aug 18, 2023 at 10:05
  • 1
    $\begingroup$ You don't really need the matrix analysis to identify the proper signs. Given $\Delta_1=\pm_1\Delta_2,\Delta_2=\pm_2\Delta_3$ you can eliminate $\Delta_2$ to get $\Delta_3=\pm_1(\pm_2\Delta_1)$, forcing the correct sign combinations. Similarly exactly $n-1$ signs are independent in $n$ dimensions. $\endgroup$ Aug 18, 2023 at 12:00

This is in a way a variation of the answer that Jean Marie wrote, but I believe it is sufficiently different in its derivation to stand on its own instead of an edit.

Let's go one dimension down, to 2d. Given 2 lines, find all the lines whose points are equidistant to both the given lines. These lines are the angular bisectors, and there are two of them.

Assume line $k$ is given by three numbers $l_k=(a_k, b_k, c_k)$ which describe the line as $a_kx + b_ky + c_k = 0$. If you have this normalized to $a_k^2+b_k^2=1$ (which you can achieve by multiplying all numbers by the same factor), then the two angular bisectors are $l_1+l_2=(a_1+a_2, b_1+b_2, c_1+c_2)$ and $l_1-l_2=(a_1-a_2, b_1-b_2, c_1-c_2)$.

2d angular bisectors, with lines and normal vectors labeled

Why? Because the vector $(a_k,b_k)$ denotes the normal of your line. If you add two normal vectors of equal length for your two lines, you end up with a normal vector pointing in the middle between the normals you started with, for reasons of symmetry. Also, if you flip all three coefficients in the equation, you get the same line, so $l_2$ and $-l_2$ are the same line and that the symmetry argument applies to the difference as well. That's also the reason why $l_2-l_1$ or $-l_1-l_2$ are not worth considering: these are lines you already have, just with all the signs in the equation flipped around. So that explains why the sum or difference yields a line which has the direction of an angular bisector. The fact that they also pass through the intersection of the original lines is because a point that satisfies a pair of equations must also satisfy any linear combination of these equations. So you don't need to explicitly compute the point of intersection to get your hands at these angular bisector lines.

OK, now back to 3d. You can play the same game there: let plane $k$ be given by $p_k=(a_k,b_k,c_k,d_k)$ representing the plane $a_kx+b_ky+c_kz+d_k=0$. Normalize so that $a_k^2+b_k^2+c_k^2=1$. Now $p_1+p_2$ is one angular bisector plane between these two planes, and $p_1-p_2$ is another. Likewise $p_2+p_3$ and $p_2-p_3$ are angular bisectors between $p_2$ and $p_3$.

In order to get a line whose points are equidistant to all three planes, you need to intersect two of these planes, e.g. intersect $p_1+p_2$ with $p_2+p_3$. There are $2\times2=4$ possible choices how you can pick which planes to intersect. Since there are many valid formalisms for how to represent a line in 3d space, I'll leave this intersection as an exercise to you.

You don't need to consider any angular bisector between $p_1$ and $p_3$ for reasons of transitivity: If a point has the same distance to $p_1$ which it has to $p_2$, and it has the same distance to $p_2$ which it has to $p_3$, then it automatically has the same distance to $p_1$ which it has to $p_3$. You can of course use that third pair of planes to check your results. Or you can try to pick your pairs which you work with in such a way as to minimize numerical errors, e.g. by avoiding both bisectors and intersections for almost parallel planes.

The benefit of this approach is that the most computationally complicated part lies in the square root needed to normalize the planes. You can likely avoid dealing with systems of equations in three variables if you can intersect your planes in a way which is simpler than that.

  • $\begingroup$ [+1] Interesting analysis. As you have done, I have added a graphical explanation. $\endgroup$
    – Jean Marie
    Aug 17, 2023 at 20:23
  • $\begingroup$ Please have a look at the Edit I have added to my answer. $\endgroup$
    – Jean Marie
    Aug 18, 2023 at 9:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .