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I would like to construct formulaic equations for a line resulting from the intersection of two given planes. However, I would like the equations to be symmetric. Something like:

$$\frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C}$$

Given two equations of planes:

$$ N_{AX}x + N_{AY}y + N_{AZ}z + D_A = 0\\ N_{BX}x + N_{BY}y + N_{BZ}z + D_B = 0 $$

we can combine them to eliminate a variable. Only two of these are needed, but to help look for the symmetry I do all three, knowing that one is a linear combination of the other two:

$$ (N_{AY}N_{BX}-N_{AX}N_{BY})y + (N_{AZ}N_{BX}-N_{AX}N_{BZ})z + (N_{BX}D_A-N_{AX}D_B) = 0\\ (N_{AX}N_{BY}-N_{AY}N_{BX})x + (N_{AZ}N_{BY}-N_{AY}N_{BZ})z + (N_{BY}D_A-N_{AY}D_B) = 0\\ (N_{AX}N_{BZ}-N_{AZ}N_{BX})x + (N_{AY}N_{BZ}-N_{AZ}N_{BY})y + (N_{BZ}D_A-N_{AZ}D_B) = 0 $$

Here it is convenient to define place-holders:

$$ \begin{align} &A = N_{AY}N_{BZ}-N_{AZ}N_{BY}\\ &B = N_{AZ}N_{BX}-N_{AX}N_{BZ}\\ &C = N_{AX}N_{BY}-N_{AY}N_{BX}\\ &D = N_{BX}D_A-N_{AX}D_B\\ &E = N_{BY}D_A-N_{AY}D_B\\ &F = N_{BZ}D_A-N_{AZ}D_B \end{align} $$

So our three equations are now:

$$ \begin{align} -Cy + Bz + D &= 0\\ Cx + -Az + E &= 0\\ -Bx + Ay + F &= 0 \end{align} $$

There are three ways to pick two of the equations and combine them to form a set of equalities. The results of the three different combinations are:

$$ \begin{align} \frac{x}{A} = \frac{y + F/A}{B} = \frac{z - E/A}{C}\\\\ \frac{x - F/B}{A} = \frac{y}{B} = \frac{z + D/B}{C}\\\\ \frac{x + E/C}{A} = \frac{y - D/C}{B} = \frac{z}{C} \end{align} $$

They correspond to a point $(x_0, y_0, z_0)$ picked respectively from the line's y-z-plane-intercept, x-z-plane-intercept, and x-y-plane-intercept. This choice results in a lack of symmetry.

Is there way to construct a point $(x_0, y_0, z_0)$ such that each component is of a similar form and the resulting equation is completely symmetric? What is the geometric significance of this point?

  • $\begingroup$ I came up with a solution and posted it but I am interested to see other ways of solving this. $\endgroup$
    – caPNCApn
    Apr 2, 2021 at 23:04

2 Answers 2


Here's an alternative derivation of OP's solution.

Let's write the equations as $$\begin{align} u\cdot X = a \tag1\\ v\cdot X = b \tag2 \end{align}$$ where $X:=(x,y,z)$, (non-parallel) $u$ and $v$ are unit normal vectors to the planes, and $a$ and $b$ are the (signed) distances from the origin to the planes in the direction of those normals.

Then $w := u\times v = (u_y v_z-u_z v_y , u_z v_x - u_x v_z, u_x v_y-u_y v_x)$, being perpendicular to $u$ and $v$, is the direction vector of the line; moreover, $$w\cdot X = 0 \tag3$$ is the plane through the origin perpendicular to the line, so that its intersection with the line (call it $P$) is closest to the origin (and is therefore the most-natural "symmetric point" to consider). Since $u$, $v$, $w$ span space, we can write $P=pu+qv+rw$ so that $$\begin{align} u\cdot(pu+qv+rw) = a &\quad\to\quad a = p + q u\cdot v \\ v\cdot(pu+qv+rw) = b &\quad\to\quad b = q + p u\cdot v \tag4\\ w\cdot(pu+qv+rw) = 0 &\quad\to\quad 0 = r \end{align}$$ and we conclude $$\begin{align} P &= \frac1{1-(u\cdot v)^2}\left(\;(a-b u\cdot v) u + (b-a u\cdot v)v\;\right) \\[4pt] &= \frac1{\sin^2\theta}\left(\;(a-b\cos\theta) u + (b-a \cos\theta)v\;\right) \tag{5} \end{align}$$ where $\theta$ is the angle between $u$ and $v$ (and a dihedral angle between the planes). Note that $(5)$ is symmetric in the elements of the planes. Thus, the line of intersection $$\frac{x-P_x}{w_x} = \frac{y-P_y}{w_y}=\frac{z-P_z}{w_z} \tag6$$

is (probably) equivalent to OP's own answer. $\square$


I hypothesized that the point I was looking for might be the point on the line that is closest to the origin, so I attempted to construct it by using one of the intercept equations:

$$x_0/A = (y_0 + F/A)/B = (z_0 - E/A)/C$$

and setting to zero the dot product of the vector from the origin to the point we want with the direction vector of the line:

$$(A,B,C)\cdot((x_0,y_0,z_0) - (0,0,0)) = 0$$

This gives us three equations in three variables:

$$ Ax_0 + By_0 + Cz_0 = 0\\ y_0 = (Bx_0 - F)/A\\ z_0 = (Cx_0 + E)/A $$

So I solved for $x_0$ and then attempted to sub in for $y_0$ and $z_0$

$$ \begin{align} x_0 &= (BF - CE)/(AA + BB + CC)\\ y_0 &= (B(BF - CE)/(AA + BB + CC) - F)/A\\ z_0 &= (C(BF - CE)/(AA + BB + CC) + E)/A \end{align} $$ But these last two equations look ugly and hard to simplify.

Since the point on the line closest to the origin does not depened on which intercept I use to find it, I can use the line equations based on the other intercepts:

$$ Ax_0 + By_0 + Cz_0 = 0\\ x_0 = (Ay_0 + F)/B\\ z_0 = (Cy_0 - D)/B\\ y_0 = (CD - AF)/(AA + BB + CC) $$

$$ Ax_0 + By_0 + Cz_0 = 0\\ x_0 = (Az_0 - E)/C\\ y_0 = (Bz_0 + D)/C\\ z_0 = (AE - BD)/(AA + BB + CC) $$

So at long last we have the point on the line closest to the origin and it is indeed symmetric

$$ x_0 = (BF - CE)/(AA + BB + CC)\\ y_0 = (CD - AF)/(AA + BB + CC)\\ z_0 = (AE - BD)/(AA + BB + CC) $$


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