Intersection of Two Planes
John Krumm
Microsoft Research
Redmond, WA, USA
5/15/00
Everyone knows that the intersection of two planes in 3D is a line, and it’s easy to
compute the line’s parameters. But, the cookbook formulae for the line are not
necessarily the best nor most intuitive way of representing the line. This short note gives
the derivation and the formula that I came up with when I was faced with this problem
and couldn't find a solution l liked anywhere. It is also a simple example of how to use
Lagrange multipliers.

The line of intersection is normally represented as a point on the line p  ( p x , p y , p z )

and a direction vector n  (n x , n y , n z ) emanating from this point. While the direction

vector, n , is easy to compute and obvious to represent (can be normalized to unit length),

the point on the line, p , can legitimately be any point on the line. Mathematically it
doesn't matter what the point is, as long as it's on the line. But it is easier to understand
the line if the point is somehow canonical or somehow close to the other points we are
concerned with. The formula below gives the point on the line that is closest to some


other point p o that can be specified by the user. The point p o could be the origin, giving
a canonical representation of the line. Or it could be some other point, like the midpoint
of a line connecting points on the two planes, giving a point on the line that is nearby the
other points in the problem's representation. At the end of this note I give examples of

both kinds of choices for p o .

The two planes are represented by their normal vectors, n1  (n1x , n1 y , n1z ) and


n2  (n2 x , n2 y , n2 z ) , and points on the two planes, p1  ( p1x , p1 y , p1z ) and

p 2  ( p 2 x , p 2 y , p 2 z ) . The direction vector of the line of intersection is the cross product of
  
the two normal vectors, i.e. n  n1  n2 . This can be normalized for a canonical
representation, but the normalization is not necessary for what follows.

We will compute the point on the line, p , as the point on the line that is closest to

some arbitrarily chosen point p o  ( p 0 x , p 0 y , p 0 z ) . This point could be the origin


( po  (0,0,0) ) or any other point. In the second example below we choose p o to be the


midpoint of a line segment connecting p1 and p 2 .

The point p must be on both planes, so we have two constraints:
  
(1)
( p  p1 )  n1  0
  
( p  p 2 )  n2  0
(2)


The point p should also be as close as possible to the point p o , so we want to minimize
the distance between the two. The distance is
 
p  p0
2
  p x  p0 x    p y  p0 y    p z  p0 z 
2
2
2
(3)
This is a textbook example of a problem for Lagrange multipliers, with one objective
function (Equation (3)) and two constraints (Equations (1) and (2)). The function w
containing the constraints and objective function is
  2
  
  
w  p  p0   ( p  p1 )  n1   ( p  p 2 )  n2
 ( p x  p0 x ) 2  ( p y  p0 y ) 2  ( p z  p0 z ) 2 
(4)
 
p x n1x  p y n1 y  p z n1z  p1  n1 
 
p x n2 x  p y n2 y  p z n2 z  p 2  n2
Here  and  are the two Lagrange multipliers (one for each constraint). The expanded
form of w has been written to make the next step easier, which is the partial derivatives.
Continuing the recipe for Lagrange multipliers, we compute the partial derivatives and
set them to zero.
w
 2 p x  p 0 x   n1 x  n2 x  0
p x
w
 2 p y  p 0 y   n1 y  n2 y  0
p y
w
 2 p z  p 0 z   n1z  n2 z  0
p z
(5)
w
 
 p x n1 x  p y n1 y  p z n1 z  p1  n1  0

w
 
 p x n2 x  p y n2 y  p z n2 z  p 2  n2  0


We are interested in solving these equations for p  ( p x , p y , p z ) , but we will also
have to solve for the Lagrange multipliers. Equation (5) represents five linear equations
in five unknowns. In matrix form, the five equations are
0
0 n1x n2 x  p x   2 p0 x 
2

0
  
2
0
n
n
2
p
p




1
y
2
y
0
y
y





0
0
2 n1z n2 z  p z  2 p0 z 
(6)

     
0     p1  n1 
 n1x n1 y n1z 0
 
n2 x n2 y n2 z 0
0     p2  n2 

T
Solving this matrix equation for the unknown vector  px , px , pz ,  ,   gives the point on

the line closest to p o . We don’t care about the values of the two Lagrange multipliers,
but we have to compute them anyway.
Example 1 – Simple
This is a simple example just to see if the solution confirms what we know is the right
answer. Suppose one of the two planes corresponds to the xy plane. We can choose


n1  0,0,3 and p1  7,11,0 . Suppose that the second plane cuts through the x axis at
1,0,0 , cuts through the y axis at 0,1,0 and is parallel to the z axis. A point on this


plane is p  1,0,0 and its normal is n  5,5,0 . This is illustrated in Figure 1. We will
2
2

find the point of intersection closest to the origin, thus po  (0,0,0) .
The resulting line of intersection should have a direction of  1,1,0 . The computed
  
direction of the line of intersection is n  n1  n2   15,15,0 , which is along the
direction we expected. From the figure we can tell that the point on the line of
intersection closest to the origin is 0.5,0.5,0 . The matrix equation becomes
 2 0 0 0 5  p x   0 
0 2 0 0 5  p   0 

 y   
0 0 2 3 0  p z    0 
(7)

   
0 0 3 0 0     0 
5 5 0 0 0    5 
The solution is  p x , p x , p z , ,    0.5,0.5,0,0,0.2 , which confirms our intuition for
the point on the line closest to the origin.
T
y
1.2
1
0.8
0.6
0.4
0.2
x
-0.2
0.2
0.4
0.6
0.8
1
1.2
-0.2
Figure 1: Plane 1 is the xy plane, and plane 2 cuts across to make the diagonal line.
Example – Two Planar Patches
This is an example that might come from a computer graphics application that uses
polygons to model a 3D object. As shown in Figure 2, there are two triangular polygons
that share a common edge. We want to find the line that represents this edge. The vertices
of the two polygons are 0,0,0, 4,1,1, 4,1,1 for the first one and
0,0,0, 4,1,1, 4,1,1 for the second one. The normals of the two planes are (from


cross products of their vertices) n1  2,8,0 and n2  2,0,8 . Points on the two planes


are (from the means of their vertices) p1  8 ,  2 ,0 and p 2  8 ,0,  2 . We want
3
3
3
3
to find a point on the line of intersection that is close to the two polygons, so we will pick







p o as the midpoint of p1 and p 2 :

a line segment connecting p1 and





po   p1  p 2  / 2  8 ,  1 ,  1 . The figure shows
3
3
3


p 2 , and p o is the midpoint of this line segment.
The direction vector of the line of intersection is
1
  
n  n1  n2  64,16,16   4,1,1 , which points along the intersection as it
16
should. Plugging in everything to the matrix equation, we have
2 0 0 2 2 p x   16 3 
0 2 0 8 0   p    2 

 y   3 
0 0 2 0 8   p z     2 

   3 
2
8
0
0
0

    0 
2 0 8 0 0   

 0 
The solution is  p x , p x , p z ,  ,   
T
(8)
1
68,17,17,2,2  2.52,0.63,0.63,0.07,0.07  .
27
Figure 2: Two planar patches intersect along a line. We find a point on the line that is closest to the
midpoint of a line connecting the two patches.