Intersection of 3 Planes

This example is the same as the previous example except that P3 has changed.
  P1    x  - y + 4z = 5   => n1 = (1,-1,4)
  P2  3x + y +   z = 3   => n2 = (3, 1,1)
  P3  5x - y +  9z = 13 => n3 = (5,-1,9)

As before, L12 has  m = (-5,11,4) and contains point (2,3,0). This time the point lies on P3

The 3 planes do have a common intersection
  r = (2,3,0) + t(-5,11,4)

plane8a.gif (1979 bytes)
Now consider the following 3 planes.

  P1   x  +  y  +  z = 1  => n1 = (1,1,1)
  P2   x + 2y + 3z = 3  => n2 = (1,2,3)
  P3   x + 2y + 5z = 9  => n3 = (1,2,5)

Analysis of normal vectors
  n1 x n2 = (1,-2,1)
L12  has direction vector
  m = (1,-2,1)
  m n3 = 2
L12  intersects P3 in a point.

For a point on L12  let  z = 0
     x  +  y = 1
     x + 2y = 3
    ------------
    y = 2 ,  x = -1 ,  z = 0
L12  has eq'n r = (-1,2,0) + t(1,-2,1)

Sub  L12  into P3  
   (-1+t) + 2(2-2t) + 5t = 9
   t = 3
The 3 planes intersect at point
  P(2, -4, 3) 

 

 

plane7a.gif (2186 bytes)

Summary:
    n1 x n2 = direction vector for the line of intersection of   P1 and P2
    if  m n3   =  0   either no solution or a line of intersection
    if  m n3 <> 0   then a unique intersection point.