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) 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) Summary:     n1 x n2 = m  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.