Make your own free website on Tripod.com







The Line of Intersection between two Planes

Problem:

Find theline of intersection between the two planes given by the vector equations r1.[3, 4, 0]  = 5 and  r2.[1, 2, 3]  = 6

A diagram of this is shown on the right. [Graphics:Images/index_gr_1.gif] is a normal vector to Plane 1

[Graphics:Images/index_gr_2.gif] is a normal vector to Plane 2.

By simple geometrical reasoning;
the line of intersection is perpendicular to both normals.
To find the position vector, r, of any point on the line of intersection;

        find a vector, v, to which the line is parallel,


       find the position vector, a, of specific apoint on the line,

       then;      r  = a  +  tv, is the required result.
To find v;

                  the cross product of two vectors gives a third vector which is perpendicular to both vectors.
                  v is perpendicular to both [Graphics:Images/index_gr_3.gif] and [Graphics:Images/index_gr_4.gif],
                  therefore, v will beparallel to   [Graphics:Images/index_gr_5.gif] × [Graphics:Images/index_gr_6.gif],  
                                
                                   
                  From the equations to the planes,
                  
                       [Graphics:Images/index_gr_7.gif].[3, 4, 0]  = 5       and,     [Graphics:Images/index_gr_8.gif].[1, 2, 3]  = 6
                       
                       [Graphics:Images/index_gr_9.gif]  =   [3, 4, 0]       and        [Graphics:Images/index_gr_10.gif]  =  [1, 2, 3]
                       
                       
                     therefore    [Graphics:Images/index_gr_11.gif]× [Graphics:Images/index_gr_12.gif]  = [3, 4, 0] × [1, 2, 3]

                      = [Graphics:Images/index_gr_13.gif]
                                    
                       = [Graphics:Images/index_gr_14.gif]i     -   [Graphics:Images/index_gr_15.gif]j    +  [Graphics:Images/index_gr_16.gif]k
                                    

                        =  [12,   -9,  9]                                             1.


To find a;

                  at any point on the line the position vector, [Graphics:Images/index_gr_17.gif],  =   [Graphics:Images/index_gr_18.gif]
                  let the point be (x, y, z), then;
                   
                   From the equations to the planes,
                                
                  (x, y, z) . [3, 4, 0]  = 5    and   (x, y, z) . [1, 2, 3]  = 6               
                
                                    3x  +  4y   =  5   and    x  +  2y  +  3z  =  6
                  
                  Let the point have an x co-ordinate = 0 then;
                  
                                              y  = [Graphics:Images/index_gr_19.gif]     and   z  =  [Graphics:Images/index_gr_20.gif]
                                              
                   a is therefore [ 0,  [Graphics:Images/index_gr_21.gif],  [Graphics:Images/index_gr_22.gif]]                            2.                         
                      
Substituting result 1. and 2. into   r  = a  +  tv  gives;

                              r  = [ 0,  [Graphics:Images/index_gr_23.gif],  [Graphics:Images/index_gr_24.gif]]  +  t [12,   -9,  9]
                              
        or, more simply,   r  = [ 0,  [Graphics:Images/index_gr_25.gif],  [Graphics:Images/index_gr_26.gif]]  +  t [4,   -3,  3]        
        
        
        since the vector     [12,   -9,  9]  is parallel to   [4,   -3,  3]                    
 

HOME