The Perpendicular Distance between two Skew Lines

Problem:

Find the perpendicular distance between the line passing through the the point (1, -1, 1) which is parallel to the vector u =[1, 3, 0] and the line passing through the point (1, 1, 3) which is parallel to the vector v = [1, 1, 0].

A diagram of this is shown on the right. A is the given point through which the first line passes

.B is the given point through which the second line passes.

MN is the common perpendicular to both given lines.

u and v are the vectors to which the respective lines are parallel.
A line is drawn joining the points A and B. Since MN is the common perpendicular, the length MN is the required length.

Since points A and B are on lines perpendicular to MN,
MN =  the projection of  AB onto MN;
    Therefore,   MN = AB . [Graphics:Images/index_gr_1.gif]

AB =  [1, 1, 3] - [1, -1, 1]
       =  [0, 2, 2]

                                Now, the cross product of two vectors gives a third vector which is perpendicular to both vectors.
                                MN is perpendicular to both u and v,
                                Therefore, MN will beparallel to   u × v,  but we can not be sure whether  u × v is directed M to N or N to M
                                
                                Therefore,   [Graphics:Images/index_gr_2.gif]  = +/- [Graphics:Images/index_gr_3.gif]
                                    
                                                                        u × v  = [1, 3, 0] × [1, 1, 0]

                                    = [Graphics:Images/index_gr_4.gif]
                                    
                                    = [Graphics:Images/index_gr_5.gif]i     -   [Graphics:Images/index_gr_6.gif]j    +  [Graphics:Images/index_gr_7.gif]k
                                    

                                    =  [0, 0, -2]  
                    
                                      | u × v |  = 2
                  
                                Therefore,  [Graphics:Images/index_gr_8.gif]  = +/-[Graphics:Images/index_gr_9.gif]
                                
                                                           =  +/-[0, 0,-1]
                                
We can now finaly calculate MN using;
             MN = AB . [Graphics:Images/index_gr_10.gif]
                     = [0, 2, 2] . +/-[0, 0, -1]
                     
                     =  +/-( 0 + 0 + -2)
                     
                     =  +/- 2
                     
                     Since MN is a length we must take the positive result;
                    
The required distance is therefore 2 units                                                 
               
                                                  
 

HOME