The Point of intersection of two Lines

Problem:

Find the point of intersection of the line having the position vector equation r1 = [0, 0, 1]  +  t[1, -1, 1] with the line having the position vector equation r2 = [4, 1, 2]  +  s[-6, -4, 0].
A diagram of this is shown on the right. O is the origin.

P is the point of intersection of the two lines.

[Graphics:Images/index_gr_1.gif] and  [Graphics:Images/index_gr_2.gif] are the position vectors of any point on the respective lines.
The position vectors are shown drawn at the point of intersection of the two lines where [Graphics:Images/index_gr_3.gif] =  [Graphics:Images/index_gr_4.gif]. Since, at the point of intersection, the two position vectors are identical it follows that;

            [0, 0, 1] + t[1, -1, 1] = [4, 1, 2]   + s[-6, -4, 0]
            
From which t and s can be solved.
Substituting for t in [Graphics:Images/index_gr_5.gif] or, s, in [Graphics:Images/index_gr_6.gif] will give the required point. Solving for the value of s and t which satisfies;

                                                                  [0, 0, 1] + t[1, -1, 1] = [4, 1, 2]  + s[-6, -4, 0]
                                                                  
                                                         
                            Rearranging;
                                                              t[1, -1, 1] - s[-6, -4, 0 ]  =  [4, 1, 2] -  [0, 0, 1]     
                                                        
                                                       [ t -  -6s,     -t  -  -4s,       t ]  =     [4, 1, 1]       
                                                        
                                                        
                               Equating components;
                                                                                    t +  6s  = 4                         1.
                                                                                  -t  +  4s  = 1                         2.
                                                                                              t  = 1                        3.            

                              Equations 2. and 3. inply;            t  =  1           s  =  [Graphics:Images/index_gr_7.gif]
                        
                        
Substituting in the equations for [Graphics:Images/index_gr_8.gif] = [0, 0, 1] + t [1, -1, 1]                     or,           [Graphics:Images/index_gr_9.gif]= [4, 1, 2]+ [Graphics:Images/index_gr_10.gif][-6, -4, 0]    gives;

                                                         = [0, 0, 1] + 1 [1, -1, 1]                                        = [4, 1, 2]+ [Graphics:Images/index_gr_11.gif][-6, -4, 0]
                                                         
                                                         = [0  +  1,   0  +  -1,   1  +  1]                                = [4 +  -3,    1  +   -2,    2  +  0]
                                                         
                                                         = [ 1,-1,  2]                                                            = [1,  -1,  2]
                                                         
  
  The two lines therefore intersect at the point (1,  -1,  2)                                                                           
                                                                             
 

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