Equation of the line of the shortest distance between the lines 1x=−1y=1z and 0x−1=−2y+1=1z is:
Solutions · 1
Official solution
d1=(1,−1,1), d2=(0,−2,1); common perpendicular direction =d1×d2=(1,−1,−2).
Let F1=(t,−t,t) on line 1, F2=(1,−1−2s,s) on line 2, with F1F2 perpendicular to both directions. Solving gives t=32,s=0, so F2=(1,−1,0).
Line of shortest distance through F2(1,−1,0) with direction (1,−1,−2): 1x−1=−1y+1=−2z.