Question

Find the shortest distance between the lines x+1=2y=-12z and x=y+2=6z-6.

Answer 1

The calculations in this are very tough and tedious .. 
Ans is approximately 4 units  The shortest perpendicular distance between the two lines.

L1:  (x+1) / 1 =  y / (1/2) = z / (-1/12) = t       --- (1)
Point P on L1:   [ t-1, t/2, -t/12 ]                   -- (2)
Direction ratios:   (1 ,  1/2,  -1/12 )            ---(3)

L2:  x/1 = (y+2) / 1  = (z-1) / (1/6)             -- (4)
Point Q on L2:   [s , s-2, 1+ s/6 ]             --- (5)
Direction ratios:    (1,  1,  1/6)                   ---(6)

L3:   PQ
   Direction ratios of PQ: ( s - t+1,  s-2 -t/2 ,  1+s/6 + t/12 )       ----(7)

Dot product of direction ratios of PQ & L1 = 0
      1(s-t+1) +1/2* (s-2-t/2) -1/12 *(1+s/6+ t/12) = 0     --- (8)
Dot product with L2:
      1 (s-t+1) + 1 (s-2-t/2) + 1/6* (1+s/6+t/12) = 0     -- (9)

Solving (8)  &(9):    t = 2832/775     s = 2394/775

Substitute these values in (2) & (4) to get  P & Q.

Find the distance by : √[(Px-Qx)² + (Py-Qy)² + (Pz-Qz)² ]   = 4   approx

Answer 2

Hey!!

Please check the below attached file!!

Thanks!!