Let L1 be the line passing through the point P14 11 3 with d

Let L₁ be the line passing through the point P₁=(4, 11, 3) with direction vector d₁=[0, 3, 2]^T, and let L₂ be the line passing through the point P₂=(1, 11, 14) with direction vector d₂=[2, 3, 0]^T.
Find the shortest distance d between these two lines, and find a point Q₁ on L₁ and a point Q₂ on L₂ so that d(Q₁,Q₂) = d. Use the square root symbol \'\' where needed to give an exact value for your answer.

d=...

Q1=(...)

Q2=(...)

Solution

The parametric equation of the line L₁ is r = (-4,-11,-3) + s( 0,-3,-2). Also, the parametric equation of the Line L₂ is r = (1,-11, 14)+t( -2,-3,0). An arbitrary point P on the line L₁ is (-4, -11-3s, -3-2s) and an arbitrary point Q on the line L₂ is ( 1-2t, -11-3t, 14). Then, the direction vector of the line PQ is( 1-2t+4, -11-3t+11+3s, 14+3+2s) i.e. ( 5-2t, 3s-3t, 17+2s). Now, let PQ is perpendicular to the lines L₁ and L₂. Then the distance PQ is the shortest distance between the lines L₁ and L₂.

Then, we have (5-2t,3s-3t,17+2s).(0,-3,-2)=0 or,-3(3s-3t)-2(17+2s)=0 or, 9t-9s-34-4s=0 or, 9t-13s = 34

and(5-2t, 3s-3t, 17+2s).(-2,-3,0)= 0 or, -2(5-2t)-3(3s-3t) = 0 or, -10+4t-9s+9t = 0 or, 13t-9s= 10.

On solving the above two equations, we get t = -2 and s = -4.

Hence the point P is ( -4, 1,5) and the point Q is ( 5, -5,14). Therefore, the vector PQ is ( 9,-6,9).Then the shortest distance between the lines L₁ and L₂ is ||PQ || = [ 9²+ (-6)²+ 9²] = ( 81+36+81) = 198