Let L1 be the line passing through the point P13 3 2 with di

Let L_1 be the line passing through the point P_1=(-3, 3, 2) with direction vector dvector=[3, 3, -1]^T, and let L_2 be the line passing through the point P_2=(4, -2, -2) with the same direction vector. Find the shortest distance d between these two lines, and find a point Q_1 on L_1 and a point Q_2 on L_2 so that d(Q_1,Q_2)=d. Use the squareroot symbol \'Squareroot\' where needed to give an exact value for your answer d = 0 Q_1 = (0, 0, 0) Q_2 = (0, 0, 0)

Solution

The line L₁ is given by (-3+3t, 3+3t, 2-t)

The distance of point P₂ from the point (-3+3t, 3+3t, 2-t) on the line L₁ is given by:

d ² = (4+3-3t)² + (-2-3-3t)² + (-2-2+t)² = (7-3t)² + (-5-3t)² + (-4+t)²

Take derivative on both sides , we get :

2d d(d)/dt = -6(7-3t) -6(-5-3t)+2(-4+t) = 0

=> -42+18t+30+18t-8+2t = 0

=> 38t = 20

=> t = 20/38 = 10/19

d² = (7-30/19)² + (-5-30/19)² + (-4+10/19)²

=> d² = (10609)/19² + (15625)/19² + (7396)/19²

=> d² = 33630/19²

=> d = sqrt(33630)/19

The shortest distance = sqrt(33630)/19

Now this occurs when t = 10/19

Q₁ = (-3+30/19, 3+30/19, 2-10/19) = (-27/19, 87/19, 28/19)

Q₂ = (4+30/19, -2+30/19, -2-10/19) = (106/19, -8/19, -48/19)