If the point ( 1, , ) lies on the line of the shortest distance between the lines x+2 -3 = y-2 4 = z-5 2 and x+2 -1 = y+6 2 , z=1 , then + =

Step-by-step solution

The shortest distance between two skew lines can be found by identifying points P and Q on each line such that the connecting vector PQ is perpendicular to both direction vectors. Given lines in symmetric form: L₁: x+2 -3 = y-2 4 = z-5 2 with direction vector d₁ = (-3, 4, 2) L₂: x+2 -1 = y+6 2 , z=1 with direction vector d₂ = (-1, 2, 0) Parametrize L₁ as (-2-3 , 2+4 , 5+2 ) and L₂ as (-2- , -6+2 , 1) . The connecting vector is PQ = (3 - , -8 - 4 + 2 , -4 - 2 ) . Requiring PQ d₁ = 0 yields 29 - 11 = -40 Requiring PQ