how to find the vector equation of the line of shortest distance between two skew lines?? Share with your friends Share 6 Manbar Singh answered this Consider an example.Find the vector equation of a line of shortest distance between the 2 skew linesx-33 = y-8-1 = z-3 and x+3-3 = y+72 = z-64Here is the solution.The given equations of the lines are :x-33 = y-8-1 = z-3 = psay .......1and x+3-3 = y+72 = z-64 = qsay ..........2Now, coordinates of any general point P on line 1 is P3p+3,-p+8,p+3.coordinates of any general point Q on line 2 is Q-3q-3,2q-7,4q+6Now, if PQ is the required line of shortest distance, then PQ will be perpendicular to both line 1 and line 2.We know that direction ratios of a line segment joining points x1,y1,z1 and x2,y2,z2 isx2-x1, y2-y1,z2-z1Now, direction ratios of PQ are : -3q-3p-6,2q+p-15,4q-p+3Now, Direction ratios of line 1 are :3, -1, 1 Direction ratios of line 2 are :-3, 2, 4Since PQ is ⊥ to line 1, then3-3q-3p-6 - 12q+p-15+14q-p+3 = 0⇒-11p - 7q = 0 .......3Since PQ is ⊥ to line 2, then-3-3q-3p-6 +22q+p-15+44q-p+3 = 0⇒7p + 29q = 0 ........4From 3 and 4, we getp = q = 0Putting p = 0 and q = 0 in the points P and Q, we getcoordinates of P are P3,8,3coordinates of Q are Q-3,-7,6Direction ratios of PQ are : -6, -15, 3Now, cartesian equation of the line passing through point x1,y1,z1 and having direction ratios a,b,c isx-x1a = y-y1b = z-z1cSo, equations of line PQ passing through P3,8,3 and having Direction ratios -6, -15, 3 is,x-3-6 = y-8-15 = z-33⇒x-32 = y-85 = z-3-1 = λIn vector form , we getr→ = 3i^ + 8j^ + 3k^ + λ2i^ + 5j^ - k^ Hope this would have cleared your doubt. 6 View Full Answer
