Let L1 be the line passing through the point P1 4 2 1 with
Solution
We know that the equation of a line through the point A (x1, y1, z1) and the direction vector (l, m, n) is
(x -x1)/l = (y-y1)/m = (z-z1)/n = k (say)
Then a on this line at a distance ‘k’ from the point A (x1, y1, z1) is P (x1 + l k, y1 + m k, z1 + n k).
In view of the above, the cartesian equations of the lines L1 and L2 are (x-4)/2 = (y-2)/-3 = (z-1)/2 = r1 (say) …(1)
and (x +3)/2=( y+5)/-3 = (z-5)/2 = r2 (say)…(2).
Then a point Q1 on the line L1 is (4+2r1,2-3r2,1+2r1) and a point Q2 on the line L2 is (-3+2r2,-5-3r2,5+2r2)
If Q1Q2 is the line of shortest distance, Its direction ratios will be [(4+2r1+3-2r2),(2-3r1+5+3r2), (1+2r1 – 5-2r2 )]
or, (7+2r1 -2r2 , 7-3r1 +3r2 , -4+2r1 -2r2 ).This line Q1Q2 is perpendicular to both the lines L1and L2 so that (2,-3,2).( 7+2r1 -2r2 , 7-3r1 +3r2 , -4+2r1 -2r2) = 0, or, 2(7+2r1 -2r2) -3(7-3r1 +3r2) +2(-4+2r1 -2r2) = 0 or, 14+ 4r1 -4r2 -21+9r1 -9r2 -8 +4r1 -4r2 = 0 or, -15 + 17r1 – 17r2 =0 or, 17(r1 –r2 ) = 15 or r2 = r1 - 15/17.Then the direction ratios of the line Q1Q2 are ( 7+2r1 -2r1 +30/17, 7-3r1 +3r1 -45/17, -4+2r1 -2r1+30/17) or, (149/17, 74/17, -38/17) The norm of Q1 Q2 is [ (149/17)2 + (74/17)2 + (-38/17)2 ] = (1/17) ( 22201+5476 + 1444) = (1/17)29121 . Thus the shortest distance between the lines L and L is (29121)/17. The point Q1 on the line L1 is (4+2r1,2-3r2,1+2r1) and the point Q2 on the line L2 is (-3+2r2,-5-3r2,5+2r2) = (-3+2r1 -30/17, -5-3r1 +45/17, 5+2r1 -30/17) =( -81/17 +2r1 , -40/17-3r1 , 55/17+2r1 ) where r1 is an arbitrary real number.
![Let L_1 be the line passing through the point P_1 = (4, 2, 1) with direction vector 3=[2, -3, 2]^T, and let L_2 be the line passing through the point P_2 = (-3 Let L_1 be the line passing through the point P_1 = (4, 2, 1) with direction vector 3=[2, -3, 2]^T, and let L_2 be the line passing through the point P_2 = (-3](/WebImages/34/let-l1-be-the-line-passing-through-the-point-p1-4-2-1-with-1101169-1761581850-0.webp)