Find the length and the equations of the line of shortest distance between the lines given by:
and 
Given : Cartesian equations of lines
![]()
![]()
Formulae :
1. Condition for perpendicularity :
If line L1 has direction ratios (a1, a2, a3) and that of line L2 are (b1, b2, b3) then lines L1 and L2 will be perpendicular to each other if
![]()
2. Distance formula :
Distance between two points A≡(a1, a2, a3) and B≡(b1, b2, b3) is given by,
![]()
3. Equation of line :
Equation of line passing through points A≡(x1, y1, z1) and B≡(x2, y2, z2) is given by,
![]()
Answer :
Given equations of lines
![]()
![]()
Direction ratios of L1 and L2 are (-1, 2, 1) and (1, 3, 2) respectively.
Let, general point on line L1 is P≡(x1, y1, z1)
x1 = -s+3 , y1 = 2s+4 , z1 = s-2
and let, general point on line L2 is Q≡(x2, y2, z2)
x2 = t+1 , y2 = 3t – 7 , z2 = 2t - 2
![]()
![]()
![]()
Direction ratios of
are ((t + s – 2), (3t – 2s – 11), (2t – s))
PQ will be the shortest distance if it perpendicular to both the given lines
Therefore, by the condition of perpendicularity,
-1(t + s – 2) + 2(3t – 2s – 11) + 1(2t – s) = 0 and
1(t + s – 2) + 3(3t – 2s – 11) + 2(2t – s) = 0
⇒ - t – s + 2 + 6t – 4s – 22 + 2t – s = 0 and
t + s – 2 + 9t – 6s – 33 + 4t – 2s = 0
⇒ 7t – 6s = 20 and
14t - 7s = 35
Solving above two equations, we get,
t = 2 and s = -1
therefore,
P ≡ (4, 2, -3) and Q ≡ (3, -1, 2)
Now, distance between points P and Q is
![]()
![]()
![]()
![]()
Therefore, the shortest distance between two given lines is
![]()
Now, equation of line passing through points P and Q is,
![]()
![]()
![]()
![]()
Therefore, equation of line of shortest distance between two given lines is
![]()