Find the length and the equations of the line of shortest distance between the lines given by: (x + 1)/2 = (y - 1)/1 = (z - 9)/-3 and (x - 3)/2

Question

Find the length and the equations of the line of shortest distance between the lines given by:

\(\cfrac{\text x+1}2=\cfrac{y-1}1=\cfrac{z-9}{-3}\) and \(\cfrac{\text x-3}2=\cfrac{y+15}{-7}=\cfrac{z-9}5.\)

(x + 1)/2 = (y - 1)/1 = (z - 9)/-3 and (x - 3)/2 = (y + 15)/-7 = (z - 9)/5

Answer 1

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,

d = \(\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}\)

3. Equation of line :

Equation of line passing through points A ≡ (x₁, y₁, z₁) and B ≡ (x₂, y₂, z₂) is given by,

Given equations of lines

Direction ratios of L1 and L2 are (2, 1, -3) and (2, -7, 5) respectively.

Let, general point on line L1 is P≡(x₁, y₁, z₁)

x₁ = 2s-1 , y₁ = s+1 , z₁ = -3s+9

and let, general point on line L2 is Q ≡ (x₂, y₂, z₂)

x₂ = 2t+3 , y₂ = -7t – 15 , z₂ = 5t + 9

Direction ratios of \(\overline{PQ}\) are ((5t - 2s + 10), (-7t – s – 16), (5t + 3s))

PQ will be the shortest distance if it perpendicular to both the given lines

Therefore, by the condition of perpendicularity,

2(5t - 2s + 10) + 1(-7t – s – 16) - 3(5t + 3s) = 0 and

2(5t - 2s + 10) – 7(-7t – s – 16) + 5(5t + 3s) = 0

⇒ 10t – 4s + 20 - 7t – s - 16 - 15t – 9s = 0 and

10t - 4s + 20 + 49t + 7s + 112 + 25t + 15s = 0

⇒ -12t – 14s = -4 and

84t + 18s = -132

Solving above two equations, we get,

t = -2 and s = 2

therefore, P ≡ (3, 3, 3) and Q ≡ (-1, -1, -1)

Now, distance between points P and Q is

Therefore, the shortest distance between two given lines is

d = 4√3 units

Now, equation of line passing through points P and Q is,

Therefore, equation of line of shortest distance between two given lines is

x = y = z.