Question

shortest distance between two lines

x = 1 + t, y = 1 + 6t, z = 2t, t ∈ R and

x = 1 + 2k, y = 5 + 15k, z = – 2 + 6k, k ∈ R is ____ 

(A) 4

(B) 6

(C) 2

(D) 1

Answer 1

The correct option (C) 2

Explanation:

x = 1 + t, y = 1 + 6t, z = 2t, and

x = 1 + 2k, y = 5 + 15k, z = – 2 + 6k

∴ x – 1 = [(y – 1)/6] = (z/2) ⇒ vector a = (1, 1, 0) and vector ℓ = (1, 6, 2)

also [(x – 1)/2] = [(y – 5)/15] = [(z + 2)/6] ⇒ vector b = (1, 5, – 2) and

vector m = (2, 15, 6)

∴ vector l × vector m

and vector b – a = (0, 4, – 2), |ℓ × m| = √(36 + 4 + 9) = 7

Hence

vector (b – a) ∙ (ℓ × m) = (0, 4, – 2) ∙ (6, – 2, 3) = – 14

∴ shortest distance = [{|(b – a) ∙ (ℓ × m)|}/(|ℓ × m|)] 

= [(|– 14|)/7] = 2 unit.