# The equations of the lines of shortest distance between the lines …

Q: The equations of the lines of shortest distance between the lines $\large \frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$ and $\large \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ are

(A) 3(x – 21) = 3y + 92 = 3z – 32

(B) $\large \frac{x-62/3}{1/3} = \frac{y+31}{1/3} = \frac{z-31}{1/3}$

(C) $\large \frac{x-21}{1/3} = \frac{y + 92/3}{1/3} = \frac{z-32/3}{1/3}$

(D) $\large \frac{x-2}{1/3} = \frac{y+3}{1/3} = \frac{z-1}{1/3}$

Sol. Let P(2r1 , – 3r1, r1) and Q(3r2 + 2, – 5r2 + 1, 2r2 – 2) be the points on the given lines so that PQ is the line of shortest distance

d.r.s of PQ 2r1 – 3r2 – 2 , – 3r1 + 5r2 – 1, r1 – 2r2 + 2

Since it is perpendicular to given lines

2(2r1 – 3r2 – 2) – 3(3r1 + 5r2 – 1) + (r1 – 2r2 + 2) = 0

and (2r1 – 3r2 – 2) – 5(– 3r1 + 5r2 – 1) + 2(r1 – 2r2 + 2) = 0

r1 = 31/3, r2 = 19/3

P is (62/3 , -31 , 31/3) and Q is (21 , -92/3 , 32/3)

d.r.s of PQ is $(\frac{1}{3} , \frac{1}{3} , \frac{1}{3})$

Hence (A), (B) and (C) are the correct answers.