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Question: If the line \(\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 1 } { 4 }\) and \(\frac { ...

If the line x−12=y+13=z−14\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 1 } { 4 } and x−31=y−k2=z1\frac { x - 3 } { 1 } = \frac { y - k } { 2 } = \frac { z } { 1 }

intersect, then k =

A

2/9

B

9/2

C

0

D

–1

Answer

9/2

Explanation

Solution

We have, x−12=y+13=z−14=r1\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 1 } { 4 } = r _ { 1 } (Let)

x=2r1+1,y=3r1−1,z=4r1+1x = 2 r _ { 1 } + 1 , y = 3 r _ { 1 } - 1 , z = 4 r _ { 1 } + 1 i.e. point is

(2r1+1,3r1−1,4r1+1)\left( 2 r _ { 1 } + 1,3 r _ { 1 } - 1,4 r _ { 1 } + 1 \right) and x−31=y−k2=z1=r2\frac { x - 3 } { 1 } = \frac { y - k } { 2 } = \frac { z } { 1 } = r _ { 2 } (Let)

i.e. point is (r2+3,2r2+k,r2)\left( r _ { 2 } + 3,2 r _ { 2 } + k , r _ { 2 } \right).

If the lines are intersecting, then they have a common point.

⇒ 2r1+1=r2+3,3r1−1=2r2+k,4r1+1=r22 r _ { 1 } + 1 = r _ { 2 } + 3,3 r _ { 1 } - 1 = 2 r _ { 2 } + k , 4 r _ { 1 } + 1 = r _ { 2 }

On solving, r1=−3/2,r2=−5r _ { 1 } = - 3 / 2 , r _ { 2 } = - 5

Hence, k = 9/2.