Three Dimensional Geometry Question 375
Question: If the lines $ \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} $ and $ \frac{x-3}{1}=\frac{y-k}{1}=\frac{z}{1} $ intersect, then k =
[IIT Screening 2004]
Options:
A) $ \frac{2}{9} $
B) $ \frac{9}{2} $
0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Any point on $ \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} $ $ =\lambda $ is,
$ (2\lambda +1,,3\lambda -1,,4\lambda +1);\lambda \in R $
Any point on $ \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} $ $ =\mu $ is,
$ (\mu +3,,2\mu +k,,\mu );,\mu \in R $
The given lines intersect if and only if the system of equations (in $ \lambda , $ and $ \mu $ )
$ 2\lambda +1=\mu +3 $ …..(i)
$ 3\lambda -1=2\mu +k $ …..(ii)
$ A+2B+3C=0 $ …..(iii)
has a unique solution.
Solving (i) and (iii), we get $ \lambda =\frac{-3}{2},,\mu =-5 $
From (ii), we get $ \frac{-9}{2}-1=-10+k\Rightarrow k=\frac{9}{2} $ .
BETA
