Question

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then k =

• A. $\frac{2}{9}$
• B. $\frac{9}{2}$
• C. \N
• D. 3

Answer 1

The correct option is B $\frac{9}{2}$

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 $\mu )$
$2\lambda +1=\mu +3....\left(i\right)$
$3\lambda -1=2\mu +k.....\left(ii\right)$
$4\lambda +1=\mu ....\left(iii\right)$
has a unique solution.
Solving (i) and (iii), we get $\lambda =\frac{-3}{2},\mu =-5$
From (ii), we get $\frac{-9}{2}-1=-10+k\Rightarrow k=\frac{9}{2}$.