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 the value of $k$ is.

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

Answer 1

Answer: (B)

Since, the lines intersect, therefore they must have a point in common, i.e.
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$
and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$
$\Rightarrow x=2\lambda +1,y=3\lambda -1$
$z=4\lambda +1$
and $x=\mu +3,y=2\mu +k,z=\mu$ are same.
$\Rightarrow 2\lambda +1=\mu +3$
$3\lambda -1=2\mu +k$
$4\lambda +1=\mu$
On solving Ist and IIIrd terms, we get, $\lambda =-\frac{3}{2}$
and $\mu =-5$
$\therefore k=3\lambda -2\mu -1$
$\Rightarrow k=3\left(-\frac{3}{2}\right)-2(-5)-1=\frac{9}{2}$
$\therefore k=\frac{9}{2}$