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 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 $x=\mu +3$,
$y=2\mu +k$,
$z=\mu$
$\Rightarrow 2\lambda +1=\mu +3,3\lambda -1=2\mu +k,4\lambda +1=\mu$
On solving we get,
$\lambda =-\frac{3}{2}$ and $\mu =-5$
$\therefore k=3\lambda -2\mu -1=3\left(-\frac{3}{2}\right)-2\left(-5\right)-1$
$\Rightarrow k=\frac{9}{2}$