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$
$\therefore k=3 \lambda-2 \mu-1$
$\Rightarrow k=3\left(-\frac{3}{2}\right)-2(-5)-1=\frac{9}{2}$
$\therefore k=\frac{9}{2}$