Question

Two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point if $k$ is equal to

• A. $\frac{2}{9}$
• B. $\frac{1}{1}$
• C. $\frac{9}{2}$
• D. $\frac{1}{6}$

Answer 1

Answer: (C)

$\frac{9}{2}$

Let,

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$

$\frac{x-3}{1}=\frac{y-k}{2}=z=\mu$

Any point on the first line is given by $(2\lambda +1,3\lambda -1,4\lambda +1)$ and for the second line is given by $(\mu +3,2\mu +k,\mu )$.

If their is a point of intersection we have

$2\lambda +1=\mu +3$.................(i)

$3\lambda -1=2\mu +k$...............(ii)

$4\lambda +1=\mu$.....................(iii)

Replacing value of $\mu$ in (i) from (iii)

we get,

$\Rightarrow \lambda =\frac{-3}{2}$

Replacing value of $\lambda$ in (iii)

we get,

$\mu =-5$.

Replacing value of $\lambda$ and $\mu$ in (ii)

$\Rightarrow 3\times \frac{-3}{2}-1=2\times -5+k$

$\Rightarrow k=\frac{9}{2}$.