Question

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{4}$ intersect each other, then find the value of '$k$'.

Answer 1

The given lines are $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{4}$
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ .......... (i)
$\therefore P(2\lambda +1,3\lambda -1,4\lambda +1)$ is any point on (i)
Also, let $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{4}=\mu$ ......... (ii)
$\therefore Q(\mu +3,2\mu +k,4\mu )$ is any point on (ii)
The two lines will intersect, if
$2\lambda +1=\mu +3$
$3\lambda -1=2\mu +k$
$4\lambda +1=4\mu$ ........ (iii)
$\mu =2\lambda -2$ ........ (iv)
$3\lambda -2\mu =k+1$ .......... (v)
Now, substituting (iv) in (iii),
$4\lambda +1=4(2\lambda -2)$
$4\lambda +1=8\lambda -8$
$4\lambda =9$
$\lambda =\frac{9}{4}$
Substituting the value of $\lambda$ in (iv),
$\mu =2\left(\frac{9}{4}\right)-2$
$\Rightarrow \mu =\frac{9}{2}-2$
$\Rightarrow \mu =\frac{5}{2}$
Substititing the value of $\lambda$ and $\mu$ in equation (v), we get
$3\lambda -2\mu =k+1$
$\Rightarrow 3\left(\frac{9}{4}\right)-2\left(\frac{5}{2}\right)=k+1$
$\Rightarrow \frac{27}{4}-5=k+1$
$\Rightarrow \frac{27}{4}-6=k$
$\Rightarrow k=\frac{3}{4}$
Thus, $k=\frac{3}{4}$