Question

If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then the vale of $k$ is

Answer 1

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

$x=2\lambda +1,y=3\lambda -2,z=4\lambda +1$ $x=\mu +3,y=2\mu +k,z=\mu$

$2\lambda +1=\mu +3$

$2\lambda -\mu =2----\left(1\right)4\lambda +1=\mu$

$4\lambda -\mu =-1---2$

$1-2$

$2\lambda -\mu -4\lambda +\mu =2+1$

$-2\lambda =3$

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

$\mu =4^2\left(\frac{-3}{x}\right)+1=-6+1$

$\mu =-5$

$3\lambda -2=2\mu +k$

$3\left(\frac{-3}{2}\right)-2=2\left(5\right)+k$

$\frac{-9}{2}-2+10=k\Rightarrow k=8-\frac{9}{2}=\frac{7}{2}$

$k=\frac{7}{2}$