Question

Find the equation of the line which intersects the lines $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ and $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and passes through the point (1, 1, 1).

Answer 1

Let $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}=\lambda$ $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\mu .$
So, the coordinates of random points on these lines are $P(\lambda -2,2\lambda +3,4\lambda -1)$ and
$Q(2\mu +1,3\mu +2,4\mu +3)$. Let A(1, 1, 1).
Now d.r.'s of line AP are $\lambda -3,2\lambda +2,4\lambda -2)$ and, d.r.'s of line AQ are $2\mu ,3\mu +1,4\mu +2.$
Therefore $\frac{\lambda -3}{2\mu }=\frac{2\lambda +2}{3\mu +1}=\frac{4\lambda -2}{4\mu +2}=k$ say.
$\Rightarrow \lambda -3=2k\mu ,2\lambda +2=k(3\mu +1),2\lambda -1=k(2\mu +1)$
$\Rightarrow \frac{\lambda -3}{2}=k\mu ,2\lambda +2=3\mu k+k,2\lambda -1=2\mu k+k$
$\therefore 2\lambda +2=3\times \frac{\lambda -3}{2}+k,2\lambda -1=2\times \frac{\lambda -3}{2}+k\Rightarrow \frac{\lambda +13}{2}=k,2\lambda -1=\lambda -3+k$
$\therefore \lambda +2=\frac{\lambda +13}{2}\Rightarrow 2\lambda +4=\lambda +13\Rightarrow \lambda =9$
Also $\frac{9-3}{2\mu }=\frac{18+2}{3\mu +1}\Rightarrow \frac{3}{2\mu }=\frac{10}{3\mu +1}\Rightarrow \mu =\frac{3}{11}$
Therefore the d.r.'s of the required line through point A are 6, 20, 34 i.e., 3, 10, 17.
Hence the equation is: $\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}.$