Question

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

Answer 1

$\begin{array}{c}Here,given.......\\ anylinepassesthroughpoint(1,1,1)\\ so,\\ \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}\\ andlineintersetthrough:\\ \frac{x-1}{2}=\frac{x-1}{3}=\frac{x-1}{4}\\ Now,determenentmustbepositive:\\ \left|\begin{array}{c}012\\ abc\\ 234\end{array}\right|=0or,\left|\begin{array}{c}abc\\ 012\\ 234\end{array}\right|=0\\ \Rightarrow -(4a-2c)+2(3a-2b)=0\\ \Rightarrow -4a+2c+6a-4b=0\\ \Rightarrow 2a-4b+2c=0\\ \therefore a-2b+c=0-----------\left(i\right)\\ Again,intersectanotherline...........\\ \frac{x-(-2)}{1}=\frac{x-3}{2}=\frac{x-(-1)}{4}\\ Now,determenentmustbesatisfiied:\\ \left|\begin{array}{c}-2-13-1-1-1\\ abc\\ 124\end{array}\right|=0\\ \Rightarrow (-3)(4b-2c)-2(4a-c)+(-2)(2a-b)=0\\ \Rightarrow -12b+6c-8a+2c-4a+2b=0\\ \Rightarrow -12a-10b+8c=0\\ \Rightarrow 6a+5b-4c=0-----------\left(ii\right)\\ fromequ\left(i\right)\&\left(ii\right),wegettheratio:\\ \frac{a}{3}=\frac{b}{10}=\frac{c}{17}\\ then,therequiredlinewill:\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}\end{array}$