STD 12 - 11. three dimension geometry — Mathematics STD 12 Science — Question

Eliminationg $n$ from both the equations, we have 

 ${l^2} + {m^2} - {\left( {l + m} \right)^2} = 0$

$ \Rightarrow {l^2} + {m^2} - {l^2} - {m^2} - 2ml = 0$

$ \Rightarrow 2lm = 0$

$ \Rightarrow lm = 0$

$ \Rightarrow l = 0\,\,\,\,or\,\,\,m = 0$

If $l=0$, we have $m+n=0$ and ${m^2} - {n^2} = 0$

$ \Rightarrow l = 0,m = \lambda ,n =  - \lambda $

If $m=0$, we have $l+m=0$ and ${l^2} - {m^2} = 0$

$ \Rightarrow l =  - \lambda ,m = 0,n = \lambda $

So, the vector parallel to these given lines

are $\vec a = \hat j - \hat k\,$ and $\,\,\vec b =  - \hat i + \hat k$

If angle between the lines is $'\theta ',$ then

$\cos \theta  = \frac{{\left| {\vec a.\vec b} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}} = \frac{1}{{\sqrt 2 .\sqrt 2 }}$

$ \Rightarrow \cos \theta  = \frac{1}{2}$

$\therefore \theta  = \frac{\pi }{3}$