Find the equation of $2$ lines through the origin which intersect the line
$\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}$ at an angle of $\displaystyle \frac{\pi}{3}$
$\bf{My\; Try::}$ Let equation of line be $\displaystyle \frac{x-0}{a} = \frac{y-0}{b} = \frac{z-0}{c}$
Where $<a,b,c>$ be the direction cosine of line parallell to that line.
and Given Line is $\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}$
Where $<2,1,1>$ be the direction cosine of line parallell to that line
and Given $\displaystyle \frac{\pi}{3}$ be the angle between $<a,b,c>$ and $<2,1,1>$
So $\displaystyle \cos \frac{\pi}{3} = \frac{2a+b+c}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{6}}\Rightarrow \frac{1}{2}=\frac{2a+b+c}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{6}}$
Now how can i solve it, Help required, Thanks
you must set $$a=3+2t$$ $$b=3+t$$ $$c=t$$ with a real number $t$ since the second line intersect the line above.