Question

Find the equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi }{3}$ each.

Answer 1

Given equation of the line is $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\lambda$ ...(i)

So, DR's of the line are 2,1,1 and DC's of the given lines are $\frac{2}{\sqrt{6}}\frac{1}{\sqrt{6}}\frac{1}{\sqrt{6}}.$
Also, the required lines make angle $\frac{\pi }{3}$ with the given line. From Eq.(i). $x=(2\lambda +3)$, $y=(\lambda +3)$ and $z=\lambda$ $\because cos\theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\therefore cos\frac{\pi }{3}=\frac{(4\lambda +6)+(\lambda +3)+\left(\lambda \right)}{\sqrt{6}\sqrt{(2\lambda +3{)}^2+(\lambda +3{)}^2+{\lambda }^2}}$ $\Rightarrow \frac{1}{2}=\frac{6\lambda +9}{\sqrt{6}\sqrt{(4{\lambda }^2+9+12\lambda +{\lambda }^2+9+6\lambda +{\lambda }^2}}\Rightarrow \frac{\sqrt{6}}{2}=\frac{6\lambda +9}{\sqrt{6{\lambda }^2+18\lambda +18}}\Rightarrow 6\sqrt{({\lambda }^2+3\lambda +3)}=2(6\lambda +9)$ $\Rightarrow 36({\lambda }^2+3\lambda +3)=36\lambda (4{\lambda }^2+9+12\lambda )$ $\Rightarrow {\lambda }^2+3\lambda +2=0\Rightarrow \lambda (\lambda +2)+1(\lambda +2)=0\Rightarrow (\lambda +1)(\lambda +2)=0\therefore \lambda =-1,-2$ So, the DC's are 1,2,-1 and -1,1,-2.
Also, both the required lines passes through origin.
So, the equations of required lines are $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}and\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$.