# Equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle of $\frac{\pi}{3}$

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 an angle of $\frac{\pi}{3}$.

Let the direction ratios of the two required lines be $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$.
Therefore the two equations are
$\frac{x-0}{a_1}=\frac{y-0}{b_1}=\frac{z-0}{c_1}$ and $\frac{x-0}{a_2}=\frac{y-0}{b_2}=\frac{z-0}{c_2}$
As these lines are making an angle of $\frac{\pi}{3}$ with $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$.
So,$\cos\frac{\pi}{3}=\frac{2a_1+b_1+c_1}{\sqrt6\sqrt{a_1^2+b_1^2+c_1^2}}$
And $\cos\frac{\pi}{3}=\frac{2a_2+b_2+c_2}{\sqrt6\sqrt{a_2^2+b_2^2+c_2^2}}$
But i am stuck here.How i can solve three variables with one equation.The book gives answer as $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$
So we have$$\frac{x-3}{2}=\frac{y-3}{1}=\frac z1=\lambda$$ $$\Rightarrow \underline{r}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}2\lambda+3\\ \lambda+3\\ \lambda\end{matrix}\right)$$
• It is not possible for me to write the equation of line in one unknown parameter.As $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$,where $t$ is a parameter.So $x=at,y=bt,z=ct$.It is still in many variables.How can i make a equation in one parameter.@David Quinn – diya Nov 16 '15 at 12:05