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

Toolbox:
• Eqn. of a line is $\large\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}=\lambda$ where $(a_1,a_2,a_3)$ is a point on the line and $(b_1,b_2,b_3)$ is d.r. of the line.
• Angle between two lines whose d.r. are $d_1$ and $d_2$ is $cos^{-1}\bigg(\large\frac{d_1.d_2}{|d_1||d_2|}\bigg)$.
Let the equation of the line be
$\large\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}=\lambda$
But given that it passes through origin.
$\Rightarrow (a_1,a_2,a_3)=(0,0,0)$
$\Rightarrow$ the eqn. is $\large\frac{x}{b_1}=\frac{y}{b_2}=\frac{z}{b_3}=\lambda$.......(i)
Let $\large\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\mu$.........(ii)
point of intersection of (i) and (ii) say Q is given by
$Q(2\mu+3,\mu+3,\mu)$
d.r.of OQ=$(2\mu+3,\mu+3,\mu)$
Since (i) passes through origin O,Line (i)= Line OQ
$\Rightarrow(b_1,b_2,b_3)=(2\mu+3,\mu+3,\mu)$
$\Rightarrow$ d.r. of (i)$= d_1=(b_1,b_2,b_3)=(2\mu+3,\mu+3,\mu)$
d.r.of (ii) $=d_2=(2,1,1)$
$d_1.d_2=2(2\mu+3)+1(\mu+3)+1.\mu=6\mu+9$
$|d_1|=\sqrt{(2\mu+3)^2+(\mu+3)^2+\mu^2}=\sqrt{6\mu^2+18\mu+18}$
$|d_2|=\sqrt{4+1+1}=\sqrt 6$
Also given that (i) and (ii) intersect at angle $\large\frac{\pi}{3}$
$\Rightarrow cos\large\frac{\pi}{3}$ $=\bigg(\large\frac{d_1.d_2}{|d_1||d_2|}\bigg)$
$\Rightarrow\large\frac{1}{2}=\frac{6\mu+9}{\sqrt {6\mu^2+18\mu+18}\sqrt 6}$
$\Rightarrow 12\mu+18=6(\sqrt {\mu^2+3\mu+3})$
$\rightarrow 2\mu+3=\sqrt {\mu^2+3\mu+3}$
squaring both the sides
$4\mu^2+9+12\mu=\mu^2+3\mu+3$
$\Rightarrow 3\mu^2+9\mu+6=0\:\:or\:\:\mu^2+3\mu+2=0$
$\Rightarrow \mu=-1\:\:or\:\:-2$
$\Rightarrow d_1=(1,2,-1)\:\:or\:\:(-1,1,-2)$
Equation of the required lines are
$\large\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda$ and
$\large\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda$
answered May 20, 2013
edited Feb 11, 2014