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 line:-

$\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=x$

$\therefore x=2r+3$

$y=r+3$

$z=r$

So, $(2r+3,r+3,r)$ is the direction ratio of two lines that intersect at $\frac{\pi }{3}$ with given line & passes through $(0,0)$.

$\therefore$ angle between the line & unknown lines is $\frac{\pi }{3}$

Direction ratio of line is $(2,1,1)$

$|a|=\sqrt{2^2+1^2+1^2}=\sqrt{6}$

$|b|=\sqrt{(2r+3{)}^2+(r+3{)}^2+r^2}=\sqrt{6r^2+18+18}$

$\cos \frac{\pi }{3}=\frac{a\cdot b}{|a||b|}$

$=\frac{1}{2}=\frac{4r+6+r+3+r}{\sqrt{6}\sqrt{6r^2+18r+18}}$

$=\frac{1}{2}=\frac{6r+9}{6\sqrt{r^2+3r+3}}$

$\therefore \sqrt{r^2+3r+3}=3r+3$

$=r^2+3r+3=4r^2+9+12r$

$=0=3r^2+6+9r$

$=0=r^2+3r+2$

$\therefore (r+1)(r+2)=0$

So, direction ratios are $(-1,1,-2)$ & $(1,2,-1)$

lines are $\frac{x-0}{-1}=\frac{y-0}{1}=\frac{z-a}{-2}$ & $\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{-1}$.