Question

Find the equation of line passing through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z-0}{1}$ at angle $\frac{\pi }{3}$.

• A. $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$
• B. $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$
• C. $\frac{x}{1}=\frac{y}{3}=\frac{z}{2}$
• D. Both $(a)$ and $(b)$

Answer 1

Answer: (D)

Given line is $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z-0}{1}=t\text{(say)}\dots \left(i\right)$
Any point on the line $\left(i\right)$ is $P\left(3+2t,3+t,t\right)$
Direction ratios of $OP$ are
$<3+2t-0$, $3+t-0$, $t-0>$
i.e. $<3+2t$, $3+t$, $t>$.
Since, $OP$ makes angle $\frac{\pi }{3}$ with $\left(i\right)$, we get
$cos\frac{\pi }{3}=\frac{|2\cdot \left(3+2t\right)+1\cdot \left(3+t\right)+1\cdot t|}{\sqrt{2^2+1^2+1^2}\sqrt{{\left(3+2t\right)}^2+{\left(3+t\right)}^2+t^2}}$
$\Rightarrow \frac{1}{2}=\frac{|6t+9|}{\sqrt{6}\sqrt{6t^2}+18t+18}$
$\Rightarrow 6\left(6t^2+18t+18\right)=4\left(36t^2+108t+81\right)$
$\Rightarrow t^2+3t+2=0$
$\Rightarrow t=-1$ or $-2$.
Substituting these values of $t$, the two possible positions of $P$ are $(1,2,-1)$ and $(-1,1,-2)$. Hence, the required lines are
$\frac{x-0}{1-0}=\frac{y-0}{2-0}=\frac{z-0}{-1-0}$ and $\frac{x-0}{-1-0}=\frac{y-0}{1-0}=\frac{z-0}{-2-0}$
i.e. $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$.