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Q.
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 $\frac{\pi}{3}$.
Three Dimensional Geometry
Solution:
Given line is $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z-0}{1}=t\,\text{(say)}\,\ldots\left(i\right)$
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
$cos \frac{\pi}{3}=\frac{\left|2\cdot\left(3+2t\right)+1\cdot\left(3+t\right)+1\cdot t\right|}{\sqrt{2^{2}+1^{2}+1^{2}}\sqrt{\left(3+2t\right)^{2}+\left(3+t\right)^{2}+t^{2}}}$
$\Rightarrow \frac{1}{2}=\frac{\left|6t+9\right|}{\sqrt{6}\sqrt{6t^{2}}+18t+18}$
$\Rightarrow 6\left(6t^{2} + 18t+18\right) = 4 \left(36t^{2}+ 108t+ 81\right)$
$\Rightarrow t^{2}+3t+2=0$
$\Rightarrow t=-1$ 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 $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$.