The equation of a line in xz plane equally inclined with x and z axes which is at a unit distance fr

Question

The equation of a line in xz plane equally inclined with x and z axes which is at a unit distance from the line $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z}{1}$ is :

Professor PC Thomas Classes · Answer

Let the required line passes through the point $(h, 0,0)$. Then its equation is
$
\frac{x-h}{1}=\frac{y-0}{0}=\frac{z-0}{1} 	ext { or, } \frac{x-h}{1}=\frac{y-0}{0}=\frac{z-0}{-1}
$
Let MN be a line perpendicular to given line and required line and its direction cosines be l : $\mathrm{m}: \mathrm{n}$, then
$
\begin{aligned}
& l 	imes 1+m 	imes 1+n 	imes 1=0 \
& l 	imes 1+m 	imes 0+n 	imes 1=0
\end{aligned}
$
Solving we get $l=-n$ and $m=0 \Rightarrow l=\frac{1}{\sqrt{2}}, n=-\frac{1}{\sqrt{2}}, m=0$
Let point A be $(\mathrm{h}, 0,0)$ and B be $(1,1,0)$ then projection of AB on the line MN is
$l(h-1)+m(0-1)+n(0-0)=1$\ $\Rightarrow \frac{1}{\sqrt{2}}(h-1)+0-\left(\frac{1}{\sqrt{2}} 	imes 0ight)=1 \Rightarrow h-1=\sqrt{2} \Rightarrow h=1+\sqrt{2}$ \ $\Rightarrow \quad \frac{x-1-\sqrt{2}}{1}=\frac{y-0}{0}=\frac{z-0}{1}$ or, $\frac{x-h}{1}=\frac{y-0}{0}=\frac{z-0}{-1} \Rightarrow l+m+n=0$ and $-l+0+n=0$ \ Solving, we get $n=\frac{1}{\sqrt{6}}, 1=\frac{1}{\sqrt{6}}$ and $m=-\frac{2}{\sqrt{6}}$\ $\Rightarrow \frac{1}{\sqrt{6}}(h-1)-\frac{2}{\sqrt{6}}(0-1)+\frac{1}{\sqrt{6}}(0-0)=1 \Rightarrow h=\sqrt{6}-1$\ $	herefore \quad \frac{x-\sqrt{6}+1}{1}=\frac{y-0}{0}=\frac{z}{-1}$.\ Hence $\frac{x-\sqrt{6}+1}{1}=\frac{y-0}{0}=\frac{z}{-1}$ and $\frac{x-1-\sqrt{2}}{1}=\frac{y-0}{0}=\frac{z-0}{1}$ are the required lines.