Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1} \quad$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}…
Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1} \quad$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4} \quad$ intersect at the point $\mathrm{R}$. Then reflection of $\mathrm{R}$ in the $x y$-plane has co-ordinates
$(2,-4,-7)$
$(2,-4,7)$
$(-2,4,7)$
$(2,4,7)$
Solution
Let $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$
$\Rightarrow x=3+\lambda, y=3 \lambda-1, \mathrm{z}=-\lambda+6$
Let $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$
$\Rightarrow x=7 \mu-5, y=-6 \mu+2, \mathrm{z}=4 \mu+3$
Both the given lines intersect each other.
$\begin{aligned}
& \text { So, } \lambda+3=7 \mu-5 \\
& \Rightarrow 7 \mu-\lambda=8...(i)
\end{aligned}$
Also, $3 \lambda-1=-6 \mu+2$
$\Rightarrow 6 \mu+3 \lambda=3$...(ii)
From (i) and (ii), we get
$\begin{aligned}
& \mu=1, \lambda=-1 \\
& \text { i.e., } x=2, y=-4, \mathrm{z}=7
\end{aligned}$
$\therefore \quad$ Co-ordinates of the intersection of the given lines are $\mathrm{R}(2,-4,7)$
Hence, reflection of $\mathrm{R}$ in the $x y$-plane is $(2,-4,-7)$.