Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$ intersect at the point $R$. The reflection of $R$ in the $x y$ - plane has coordinates:

Question

Aicharya · Accepted Answer

Let the coordinate of $P$ with respect to line

$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$ $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$ $L_{1}=(\lambda+3,3 \lambda-1,-\lambda+6)$ and coordinate of $P$ w.r.t. line $L_{2}=(7 \mu-5,-6 \mu+2,4 \mu+3)$ $\therefore \quad \lambda-7 \mu=-8,3 \lambda+6 \mu=3, \lambda+4 \mu=3$ From above equation : $\lambda=-1, \mu=1$ $\therefore \quad$ Coordinate of point of intersection $R=(2,-4,7)$ Image of $R$ w.r.t. $x y$ plane $=(2,-4,-7)$

Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$…

Solution