Question

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 $xy$- plane has coordinates :

• A. $(2,-4,-7)$
• B. $(2,4,7)$
• C. $(2,-4,-7)$
• D. $(-2,4,7)$

Answer 1

The correct option is A $(2,-4,-7)$

Consider the line $L_1$
$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda \Rightarrow x=\lambda +3,y=3\lambda -1,z=-\lambda +6$

Now, consider the line $L_2$
$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$
$\Rightarrow x=7\mu -5,y=-6\mu +2,z=4\mu +3$

Both the lines $L_1$ and $L_2$ intersects each other.
So, $\lambda +3=7\mu -5$
$\Rightarrow 7\mu -\lambda =8\cdots \left(1\right)$
and $3\lambda -1=-6\mu +2$
$\Rightarrow 6\mu +3\lambda =3\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$
$\lambda =-1,\mu =1$
$\therefore x=2,y=-4,z=7$

$\therefore$ Coordinates of the intersection of $L_1$ and $L_2$ is $R(2,-4,7)$
Hence, reflection of $R$ in the $xy$-plane is $(2,-4,-7)$