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 C $(2,-4,-7)$

$L_1:\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$

$L_2:\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$

Let $L_1:\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$

$x=\lambda +3,y=3\lambda -1,z=-\lambda +6$--(1)

$(\lambda +3,3\lambda -1,-\lambda +6)$ is a point on $L_1$

If it lies on $L_2$ then

$\frac{(\lambda +3)+5}{7}=\frac{(3\lambda -1)-2}{-6}$

$=>\lambda =-1$

Put $\lambda =-1$ in (1)

$(-1+3,-3-1,1+6)$is the point of intersection.

Reflection of Point P is (2, -4, -7) in the xy-plane

Hence, Option C is correct.