Two lines frac{x-3}{1}=frac{y+1}{3}=frac{z-6} | vedclass

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(A) Let $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$. Then $x = \lambda+3, y = 3\lambda-1, z = 6-\lambda$. Let $\frac{x+5}{7}=\frac{y-2}{-6}=\

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$(2,-4,-7)$

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(A) Let $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$. Then $x = \lambda+3, y = 3\lambda-1, z = 6-\lambda$. Let $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$. Then $x = 7\mu-5, y = -6\mu+2, z = 4\mu+3$. Since the lines intersect,we equate the coordinates: $\lambda+3 = 7\mu-5 \Rightarrow \lambda - 7\mu = -8$ $(i)$ $3\lambda-1 = -6\mu+2 \Rightarrow 3\lambda + 6\mu = 3 \Rightarrow \lambda + 2\mu = 1$ (ii) Subtracting $(i)$ from (ii): $9\mu = 9 \Rightarrow \mu = 1$. Substituting $\mu=1$ in (ii): $\lambda + 2(1) = 1 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the first line equations: $x = -1+3 = 2, y = 3(-1)-1 = -4, z = 6-(-1) = 7$. Thus,the point of intersection is $R(2, -4, 7)$. The reflection of a point $(x, y, z)$ in the $xy$-plane is $(x, y, -z)$. Therefore,the reflection of $R(2, -4, 7)$ in the $xy$-plane is $(2, -4, -7)$.

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

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