3D Geometry 3 Question 18
####18. 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
(2019 Main, 11 Jan II)
(a) $(2,-4,-7)$
(b) $(2,-4,7)$
(c) $(-2,4,7)$
(d) $(2,4,7)$
Show Answer
Answer:
(a)
Solution:
- Let $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=a$ (say). Then, any point on this line is of the form $P(a+3,3 a-1,-a+6)$
Similarly, any point on the line. $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=$ $b$ (say), is of the form $Q(7 b-5,-6 b+2,4 b+3)$
Now, if the lines are intersect, then $P=Q$ for some $a$ and $b$.
$ \begin{array}{rlrl} \Rightarrow a+3 & =7 b-5 \\ 3a - 1 = -6b + 2 \\ \text { and } -a+6 & =4 b+3 \end{array} $
$ \Rightarrow a-7 b=-8, a+2 b=1 \text { and } a+4 b=3 $
On solving $a-7 b=-8$ and $a+2 b=1$, we get $b=1$ and $a=-1$,
which also satisfy $a+4 b=3$
$ \therefore \quad P=Q \equiv(2,-4,7) \quad \text { for } a=-1 \text { and } b=1 $
Thus, coordinates of point $R$ are $(2,-4,7)$
and reflection of $R$ in $x y$-plane is $(2,-4,-7)$
