Question:
Two lines $\quad \frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1} \quad$ 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 :-
Correct Option: , 3
Solution:
Point on $\mathrm{L}_{1}(\lambda+3,3 \lambda-1,-\lambda+6)$
Point on $\mathrm{L}_{2}(7 \mu-5,-6 \mu+2,4 \mu+3)$
$\Rightarrow \lambda+3=7 \mu-5$ ......(i)
$3 \lambda-1=-6 \mu+2$ ......(ii) $\Rightarrow \lambda=-1, \mu=1$
point $R(2,-4,7)$
Reflection is $(2,-4,-7)$
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