Two lines `(x-3)/1 = (y+1)/3 = (z-6)/(-1)` and `(x+5)/7 = (y-2)/(-6) = (z-3)/4` intersect in 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)`A. (2, -4, -7)B. (2, -4, 7)C. (-2,4,7)D. (2, 4, 7)

Two lines `(x-3)/1 = (y+1)/3 = (z-6)/(-1)` and `(x+5)/7 = (y-2)/(-6) =

1.	Two lines `(x-3)/1 = (y+1)/3 = (z-6)/(-1)` and `(x+5)/7 = (y-2)/(-6) = (z-3)/4` intersect in 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)`A. (2, -4, -7)B. (2, -4, 7)C. (-2,4,7)D. (2, 4, 7)
Answer» Correct Answer - A Let `(x-3)/(1)=(y+1)/(3)=(z-6)/(-1)=a` (say). Then any point on this line is of the from `P(a+3, 3a-1,-a+6)` Similarly, any point on the line. `(x+5)/(7)=(y-2)/(-6)=(z-3)/(4)=b`(say), is of the from `Q(7b-5,-6b+2,4b+3)` Now, if the lines are intersect, then P=Q for some a and b. `implies" "a+3=7b-5` `3a-1=-6b+2` and`" "-a+6=4b+3` `impliesa-7b=-8,a+2b=1" and "a+2b=3` On solving `a-7b=-8" and "a+2b=1`, we get b=1 and a=1, which also satisfy `a+4b=3` `:.P=Q-=(2, -4,7)" for "a=-1"and"b=1` Thus, coordinates of point R are (2, -4, 7) and reflection of R in xy-plane is (2, -4, -7)

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