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Question

The reflection of the point P(1,0, 0) in the line x−12=y+1−3=x+108 is

A
(3, -4, -2)
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B
(5, - 8, - 4)
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C
(1, - 1, - 10)
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D
(2, - 3, - 8)
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Solution

The correct option is B (5, - 8, - 4)
Coordinates of any point Q on the given line are
(2r +1, - 3r - 1, 8r - 10).
So the direction cosines of PQ are 2r, - 3r - 1, 8r - 10.
Now PQ is perpendicular to the given line.
If 2(2r) -3 (-3r-1) + 8 (8r- 10) = 0
77r - 77 = 0 r = 1
And the coordinates of Q, the foot of the perpendicular from P on the line are (3, - 4, - 2).
Let R (a, b, c) be the reflection of P in the given lines than Q is the mid- point of PR
a+12=3,b2=4,c2=2
a=5,b=8,c=4
And the coordinate of the required point are (5, - 8, - 4).

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