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Question

The reflection of the point P(1,0,0) in the line x12=y+13=z+108 is
  1. (3,4,2)
  2. (5,8,4)
  3. (1,1,10)
  4. (2,3,8)

A
(5,8,4)
B
(1,1,10)
C
(3,4,2)
D
(2,3,8)
Solution
Verified by Toppr

Coordinates of any point Q on the given line are (2r+1,3r1,8r10)
So, the direction cosines of PQ are 2r,3r1,8r10
Now, PQ is perpendicular to the given line, if
2(2r)3(3r+1)+8(8r10)=077r77=0r=1
So, 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 line
Then, Q is the mid-point of PR
a+12=3,b2=4,c2=2a=5,b=8,c=4
So, the coordinates of the required point are (5,8,4)

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