The square of the distance of the image of the point (6, 1, 5) in the line x - 13 = y2 = z - 24, from the origin is .
Answer
62
Solution
Let M be the foot of the perpendicular from P(6, 1, 5): M(3 + 1, 2 , 4 + 2).
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Since PM is perpendicular to the line:
3(3 - 5) + 2(2 - 1) + 4(4 - 3) = 0
29 - 29 = 0 = 1, so M(4, 2, 6)
M is the mid-point of P and its image I, so
I = (2 4 - 6, 2 2 - 1, 2 6 - 5) = (2, 3, 7)
OI2 = 22 + 32 + 72 = 62
Answer: 62