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Question
If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.
Solution
The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.
Let the coordinates of point P be (x, y, z).
Using the distance formula, we obtain
PA2 = (x - 3)2 + (y - 4)2 + (z - 5)2
= x2 + 9 - 6x + y2 + 16 - 8y + z2 + 25 - 10z
= x2 - 6x + y2 - 8y + z2 -10z + 50
PB2 = (x + 1)2 + (y - 3)2 + (z + 7)2
= x2 + 2x + y2 - 6y + z2 + 14z + 59
Now, if PA2 + PB2 = k2, then
(x2 - 6x + y2 - 8y + z2 - 10z + 50) + (x2 + 2x + y2 - 6y + z2 + 14z + 59) = k2
= 2x2 + 2y2 + 2z2 - 4x - 14y + 4z + 109 = k2
= 2(x2 + y2 + z2 - 2x - 7y + 2z) = k2 - 109
= x2 + y2 + z2 - 2x - 7y + 2z = `(k^2 - 109)/2`
Thus, the required equation is x2 + y2 + z2 - 2x - 7y + 2z = `(k^2 - 109)/2`.
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