Find the equation of the set of points P such | vedclass

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Question

(A) Let the coordinates of point $P$ be $(x, y, z)$.$PA^{2} = (x-3)^{2} + (y-4)^{2} + (z-5)^{2} = x^{2} - 6x + 9 + y^{2} - 8y + 16 + z^{2} - 10z + 25

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$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z = 2k^{2} - 109$

Vedclass · Answer

(A) Let the coordinates of point $P$ be $(x, y, z)$.$PA^{2} = (x-3)^{2} + (y-4)^{2} + (z-5)^{2} = x^{2} - 6x + 9 + y^{2} - 8y + 16 + z^{2} - 10z + 25 = x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50$.$PB^{2} = (x+1)^{2} + (y-3)^{2} + (z+7)^{2} = x^{2} + 2x + 1 + y^{2} - 6y + 9 + z^{2} + 14z + 49 = x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59$.Given $PA^{2} + PB^{2} = 2k^{2}$,we have:$(x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50) + (x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59) = 2k^{2}$.Combining like terms:$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z + 109 = 2k^{2}$.Thus,$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z = 2k^{2} - 109$.

Find the equation of the set of points $P$ such that $PA^{2} + PB^{2} = 2k^{2}$,where $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$,respectively.

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