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

Vedclass · Accepted Answer

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

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.

Similar Questions

Let $A, B, C$ be the feet of the perpendiculars from a point $P$ on the $xy, yz,$ and $zx$-planes respectively. Find the coordinates of $A, B, C$ for the following points $P$:
$(3, 4, 5), (-5, 3, 7), (4, -3, -5)$

Let $A, B, C$ be the feet of perpendiculars from a point $P$ on the $x, y, z$-axes respectively. Find the coordinates of $A, B$ and $C$ for the following points $P$:
$(i)$ $(3, 4, 2)$
$(ii)$ $(-5, 3, 7)$
$(iii)$ $(4, -3, -5)$

Name the octant in which each of the following points lies:
$(i) (1, 2, 3)$
$(ii) (4, -2, 3)$
$(iii) (4, -2, -5)$
$(iv) (4, 2, -5)$
$(v) (-4, 2, 5)$
$(vi) (-3, -1, 6)$
$(vii) (2, -4, -7)$
$(viii) (-4, 2, -5)$

The point $(1, -3, 4)$ lies in the octant:

In the figure,if $P$ is $(2, 4, 5)$,find the coordinates of $F$.

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