Given P(3,0,-5) & Q(-9,6,2), find a rectangular equation of the perpendicular bisector of segement PQ.?

Question

So far I got the parametric equation of segment PQ is x=3-(12/sqrt229)d, y=(6/sqrt229)d, z=-5+(7/sqrt229)d so the perpendicular bisector of this segment is 12x-6y-7z=d so d=sqrt229 so 12x-6y-7z=229 but I don't even know if i did this right or how to find the answer to this question.

please help

Answer 1

Given P(3,0,-5) and Q(-9,6,2), find a rectangular equation of the perpendicular bisector of segement PQ.

In three dimensions, the perpendicular bisector of a line segment is a plane.

Define midpoint M and vector n.

M = (P + Q)/2 = [(3-9)/2, (0+6)/2, (-5+2)/2] = (-3, 3, -3/2)

n = PQ = Q - P = <-9-3, 6-0, 2-(-5)> = <-12, 6, 7>

M is a point on the plane and n is the normal vector to the plane. From these we can write the equation of the plane.

-12(x -(-3)) + 6(y - 3) + 7(z - (-3/2)) = 0
-12(x + 3) + 6(y - 3) + 7(z + 3/2) = 0
-12x - 36 + 6y - 18 + 7z + 21/2 = 0
-12x + 6y + 7z + 87/2 = 0

To get all integer coeficients multiply by -2.

24x - 12y - 14z + 87 = 0