Question

Find an equation of the set of all points equidistant from the points a(−3, 4, 3) and b(6, 1, −2). describe the set.

Answer 1

Let

P-------> the set of points equally distant from the two given points

we know that

Distance formula in three dimensions for the distance  between two points is equal to

$d=\sqrt{(x-x1)^{2}+(y-y1)^{2}+(z-z1)^{2} }$

P(x,y,z) A(-3,4,3) B(6,1,-2)

in this problem

$\left|AP\right| =\left|BP\right|$

1) Find the distance AP

$d=\sqrt{(x+3)^{2}+(y-4)^{2}+(z-3)^{2}}$

2) Find the distance BP

$d=\sqrt{(x-6)^{2}+(y-1)^{2}+(z+2)^{2}}$

3) Equate distance AP and distance BP

$\sqrt{(x+3)^{2}+(y-4)^{2}+(z-3)^{2}}=\sqrt{(x-6)^{2}+(y-1)^{2}+(z+2)^{2}} \\ \\ x^{2}+6x+9+y^{2} -8y+16+z^{2} -6z+9= x^{2}-12x+36+y^{2} -2y+1+z^{2} +4z+4 \\ \\6x-8y-6z+34= -12x-2y+4z+41 \\ \\18x-6y-10z=7$

This is a plane, equidistant from each of A and B, which is perpendicular

to the line which joins A and B

therefore

the answer is

The equation is equal to

$18x-6y-10z=7$