Plane in 3-space not parallel to xy- zx- or zy- plane

1. Oct 5, 2014

Duderonimous

1. The problem statement, all variables and given/known data
Find an equation of the set of all points equidistant from the points A(-1,5,3) and B(6,2,-2). Describe the set.

2. Relevant equations
$|A_2 B_1|/2$ = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}/2$
This would be the distance to the midpoint along the line connecting A and B
$(\frac{x_ 1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$
Coordinate of midpoint between A and B
3. The attempt at a solution

For $|A_2 B_1|/2$ I got $\frac{\sqrt{83}}{2}$
and the coordinates of the mid point are
$(\frac{5}{2},\frac{7}{2},\frac{1}{2})$
I can visualize that this set of points would be a plane bisecting a line connecting A and B at its mid point.

The books answer is $14x-6y-10z=9$
Not sure how they got this. Any help appreciated thanks.

2. Oct 5, 2014

Staff: Mentor

Do you know about the dot product for vectors? If so, the vector that joins A and B is perpendicular to every vector that lies in the plane. That means that the dot product of these vectors will be zero.

3. Oct 5, 2014

LCKurtz

Or another way, if you let $P=(x,y)$ you could try setting $d^2(A,P) = d^2(B,P)$. Using the squares eliminates the square roots in the distance formulas.