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

[Duderonimous]

Homework Statement

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.

Homework 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

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.

 

[Mark44]

Homework Statement

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.

Homework 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

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.

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.

 

[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.