3d graphing: equidistant

[jaredmt]

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

dot product
cross product

The Attempt at a Solution

the book found the line: 14x - 6y - 10z = 9 which is perpendicular and i guess cuts in half way between the two dots.
i know that u take 2 lines and multiply them using dot product and it needs to equal 0 for them to be perpendicular. but i only know how to do that with 2 lines with endpoints... idk how to do this since i have 2 end points and i need to figure out an equation.

if any1 can explain how the book got that equation or even explain how to get any correct answer, that would definitely help. thanks

 

[Defennder]

That looks more like a plane in 3D space rather than a line. It's easy to see why it's a plane and not a line, considering two points in space and a line joining them. A plane situated equidistant from the two points and normal to the connecting line would represent the locus of points which are equidistant from the 2 points. You can't represent a line in 3D space with a single equation unless it's a vector equation.

A plane is associated with a normal vector to it and an arbitrary point on the plane. Can you see how to get both?

 

[jaredmt]

ok I am not sure what calculations I am supposed to make to get this answer. I am a little clueless right now

does it have anything to do with vector project or scalar project? idk what formulas I am supposed to be using on this 1

 

[Dick]

If you want to take the direct route just write an equation that says that the distance squared from a point (x,y,z) to (-1,5,3) equals the distance to (6,2,-2) and simplify it.