Equation of Plane- Equidistant with 2 Points

emma3001
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Homework Statement



Find the equation of a plane, every point of which is equidistant from the points A(1, 1, 0) and B(5, 3, -2)


The Attempt at a Solution



I am quite stuck... I wasn't sure if I could find vectors AP and BP and then find their magnitudes using square root x^2 + y^2 + z^2
 
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It will help if you know the relationship between the components of a normal vector to a plane and the coefficients in the equation for the plane:

the plane ax + by + cz + d = 0

has the normal vector <a, b, c>.

Construct a vector between point A and B (which order doesn't matter). If you make this the normal vector to your plane, you will have an essential requirement to meet the condition for equidistance. For all the points in the plane to be equally distant from A and B, you now make sure your perpendicular plane contains the midpoint between A and B.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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