Equation of Plane- Equidistant with 2 Points


by emma3001
Tags: equation, equidistant, plane, points
emma3001
emma3001 is offline
#1
Mar18-08, 07:00 PM
P: 42
1. The problem statement, all variables and given/known data

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


3. 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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dynamicsolo
dynamicsolo is offline
#2
Mar18-08, 09:15 PM
HW Helper
P: 1,664
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.


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