Thread Closed

Equation of Plane- Equidistant with 2 Points

 
Share Thread
Mar18-08, 07:00 PM   #1
 

Equation of Plane- Equidistant with 2 Points


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
PhysOrg.com science news on PhysOrg.com

>> Leading 3-D printer firms to merge in $403M deal (Update)
>> LA to give every student an iPad; $30M order
>> CIA faulted for choosing Amazon over IBM on cloud contract
Mar18-08, 09:15 PM   #2
 
Recognitions:
Homework Helper Homework Help
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.
Thread Closed

Similar discussions for: Equation of Plane- Equidistant with 2 Points
Thread Forum Replies
Equidistant from 2 points Calculus & Beyond Homework 2
Equation of a Plane with 2 Points and Perpendicular Calculus & Beyond Homework 3
Set of points equidistant from two points Calculus & Beyond Homework 1
Equation of the plane EQUIDISTANT from two points Calculus & Beyond Homework 4
equation of a plane containing the three points Introductory Physics Homework 0