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3-D Planes

  • Thread starter DiamondV
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Homework Statement


The method that we are taught on how to determine the equation of a plane is as follows when given 3 coplanar points:
1.
Determine the vectors
2.
Find the cross product of the two vectors.
3.
Substitute one point into the Cartesian equation to solve for d.


Homework Equations




The Attempt at a Solution



I know how to do this but my issue is with the intuition behind it, by getting the cross product of two vectors on the plane we are essentially getting the normal vector of the entire plane. we then take the coefficents of this vector and put it into sort of an equation like this x+y+z=d, then sub a point into this to find d and thats how you get the equation of the plane. I mean what exactly is happening here? How does the coefficents of the x, y and z components(or magnitude of x,y,z, basically whatever is front of the x, y,z) give us the equation of the plane?
 

Answers and Replies

  • #2
andrewkirk
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The cross product of two vectors in the plane will be a vector ##\vec v=(a,b,c)## that is normal to the plane. Let ##D=(d,e,f)## be any point in the plane. Then for any other point ##P=(x,y,z)## in the plane, the vector ##\vec u## from D to P lies in the plane and hence must be perpendicular to ##\vec v##.

So we have

$$0=\vec u\cdot \vec v =(x-d,y-e,z-f)\cdot(a,b,c)=a(x-d)+b(y-e)+c(z-f)=ax+by+cz-(ad+be+cf)$$

So the equation of the plane is

$$ax+by+cz=(ad+be+cf)$$
 

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