Equation of a Circle in 3-space

[Philrock]

Given the circle's center at x1, y1, z1, and the 3 direction cosines of the normal to plane of the circle, what is the equation of the circle?

 

[JStephen]

I assume you also have the radius of the circle?

Anyway, a circle in space would be the intersection of a plane and a (hollow) sphere. The sphere equation is easy. Work out the plane equation from the direction cosines (I don't even remember what a direction cosine is!) and the circle center point, substitute one equation in the other.

 

[prex]

As suggested by JStephen you'll have not a single equation, but a system of two equations:

(x-x₁)²+(y-y₁)²+(z-z₁)²=r²

and

(x-x₁)cos(xn)+(y-y₁)cos(yn)+(z-z₁)cos(zn)=0

prex

http://www.xcalcs.com

Online tools for structural design

 

[zekeman]

Take the crossproduct of the unit vector cosA,cosB,cosC,(A,B,C the direction angles) with the vector x-x1,y-y1,z-z1. ;then take the self dot product and set it equal to R^2.
For the crossproduct, I get [cosB(z-z1)-cosC(y-y1)],[cosC(x-x1)-cosA(z-z1)],[cosA(y-y1)-cosB(x-x1)]
This vector is in the direction parallel to the plane of the circle and has a magnitude of R. Taking the self dot product equal to R^2 should yield the equation, namely

[[cosB(z-z1)-cosC(y-y1)]^2+[cosC(x-x1)-cosA(z-z1)]^2+[cosA(y-y1)-cosB(x-x1)]^2=R^2

 

[prex]

zekeman
a single equation in 3D necessarily represents a surface. I didn't check your way of reasoning, but your equation should be a cylinder: you still need to intersect it with a plane to get a circle (hence again a system of two equations).

prex

http://www.xcalcs.com

Online tools for structural design

 

[Tunalover]

Philrock-
I find it hard to believe that someone would have a problem like this in the "real world."

This sounds like a homework problem! If it is, please refrain from using this forum to do your homework for you!


Tunalover

 

[zekeman]

Prex,
You are correct. Mine is the eq of the cylinder perpendicular to the circle in space and, indeed I would need the eq of the intersecting plane to complete the answer.
However, a discussion of the more general closed planer curve problem rotated in space, namely
F(x,y)=0
would be more interesting, since it would not involve a familiar intersection of a plane and a sphere or cylinder.
Try it in your spare time!

 

[Philrock]

Thanks very much for all the help!

Tunalover - this is indeed a real world problem - vehicle steering geometry - king pin inclination, caster, tracking while turning, etc.