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Possible Duplicate:
Parametric Equation of a Circle in 3D Space?

I know that, for example, if a circle is on a plane with counter-clockwise orientation, and with center $(a,b)$ and radius $R$, it has parametrization

$r(t)=(a + R \cos{t};b + R \sin{t}) \quad 0 \leq t \leq 2\pi$

and with clockwise orientation

$r(t)=(a + R \sin{t},b + R \cos{t}).$

Also, I know of forms of circle parametrization if it lies on horizontal plane $z=c$ and center $O(a,b,c)$, or if it is located on the plane $x=c$, I know how to parametrize the circle in this case. I am interested in what happens if the circle does not lie in any plane parallel to the coordinate planes?

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    @Théophile maybe ,because my native language is georgian and not english,you are right2012-07-27

2 Answers 2

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Let $\mathbf{u},\mathbf{v}$ be any two orthonormal vectors in $\mathbb{R}^n$, let $\mathbf{a} \in \mathbb{R}^n$ and let $R > 0$ be a positive real number. Then the circle of radius $R$ with centre $\mathbf{a}$ lying in the plane through $\mathbf{a}$ which is parallel to $\mathbf{u}$ and $\mathbf{v}$ is given by

$\mathbf{r}(t) = \mathbf{a} + (R\cos t)\mathbf{u} + (R\sin t)\mathbf{v}$

where $\mathbf{r}(t)$ denotes the locus of the points on the circle.

So given a plane $\Pi \subseteq \mathbb{R}^3$, calculate $\mathbf{u}$ and $\mathbf{v}$ and substitute into the above.

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    thanks very much for answer2012-07-27
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(Clive beat me by a few minutes, but I leave my answer here anyway, as it addresses aspects not covered by Clive.)

Assume your circle $\gamma$ has center ${\bf p}\in{\mathbb R}^3$. The plane $\Pi$ in which the circle lies has to be given somehow. If the plane is given in the form ${\bf n}\cdot{\bf x}=\rho$, where $\rho={\bf n}\cdot {\bf p}$, then we first have to provide ourselves with two mutually orthogonal unit vectors ${\bf e}_i$ which "span $\Pi$" or, to be exact, are orthogonal to ${\bf n}$. The first of these vectors can be obtained by normalizing the vector ${\bf e}_1':=(-n_2,n_1,0)$ (which is orthogonal to ${\bf n}$). Normalizing means replacing the vector ${\bf e}_1'$ by the unit vector ${\bf e}_1:={{\bf e}_1'\over\bigl|{\bf e}_1'\bigr|}$ which points in the same direction. The second vector ${\bf e}_2$ is then obtained by normalizing the vector ${\bf e}_2':={\bf n}\times{\bf e}_1$.

After these preparatory steps it is easy to write down a parametric representation of $\gamma$: $\gamma:\quad t\mapsto{\bf p}+R\cos(t){\bf e}_1+R\sin(t){\bf e}_2\qquad(0\leq t\leq 2\pi)\ .$

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    sorry normalize vector $e_1=(-n_2,n_1,0)$ what does mean?which vector we are normalizing2012-07-27