A circle on the plane

Question

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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?

"i am interested what if f the circle does not lie in any plane parallel to the coordinate planes" - then you use an appropriate rotation matrix on the parametric equations for a circle. — J. M. ain't a mathematician, Jul 27 '12 at 13:36
generally no,i now some kind of rotation matrices,but how they are built up,i dont know — dato datuashvili, Jul 27 '12 at 13:44
I did no such thing; why did you automatically assume that I was your downvoter? In any event, I am advicing you to read up a bit on rotation matrices first, and then we can talk. — J. M. ain't a mathematician, Jul 27 '12 at 13:49
so it means ,that someone can't understand question and just does it without any thinking,no you can tell me ,dont worry if i would not understand it.please tell me how can i solve this task?use any concrete example if you would like — dato datuashvili, Jul 27 '12 at 13:53
@dato: You may have been downvoted because of the unclear writing. It would help to use (more) punctuation so that your question is not a single sentence. — Théophile, Jul 27 '12 at 14:01
@Théophile maybe ,because my native language is georgian and not english,you are right — dato datuashvili, Jul 27 '12 at 14:06

score 10 · Accepted Answer · answered Jul 27 '12 at 13:59

10

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.

answered Jul 27 '12 at 13:59

Clive Newstead

64,925

$u$ and $v$ we can calculate using this parametrizations,which i have written above or? – dato datuashvili Jul 27 '12 at 14:04
@dato: I don't really understand what you're asking. You can calculate $\mathbf{u}$ and $\mathbf{v}$ by looking at the specification of the plane you want your circle to be in. There are a number of ways of specifying a plane, which I'm sure you've come across already. – Clive Newstead Jul 27 '12 at 14:06
aa ok ok thanks very much,just one question $a$ in this case is a scalar right?i dont want to mix,or it is vector? – dato datuashvili Jul 27 '12 at 14:08
1

It's a vector. It couldn't possibly be a scalar, because you can't add a scalar to a vector. Here, $\mathbf{a}$ is the centre of the circle, so you could write, say, $\mathbf{a} = (a,b,c)$ if you wanted to. (At least, you could when working in 3 dimensions.) – Clive Newstead Jul 27 '12 at 14:13
thanks very much for answer – dato datuashvili Jul 27 '12 at 14:14

Christian Blatter · Answer 2 · 2012-07-27T14:27:32.763

(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 $\gamma$: $$\gamma:\quad t\mapsto{\bf p}+R\cos(t){\bf e}_1+R\sin(t){\bf e}_2\qquad(0\leq t\leq 2\pi)\ .$$

sorry normalize vector $e_1=(-n_2,n_1,0)$ what does mean?which vector we are normalizing — dato datuashvili, Jul 27 '12 at 14:23

A circle on the plane

2 Answers2