How to parametrise a circle in a plane

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Solution 1

The unit sphere has equation $x^2+y^2+z^2=1$, its intersection with the plane is the circle you are looking for. So you have $$y=-x$$ and $$2x^2+z^2=1$$. You can now use polar coordinates so put $x=-y={1 \over \sqrt 2} \cos t$ and $z=\sin t$ so that $$2{1 \over 2} \cos^2 t+ \sin^2 t=1$$ is identically verified.

Solution 2

Here’s a general solution:

In $\mathbb R^2$, if $\vec u$ and $\vec v$ are any pair of orthogonal unit vectors, then $\vec c+r\vec u\cos t+r\vec v\sin t$ is a parameterization of a circle of radius $r$ centered at $\vec c$. The same parameterization will work in any Euclidean space, so your problem is reduced to finding a pair of orthogonal unit vectors parallel to the plane of the circle.

In $\mathbb R^3$, a plane can be given by the equation $ax+by+cz+d=0$, which can be written as $(a,b,c)\cdot(x,y,z)+d=0$. The vector $\vec n=(a,b,c)$ is normal to the plane, and it can be easily verified that both $(b,-a,0)$ and $(c,0,-a)$ are perpendicular to this vector and so parallel to the plane. Pick either one of these vectors and normalize it to get $\vec u$. For $\vec v$, compute the cross product of whichever vector you picked with $(a,b,c)$ and normalize that.

For your specific problem, $\vec n=(1,1,0)$, so we can take $\vec u=\frac1{\sqrt2}(1,-1,0)$. For $\vec v$, $(1,-1,0)\times(1,1,0)=(0,0,2)$, so $\vec v=(0,0,1)$. One possible parameterization of the circle is therefore $$(0,0,0)+\frac1{\sqrt2}\cos t\,(1,-1,0)+\sin t\,(0,0,1)=\left(\frac1{\sqrt2}\cos t,-\frac1{\sqrt2}\cos t,\sin t\right).$$

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Updated on April 13, 2020

Comments

  • Jonahhill
    Jonahhill over 3 years

    I've been studying for an exam and in one of the exams it asks me the parametrise the unit circle in $R^3$ which lines in the plane $$x+y=0$$.

    I know how to parametrise a normal circle, but I don't really have any idea how to parametrise it in a plane like that. Any help would be appreciated, thanks 😊

    • Michael Hoppe
      Michael Hoppe over 6 years
    • hardmath
      hardmath over 6 years
      It would be clearer if the "unit circle in $\mathbb{R}^3$ which [lies] in the plane $x+y=0$" were specified. There are many such circles of unit radius, so one supposes that the circle centered on the origin is intended.
  • Michael Hoppe
    Michael Hoppe over 6 years
    Your equation doesn't describe a circle in $3$-space.
  • amd
    amd over 6 years
    Why the downvote, o anonymous detractor? Is there something incorrect or unclear?
  • N74
    N74 over 6 years
    @MichaelHoppe It seems It does: for every $t$ you have a point that is both in the unit sphere and on a plane, this should be enough... but if you are right feel free to add your answer.
  • jorgeegroj
    jorgeegroj over 6 years
    This is absolutely correct, I messed up, I'm sorry. I'll edit my answer, thank you for pointing that out!