For a general plane, what is the parametric equation for a circle laying in the plane Given a general equation for a plane through the origin
$$\vec{n}\cdot\vec{r}=0$$
With no assumptions made on $\vec{n}$ except having unit modulus, real $3\times1$ vector. How can you describe a unit circle, centred at the origin, laying in this plane?
I can only seem to find parametric equations that rely on knowing two vectors in the plane, but with no knowledge of the vector $\vec{n}$ you can't generally create two such vectors, as some component(s) of $\vec{n}$ may be zero. All the information you need to define such a circle is contained within the normal to the plane, so I am confused as to why there is not a form defined only with reference to this vector.
EDIT#1: With reference to this matrix. Can we start with in the $xy$ plane
$$(x,y,z)=(\cos(\theta),\sin(\theta),0)$$
Then rotate this about the axis ($\vec{u}$ in the link) 
$$\vec{u}=(-n_2,n_1,0)$$
about an angle $\phi$ that satisfies
$$\tan(\phi)=\frac{n_3}{\sqrt{n_1^2+n_2^2}}.$$
I claim that $\vec{u}$ is the axis of rotation as this vector is perpendicular to the normal of the plane $\vec{n}$ and lies in the $xy$ plane. Also that $\phi$ is the angle which the $xy$ plane is rotate about $\vec{u}$ by.
Therefore by substituting into the matrix linked to at the beginning of this edit, transforming $(x,y,z)=(\cos(\theta),\sin(\theta),0)$ by said matrix will give parametric coordinates for the tilted circle in terms of $\vec{n}$?
EDIT #2: I find this for the rotation matrix from the $xy$ plane to the plane with normal $\vec{n}$, from the method described above.
$$Q=\small{\left(\begin{array}{ccc} {\mathrm{n_2}}^2 - {\mathrm{n_2}}^2\, \sqrt{1 - {\mathrm{n_3}}^2} + \sqrt{1 - {\mathrm{n_3}}^2} & \mathrm{n_1}\, \mathrm{n_2}\, \left(\sqrt{1 - {\mathrm{n_3}}^2} - 1\right) & \mathrm{n_1}\, \mathrm{n_3}\\ \mathrm{n_1}\, \mathrm{n_2}\, \left(\sqrt{1 - {\mathrm{n_3}}^2} - 1\right) & {\mathrm{n_2}}^2\, \sqrt{1 - {\mathrm{n_3}}^2} + {\mathrm{n_3}}^2\, \sqrt{1 - {\mathrm{n_3}}^2} - {\mathrm{n_2}}^2 - {\mathrm{n_3}}^2 + 1 & \mathrm{n_2}\, \mathrm{n_3}\\ - \mathrm{n_1}\, \mathrm{n_3} & - \mathrm{n_2}\, \mathrm{n_3} & \sqrt{1 - {\mathrm{n_3}}^2} \end{array}\right)}$$
This is found from this MATLAB code.
EDIT #3: Using $\vec{n}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ I find this parametrically plots

 A: Let $\|n\|=1$ and choose any $p$ with $\langle n,p\rangle=0$ and $\|p\|=1$.  Then $n\times p$ satisfies $\langle n, n\times p\rangle=0$, $\|n\times p\|=1$  and 
$$c(t)=\bigl(p\cdot\cos(t), (n\times p)\cdot\sin(t)\bigr)$$
is a parametrization of the unit circle in the plane.
This may easily be extended to arbitrary planes and circles with arbitrary radii.
Michael
A: How about this. WLOG, let $\Vert n \Vert = 1$. Chose $\mu \neq n \in S^2$ and compute
$$\nu_1 = n\times \mu; \nu_2 = \nu_1 \times n$$
Then $\nu_1, \nu_2 \in A := \{x\in S^2 | n\cdot x = 0\}$
So we have our two points and only need one case:
$$n = \mu_0 \Rightarrow \text{chose } \mu = \mu_1, \text{ else } \mu = \mu_0$$
with both $\mu_0\neq \mu_1 \in S^2$ chosen prior to all.
A: For any vector $u$ that is not parallel to $n$, you can construct an orthogonal basis for the plane using the cross product: just take $e_1 = n \times u$ and $e_2 = n \times e_1$. Your problem seems to be that you want to be able to generate $e_1$ using some function of $n$. Indeed such a function is hard to find - it turns out that for any continuous map of the sphere, some vector will be mapped to a multiple of itself. (This is the "fixed-point property of the real projective plane".) Thus for this approach to work you would have to use a discontinuous map - I'm not sure whether or not this is an issue for you. (For example you could send $n$ to $(1,0,0)$ if $n=(0,0,\pm1)$, or just rotate about the $z$ axis by some fixed amount for any other $n$.)
I think your best bet is to write $n$ in spherical polar coordinates as $n = (\theta, \phi)$ for $\theta$ the azimuthal angle, and then find a rotation matrix $M$ that sends $(0,0,1)$ to $n$ - I believe an example is $$M = \left(
\begin{array}{ccc}
 -\sin (\theta ) & \cos (\theta ) \cos (\phi ) & \cos (\theta ) \sin (\phi ) \\
 \cos (\theta ) & \cos (\phi ) \sin (\theta ) & \sin (\theta ) \sin (\phi ) \\
 0 & -\sin (\phi ) & \cos (\phi ) \\
\end{array}
\right).$$
$e_1 = M(1,0,0)$ and $e_2 = M(0,1,0)$ are then an orthonormal basis for the plane. You should be able to write $\theta,\phi$ in terms of the cartesian components of $n$ to extract an explicit formula for the $e_1$; and then of course $\sin(s) e_1 + \cos(s)e_2$ will parametrise your circle.
