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
Best Answer
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