Since $\mathbf{n}\cdot \mathbf{n}=1$, the vector $K\mathbf{r}$ is perpendicular to $\mathbf{n}$:It follows that $K^2\mathbf{r}=K\mathbf{r}$. If $\mathbf{r}=t\mathbf{n}$, then $K\mathbf{r}=0$; if $\mathbf{n}\cdot \mathbf{r}=0$, then $K\mathbf{r}=\mathbf{r}$. Thus $K=I-\mathbf{n}{\mathbf{n}}^T$ is the orthogonal projection matrix onto ${\mathbf{n}}^{\perp }$: it removes exactly the component normal to the plane.

For the equally inclined plane choose $\mathbf{n}=(1,1,1)/\sqrt{3}$. Reversing its sign gives the same plane; other sign choices give equivalent reflections of the coordinate axes. One suitable orthonormal basis of this plane isTheir dot products with $\mathbf{n}$ vanish, their lengths are one, and their mutual dot product is zero. The equal first two components of ${\mathbf{e}}_1$ give its equal inclinations to $\mathbf{i},\mathbf{j}$. Because ${\mathbf{e}}_j$ is already in the projection plane, ${\mathbf{e}}_j\cdot K\mathbf{r}={\mathbf{e}}_j\cdot \mathbf{r}$. The coordinates of the projected point are thereforeFor a circle of radius $R$ centered at the origin in $z=0$, its equation $x^2+y^2=R^2$ becomes $3u^2+v^2=R^2$. This is an ellipse with semiaxes $R/\sqrt{3}$ and $R$. Translating the original center only translates its orthogonal projection, so the result applies to every circle in that plane. ConsequentlyThe general reason is the orthogonal projection of a circle between planes: the direction common to the planes retains its length, while the perpendicular direction is shortened by the cosine of their angle.

The angle-subtraction identity gives $\sqrt{3}\cos \theta +\sin \theta =2\cos (\theta -\pi /6)$. Thus $\cos (\theta -\pi /6)=1/\sqrt{2}$, so $\theta -\pi /6=\pm \pi /4+2k\pi$. Keeping the two values in the specified interval gives