= Solution
Since $\boldsymbol n\cdot\boldsymbol n=1$, the <vector> $K\boldsymbol r$ is perpendicular to $\boldsymbol n$:
$$
\boldsymbol n\cdot K\boldsymbol r=\boldsymbol n\cdot\boldsymbol r-(\boldsymbol n\cdot\boldsymbol r)(\boldsymbol n\cdot\boldsymbol n)=0.
$$
It follows that $K^2\boldsymbol r=K\boldsymbol r$. If $\boldsymbol r=t\boldsymbol n$, then $K\boldsymbol r=0$; if $\boldsymbol n\cdot\boldsymbol r=0$, then $K\boldsymbol r=\boldsymbol r$. Thus $K=I-\boldsymbol n\boldsymbol n^T$ is the <orthogonal projection matrix> onto $\boldsymbol n^\perp$: it removes exactly the component normal to the plane.
For the equally inclined plane choose $\boldsymbol n=(1,1,1)/\sqrt3$. Reversing its sign gives the same plane; other sign choices give equivalent reflections of the coordinate axes. One suitable <orthonormal basis> of this plane is
$$
\boldsymbol e_1=\frac{(1,1,-2)}{\sqrt6},\qquad \boldsymbol e_2=\frac{(-1,1,0)}{\sqrt2}.
$$
Their <dot products> with $\boldsymbol n$ vanish, their lengths are one, and their mutual <dot product> is zero. The equal first two components of $\boldsymbol e_1$ give its equal inclinations to $\boldsymbol i,\boldsymbol j$. Because $\boldsymbol e_j$ is already in the projection plane, $\boldsymbol e_j\cdot K\boldsymbol r=\boldsymbol e_j\cdot\boldsymbol r$. The coordinates of the projected point are therefore
$$
\boxed{u=\frac{x+y-2z}{\sqrt6},\qquad v=\frac{-x+y}{\sqrt2}.}
$$
For 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 $R/\sqrt3$ and $R$. Translating the original center only translates its <orthogonal projection>, so the result applies to every circle in that plane. Consequently
$$
\boxed{\frac{\text{minor axis}}{\text{major axis}}=\frac1{\sqrt3}.}
$$
The 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.
Back to article page
