I've been working through a problem and I have managed to reduce it to the following:$$x=\frac{2r}{3}\cos\theta - \frac{r}{3}\sin\theta$$ $$y=\frac{2r}{3}\sin\theta - \frac{r}{3}\cos\theta$$ $$z = -\frac{r}{3}(\cos\theta + \sin\theta)$$
I need to show this is an ellipse on the plane $x+y+z=0$ and the problem is complete. I can see this is true using software, but unfortunately I am unable to prove it algebraically.
As always, your help is much appreciated!
Using vector transformations: imagine rotating the 3D ellipse onto the x-y plane. To find the transformation, imagine rotating the normal to the x+y+z=0 plane to the z-axis. First, rotate by $\pi/4$ about the z-axis, then rotate about the x-axis by $arctan(\sqrt{2})$. Both of these rotations are unitary transformations, so they do not distort the ellipse. The first rotation matrix is $U_1=\begin{pmatrix} c_1 & -s_1 & 0 \\ s_1 & c_1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$ where $c_1=s_1=1/\sqrt{2}$. The second is $U_2=\begin{pmatrix} 1 & 0 & 0 \\ 0 & c_2 & -s_2 \\ 0 & s_2 & c_2 \\ \end{pmatrix}$ where $c_2=1/\sqrt{3}$ and $s_2=\sqrt{2/3}$. Applying the rotations to the ellipse gives $$\begin{Bmatrix} x'\\y'\\z' \end{Bmatrix} = U_2 \cdot U_1 \begin{Bmatrix} x\\y\\ z \end{Bmatrix} = \begin{Bmatrix} (x-y)/\sqrt{2}\\(x+y-2z)/\sqrt{6}\\(x+y+z)/\sqrt{3} \end{Bmatrix} =\begin{Bmatrix} r(cos\theta -sin\theta)/\sqrt{2}\\r(cos\theta +sin\theta)/\sqrt{6}\\0 \end{Bmatrix}$$ Almost there! All that remains is to show that $x'^2+3y'^2=r^2$, which is straight forward.