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!

  • $\begingroup$ $\sin^2 \theta + \cos^2 \theta = 1$. Also, multiply by $3$. $\endgroup$ – James S. Cook Nov 26 '14 at 22:49
  • $\begingroup$ I tried this approach but didn't quite manage to bring all the terms together. Will the solution be of the form $f(x,y,z) = \text{constant}$? $\endgroup$ – user211337 Nov 26 '14 at 22:55
  • $\begingroup$ To show it is on the plane just add the formulas for $x,y,z$. If they sum to zero then that suffices to show the curve is on the plane. To show it is an ellipse, that depends on how you want to characterize an ellipse. $\endgroup$ – James S. Cook Nov 26 '14 at 22:57
  • $\begingroup$ Perhaps you can also show something like $x^2+y^2=1$? (not necessarily this, but something similar) then the curve is the intersection of a cylinder and plane which you might be able to see is an ellipse? $\endgroup$ – James S. Cook Nov 26 '14 at 22:59
  • $\begingroup$ Is there a form to have the major and minor axes stated explicitly, such that it is possible to simply read them off? $\endgroup$ – user211337 Nov 26 '14 at 22:59

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.

  • $\begingroup$ A truly beautiful application of mathematics! Ingenious method as well, may I add. Many thanks good sir! $\endgroup$ – user211337 Nov 27 '14 at 12:11
  • $\begingroup$ nice work. I had used the interpretation $r^2=x^2+y^2$, but, it seems $r$ was a constant. $\endgroup$ – James S. Cook Nov 28 '14 at 10:04
  • $\begingroup$ @James Yes, I almost made the $x=r cos\theta$ substitution, but realized that that $\theta$ was a parameter. I wanted to change it $t$, or something, because using $\theta$ in this context throws off my intuition. And the same the $r$. $\endgroup$ – LouisB Nov 29 '14 at 4:50

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