Intersection between the plane $y-x=0$ and $x^2+y^2+z^2=4$

Question

When I plug in $y=x$ to the other equation I get an elipse equation. But when plotting it 3d I see the intersection curve is a circle. What am I supposed to do?

Answer 1

You are right, the equation $2x^2+z^2=4$ is the equation for an ellipse in two dimensions, and you are also right, that the intersection of a plane $y-x=0$ and a sphere $x^2+y^2+z^2=4$ is (always) a circle.

However this is not a contradiction, because $2x^2+z^2=4$ is not the equation of the intersection of the plane an the sphere in three dimensions.

The equation

$$2x^2+z^2=4, y\in\mathbb{R}$$

describes an infinitely long elliptical tube in $\mathbb{R}^3$.

Any intersection of this parallel the $y$ direction creates an ellipse, however, that's not what you are intersecting this elliptical tube with.

If intersecting the tube

$$2x^2+z^2=4, y\in\mathbb{R}$$

with $y-x=0$, you indeed get a circle, because this plane makes a $45°$ angle with the $y$ axis:

Moreover, circles in multiple dimensions cannot be described with a single equation. The system of equations

$$\begin{cases}y-x=0 \\ x^2+y^2+z^2=4\end{cases}$$

is a circle, and so is

$$\begin{cases}y-x=0 \\ 2x^2+z^2=4\end{cases}$$

and so is

$$\begin{cases}y-x=0 \\ 2y^2+z^2=4\end{cases}$$

You may recall from geometry class, that lines in $3$ dimensions were described by a system of equation, not a single equation, something like

$$\begin{cases}x=x_0+ta\\y=y_0+tb\\z=z_0+tc\end{cases}$$

so you can't expect a circle to behave differently.

Answer 2

What you need to do is a orthonormal change of basis: You set $e_1 = \left( (1,0,0) + (0,1,0) \right)/\sqrt{2}$, $e_2 = \left( (-1,0,0) + (0,1,0) \right)/\sqrt{2}$ and $e_3 = (0,0,1)$ Your matrix is indeed in the orthogonal group. $$U = \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix}R_{\pi/4} &0\\0 & 1\end{pmatrix}$$

Let us work with vectors directly: $$X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \qquad X' = UX=\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}$$ We have a good change of "repère" (sorry I only know the French word).

You first equation becomes $y' = 0$ and the second was $X^T X = 4$ so $$(U^TX')^T U^T X' = 4 = X'^T X' = 4$$ which leaves us with $x'^2 + y'^2 +z^2 = 4$. Combining the two is: $$\left\{\begin{array}{l} y' = 0\\ x'^2 + z^2 = 2^2 \end{array}\right.$$ So with an orthogonal projection orthogonal to $(-1,1,0)$ you get a circle

If you don't want to bother with matrices, just use $x = \frac{\sqrt{2}}{2}(x'-y')$ and $y = \frac{\sqrt{2}}{2}(x'+y')$

Answer 3

You know that an equation like $2x^2+z^2=4$, with unequal coefficients on the two squares, represents can ellipse. But you may have forgotten the context in which you learned this: The two axes with respect to which $x$ and $z$ are measured need to be perpendicular and the units of measurement along the two axes need to be the same. By substituting $y=x$, you've created an $x$-$z$ coordinate system (in the $x=y$ plane) for your equation. This coordinate system has perpendicular axes all right, but the units of measurement are not the same. If you increase $x$ by $1$, you have, because of setting $y=x$, also increased $y$ by $1$, so the distance you've moved is not $1$ but $\sqrt2$. On the other hand, increasing $z$ by $1$ just moves you a distance $1$.

So your equation $2x^2+z^2=4$ represents an "ellipse" in a distorted (by a factor $\sqrt2$) coordinate system. The distortion in this case exactly compensates for the eccentricity of the ellipse, so your curve would be a circle in an undistorted view (as, of course, a nontrivial intersection of a plane and a sphere has to be).

I hope this makes the problem clear. What to do about it is explained in detail in the answer from Jules Besson