What is the (parametric) intersection of a plane and a sphere?

Question

Can someone please show me how to prove that the intersection of the plane

$$x+y+z=0$$ and the sphere

$$x^2+y^2+z^2=1$$

can be expressed as

$$x(t)=\frac{\cos t-\sqrt3 \cdot\sin t}{\sqrt6}$$

$$y(t)=\frac{\cos t+\sqrt3 \cdot\sin t}{\sqrt6}$$

$$z(t)=\frac{-2\cos t}{\sqrt6}$$

Ps: Also, why am I not getting the correct notation? I am using a macbook pro (safari) if that is a concern?

Answer 1

You can solve for $z=-x-y$ and plug in $x^2+y^2+(-x-y)^2=1$, which rearranges to $2x^2+2xy+2y^2=1$. Hence for any fixed $z$ you have a rotated ellipse.

Let $u=(x+y/2)$. Then $u^2=x^2+xy+y^2/4$, so the equation becomes $2u^2+\frac{3}{2}y^2=1$. In standard form this is: $$\frac{u^2}{(\frac{1}{\sqrt{2}})^2}+\frac{y^2}{(\frac{\sqrt{2}}{\sqrt{3}})^2}=1$$

We can solve this parametrically as $u=\frac{1}{\sqrt{2}}\sin t$, $$y=\sqrt{\frac{2}{3}}\cos t$$ We can now find $$x=u-y/2=\frac{1}{\sqrt{2}}\sin t-\frac{1}{\sqrt{6}}\cos t$$ and $$z=-x-y=\frac{-1}{\sqrt{2}}\sin t +(-\frac{1}{\sqrt{6}}-\sqrt{\frac{2}{3}})\cos t$$

Alas, you posted your revisions after I'd made my choices so my solution will not agree with yours. There are three choices at the first step (solve for $z,y,x$), and two choices at the second step (replace $x$ or $y$). One of those six choices might give the result you have. :-)

Followup: A cross term ($xy$) is a rotation and can always be eliminated by a change of variables, pointing in the directions of the major and minor axes. There is a systematic way to do this; see here or here, or you can do it by the seat of your pants, like I did. I didn't directly rotate, instead I sheared, which made the computations a bit simpler.

Edit: fix sign error, thanks @Mhenni.

Edit 2: add general method

Answer 2

(Originally I had $u,v=x+y,x-y$, but after going through half the problem, I considered that $2u,2v=x+y,x-y$ would be a little more convenient.)

amanda, after your substitution, we consider $x^2+xy+y^2={1\over2}$.

Here are is a nice choice for the next substitution,

\begin{align} 2u&=x+y,\\ 2v&=x-y. \end{align}

We next consider

\begin{align} 4u^2&=x^2+y^2+2xy,\\ 4v^2&=x^2+y^2-2xy. \end{align}

To get $x^2+y^2+xy$, we need $2u^2+2v^2+(u^2-v^2)$. Thus our problem is transformed to

$$6u^2+2v^2=1.$$

Then we have parametric solutions

$$u=\sqrt{1\over6}\cos t,\ v=\sqrt{1\over2}\sin t.$$

From there, we find for instance

$$x=u+v={\cos t+\sqrt3\sin t\over\sqrt6},$$

and the rest should be straightforward.

To get the sign to match what is shown in your equations, I can think of several ways to go.

• You could simply note the symmetry in $x$ and $y$ and switch them (which suggests that if we had made the substitution $2v=y-x$ instead of $x-y$, that would have worked also).
• From $6u^2+2v^2=1$, we could have chosen $v=-\sqrt{1\over2}\sin t$.
• If we make the substitution $t=-s$, then we have $x={\cos (-s)+\sqrt3\sin (-s)\over\sqrt6}={\cos s-\sqrt3\sin s\over\sqrt6}$. Now, call it $t$ again.