Parametrization for intersection of sphere and plane

Question

Given is the sphere $x^2 + y^2 + z^2 = 4$ and the plane $x + y = 2$ in $\mathbb R^3 $.

How can I find a parametrization for the intersection of the two?

Answer 1

You are in luck because one of the equations is linear.

Since $x+y=2$ it follows that $x=2-y$. We can substitute this into the first equation:

$$x^2+y^2+z^2=4 \implies (2-y)^2+y^2+z^2=4 \implies 2y^2-4y+z^2=0\,. $$

We can "complete the square" on the first part to give:

$$2[(y-1)^2-1]+z^2=0 \implies 2(y-1)^2+z^2=2 \, . $$

Considering only the $yz$-plane then we have an ellipse in the $yz$-plane parametrised by $(y(\theta),z(\theta)) = (1+\cos\theta,\sqrt{2}\sin\theta)$. We also know that $x=2-y$, hence:

$$\gamma(\theta) = (1-\cos\theta,1+\cos\theta,\sqrt{2}\sin\theta)$$

would give a regular parametrisation of the set in question.

Answer 2

The intersection is an ellipse:

$$x^2 + (2-x)^2 + z^2 = 4 \implies 2(x- 1)^2 +z^2 = 2$$

So parametrize as follows:

$$x=1 + \cos{t}$$ $$z= \sqrt{2} \sin{t}$$

Answer 3

Try these equations.

$$\cases{ x=r \sin(s) \cos(t) \cr y=r \cos(s) \cos(t)\cr z=r \sin(t)\cr} $$

This is not a homeomorphism. The value $r$ is the radius of the sphere. It will parametrize the sphere for the right values of $s$ and $t$. This could be useful in parametrizing the ellipse. You can reparametrize if necessary to avoid any nastiness.

Answer 4

Let $\mathbf c = (1,1,0)$, and define two orthogonal unit vectors by $\mathbf u = (1,-1,0)/\sqrt 2$, and $\mathbf v = (0,0,1).$ Then Fly By Night's equations can be written as:

$$\gamma(\theta) = \mathbf c + (\sqrt 2 \cos\theta)\mathbf u + (\sqrt 2 \sin\theta)\mathbf v $$

This makes it clear that the curve is a circle with center at $\mathbf c$ and radius $\sqrt 2$.

You only asked about the parametrisation of the curve, not its shape, so this is slightly off topic, I suppose, but it might still be of interest.