Sets of Points Describing Surfaces

## Sets of Points Describing Surfaces

We have recently looked at various surfaces in $\mathbb{R}^3$ from the Geometry in Three-Dimensional Space page. We are now going to look further into this idea.

Consider the plane $x + y + z = 1$. This plane is constructed by an infinite set of points $(x, y, z)$ which satisfy the equation of the plane. If we denote this plane by the symbol $\Pi$ then the points $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ are contained within the set that defines $\Pi$. Therefore $\Pi = \{ (x, y, z) \in \mathbb{R}^3 : x + y + z = 1 \}$. In that sense, we can think of a surface in $\mathbb{R}^3$ as a set of ordered $n$-tuples which satisfy a set of equations defining the surface.

## Example 1

Describe the set of points that form a sphere centered at $(1, 2, 3)$ and with radius $4$.

We note that the general equation of a sphere is $(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$ where the center of the sphere is $(h, k, l)$ and the radius of the sphere is $r$. Therefore the equation of the sphere centered at $(1, 2, 3)$ and with radius $4$ is given by:

(1)
$$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 16$$

Therefore this sphere is defined by the set of points $\{ (x, y, z) \in \mathbb{R}^3 : (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 16 \}$.

## Example 2

Describe the set of points $(x, y, z)$ that satisfy the system of equations $\left\{\begin{matrix} x^2 + y^2 + z^2 = 1 \\ x + y + z = 1 \end{matrix}\right.$.

The first equation in the system is a sphere with radius $1$, while the second equation represents a plane. Let $S = \{ (x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \: \mathrm{and} \: x + y + z = 1 \}$. There are three possibilities for what our surface $S$ could represent.

• 1) If $S = \emptyset$ then there are no solutions to the system $\left\{\begin{matrix} x^2 + y^2 + z^2 = 1 \\ x + y + z = 1 \end{matrix}\right.$, and so our surface $S$ may be nothing at all.
• 2) Suppose that the plane intersects the sphere at only one point. Then the plane itself is tangential to the sphere, and so the set $S$ contains only point, and our surface would simply be a single point.
• 3) Suppose that the plane cuts through the sphere. Then the intersection will be a circle and $S$ will contain infinitely many points.

We note that $(1, 0, 0), (0, 1, 0), (0, 0, 1) \in S$, and so the intersection between the sphere and the plane must produce a circle.