Writing Equations in [latex]\mathbb{R}^3[/latex]

Now that you can represent points in space and calculate distances between them, you can learn to write equations for geometric objects like lines, planes, and curved surfaces in three-dimensional space.

The same equation can represent completely different geometric objects depending on the dimensional space you’re working in. Consider the simple equation [latex]x = 0[/latex]:

• In [latex]\mathbb{R}[/latex] (1D): Describes a single point
• In [latex]\mathbb{R}^2[/latex] (2D): Describes a line (the [latex]y[/latex]-axis)
• In [latex]\mathbb{R}^3[/latex] (3D): Describes a plane (the [latex]yz[/latex]-plane)

In 3D space, each coordinate plane has a simple equation:

• [latex]yz[/latex]-plane: [latex]x = 0[/latex] (contains all points [latex](0, y, z)[/latex])
• [latex]xy[/latex]-plane: [latex]z = 0[/latex] (contains all points [latex](x, y, 0)[/latex])
• [latex]xz[/latex]-plane: [latex]y = 0[/latex] (contains all points [latex](x, 0, z)[/latex])

Once you understand coordinate plane equations, you can write equations for any plane parallel to a coordinate plane. The key insight: when a plane is parallel to a coordinate plane, one coordinate remains constant for all points in that plane.

For example, if a plane is parallel to the [latex]xy[/latex]-plane, every point in that plane has the same [latex]z[/latex]-coordinate. Only the [latex]x[/latex]– and [latex]y[/latex]-coordinates vary from point to point.

Spheres

Just as a circle consists of all points in a plane equidistant from a center point, a sphere consists of all points in 3D space equidistant from a center point. The distance from the center to any point on the sphere is the radius.

The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance.

Find the standard equation of the sphere with center [latex](10, 7, 4)[/latex] and point [latex](-1, 3, -2)[/latex] as shown in Figure 4.

Show Solution

Use the distance formula to find the radius [latex]r[/latex] of the sphere:

[latex]\begin{align*} r &= \sqrt{(-1 - 10)^2 + (3-7)^2 + (-2 - 4)^2}\\ &= \sqrt{(-11)^2 + (-4)^2 + (-6)^2}\\ &= \sqrt{173} \end{align*}[/latex].

The standard equation of the sphere is

[latex](x-10)^2 + (y-7)^2+ (z-4)^2 = 173[/latex].

Let [latex]P=(-5, 2, 3)[/latex] and [latex]Q=(3, 4, -1)[/latex], and suppose line segment [latex]PQ[/latex] forms the diameter of a sphere (Figure 5). Find the equation of the sphere.

Show Solution

Since [latex]PQ[/latex] is a diameter of the sphere, we know the center of the sphere is the midpoint of [latex]PQ[/latex]. Then,

[latex]\begin{align*} C &= (\frac{-5+3}{2}, \frac{2+4}{2}, \frac{3+(-1)}{2})\\ &= (-1, 3, 1)\\ \end{align*}[/latex].

Furthermore, we know the radius of the sphere is half the length of the diameter. This gives

[latex]\begin{align*} r &= \frac{1}{2}\sqrt{(-5 - 3)^2 + (2-4)^2 + (3 - (-1))^2}\\ &= \frac{1}{2}\sqrt{64+4+16}\\ &= \sqrt{21} \end{align*}[/latex].

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Then, the equation of the sphere is [latex](x+1)^2 + (y-3)^2 + (z-1)^2 = 21[/latex].

Describe the set of points that satisfies [latex](x-4)(z-2)=0[/latex], and graph the set.

Show Solution

We must have either [latex]x-4=0[/latex] or [latex]z-2=0[/latex], so the set of points forms the two planes [latex]x=4[/latex] and [latex]z=2[/latex] (Figure 6).

Describe the set of points in three-dimensional space that satisfies [latex](x+2)^{2}+(y-1)^{2}=4[/latex], and graph the set.

Show Solution

The [latex]x[/latex]– and [latex]y[/latex]-coordinates form a circle in the [latex]xy[/latex]-plane of radius 2, centered at [latex](2, 1)[/latex]. Since there is no restriction on the [latex]z[/latex]-coordinate, the three-dimensional result is a circular cylinder of radius 2 centered on the line with [latex]x=2[/latex] and [latex]x=1[/latex]. The cylinder extends indefinitely in the [latex]z[/latex]-direction (Figure 8).