Distance in 3D
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.2: “Vectors in Three Dimensions” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-2-vectors-in-three-dimensions |
| Textbook used in class | Stewart, Calculus, Section 12.1: “Three-Dimensional Coordinate Systems” (Examples 3, 4) |
Opening Scenario
The Pythagorean theorem gives the distance between two points in the plane as $\sqrt{(\Delta x)^2 + (\Delta y)^2}$. In 3D you apply it twice: once to get the diagonal of the base rectangle, then again to lift that diagonal up to the point at height $\Delta z$. The result is the three-dimensional distance formula, and it leads directly to the equation of a sphere.
Quick Reference
Distance formula. \[ |P_1 P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \]
Equation of a sphere with center $(a, b, c)$ and radius $r$: \[ (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2. \]
A sphere is the set of all points at distance $r$ from the center.
Key Concepts
1. Deriving and Applying the Distance Formula
Example 1. Find the distance from $P_1 = (1, -2, 4)$ to $P_2 = (3, 0, -1)$. (Stewart 12.1, Example 3.)
\[ |P_1 P_2| = \sqrt{(3-1)^2 + (0-(-2))^2 + (-1-4)^2} = \sqrt{4 + 4 + 25} = \sqrt{33}. \]
Boxed answer: $\sqrt{33}$.
2. The Equation of a Sphere
A sphere with center $C = (a, b, c)$ and radius $r$ is the set of all points $P = (x, y, z)$ with $|CP| = r$: \[ (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2. \]
Example 2. Show that $x^2 + y^2 + z^2 + 4x - 6y + 2z + 6 = 0$ is the equation of a sphere, and find its center and radius. (Stewart 12.1, Example 4.)
Complete the square in each variable: \[ (x^2 + 4x + 4) + (y^2 - 6y + 9) + (z^2 + 2z + 1) = -6 + 4 + 9 + 1 = 8. \] \[ (x+2)^2 + (y-3)^2 + (z+1)^2 = 8. \]
Boxed answer: Center $(-2, 3, -1)$, radius $r = \sqrt{8} = 2\sqrt{2}$.
Recap. Completing the square in three variables simultaneously is the standard technique. Add the same constants to both sides. The right side must be positive for a real sphere.
3. Midpoint Formula
The midpoint of $P_1 = (x_1, y_1, z_1)$ and $P_2 = (x_2, y_2, z_2)$ is \[ M = \left(\frac{x_1+x_2}{2},\, \frac{y_1+y_2}{2},\, \frac{z_1+z_2}{2}\right). \] This is the 3D analogue of the 2D midpoint formula.
taking the center of the sphere directly from the expanded equation without completing the square. The equation $x^2 + y^2 + z^2 + 4x - 6y + 2z + 6 = 0$ does NOT have center $(4, -6, 2)$. Those are the coefficients of the linear terms, not the coordinates of the center. Complete the square to find the center $(-2, 3, -1)$.
Common Errors Summary
| Error | Correction |
|---|---|
| Not squaring differences before summing | Distance uses $(x_2 - x_1)^2$, not $|x_2 - x_1|$ summed; the sum is inside the square root |
| Forgetting to add the completing-square constants to both sides | For each variable $x^2 + bx$, add $(b/2)^2$ to both sides |
| Accepting a negative right side as a sphere | If $r^2 \leq 0$ after completing the square, the “sphere” is imaginary or a single point |
Common Misconceptions
the center of a sphere can be read directly from the linear coefficients of the expanded equation.
This is the concept-image-conflicts-definition error. In the expanded form $x^2 + y^2 + z^2 + 4x - 6y + 2z + 6 = 0$, the numbers $4$, $-6$, and $2$ are not the center coordinates. Completing the square in each variable reveals the center $(-2, 3, -1)$ and radius $2\sqrt{2}$. Reading the linear coefficients directly gives the wrong sign and misses the halving step.
Leveled Practice
Problem 1. Find the equation of the sphere with center $(1, -3, 2)$ and radius $4$.
Show answer
$(x-1)^2 + (y+3)^2 + (z-2)^2 = 16$.
Problem 2. Find the center and radius of $x^2 + y^2 + z^2 - 2x + 4y - 6z + 5 = 0$.
Show answer
$(x^2 - 2x + 1) + (y^2 + 4y + 4) + (z^2 - 6z + 9) = -5 + 1 + 4 + 9 = 9$.
$(x-1)^2 + (y+2)^2 + (z-3)^2 = 9$.
Center $(1, -2, 3)$, radius $3$.
Mastery Checklist
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