Teaching Vector Geometry Through a 3D STEM Project: Finding the Fourth Vertex of a Regular Tetrahedron

In modern STEM education, hands-on geometry projects bridge abstract mathematical concepts with real-world understanding. One compelling application is teaching vector geometry using 3D spatial reasoning—tasks like finding the missing vertex of a regular tetrahedron challenge students to apply coordinates, symmetry, and vector properties. A classic example involves plotting four points in 3D space to form a regular tetrahedron, where all edges are equal in length. This article explores a real classroom scenario where a STEM educator guides students through discovering the integer coordinates of the fourth vertex $D$ of a regular tetrahedron with given vertices $A(1, 0, 0)$, $B(0, 1, 0)$, and $C(0, 0, 1)$.

Understanding the Context

What Is a Regular Tetrahedron?

A regular tetrahedron is a polyhedron with four equilateral triangular faces, six equal edges, and four vertices. Requiring all edges to be equal makes this an ideal model for teaching spatial geometry and vector magnitude calculations.

Given points $A(1, 0, 0)$, $B(0, 1, 0)$, and $C(0, 0, 1)$, we aim to find integer coordinates for $D(x, y, z)$ such that
[
|AB| = |AC| = |AD| = |BC| = |BD| = |CD|.
]

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Key Insights

Step 1: Confirm Equal Edge Lengths Among Given Points

First, compute the distances between $A$, $B$, and $C$:

  • Distance $AB = \sqrt{(1-0)^2 + (0-1)^2 + (0-0)^2} = \sqrt{1 + 1} = \sqrt{2}$
    - Distance $AC = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{1 + 1} = \sqrt{2}$
    - Distance $BC = \sqrt{(0-0)^2 + (1-0)^2 + (0-1)^2} = \sqrt{1 + 1} = \sqrt{2}$

All edges between $A$, $B$, and $C$ are $\sqrt{2}$, confirming triangle $ABC$ is equilateral in the plane $x+y+z=1$. Now, we seek point $D(x, y, z)$ such that its distance to each of $A$, $B$, and $C$ is also $\sqrt{2}$, and all coordinates are integers.

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Then from (D): $ 6(0) + b = 1 $ → $ b = 1 $
From (A): $ 7(0) + 3(1) + c = 5 $ → $ 3 + c = 5 $ → $ c = 2 $
From (1): $ 0 + 1 + 2 + d = 3 $ → $ d = 0 $

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Final Thoughts

Step 2: Set Up Equations Using Distance Formula

We enforce $|AD| = \sqrt{2}$:

[
|AD|^2 = (x - 1)^2 + (y - 0)^2 + (z - 0)^2 = 2
]
[
\Rightarrow (x - 1)^2 + y^2 + z^2 = 2 \quad \ ext{(1)}
]

Similarly, $|BD|^2 = 2$:

[
(x - 0)^2 + (y - 1)^2 + (z - 0)^2 = 2
\Rightarrow x^2 + (y - 1)^2 + z^2 = 2 \quad \ ext{(2)}
]

And $|CD|^2 = 2$:

[
x^2 + y^2 + (z - 1)^2 = 2 \quad \ ext{(3)}
]

Step 3: Subtract Equations to Eliminate Quadratic Terms

Subtract (1) – (2):