Question: An educator is using a STEM project to teach vector geometry. In a 3D coordinate system, a student plots three vertices of a regular tetrahedron: $A(1, 0, 0)$, $B(0, 1, 0)$, and $C(0, 0, 1)$. Find the integer coordinates of the fourth vertex $D$ such that all edges of the tetrahedron are of equal length. - Decision Point
Teaching Vector Geometry Through a 3D STEM Project: Finding the Fourth Vertex of a Regular Tetrahedron
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 reasoningâÃÂÃÂtasks 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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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) âÃÂà(2):
