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.

February 17, 2026

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 reasoningÃÂ¢ÃÂÃÂtasks 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 $C(0, 0, 1)$.

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|.]

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.

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) ÃÂ¢ÃÂÃÂ (2):

[[(x - 1)^2 + y^2 + z^2] - [x^2 + (y - 1)^2 + z^2] = 0]

Expand:

[(x^2 - 2x + 1) + y^2 + z^2 - [x^2 + y^2 - 2y + 1 + z^2] = 0]

Simplify:

[-2x + 1 + 2y - 1 = 0 \Rightarrow -2x + 2y = 0 \Rightarrow x = y]

Now subtract (2) ÃÂ¢ÃÂÃÂ (3):

[[x^2 + (y - 1)^2 + z^2] - [x^2 + y^2 + (z - 1)^2] = 0]

Expand:

[x^2 + y^2 - 2y + 1 + z^2 - x^2 - y^2 - (z^2 - 2z + 1) = 0]

Simplify:

[-2y + 1 + 2z - 1 = 0 \Rightarrow -2y + 2z = 0 \Rightarrow y = z]

Thus, $x = y = z$. Let $x = y = z = t$, where $t$ is an integer (by problem constraint).

Step 4: Substitute into One Equation

Use equation (1) with $x = y = z = t$:

[(t - 1)^2 + t^2 + t^2 = 2\Rightarrow (t^2 - 2t + 1) + t^2 + t^2 = 2\Rightarrow 3t^2 - 2t + 1 = 2\Rightarrow 3t^2 - 2t - 1 = 0]

Solve the quadratic:

[t = rac{2 \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)} = rac{2 \pm \sqrt{4 + 12}}{6} = rac{2 \pm \sqrt{16}}{6} = rac{2 \pm 4}{6}]

So, $t = 1$ or $t = -rac{1}{3}$

Only $t = 1$ is an integer. Try $t = 1$: then $D = (1, 1, 1)$

Check distances:

$|AD| = \sqrt{(1-1)^2 + (1-0)^2 + (1-0)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$- $|BD| = \sqrt{(1-0)^2 + (1-1)^2 + (1-0)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$- $|CD| = \sqrt{(1-0)^2 + (1-0)^2 + (1-1)^2} = \sqrt{1 + 1 + 0} = \sqrt{2}$

All edges match. However, verify if all edges are equal:

$|AB| = |AC| = \sqrt{2}$- $|AD| = |BD| = |CD| = \sqrt{2}$

All six edges are $\sqrt{2}$, so vertex $D(1,1,1)$ forms a regular tetrahedron with $A$, $B$, $C$.

But wait ÃÂ¢ÃÂÃÂ geometrically, point $(1,1,1)$ lies in the same octant as $A$, $B$, $C$, but is it valid?

LetÃÂ¢ÃÂÃÂs reconsider: The tetrahedron with vertices $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$, and $D(1,1,1)$ ÃÂ¢ÃÂÃÂ is this regular?

Check distance $CD$: already $\sqrt{2}$, good.Check $AD$, $BD$, $CD$ all $\sqrt{2}$ ÃÂ¢ÃÂÃÂ yes.But is triangle $ABC$ really in a plane? Yes, but does this form a closed, symmetric tetrahedron?

Actually, in the standard regular tetrahedron with vertices at $(1,1,1)$, $(-1,-1,1)$, $(-1,1,-1)$, $(1,-1,-1)$, normalization differs. But here, $D(1,1,1)$ yields valid equal lengths.

However, note: the centroid of $A, B, C$ is $(rac{1}{3}, rac{1}{3}, rac{1}{3})$, and $D(1,1,1)$ lies far from origin ÃÂ¢ÃÂÃÂ distance from origin to $D$ is $\sqrt{3}$, while $A$ is at distance 1. Not symmetric around origin, but thatÃÂ¢ÃÂÃÂs acceptable.

But is there another integer solution?

From quadratic, only integer solution is $t=1$. The other root $t = -1/3$ is not integer.

So the only integer-coordinate vertex that satisfies equal edge lengths is $D(1,1,1)$

But wait ÃÂ¢ÃÂÃÂ does this form a valid 3D tetrahedron?

Check volume: The volume $V$ of tetrahedron with vertices $A, B, C, D$ is:

[V = rac{1}{6} \left| \det egin{bmatrix}-1 & 1 & 0 & 1 -1 & 0 & 1 & 1 -1 & 0 & 0 & 1 0 & 1 & 1 & 1\end{bmatrix}]

But simpler: vectors:

$ec{AB} = (-1,1,0)$, $ec{AC} = (-1,0,1)$, $ec{AD} = (0,1,1)$

Triple scalar product:

[ec{AB} \cdot (ec{AC} \ imes ec{AD})]

First, compute $ec{AC} \ imes ec{AD}$:

[egin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} -1 & 0 & 1 0 & 1 & 1 \end{vmatrix}= \mathbf{i}(0 \cdot 1 - 1 \cdot 1) - \mathbf{j}(-1 \cdot 1 - 1 \cdot 0) + \mathbf{k}(-1 \cdot 1 - 0 \cdot 0) = \mathbf{i}(-1) - \mathbf{j}(-1) + \mathbf{k}(-1) = (-1, 1, -1)]

Now dot with $ec{AB} = (-1,1,0)$:

[(-1)(-1) + (1)(1) + (0)(-1) = 1 + 1 = 2 e 0]

Thus, vectors are not coplanar ÃÂ¢ÃÂÃÂ volume $ rac{1}{6} |2| = rac{1}{3} e 0$: non-degenerate.

So $D(1,1,1)$ is valid.

But is there another integer point?

Suppose $D(x,y,z)$ with integer coordinates such that all pairwise distances are $\sqrt{2}$. From symmetry, $(\pm1,\pm1,\pm1)$ with odd number of minuses may work.

Try $D = (1,1,-1)$:

$|AD|^2 = (0)^2 + (1)^2 + (-1)^2 = 0 + 1 + 1 = 2$ ÃÂ¢ÃÂÃÂ- $|BD|^2 = (1)^2 + (0)^2 + (-1)^2 = 1 + 0 + 1 = 2$ ÃÂ¢ÃÂÃÂ- $|CD|^2 = (1)^2 + (1)^2 + (-2)^2 = 1 + 1 + 4 = 6 e 2$ ÃÂ¢ÃÂÃÂ

Try $D = (-1,1,1)$:

$|AD|^2 = (-2)^2 +1^2 +1^2 = 4+1+1=6$ ÃÂ¢ÃÂÃÂ

Try $D = (1,-1,1)$: $|AD|^2 = 0 +1+1=2$ ÃÂ¢ÃÂÃÂ$|BD|^2 =1+4+1=6$ ÃÂ¢ÃÂÃÂ

Try $D = (-1,-1,1)$: $|AD|^2 = (-2)^2+(-1)^2+1^2=4+1+1=6$ ÃÂ¢ÃÂÃÂ

All others fail.

Thus, the only integer-coordinate point $D$ such that $|AD|=|BD|=|CD|=\sqrt{2}$ is $D(1,1,1)$.

Pedagogical Insight: Why This Project Works

This STEM project engages students in:

Collaborative problem solving: }\ Students work in teams to derive equations and interpret geometry.- Vector applications: Students use coordinates to compute magnitudes, simulating real engineering constraints.- Critical thinking: Discovering why not all space-filling integer points work fosters deeper conceptual understanding.- Geometric intuition: Visualizing symmetry in 3D reinforces abstract vector properties.

Final Answer

The integer coordinates of the fourth vertex $D$ that form a regular tetrahedron with $A(1,0,0)$, $B(0,1,0)$, and $C(0,0,1)$ are:

[oxed{(1, 1, 1)}]