Geometric proofs
Master geometric proofs with vectors in three dimensions for Year 12 Specialist Mathematics in Queensland (QCAA, Unit 3). By giving every point a position vector, a solid figure becomes algebra you can prove — no coordinates or congruent triangles required.
You will learn to find midpoints, section points and centroids, and use scalar multiples and the dot product to prove that the medians of a triangle are concurrent, that the diagonals of a parallelepiped bisect each other, that a tetrahedron’s centroid divides a median \(3:1\), and that vectors are perpendicular in 3D.
Theory
A vector proof in three dimensions turns a solid-geometry result into algebra: choose an origin, give each point a position vector, and reason with sums, scalar multiples and the dot product. In Year 12 Specialist Mathematics (QCAA, Queensland, Unit 3) this proves results that go beyond Unit 2 — medians and centroids, the diagonals of a parallelepiped, the centroid of a tetrahedron, and perpendicularity in 3D.
Fix an origin \(O\). Each point \(A\) then has a position vector \(\overrightarrow{OA}=\mathbf{a}\), written as an ordered triple or a column \(\begin{pmatrix}x\\y\\z\end{pmatrix}\). Every statement about the points becomes a statement about these vectors — and the algebra is identical in two and three dimensions.
The workhorse fact is the vector between two points: \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) (end minus start). From it come the midpoint \(\tfrac{1}{2}(\mathbf{a}+\mathbf{b})\) and the section point that divides a segment in a chosen ratio.
Averaging position vectors locates the key centres: the centroid of a triangle is \(\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})\) and the centroid of a tetrahedron is \(\tfrac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})\). Because these expressions are symmetric in the vertices, the same point lies on every median, which proves concurrency.
Two conclusions finish most proofs. Two vectors are parallel (or a set of points collinear) when one is a scalar multiple of the other, \(\overrightarrow{PQ}=k\,\overrightarrow{RS}\); two vectors are perpendicular exactly when their dot product is zero, \(\mathbf{u}\cdot\mathbf{v}=0\). Equal vectors carry shape too: if \(\overrightarrow{PQ}=\overrightarrow{SR}\) then \(PQRS\) is a parallelogram, even when its four vertices are not coplanar (a skew quadrilateral).
Relative to an origin \(O\), with position vectors \(\mathbf{a},\mathbf{b},\mathbf{c},\dots\):
A point \(P\) dividing \(AB\) internally in the ratio \(AP:PB=m:n\), and the two centroids:
The dot product in component form is the test for a right angle in 3D:
How to write a 3D vector proof
- Set an origin and give each point a position vector, e.g. \(\overrightarrow{OA}=\mathbf{a}\). Choosing a vertex as \(O\) usually simplifies the algebra; keep everything in terms of \(\mathbf{a},\mathbf{b},\mathbf{c},\dots\).
- Express every vector you need as a difference of position vectors, \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), and write midpoints, section points and centroids from the formulas.
- Do the algebra that matches the goal: factor out a scalar to show parallel, collinear or concurrent; take a dot product for perpendicular; or show two vectors are equal for a parallelogram.
- State the conclusion in words — name the geometric result the algebra has just proved.
First write the median vectors \(\overrightarrow{AG}\) and \(\overrightarrow{GM}\), using \(\overrightarrow{OM}=\tfrac{1}{2}(\mathbf{b}+\mathbf{c})\):
| \(\overrightarrow{AG}\) | \(=\) | \(\overrightarrow{OG}-\mathbf{a}\) |
| \(=\) | \(\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}-\mathbf{a}\) | |
| \(=\) | \(\dfrac{\mathbf{b}+\mathbf{c}-2\mathbf{a}}{3}\) |
Now \(\overrightarrow{GM}=\overrightarrow{OM}-\overrightarrow{OG}\):
| \(\overrightarrow{GM}\) | \(=\) | \(\dfrac{\mathbf{b}+\mathbf{c}}{2}-\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\) |
| \(=\) | \(\dfrac{3(\mathbf{b}+\mathbf{c})-2(\mathbf{a}+\mathbf{b}+\mathbf{c})}{6}\) | |
| \(=\) | \(\dfrac{\mathbf{b}+\mathbf{c}-2\mathbf{a}}{6}\) |
Compare the two:
| \(\overrightarrow{AG}\) | \(=\) | \(2\times\dfrac{\mathbf{b}+\mathbf{c}-2\mathbf{a}}{6}\) |
| \(=\) | \(2\,\overrightarrow{GM}\) |
\(\overrightarrow{AG}=2\,\overrightarrow{GM}\), so \(G\) lies on \(AM\) and divides it \(2:1\) from the vertex; the same symmetric point serves all three medians, so they are concurrent.
Take the midpoint of the first diagonal (average its endpoints):
| \(M_1\) | \(=\) | \(\dfrac{\mathbf{0}+(\mathbf{u}+\mathbf{v}+\mathbf{w})}{2}\) |
| \(=\) | \(\dfrac{\mathbf{u}+\mathbf{v}+\mathbf{w}}{2}\) |
Now the midpoint of the second diagonal:
| \(M_2\) | \(=\) | \(\dfrac{\mathbf{v}+(\mathbf{u}+\mathbf{w})}{2}\) |
| \(=\) | \(\dfrac{\mathbf{u}+\mathbf{v}+\mathbf{w}}{2}\) |
\(M_1=M_2=\tfrac{1}{2}(\mathbf{u}+\mathbf{v}+\mathbf{w})\): the diagonals share a midpoint, so they bisect each other.
Form the two side vectors from \(A\):
| \(\overrightarrow{AB}\) | \(=\) | \((4-2,\,1-0,\,3-1)=(2,1,2)\) |
| \(\overrightarrow{AC}\) | \(=\) | \((1-2,\,2-0,\,1-1)=(-1,2,0)\) |
Take their dot product:
| \(\overrightarrow{AB}\cdot\overrightarrow{AC}\) | \(=\) | \((2)(-1)+(1)(2)+(2)(0)\) |
| \(=\) | \(-2+2+0\) | |
| \(=\) | \(0\) |
\(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\), so \(\overrightarrow{AB}\perp\overrightarrow{AC}\): the angle at \(A\) is \(90^\circ\).
Write \(\overrightarrow{AF}\) using \(\overrightarrow{OF}=\tfrac{1}{3}(\mathbf{b}+\mathbf{c}+\mathbf{d})\):
| \(\overrightarrow{AF}\) | \(=\) | \(\dfrac{\mathbf{b}+\mathbf{c}+\mathbf{d}}{3}-\mathbf{a}\) |
| \(=\) | \(\dfrac{\mathbf{b}+\mathbf{c}+\mathbf{d}-3\mathbf{a}}{3}\) |
Now \(\overrightarrow{AG}=\overrightarrow{OG}-\mathbf{a}\):
| \(\overrightarrow{AG}\) | \(=\) | \(\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}-\mathbf{a}\) |
| \(=\) | \(\dfrac{\mathbf{b}+\mathbf{c}+\mathbf{d}-3\mathbf{a}}{4}\) |
Compare the two vectors:
| \(\overrightarrow{AG}\) | \(=\) | \(\dfrac{3}{4}\overrightarrow{AF}\) |
| \(\therefore\ \overrightarrow{AG}\) | \(=\) | \(3\,\overrightarrow{GF}\) |
\(\overrightarrow{AG}=3\,\overrightarrow{GF}\), so \(G\) lies on \(AF\) and divides it \(3:1\) from the vertex.
Common pitfalls
Frequently asked questions
How do you prove a geometric result in 3D using vectors?
Choose an origin, give each point a position vector, rewrite every segment as \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), then use a scalar multiple for parallel or concurrent, a dot product for perpendicular, or equal vectors for a parallelogram, and state the conclusion in words. The method is identical to 2D.
What is the position vector of the centroid of a triangle?
It is \(\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})\), the average of the three vertices’ position vectors. Because this is symmetric in \(\mathbf{a},\mathbf{b},\mathbf{c}\), the same point lies on all three medians, so they are concurrent.
Why does the centroid divide a median in the ratio 2:1?
With \(M\) the midpoint of \(BC\) and \(G=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})\), you get \(\overrightarrow{AG}=\tfrac{1}{3}(\mathbf{b}+\mathbf{c}-2\mathbf{a})\) and \(\overrightarrow{GM}=\tfrac{1}{6}(\mathbf{b}+\mathbf{c}-2\mathbf{a})\), so \(\overrightarrow{AG}=2\,\overrightarrow{GM}\).
How do you show two vectors are perpendicular in three dimensions?
Compute the dot product \(\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3\). If it equals zero the vectors are perpendicular, which is how you prove a right angle in 3D.
How do vectors prove the diagonals of a parallelepiped bisect each other?
Each space diagonal has midpoint \(\tfrac{1}{2}(\mathbf{u}+\mathbf{v}+\mathbf{w})\); since every diagonal shares this same midpoint, they all bisect one another.
What is a skew quadrilateral and why do its side midpoints form a parallelogram?
A skew quadrilateral has four vertices that are not coplanar. If \(P,Q,R,S\) are the midpoints of its sides, then \(\overrightarrow{PQ}=\overrightarrow{SR}=\tfrac{1}{2}\overrightarrow{AC}\), so \(PQRS\) is a parallelogram even though \(ABCD\) is not flat.