Year 12 Maths - Specialist (Unit 3 & Unit 4) Vectors

Geometric proofs with vectors

Q: What is the position vector of the centroid of a triangle?

It is one third of (a plus b plus c), the average of the three vertices' position vectors. Because this is symmetric in a, b and c, the same point lies on all three medians, so they are concurrent.

Q: Why does the centroid divide a median in the ratio 2 to 1?

With M the midpoint of BC and G equal to one third of (a plus b plus c), you get vector AG equals one third of (b plus c minus two a) and vector GM equals one sixth of the same, so vector AG equals two times vector GM.

Q: How do you show two vectors are perpendicular in three dimensions?

Compute the dot product u dot v equals u1 v1 plus u2 v2 plus u3 v3. If it equals zero the vectors are perpendicular, which is how you prove a right angle in 3D.

Q: How do vectors prove the diagonals of a parallelepiped bisect each other?

Each space diagonal has midpoint one half of (u plus v plus w); since every diagonal shares this same midpoint, they all bisect one another.

20 practice questions 0 video lessons Theory + worked examples

Master geometric proofs with vectors in Year 12 VCE Specialist Mathematics. By giving every point a position vector, a solid figure becomes algebra you can prove — no coordinates or congruent triangles required. It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Vectors topic of Unit 3.

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 parallelogram bisect each other, and that vectors are perpendicular in three dimensions — a rigorous proof technique for geometry.

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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 (, , 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).

Parallelepiped with edges \(\mathbf{u},\mathbf{v},\mathbf{w}\): both space diagonals have midpoint \(\tfrac{1}{2}(\mathbf{u}+\mathbf{v}+\mathbf{w})\), so the diagonals bisect each other at \(M\).

Tetrahedron \(ABCD\): the centroid \(G=\tfrac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})\) lies on \(AF\) (\(F\) the centroid of face \(BCD\)) with \(AG:GF=3:1\).

Relative to an origin \(O\), with position vectors \(\mathbf{a},\mathbf{b},\mathbf{c},\dots\):

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a} \qquad M_{AB}=\tfrac{1}{2}(\mathbf{a}+\mathbf{b}) \]

\vec{A B} = b - a

A point \(P\) dividing \(AB\) internally in the ratio \(AP:PB=m:n\), and the two centroids:

\[ \overrightarrow{OP}=\dfrac{n\mathbf{a}+m\mathbf{b}}{m+n} \qquad G_{\triangle}=\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3} \qquad G_{\text{tet}}=\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4} \]

G_{△} = \frac{a + b + c}{3}

The dot product in component form is the test for a right angle in 3D:

\[ \mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3 \qquad \mathbf{u}\perp\mathbf{v}\iff\mathbf{u}\cdot\mathbf{v}=0 \]

u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}

The three conclusions. Parallel / collinear / concurrent: one vector is a scalar multiple of another, \(\overrightarrow{PQ}=k\,\overrightarrow{RS}\). Perpendicular: \(\mathbf{u}\cdot\mathbf{v}=0\). Parallelogram (even if skew): \(\overrightarrow{PQ}=\overrightarrow{SR}\).

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.

Example 1 — Centroid divides a median 2:1

In triangle \(ABC\), \(M\) is the midpoint of \(BC\) and \(G\) is the point on the median \(AM\) with \(\overrightarrow{OG}=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})\). Show that \(G\) lies on \(AM\) and that \(\overrightarrow{AG}=2\,\overrightarrow{GM}\).

Solution

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.

Example 2 — Diagonals of a parallelepiped bisect

A parallelepiped has a vertex at \(O\) and edge vectors \(\mathbf{u}\), \(\mathbf{v}\), \(\mathbf{w}\). Show that the space diagonal from \(O\) to \(\mathbf{u}+\mathbf{v}+\mathbf{w}\) and the space diagonal from \(\mathbf{v}\) to \(\mathbf{u}+\mathbf{w}\) bisect each other.

Solution

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.

Example 3 — A right angle in 3D (dot product)

The points are \(A(2,0,1)\), \(B(4,1,3)\) and \(C(1,2,1)\). Show that triangle \(ABC\) is right-angled at \(A\).

Solution

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 \(90^\circ\).

Example 4 — Centroid of a tetrahedron (3:1)

A tetrahedron has vertices \(A,B,C,D\). Let \(F\) be the centroid of face \(BCD\) and \(G=\tfrac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})\) be the centroid of the tetrahedron. Show that \(G\) lies on \(AF\) with \(\overrightarrow{AG}=3\,\overrightarrow{GF}\).

Solution

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

Reversing the vector between two points. \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) is end minus start, not \(\mathbf{a}-\mathbf{b}\). Getting it backwards flips every sign in the proof.

Confusing parallel with perpendicular. A scalar multiple \(\overrightarrow{PQ}=k\,\overrightarrow{RS}\) proves parallel (or concurrent); a zero dot product \(\mathbf{u}\cdot\mathbf{v}=0\) proves perpendicular. They are different tests.

Using the wrong centroid fraction. A triangle's centroid divides a median \(2:1\) (the \(\tfrac{1}{3}\) average of three vertices); a tetrahedron's divides the vertex-to-face segment \(3:1\) (the \(\tfrac{1}{4}\) average of four). Do not mix them.

Stopping at the algebra. A vector proof is not finished until you state the geometric conclusion in words — "so the diagonals bisect each other", not just a line of vectors.

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.

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