Geometric proofs using vectors
Master geometric proofs using vectors for Year 11 Specialist Mathematics in Queensland (QCAA). By giving every point a position vector, a diagram becomes algebra you can prove — no coordinates or congruent triangles required.
You will learn to write the vector between two points, find midpoints and section points, and use scalar multiples and the scalar product to prove results such as the midpoint theorem, that a rhombus has perpendicular diagonals, and that the angle in a semicircle is a right angle.
Theory
A vector proof turns a geometry result into algebra: choose an origin, give each point a position vector, and reason with sums, scalar multiples and the scalar product. In Year 11 Specialist Mathematics (QCAA, Queensland) this proves classic results about triangles, parallelograms and circles without coordinates or congruent triangles.
Fix an origin \(O\). Each point \(A\) then has a position vector \(\overrightarrow{OA}=\mathbf{a}\). Every geometric statement about the points becomes a statement about these vectors.
The single most-used 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 dividing \(AB\) in a chosen ratio.
Three ideas convert a picture into a conclusion. Two vectors are parallel when one is a scalar multiple of the other, \(\overrightarrow{PQ}=k\,\overrightarrow{RS}\); three points are collinear when two vectors along them are parallel and share a point; and two vectors are perpendicular exactly when their scalar product is zero, \(\mathbf{u}\cdot\mathbf{v}=0\).
Equal vectors also carry shape: if \(\overrightarrow{PQ}=\overrightarrow{SR}\) then \(PQRS\) is a parallelogram. Combining these, vectors prove that the diagonals of a parallelogram bisect each other, that a rhombus has perpendicular diagonals, and that the angle in a semicircle is a right angle.
Relative to an origin \(O\), with \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\):
A point \(P\) dividing \(AB\) internally in the ratio \(AP:PB=m:n\):
The scalar-product tools that finish perpendicularity and length proofs:
How to write a vector proof
- Set an origin and give each point a position vector, e.g. \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OB}=\mathbf{b}\). Choosing a vertex as \(O\) usually simplifies the algebra.
- Express every vector you need as a difference of position vectors, \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), and write midpoints and section points from the formulas.
- Do the algebra that matches the goal: factor out a scalar to show parallel/collinear, take a scalar 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.
Write the two half position vectors, then subtract to reach \(\overrightarrow{PQ}\):
| \(\overrightarrow{OP}\) | \(=\) | \(\tfrac{1}{2}\mathbf{a}\) |
| \(\overrightarrow{OQ}\) | \(=\) | \(\tfrac{1}{2}\mathbf{b}\) |
| \(\overrightarrow{PQ}\) | \(=\) | \(\overrightarrow{OQ}-\overrightarrow{OP}\) |
| \(=\) | \(\tfrac{1}{2}\mathbf{b}-\tfrac{1}{2}\mathbf{a}\) | |
| \(=\) | \(\tfrac{1}{2}(\mathbf{b}-\mathbf{a})\) | |
| \(=\) | \(\tfrac{1}{2}\overrightarrow{AB}\) |
\(\overrightarrow{PQ}=\tfrac{1}{2}\overrightarrow{AB}\), so \(PQ\) is parallel to \(AB\) and half its length.
First build diagonal \(\overrightarrow{OB}\), then take its midpoint:
| \(\overrightarrow{OB}\) | \(=\) | \(\overrightarrow{OA}+\overrightarrow{AB}\) |
| \(=\) | \(\mathbf{a}+\mathbf{c}\) | |
| \(M_{OB}\) | \(=\) | \(\tfrac{1}{2}(\mathbf{a}+\mathbf{c})\) |
Now the midpoint of the other diagonal \(AC\):
| \(M_{AC}\) | \(=\) | \(\tfrac{1}{2}(\overrightarrow{OA}+\overrightarrow{OC})\) |
| \(=\) | \(\tfrac{1}{2}(\mathbf{a}+\mathbf{c})\) |
Both midpoints are \(\tfrac{1}{2}(\mathbf{a}+\mathbf{c})\), so the diagonals bisect each other.
Take the scalar product of the two diagonals and expand:
| \((\mathbf{a}+\mathbf{c})\cdot(\mathbf{c}-\mathbf{a})\) | \(=\) | \(\mathbf{c}\cdot\mathbf{c}-\mathbf{a}\cdot\mathbf{a}\) |
| \(=\) | \(|\mathbf{c}|^2-|\mathbf{a}|^2\) |
The diagonals are perpendicular when this scalar product is zero:
| \(|\mathbf{c}|^2-|\mathbf{a}|^2\) | \(=\) | \(0\) |
| \(|\mathbf{c}|^2\) | \(=\) | \(|\mathbf{a}|^2\) |
| \(|\mathbf{a}|\) | \(=\) | \(|\mathbf{c}|\) |
The diagonals meet at right angles exactly when \(|\mathbf{a}|=|\mathbf{c}|\); that is, when the sides are equal, so the parallelogram is a rhombus.
Write \(\overrightarrow{PA}\) and \(\overrightarrow{PB}\) as differences of position vectors:
| \(\overrightarrow{PA}\) | \(=\) | \(\mathbf{a}-\mathbf{p}\) |
| \(\overrightarrow{PB}\) | \(=\) | \(-\mathbf{a}-\mathbf{p}\) |
Take their scalar product and use \(\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2\):
| \(\overrightarrow{PA}\cdot\overrightarrow{PB}\) | \(=\) | \((\mathbf{a}-\mathbf{p})\cdot(-\mathbf{a}-\mathbf{p})\) |
| \(=\) | \(-\mathbf{a}\cdot\mathbf{a}+\mathbf{p}\cdot\mathbf{p}\) | |
| \(=\) | \(|\mathbf{p}|^2-|\mathbf{a}|^2\) | |
| \(=\) | \(r^2-r^2\) | |
| \(=\) | \(0\) |
\(\overrightarrow{PA}\cdot\overrightarrow{PB}=0\), so \(PA\perp PB\): the angle in a semicircle is a right angle.
Common pitfalls
Frequently asked questions
How do you prove a geometry result 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, a scalar product for perpendicular, or equal vectors for a parallelogram, and state the conclusion.
What is the position vector of a midpoint?
The midpoint of \(AB\) has position vector \(\tfrac{1}{2}(\mathbf{a}+\mathbf{b})\) — the average of the position vectors of the two endpoints.
How do you show two vectors are parallel?
Show that one is a scalar multiple of the other, \(\overrightarrow{PQ}=k\,\overrightarrow{RS}\). If they also share a point, the three points are collinear.
How do you prove the angle in a semicircle is 90 degrees with vectors?
Put the centre at \(O\) with diameter ends \(\mathbf{a}\) and \(-\mathbf{a}\) and \(P=\mathbf{p}\) on the circle. Then \(\overrightarrow{PA}\cdot\overrightarrow{PB}=|\mathbf{p}|^2-|\mathbf{a}|^2=0\), so \(PA\perp PB\).
Why do the diagonals of a rhombus meet at right angles?
With sides \(\mathbf{a}\) and \(\mathbf{c}\) the diagonals are \(\mathbf{a}+\mathbf{c}\) and \(\mathbf{c}-\mathbf{a}\); their scalar product is \(|\mathbf{c}|^2-|\mathbf{a}|^2\), which is zero exactly when \(|\mathbf{a}|=|\mathbf{c}|\).
How do you prove three points are collinear using vectors?
Form two vectors along the points, such as \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\); if \(\overrightarrow{AC}=k\,\overrightarrow{AB}\) they are parallel and share \(A\), so \(A,B,C\) are collinear.