Prove geometric results using vectors, including properties of triangles, parallelograms and the diagonals of quadrilaterals, by expressing points as position vectors

What this dot point is asking

NESA wants you to prove geometric theorems using vectors rather than coordinate or Euclidean methods. You represent each point by a position vector, express segments as differences of position vectors, and translate geometric facts (parallel, collinear, equal length, perpendicular, midpoint) into vector statements. The proof then follows from vector algebra and the scalar product. The appeal is that one or two lines of algebra often replace a page of coordinate work, and the result holds for any configuration, including in three dimensions where coordinate geometry becomes unwieldy.

Position vectors and segments

Fix an origin $O$. Every point $P$ then corresponds to a position vector $\mathbf{p} = \overrightarrow{OP}$. The directed segment from $A$ to $B$ is

This single identity is the engine of vector geometry. Two segments $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are parallel exactly when $\mathbf{b} - \mathbf{a} = \lambda(\mathbf{d} - \mathbf{c})$ for some scalar $\lambda$. They are equal in length when $|\mathbf{b} - \mathbf{a}| = |\mathbf{d} - \mathbf{c}|$, and perpendicular when $(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{d} - \mathbf{c}) = 0$.

The four-step tactic (use this every time)

Vector proofs all follow the same shape. Treat it as a fixed procedure:

1. Assign a position vector to every labelled point (and, if it helps, put the origin at a useful vertex so one or more vectors become $\mathbf{0}$).
2. Express the displacement vectors the question is about, using $\overrightarrow{XY} = \mathbf{y} - \mathbf{x}$.
3. Compute the algebraic relation that encodes the geometric claim: an equality, a scalar multiple, or a zero scalar product.
4. Conclude by translating that relation back into the geometric statement, in words.

The skill is matching the geometric claim to the right algebraic relation in step 3. Here is the dictionary.

The midpoint and section formulas

The midpoint $M$ of segment $AB$ has position vector

More generally, the point $P$ dividing $AB$ internally in the ratio $m:n$ (so $AP:PB = m:n$) has position vector

This follows because $\overrightarrow{AP} = \frac{m}{m+n}\overrightarrow{AB}$, so $\mathbf{p} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a})$, which simplifies to the stated form. The cross-over of the labels is worth memorising: the point nearer $B$ (large $m$) carries the weight $m$ on $\mathbf{b}$. The midpoint is just the case $m:n = 1:1$.

Collinearity

Three points $A$, $B$, $C$ are collinear when one displacement is a scalar multiple of another and they share a point: $\overrightarrow{AB} = \lambda\,\overrightarrow{AC}$. The shared vertex $A$ is what upgrades "parallel" to "on the same line", since two parallel vectors that share an endpoint must lie along one line through that endpoint.

Using the scalar product for angles and lengths

When perpendicularity or equal lengths appear, the scalar product is decisive. The key identities are

The first turns a length statement into an algebraic one (expand the square with $\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2$); the second is the perpendicularity test. Together they prove, for example, that a triangle inscribed in a semicircle has a right angle at the circumference, by placing the centre at the origin and showing two radii-based vectors have zero scalar product.

Proofs about three-dimensional figures

In Extension 2 the same machinery proves properties of solids: cubes, rectangular boxes and pyramids. The move is identical, only now the position vectors are three-component. The natural setup for a box is to put one vertex at the origin and call the three edges from it $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$; every other vertex is then a sum of these, and the space diagonal is $\mathbf{a} + \mathbf{b} + \mathbf{c}$.

With the edges mutually perpendicular and equal in length (a cube), $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{c} = \mathbf{c}\cdot\mathbf{a} = 0$ and $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}|$. These two facts collapse most cube proofs to a couple of lines, as the worked example shows.

How exam questions ask about vector proofs

• "Show that $X$, $Y$, $Z$ are collinear": show $\overrightarrow{XY} = \lambda\,\overrightarrow{XZ}$, then state they share the point $X$.
• "Show that $P$ lies on the line through $A$ and $B$": write $\overrightarrow{OP} = \overrightarrow{OA} + t\,\overrightarrow{AB}$ for some scalar $t$.
• "Prove that $PQRS$ is a parallelogram": show one pair of opposite sides are equal vectors, $\overrightarrow{PQ} = \overrightarrow{SR}$.
• "Prove the diagonals bisect each other": show the midpoints of the two diagonals have the same position vector.
• "$P$ divides $AB$ in the ratio $m:n$; find its position vector": apply the section formula $\frac{n\mathbf{a} + m\mathbf{b}}{m+n}$.
• "Show that the angle ... is a right angle" or "... edges are perpendicular": show the relevant scalar product is zero, expanding with $\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2$ where needed.