Year 12 Maths Extension 1 (2027) Vectors

Vectors in geometry (proofs)

Q: How do vectors prove geometry results?

Write points as position vectors, then use the midpoint and section formulas, the parallel condition, or the dot product.

Q: What is the section formula?

The point dividing AB in ratio m to n is (n a plus m b) over m plus n.

Q: How do you show two lines are perpendicular?

Show the dot product of their direction vectors is zero.

Q: How do you show equal lengths?

Compare the dot product of each vector with itself, which equals its magnitude squared.

20 practice questions 2 video lessons Theory + worked examples

Use vectors to prove geometric results in NSW Year 12 Mathematics Extension 1. Representing points and sides as vectors turns statements about shapes into clean algebra, without relying on a specific diagram.

You will learn to prove properties of triangles, parallelograms and other figures using position vectors, the dot product, and tests for parallel and perpendicular lines — an elegant vector proof approach valued in the HSC Extension 1 exam.

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Theory

Vectors give clean geometry proofs: set an origin, use the midpoint/section formulas, the parallel condition $\vec{A B} = k \vec{C D}$ , and the dot product for perpendicularity and length. This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-02.

Vectors give clean proofs of geometry results. Set an origin $O$ , write each point as a position vector $\vec{O A} = a$ , then $\vec{A B} = b - a$ , and use three tools: the midpoint/section formulas, the parallel condition, and the dot product.

Midpoint of $A B$ : $\frac{1}{2} (a + b)$ . The point dividing $A B$ in ratio $m : n$ is $\frac{n a + m b}{m + n}$ .

Collinear/parallel: $\vec{A B} = k \vec{C D}$ . Perpendicular: $\vec{A B} \cdot \vec{C D} = 0$ . Length: $| a |^{2} = a \cdot a$ .

NESA link. Part of the Year 12 Introduction to vectors focus area, outcome ME1-12-02 ("operates with 2D and 3D vectors and uses 2D vectors to solve problems involving motion in two dimensions") with MAO-WM-01.

The point dividing

A B

in ratio

m : n

\frac{n a + m b}{m + n}

Angle in a semicircle:

\vec{A P} \cdot \vec{B P} = 0

midpoint = \frac{1}{2} (a + b), ratio m : n \Rightarrow \frac{n a + m b}{m + n} .

midpoint = (a+b)/2; ratio m:n gives (n a + m b)/(m+n)

\vec{A B} = k \vec{C D} (parallel), \vec{A B} \cdot \vec{C D} = 0 (⊥), | a |^{2} = a \cdot a .

parallel: AB = k CD; perpendicular: AB.CD = 0; length squared: a.a

Strategy. To prove two segments bisect, show equal midpoints; perpendicular, show the dot product is $0$ ; equal length, compare $a \cdot a$ .

How to prove with vectors

Choose a convenient origin (a vertex or centre) to simplify the algebra.
Write each point as a position vector and each segment as head minus tail.
Apply the right tool: midpoint for bisection, dot product for perpendicularity or length.
Conclude with a sentence linking the algebra to the geometric claim.

Example 1 — Section point

P

divides

A B

with

A P : P B = 3 : 1

. If

A

B

have position vectors

a

b

, find

\vec{O P}

Solution

\vec{O P}

=

a + \frac{3}{4} (b - a) = \frac{1}{4} a + \frac{3}{4} b

OP = (1/4)a + (3/4)b

$\vec{O P} = \frac{1}{4} a + \frac{3}{4} b$ .

Example 2 — Rhombus diagonals

In a rhombus with

\vec{O A} = a

\vec{O C} = c

and

| a | = | c |

, prove the diagonals

a + c

and

c - a

are perpendicular.

Solution

(a + c) \cdot (c - a)

=

| c |^{2} - | a |^{2} = 0

(a+c).(c-a) = |c|^2 - |a|^2 = 0, so perpendicular

So the diagonals are perpendicular.

Example 3 — Angle in a semicircle

A

B

are ends of a diameter (centre

O

), and

P

is on the circle. Prove

∠ A P B = 90^{\circ}

Solution

Take $O$ as origin, so $\vec{O A} = a$ , $\vec{O B} = - a$ , $| \vec{O P} | = | a |$ .

\vec{A P} \cdot \vec{B P}

=

(p - a) \cdot (p + a) = | p |^{2} - | a |^{2} = 0

AP.BP = |p|^2 - |a|^2 = 0, so angle APB = 90 degrees

So $∠ A P B = 90^{\circ}$ .

Example 4 — Centroid

In triangle

O A B

M

is the midpoint of

A B

and

G

lies on

O M

with

O G : G M = 2 : 1

. Show

\vec{O G} = \frac{1}{3} (a + b)

Solution

$\vec{O M} = \frac{1}{2} (a + b)$ .

\vec{O G}

=

\frac{2}{3} \vec{O M} = \frac{2}{3} \cdot \frac{1}{2} (a + b) = \frac{1}{3} (a + b)

OG = (1/3)(a + b)

So $\vec{O G} = \frac{1}{3} (a + b)$ .

Common pitfalls

Origin choice. A poorly chosen origin makes the algebra messy — use a vertex or centre.

Section formula order. Ratio

m : n

gives

\frac{n a + m b}{m + n}

— the weights cross over.

Proving perpendicular. Show the dot product is

0

; for equal lengths compare

a \cdot a

No conclusion. Always finish with a sentence linking the algebra to the geometric claim.

Frequently asked questions

How do vectors prove geometry results?

Write points as position vectors, then use the midpoint/section formulas, the parallel condition, or the dot product.

What is the section formula?

The point dividing $A B$ in ratio $m : n$ is $\frac{n a + m b}{m + n}$ .

How do you show two lines are perpendicular?

Show the dot product of their direction vectors is $0$ .

How do you show equal lengths?

Compare $a \cdot a = | a |^{2}$ for each.

Why choose the origin carefully?

A vertex or centre often makes several position vectors simple, cutting the algebra.

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Projection of a vector & perpendicular component

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Motion in vector form 2D (velocity/accel, relative velocity, crosswind)

Vectors in geometry (proofs)

📖 Theory

How to prove with vectors

Common pitfalls

Frequently asked questions

Theory