Vector geometry proofs | Year 11

01 / LEARN

Understand the idea.

Start with a useful choice of origin

A vector proof uses algebra to establish a geometric fact for every figure satisfying the assumptions. Choose an origin and name the position vectors, then express derived points using those same vectors. A midpoint, for example, has the average of its endpoint positions.

Use the diagram to choose an approach, rather than as evidence. State any needed conditions: distinct points give non-zero displacements, and a non-degenerate triangle or parallelogram rules out collapsed figures where a direction or angle could be missing.

Match algebra to a geometric conclusion

Match the conclusion to the calculation. A scalar-multiple relation establishes parallel directions; using a common starting point then establishes collinearity. A zero dot product establishes perpendicularity when both vectors are non-zero.

Equal directed vectors establish both equal lengths and matching direction. Equal magnitudes establish only equal lengths. Finish a proof by naming the geometric implication, including any property of a triangle or parallelogram needed to reach the stated result.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

The triangle midpoint result

In a non-degenerate triangle OAB, let OA=a and OB=b. M and N are the midpoints of OA and OB. Prove that MN is parallel to AB and half its length.

Worked solution

1
Express the midpoint positions.
$\vec{O M} = \frac{1}{2} a, \vec{O N} = \frac{1}{2} b$
M and N each lie halfway from the chosen origin to the respective vertex.
2
Subtract positions to find the joining vector.
$\begin{aligned} \vec{M N} & = \vec{O N} - \vec{O M} \\ = \frac{1}{2} b - \frac{1}{2} a \\ = \frac{1}{2} (b - a) . \end{aligned}$
A displacement is its endpoint position minus its starting position.
3
Compare with the triangle side.
$\vec{A B} = b - a, \vec{M N} = \frac{1}{2} \vec{A B}$
Both displacements have been expressed using the same original vectors.
4
State both geometric conclusions.
$M N ∥ A B, ∣ \vec{M N} ∣ = \frac{1}{2} ∣ \vec{A B} ∣$
The side AB is non-zero; its positive scalar multiple has the same direction and half the magnitude.

Answer

MN is parallel to AB and $∣ M N ∣ = \frac{1}{2} ∣ A B ∣$ .

EXAMPLE 02

A rectangle from vector sides

Adjacent sides of a non-degenerate parallelogram are a=(3,1) and b=(-2,6). Prove that it is a rectangle.

Worked solution

1
Test the adjacent side directions.
$a \cdot b = (3, 1) \cdot (- 2, 6) = 3 (- 2) + 1 (6) = 0$
The dot-product test detects a right angle between the adjacent side vectors.
2
Establish a genuine right angle.
$∣ a ∣ = \sqrt{10} > 0, ∣ b ∣ = \sqrt{40} > 0$
Both adjacent sides are non-zero, so the zero dot product proves perpendicularity.
3
Use the parallelogram assumption.
The parallelogram has one right angle. Adjacent angles in a parallelogram are supplementary and opposite angles are equal, so all four angles are $90^{\circ}$ . It is therefore a rectangle.

Answer

The parallelogram is a rectangle.

03 / YOUR TURN

Try it yourself.

Check your understanding

If AB=DC as directed vectors in a quadrilateral with vertices A, B, C, D in order, what can you conclude when it is non-degenerate?

Show answer & reasoning

It is a parallelogram.

Writing positions as $a, b, c, d$ , the condition gives $b - a = c - d$ . Rearranging gives $d - a = c - b$ , so the other directed opposite sides also agree. The non-degenerate quadrilateral has both pairs of opposite sides equal and parallel, hence is a parallelogram.

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6 worksheet questions · 19 marks

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Read the questions online

$A$ , $B$ have position vectors $a$ , $b$ . Write the midpoint $M$ position vector.
2 marks
$A$ , $B$ , $C$ have position vectors $a$ , $b$ , $c$ . $M$ and $N$ are midpoints of $A B$ and $A C$ . Show that $M N = \frac{1}{2} B C$ .
3 marks
In a parallelogram with adjacent side vectors $a$ and $b$ , show that the diagonals have the same midpoint.
3 marks
For non-zero perpendicular $a$ and $b$ , prove $∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2}$ .
3 marks
A parallelogram has $∣ a ∣ = ∣ b ∣$ . Prove that its diagonal directions $a + b$ and $a - b$ are perpendicular, assuming neither diagonal is zero.
4 marks
A quadrilateral's diagonals have the same midpoint. Use position vectors $a$ , $b$ , $c$ , $d$ to show $A B = D C$ .
4 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$A$ , $B$ have position vectors $a$ , $b$ . Write the midpoint $M$ position vector.
2 marks
Average the two position vectors.
$\frac{1}{2} (a + b)$ .
$A$ , $B$ , $C$ have position vectors $a$ , $b$ , $c$ . $M$ and $N$ are midpoints of $A B$ and $A C$ . Show that $M N = \frac{1}{2} B C$ .
3 marks
$O M = \frac{a + b}{2}$ and $O N = \frac{a + c}{2}$ .
Subtract $O M$ from $O N$ .
$\vec{M N} = \frac{1}{2} (c - b)$ .
In a parallelogram with adjacent side vectors $a$ and $b$ , show that the diagonals have the same midpoint.
3 marks
The opposite vertex has position $a + b$ .
The other diagonal joins positions $a$ and $b$ .
Both midpoints are $\frac{1}{2} (a + b)$ when one vertex is the origin.
For non-zero perpendicular $a$ and $b$ , prove $∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2}$ .
3 marks
$(a + b) \cdot (a + b) = ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2}$ .
The middle term is zero.
Expand the squared length and use $a \cdot b = 0$ .
A parallelogram has $∣ a ∣ = ∣ b ∣$ . Prove that its diagonal directions $a + b$ and $a - b$ are perpendicular, assuming neither diagonal is zero.
4 marks
Expand to $∣ a ∣^{2} - ∣ b ∣^{2}$ .
Equal adjacent lengths make the difference zero.
$(a + b) \cdot (a - b) = 0$ .
A quadrilateral's diagonals have the same midpoint. Use position vectors $a$ , $b$ , $c$ , $d$ to show $A B = D C$ .
4 marks
Equal diagonal midpoints give $a + c = b + d$ .
Rearrange to get the two equal directed side vectors.
$b - a = c - d$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: vectors in the plane. Covers magnitude, components, geometry, scalar products, projections and proofs; follows the approved Specialist preparation structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Start with a useful choice of origin

Match algebra to a geometric conclusion

See the method in action.

The triangle midpoint result

Express the midpoint positions.

Subtract positions to find the joining vector.

Compare with the triangle side.

State both geometric conclusions.

A rectangle from vector sides

Test the adjacent side directions.

Establish a genuine right angle.

Use the parallelogram assumption.

Try it yourself.

Check your understanding

Keep practising