Vector geometry proofs | Year 11

01 / LEARN

The idea, step by step

Start with a useful choice of origin

A vector proof begins by naming vectors clearly. Choosing one vertex as the origin often reduces the number of terms. Express derived points, such as midpoints, using the same named vectors before comparing displacements.

A diagram suggests the relationship but cannot prove it. Show the scalar-multiple or dot-product calculation and then state the geometric fact it establishes. Use distinct points and a non-degenerate figure when those are required.

Match algebra to a geometric conclusion

To prove parallelism, show that one non-zero displacement is a scalar multiple of another. To prove collinearity, do the same with displacements from a common point. To prove perpendicularity, show the relevant non-zero vectors have dot product zero.

Equal vector displacements give both equal lengths and parallel directions. Equal magnitudes by themselves do not give parallelism. A complete argument says which fact is being used, rather than leaving the reader to infer it from a calculation.

A common trap

Writing down equal-looking lengths from a sketch as if that were a proof.

Instead: Express the relevant vectors algebraically and justify each implication, including why the figure is non-degenerate.

02 / WORKED EXAMPLES

See the thinking

Example 1

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.

OM= $\frac{1}{2} a$ and ON= $\frac{1}{2} b$ .
MN= $\frac{1}{2} b - \frac{1}{2} a = \frac{1}{2} (b - a) = \frac{1}{2}$ AB.
The positive scalar multiple gives parallel directions and magnitude ratio one half.

MN is parallel to AB and

∣ M N ∣ = \frac{1}{2} ∣ A B ∣

Example 2

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.

$a \cdot b = 3 (- 2) + 1 (6) = 0$ .
Both adjacent sides are non-zero, so their included angle is a right angle.
A parallelogram with a right angle is a rectangle.

The parallelogram is a rectangle.

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

It is a parallelogram.

The opposite sides AB and DC are equal in length and have the same direction; equivalently the position equation gives equal opposite-side displacements.

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF Worked solutions

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 AB and AC. Show that MN= $\frac{1}{2}$ BC.
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 AB=DC.
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 AB and AC. Show that MN= $\frac{1}{2}$ BC.
3 marks
OM= $(a + b) / 2$ and ON= $(a + c) / 2$ .
Subtract OM from ON.
$\vec{M N} = \frac{1}{2} (c - b)$ .
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 AB=DC.
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 ↗

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