GCSE · Geometry and measures

Vectors and Vector Proof

A vector is a movement: how far across and how far up, in one package. Vectors let you add journeys, stretch them, reverse them, and at Higher tier prove geometric facts (that three points lie on a straight line, or that two lines are parallel) with algebra instead of a protractor.

GCSE subject content G24: describe translations as 2D vectors. G25: apply addition and subtraction of vectors, multiplication of vectors by a scalar, and diagrammatic and column representations of vectors; (Higher) use vectors to construct geometric arguments and proofs.

1 · What a vector is, and how to write it

You met vectors at KS3 as the way to describe a translation (a slide; see Transformations). A vector has a size and a direction, and it doesn't care where it starts. We write it three ways:

As a column vector: top number = movement right (negative = left), bottom number = movement up (negative = down). 3−2 means 3 right, 2 down.
As an arrow on a diagram: the arrowhead shows the direction.
As a letter: printed in bold, a, and handwritten underlined, a. The vector from point A to point B is written AB.

Drag the tail or the head of the arrow. Notice that moving the whole arrow doesn't change the vector: only its length and direction matter.

Column vector

In words

Common mistakes: writing the column vector upside down (it's always across on top, up underneath, like coordinates); and getting AB and BA confused: BA = −AB, same length, opposite direction.

2 · Adding vectors: one journey after another

To add a + b, do a and then b: put the tail of b on the head of a ("tip to tail"). The resultant goes straight from the start to the finish. In columns, just add the tops and add the bottoms. Doing b first and then a lands in the same place (the dashed route), so a + b = b + a.

Drag the heads of a (teal) and b (orange).

3 · Subtracting and multiplying by a number

Multiplying by a scalar (an ordinary number) k stretches the vector: 2a is twice as long in the same direction, ½a half as long. A negative number flips the direction: −a is a reversed. Subtracting is adding the reverse: a − b = a + (−b).

show k

4 · Magnitude: how long is a vector? Higher

The magnitude of a vector is its length, written |a|. The column vector gives a right-angled triangle (across and up), so Pythagoras finds the length. This isn't listed separately in the DfE content, but it's standard on the Higher route and simply uses Pythagoras.

a = (x, y) as a column → |a| = √(x² + y²)

5 · Parallel vectors

Two vectors are parallel when one is a multiple of the other. 6−9 = 3 × 2−3, so they're parallel. To check, divide matching components: if both give the same number, they're parallel. A negative multiple is still parallel, just pointing the opposite way.

6 · Paths in a diagram Higher

In geometry questions you're given two vectors, usually a and b, along two sides of a shape, and asked to write other lines in terms of them. The rule: find a route from the start to the end along lines you know, and add up the vectors on the way, using a minus sign when you travel backwards along an arrow.

Here OABC is a parallelogram with OA = a and OC = b. M is the midpoint of AB, and P is on OB with OP : PB = 1 : 2. Build a route by clicking points.

7 · Midpoints and ratios along a line Higher

If P lies on AB with AP : PB = m : n, then P is m/(m + n) of the way from A to B. So AP = m/(m+n) × AB, and to get the position vector of P go to A first: OP = OA + AP. For the midpoint (1 : 1) this gives OM = ½(a + b).

m n

Drag A and B.

Trap: a ratio 1 : 2 means P is one third of the way along, not one half. Add the parts: 1 + 2 = 3, so the fraction is 1/3.

8 · Vector proof: collinear points and parallel lines Higher

Two key facts:

Parallel lines: if PQ = k × RS for a number k, then PQ is parallel to RS (and k tells you how many times longer).
Collinear points (on one straight line): if PQ = k × PR, the lines are parallel and share the point P, so P, Q and R lie on a straight line. You must mention both facts.

Step through a full proof using the parallelogram from section 6: show that C, P and M are collinear.

Worked examples

a = (3, −1) and b = (−2, 4) as column vectors. Find a + b and 2a − b.

a + b = 3 + (−2)−1 + 4 = 13

2a = 6−2, so 2a − b = 6 − (−2)−2 − 4 = 8−6. (Subtracting a negative: 6 − (−2) = 8.)

Find the magnitude of the vector (5, −12). Higher

|v| = √(5² + (−12)²) = √(25 + 144) = √169 = 13.

Squaring removes the minus sign: (−12)² = 144.

The vector (k, 6) is parallel to (2, 3). Find k.

Parallel means (k, 6) = t × (2, 3) for some number t.

Bottom: 6 = 3t, so t = 2. Top: k = 2t = 4.

OABC is a parallelogram, OA = a, OC = b, M is the midpoint of AB. Find OM and CM. Higher

AB is parallel and equal to OC (opposite sides of a parallelogram), so AB = b and AM = ½b.

OM = OA + AM = a + ½b.

CM = CO + OM = −b + a + ½b = a − ½b.

OA = a, OB = b, and P is on AB with AP : PB = 1 : 3. Find OP. Higher

AB = AO + OB = −a + b = b − a.

P is 1/(1 + 3) = ¼ of the way from A: AP = ¼(b − a).

OP = a + ¼(b − a) = ¾a + ¼b.

Proof: in the parallelogram OABC above, P is on OB with OP : PB = 2 : 1. Show that C, P and M are collinear. Higher

OB = a + b, so OP = ⅔(a + b).

CP = CO + OP = −b + ⅔a + ⅔b = ⅔a − ⅓b = ⅓(2a − b).

CM = a − ½b (previous example) = ½(2a − b).

So CM = 3/2 × CP: the lines CP and CM are parallel, and they share the point C. Therefore C, P and M are collinear.

Check it matters: with OP : PB = 1 : 2 instead, CP = ⅓a − ⅔b, which is not a multiple of 2a − b, so the points would not be collinear. Always finish with both reasons: "multiple, so parallel" and "common point".

Quick check

In the quiz, column vectors are written sideways as (top, bottom) so they fit on a button.

What you've learned

A vector has size and direction. Column vector: across on top, up underneath. Bold a in print, underlined a by hand, AB from A to B, and BA = −AB.
Add tip to tail (add the components); subtract by adding the reverse; multiplying by k stretches, and a negative k reverses.
Higher: magnitude |a| = √(x² + y²) by Pythagoras.
Vectors are parallel when one is a multiple of the other.
Higher: write paths by going round known sides (minus for going backwards). A point dividing AB in the ratio m : n is m/(m + n) of the way along AB = b − a.
Higher proof: PQ = kRS means parallel; PQ = kPR means parallel with a common point, so collinear.