Geometry · GCSE Maths
Vectors
GCSE Maths vectors: add column vectors, find a resultant or midpoint, and write a proof-style “show that” argument with directed segments such as AB = b − a.
A vector has magnitude and direction. AB = b − a if a and b are position vectors of A and B. To show two vectors are parallel, show one is a scalar multiple of the other.
The important bits
What you need to know
- 1
A column vector (3 / −2) means 3 right and 2 down. Add vectors by adding components: (3 / −2) + (1 / 4) = (4 / 2). Subtract by subtracting components.
- 2
Multiplying by a scalar stretches or reverses: 3(2 / −1) = (6 / −3). A negative scalar reverses direction. Parallel vectors are scalar multiples, including a negative multiple for opposite directions.
- 3
If a and b are position vectors from a common origin O, then vector AB = b − a, and vector BA = a − b. Midpoint M of AB has position vector (a + b) / 2.
- 4
The resultant of two vectors is their sum, nose to tail. The resultant of a walk 3 east then 4 north is the hypotenuse 5 if you need magnitude, or the column (3 / 4) if you need the vector.
- 5
Magnitude of (x / y) is √(x² + y²). A unit vector in the same direction is the vector divided by its magnitude.
- 6
Proof-style questions: express each path in terms of a and b, simplify, and compare. To show M is the midpoint, show OM = (OA + OB) / 2, or show AM = MB as vectors.
- 7
To show that three points are collinear, show that two segments along the line are parallel (scalar multiples) and share a point.
- 8
Vector geometry uses the same letters as geometry proofs: write a reason, “AB = b − a”, then the next line. A correct column without a path often misses the “show that” mark.
Quotations worth analysing
Short evidence. Real method.
“AB = b − a”
This is the line that turns a diagram into algebra. Students who write AB = a − b have reversed the journey and will not show parallelism correctly.
“Show that”
Start from the given position vectors, write each required vector as a combination of a and b, and finish with a sentence: “this is a scalar multiple of …, so parallel”.
“Parallel vectors are scalar multiples”
PQ = k × RS, with k a number (possibly negative or a fraction). Equal vectors have k = 1 and the same direction. Opposite direction is k < 0.
Go deeper
Column arithmetic is the easy third of the paper
Add, subtract, multiply by a scalar, find magnitude. (4 / −1) + (−2 / 5) = (2 / 4). 3(4 / −1) − 2(−2 / 5) = (12 / −3) + (4 / −10) = (16 / −13). Magnitude of (3 / −4) is 5. These are method-mark friendly if you write the columns lined up. The resultant of two journeys is the single column that starts where they started and ends where they finished. On a vector diagram, draw them nose to tail; the resultant is the shortcut. A question that asks for “the vector that takes A to B” wants AB, not BA, and not the magnitude unless it says length or distance. If it asks for a bearing or an angle, you have left pure column work and need tan⁻¹ of the components, watching which quadrant the vector sits in.
Go deeper
Show that: write a path, then simplify
O is the origin, OA = a, OB = b, P is the midpoint of OA, Q divides OB in the ratio 1:2 (OQ = (1/3)b). Show that PQ is parallel to a line from A to a point R on OB with OR = (2/3)b, or similar. Method: write every vector from the letters. PQ = Q − P = (1/3)b − (1/2)a. AR = R − A = (2/3)b − a. These two are not obviously multiples. A typical successful item is: M midpoint of AB, show OM = (a + b)/2. OM = OA + (1/2)AB = a + (1/2)(b − a) = a + (1/2)b − (1/2)a = (1/2)a + (1/2)b. That chain is the “show that”. Each equality needs a reason: “M midpoint so AM = (1/2)AB” or “triangle law”. Finish with “therefore M has position vector (a + b)/2”. The last sentence converts algebra back into geometry.
Go deeper
Midpoints, ratios and collinearity
Section formula: a point dividing AB in the ratio m:n has position vector (n a + m b) / (m + n). For a midpoint, m = n = 1, so (a + b)/2. For 2:1 towards B, the position is (1 a + 2 b) / 3. Check the way round: if the first part is 0, the point should be at A. Collinear A, P, B: show AP = k AB for some k. If 0 < k < 1, P is between A and B; if k = 2, P is beyond B. Parallel is not collinear: AB = 2 CD can sit on a different line; you need a shared point. To show a parallelogram, show PQ = SR as vectors — same magnitude and direction, which is stronger than parallel.
See the idea in action
OA = a, OB = b. M is the midpoint of AB. N is the point on OA such that ON = (1/3)a. Find MN in terms of a and b, and show that MN is not parallel to OB unless a is a multiple of b. Step 1: AB = b − a. M midpoint, so OM = a + (1/2)(b − a) = (1/2)a + (1/2)b. Step 2: ON = (1/3)a, so position vector of N is (1/3)a. Step 3: MN = OM − ON = (1/2 a + 1/2 b) − (1/3 a) = (1/6)a + (1/2)b. Step 4: MN = (1/6)a + (1/2)b, which is a multiple of b alone only if the a-component is 0, i.e. if a is the zero vector. So MN is not parallel to b in general. Check: if you only needed MN, the three lines OM, ON, subtract are the whole method.
Exam technique
Turn knowledge into marks
For “show that”, start from position vectors, write AB = b − a as an early line, and end with a sentence: parallel because one is a scalar multiple of the other.
Common mistakes
Do not give these marks away
- 01
Writing AB = a − b, which is the journey from B to A, then claiming a midpoint or a parallelogram that faces the wrong way.
- 02
Showing two vectors are parallel but forgetting the concluding sentence the mark scheme lists as “so PQ is parallel to SR”.
- 03
Adding magnitudes instead of adding columns, or giving a length when the question asked for a vector.
If OA = a and OB = b, then vector AB is
Ab − a
Ba − b
Ca + b
D(a + b) / 2
Show the answer
b − a. AB = (position of B) − (position of A) = b − a. a − b is BA. a + b is not a side of the triangle. (a + b)/2 is the position vector of the midpoint of AB.
Quick questions
If this is the bit you searched
How do you add column vectors?
Add the top numbers and add the bottom numbers separately. (3 / −2) + (1 / 4) = (4 / 2). Subtract by subtracting components.
What does AB = b − a mean?
If a and b are the position vectors of A and B from O, the journey from A to B is the position of B minus the position of A.
How do I show two vectors are parallel?
Show that one is a scalar multiple of the other, including a negative scalar for opposite directions. Then write that conclusion in words.
How do you find the midpoint using vectors?
The midpoint M of AB has position vector (a + b) / 2. Equivalently, OM = OA + (1/2)AB = a + (1/2)(b − a).