Vector notation, magnitude, direction, position vectors, vector arithmetic in 2D.
Answer at least 3 of 3 correctly to complete this section.
At GCSE, you may have met vectors briefly — adding arrows and perhaps some simple column vector work. At A-Level, vectors become a fundamental tool for geometry. Instead of working with coordinates and equations of lines, you work directly with directed quantities that have both magnitude and direction.
Vectors let you prove geometric results elegantly: that three points lie on the same line, that a point is a midpoint, that two lines are parallel. These proofs are a favourite of examiners and appear on almost every paper.
A 2D vector can be written as a column vector:
a=(a1a2)
This represents a displacement of a1 in the x-direction and a2 in the y-direction.
The standard unit vectors are:
i=(10),j=(01)
So (3−2)=3i−2j.
Drag points A and B to explore position and displacement vectors.
The magnitude (or modulus) of a vector a=(a1a2) is its length:
∣a∣=a12+a22
This comes directly from Pythagoras’ theorem.
A unit vector has magnitude 1. To find the unit vector in the direction of a:
a^=∣a∣a
You divide each component by the magnitude.
Find the magnitude of a=(3−4) and the unit vector in the direction of a.
∣a∣=32+(−4)2=9+16=25=5
a^=51(3−4)=(3/5−4/5)
Check: ∣a^∣=(3/5)2+(−4/5)2=9/25+16/25=25/25=1 ✓
Answer at least 3 of 3 correctly to complete this section.
The position vector of a point A is the vector from the origin O to A. If A=(3,5), then its position vector is:
OA=a=(35)
The vector from A to B is:
AB=b−a
Think of it as: “to go from A to B, go back to the origin (−a), then out to B (+b).”
Direction matters: AB=b−a but BA=a−b=−AB. Getting this the wrong way round is one of the most common errors.
Addition: (a1a2)+(b1b2)=(a1+b1a2+b2)
Subtraction: (a1a2)−(b1b2)=(a1−b1a2−b2)
Scalar multiplication: k(a1a2)=(ka1ka2)
Points A and B have position vectors a=(21) and b=(85). Find AB and ∣AB∣.
AB=b−a=(85)−(21)=(64)
∣AB∣=62+42=36+16=52=213
Find the position vector of the midpoint M of A and B where a=(21) and b=(85).
The position vector of the midpoint is:
m=21(a+b)=21((21)+(85))=21(106)=(53)
So M=(5,3).
Answer at least 3 of 3 correctly to complete this section.
Two vectors are parallel if one is a scalar multiple of the other:
b=kafor some scalar k
If k>0, they point in the same direction. If k<0, they point in opposite directions.
Three points A, B, C are collinear (lie on the same straight line) if:
Both conditions are needed. Two vectors can be parallel without the points being on the same line.
Exam trap: Showing that AB is parallel to CD does NOT prove collinearity — you need the vectors to share a point. For example, AB=2AC proves A, B, C are collinear because they share point A.
OABC is a quadrilateral. OA=a, OC=c, and AB=c. Prove that CB is parallel to OA.
First, find OB:
OB=OA+AB=a+c
Now find CB:
CB=CO+OB=−c+(a+c)=a
Since CB=a=OA, the vectors are equal (same direction, same magnitude). In particular, CB is parallel to OA. In fact, OABC is a parallelogram. □
Points A, B, and C have position vectors a=(12), b=(48), and c=(612). Show that A, B, C are collinear.
AB=b−a=(36)
AC=c−a=(510)
Now check: is AC a scalar multiple of AB?
AC=(510)=35(36)=35AB
Since AC=35AB and both vectors start from A, the points A, B, C are collinear. □
Answer at least 3 of 3 correctly to complete this section.
Point P divides AB in the ratio 2:1, where a=(13) and b=(79). Find the position vector of P.
P divides AB in the ratio 2:1, so:
AP=32AB
AB=b−a=(66)
OP=a+AP=a+32AB=(13)+32(66)=(13)+(44)=(57)
In triangle OAB, M is the midpoint of AB. Given OA=a and OB=b, find OM and show that M lies on the line from O through the centroid.
OM=OA+AM=a+21AB=a+21(b−a)=21a+21b=21(a+b)
This is the standard result: the position vector of the midpoint of AB is 21(a+b).
| Question type | Approach |
|---|---|
| ”Find AB“ | Calculate b−a |
| “Find the midpoint of AB“ | Position vector =21(a+b) |
| “Show X, Y, Z are collinear” | Show XY=kXZ for some scalar k |
| “Prove AB is parallel to CD“ | Show AB=kCD |
| ”P divides AB in ratio m:n” | OP=a+m+nmAB |
Answer at least 4 of 5 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Equation of a straight line, gradient, midpoint, distance between two points, parallel and perpendicular lines.
Extending vector operations to three dimensions, distance in 3D.
Using vectors to prove geometric properties, collinearity, ratios on line segments.