A vector is a quantity with both magnitude and direction, such as displacement, velocity or force. A scalar has magnitude only. This sub-topic sets out the notation used throughout H2 Vectors and the basic operations everything else relies on. These are addition, subtraction, scalar multiples, position vectors, magnitude and unit vectors, and each works the same way in 2D and 3D.
Every later result in H2 Vectors, on products of vectors, lines and planes, is built from position vectors, collinearity and the ratio theorem. A-level vectors questions very often open with a ratio-theorem or collinearity part, and the written justification is often where the marks are.
What these notes cover
- •Scalars, vectors and notation
- •Equal, negative and zero vectors
- •Addition: triangle and parallelogram laws
- •Subtraction and scalar multiplication
- •Laws of vector algebra and route-finding
- •Position, displacement and direction vectors
- •Magnitude and unit vectors
- •Parallel vectors and collinearity
- •Comparing coefficients
- •The ratio theorem and lines produced
- •Centroid and parallelogram applications
- •Common mistakes and exam technique
Key definitions
- Vector
- A quantity with both magnitude and direction, such as displacement, velocity or force. Geometrically it is a directed line segment: its length represents the magnitude and its arrowhead shows the direction.
- Position vector
- A vector that starts at the origin O, so it is tied to one particular point. The point A(1, −2, 4) has position vector OA→ = i − 2j + 4k. A point is a location and its position vector is an arrow, so a point never equals a vector.
- Direction vector
- Any non-zero vector that points the way you need. Only its direction matters, so any non-zero scalar multiple does the same job and it may be rescaled freely. A position vector must never be rescaled.
- Unit vector
- A vector with magnitude exactly 1, found by dividing a non-zero vector by its own magnitude.
- Collinear points
- Points that lie on one straight line. Showing this needs two parallel vectors and a common point.
Key formulas
Displacement vector
End minus start: the point where the arrow finishes comes first. Reversing the arrow flips the sign, so BA→ = −AB→.
Magnitude
For a = xi + yj + zk. In 2D, drop the z term. Every component is squared, so negative components still add to the length. The distance between two points is AB = |b − a|.
Unit vector
a must be non-zero. A vector of length L parallel to a is ±Lâ. If the question says 'in the direction of a', only +Lâ is accepted.
Collinearity test
For some scalar k. The equation shows that AB→ is parallel to AC→, and A is a common point, so A, B and C lie on one line. Two non-zero vectors are parallel when a = kb with k ≠ 0.
Ratio theorem
P lies on the line segment AB with AP : PB = λ : μ. The weights cross over: a takes μ and b takes λ. With λ = μ = 1, the midpoint of AB is (a + b) ÷ 2.
Comparing coefficients
This holds only when a and b are non-zero and non-parallel. State that condition before comparing. It is mainly used to find where two lines meet.
Worked example
The points A, B and C have coordinates (1, −2, 3), (3, 1, 2) and (7, 7, 0) respectively. Show that A, B and C are collinear, and find the ratio AB : BC.
- 1Use two displacement vectors that share the point A: AB→ = OB→ − OA→ = (3i + j + 2k) − (i − 2j + 3k) = 2i + 3j − k.
- 2Similarly, AC→ = OC→ − OA→ = (7i + 7j) − (i − 2j + 3k) = 6i + 9j − 3k = 3(2i + 3j − k).
- 3So AC→ = 3AB→, which means AC→ is parallel to AB→.
- 4Since AC→ is parallel to AB→ and A is a common point, A, B and C are collinear.
- 5Since k = 3 > 0, B lies between A and C, and BC→ = AC→ − AB→ = 3AB→ − AB→ = 2AB→.
- 6Hence AB : BC = 1 : 2.
Watch out: Parallel alone does not prove collinearity, so the conclusion must state that A is a common point. Here AB : AC = 1 : 3, not 1 : 2, so always check which ratio the question asks for.
Common mistakes
Displacement the wrong way round
Wrong: AB→ = a − b
Right: AB→ = OB→ − OA→ = b − a. The rule is always 'end minus start'.
Ratio theorem weights not crossed over
Wrong: AP : PB = 2 : 3 gives OP→ = (2a + 3b) ÷ 5
Right: OP→ = (3a + 2b) ÷ 5. The coefficients cross over. P is nearer A, so a gets the bigger weight.
Comparing coefficients with no justification
Wrong: λa + 2b = 3a + μb, so λ = 3 and μ = 2, with no reason given
Right: First write 'since a and b are non-zero and non-parallel'. Without this line the comparison is not justified and the mark is withheld.
Questions students ask
How do you show three points are collinear using vectors?
Show that AB→ = k AC→ for some scalar k, so the two vectors are parallel. Then state that A is a common point. Parallel alone only proves the lines are parallel, because they could be two separate lines. The common point is what puts all three points on the same line.
What is the ratio theorem in H2 Math vectors?
If P lies on the line segment AB with AP : PB = λ : μ, then OP→ = (μa + λb) ÷ (λ + μ). The weights cross over, so the coefficient of a is μ, which is the part of the ratio next to B. Taking λ = μ = 1 gives the midpoint (a + b) ÷ 2.
How do you find a unit vector in 3D?
Divide the vector by its own magnitude: â = a ÷ |a|, where |a| = √(x² + y² + z²). For AB→ = 2i + 6j − 3k, |AB→| = √49 = 7, so the unit vector is (2i + 6j − 3k) ÷ 7. Check it by squaring the components and adding; the total must be 1.
What does 'AB produced' mean in vectors?
'AB produced' means the line AB extended beyond B, and 'BA produced' means the line extended beyond A. The ratio theorem as stated needs the point to lie between the two endpoints. So sketch the line, identify the middle point, apply the ratio theorem to that point, and rearrange for the unknown vector.
In the full PDF
- •Labelled diagrams, including the i, j, k cuboid and the intersecting-lines set-up with O, A, B, C, D and X
- •Three further full worked examples: magnitude and unit vectors in 3D, parallel vectors and collinear points, and an exam-style intersecting-lines question
- •Worked examples on finding where two lines meet by comparing coefficients and on a point on a line produced
- •Geometrical applications: the centroid of a triangle, the diagonals of a parallelogram and the parallelogram condition
- •The remaining common mistakes and a six-point exam technique checklist
- •A summary of every key result in one place