Topic 10 covers vectors in two and three dimensions: column and i, j, k notation, magnitude and direction, adding and scaling vectors, unit vectors, position vectors and the distance between two points in 3D, testing for parallel vectors and collinear points, dividing a line in a given ratio, and using vectors to solve geometric problems.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
Vectors · notation
Vectors in two and three dimensions
A vector has magnitude and direction; a scalar has magnitude only. In 3D we use the mutually perpendicular unit vectors i, j, k along the x, y and z axes.
A 3D vector and its magnitude — Pythagoras, applied twice.
a = xi + yj + zk · |a| = √(x² + y² + z²)the column form is the same thing written vertically
Exam trap: the magnitude is never negative — square the components, then take the positive root. |2i − 3j + 6k| = √(4 + 9 + 36) = 7, not √(4 − 9 + 36).
Calculate
Your turn — magnitude in 2D
1Find the magnitude of the vector 3i − 4j.
Set-up
|3i − 4j| = √(9 + 16) = √25 = 5.
Hint: √(3² + (−4)²) = √25.
Calculate
Your turn — magnitude in 3D
2Find the magnitude of the vector a = 2i − 3j + 6k.
Set-up
|a| = √(4 + 9 + 36) = √49 = 7.
Hint: √(2² + 3² + 6²) = √49.
Quick check
Magnitude in 3D
?What is |xi + yj + zk|?
Vectors · algebra
Adding, scaling and unit vectors
Adding: add the components. (2i + j) + (3i − 4j) = 5i − 3j. Geometrically it is the triangle (or parallelogram) law.
Scaling: λa has the same direction as a (opposite if λ < 0) and magnitude |λ||a|.
Parallel: b is parallel to a exactly when b = λa for some scalar λ. So 4i − 2j + 6k = 2(2i − j + 3k) is parallel to 2i − j + 3k.
Unit vector: divide by the magnitude — â = a / |a|, which has magnitude 1.
Worked example — the unit vector in the direction of a = 2i − 3j + 6k
3Find the i-component of the unit vector in the direction of a = 2i − 3j + 6k. (Give a decimal to 4 d.p., or the fraction 2/7.)
Set-up
â = (1/7)(2i − 3j + 6k), so the i-component is 2/7 = 0.2857 (4 d.p.).
Hint: Divide each component by |a| = 7.
Quick check
Parallel vectors
?Which vector is parallel to 2i − j + 3k?
Vectors · direction
Magnitude and direction form
In 2D you can convert between component form and magnitude–direction form.
|v| = √(x² + y²) · tan θ = y/xθ is measured anticlockwise from the positive x-axis — sketch it to get the correct quadrant
Worked example — v = 3i + 4j
|v| = √(9 + 16) = 5.
θ = arctan(4/3) = 53.1° (1 d.p.) above the positive x-axis. Both components are positive, so the vector really is in the first quadrant. ✓
Exam trap: arctan only ever returns an angle between −90° and 90°. For v = −3i − 4j the calculator still says 53.1°, but the true direction is 53.1 + 180 = 233.1°. Always sketch.
Calculate
Your turn — direction
4The vector v = 3i + 4j makes an angle θ with the positive x-axis. Find θ in degrees to 1 decimal place.
Set-up
tan θ = 4/3 = 1.3333
θ = arctan(1.3333) = 53.130…° = 53.1° (1 d.p.), in the first quadrant.
°
Hint: tan θ = 4/3.
Sort it
Sort by magnitude
Tap a vector, then its magnitude. Square the components and add — no calculator needed.
5️⃣ Magnitude 5
7️⃣ Magnitude 7
1️⃣3️⃣ Magnitude 13
Vectors · position
Position vectors and distance
The position vector of A is OA = a, measured from the origin. Then:
AB = b − a“destination minus start” — the single most useful line in the topic
The distance between two points in 3D is the magnitude of that vector:
d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
Worked example — A(1, 2, 3) and B(4, 6, 15)
AB = b − a = (4 − 1)i + (6 − 2)j + (15 − 3)k = 3i + 4j + 12k.
|AB| = √(9 + 16 + 144) = √169 = 13.
Exam trap: AB = b − a, not a − b. Getting it backwards gives a vector of the right length pointing the wrong way — fine for a distance, fatal for a direction or a ratio.
Calculate
Your turn — distance in 3D
5A is the point (1, 2, 3) and B is (4, 6, 15). Find the exact distance AB.
The point P dividing AB in the ratio m : n has OP = a + [m/(m + n)](b − a).
Collinear points: A, B, C are collinear if AB is a scalar multiple of BC (parallel and sharing the point B).
Worked example — A(1, −2, 4), B(5, 2, −4), and P divides AB in the ratio 1 : 3
AB = b − a = 4i + 4j − 8k.
OP = a + ¼(AB) = (1 + 1)i + (−2 + 1)j + (4 − 2)k = 2i − j + 2k, so P is (2, −1, 2).
|OP| = √(4 + 1 + 4) = √9 = 3.
Exam trap: the ratio 1 : 3 means P is one quarter of the way from A to B (1 part out of 1 + 3 = 4), not one third. Count the total parts.
Calculate
Your turn — ratio and magnitude
6A is (1, −2, 4) and B is (5, 2, −4). The point P divides AB in the ratio 1 : 3. Find the distance OP from the origin.
Set-up
b − a = 4i + 4j − 8k, so ¼(b − a) = i + j − 2k.
OP = (i − 2j + 4k) + (i + j − 2k) = 2i − j + 2k, so P = (2, −1, 2).
|OP| = √(4 + 1 + 4) = √9 = 3.
Hint: OP = a + ¼(b − a). Find P first, then its magnitude.
Quick check
Collinear points
?How do you show that the points A, B and C are collinear?
Match it
Vector flashcards
Tap a card on the left, then its meaning.
Object
Meaning
Vectors · in context
Using vectors in context
Vectors are the language of mechanics (Paper 3): forces, velocities and displacements all add as vectors.
Worked example — resultant force
Two forces act on a particle: F₁ = (3i + 4j) N and F₂ = (−i + 2j) N.
Resultant R = F₁ + F₂ = 2i + 6j N.
|R| = √(4 + 36) = √40 = 6.32 N, at arctan(6/2) = 71.6° from the i-direction.
A particle in equilibrium has resultant force 0, so the i-components sum to zero and the j-components sum to zero — two equations.
s = ut + ½at² works exactly the same way with vector u and a.
Not on 9MA0: the scalar (dot) product and the vector equation of a line belong to Further Mathematics. If a question seems to need the angle between two general vectors, you are looking at the wrong specification.
Recap
The big ideas to take away
Notation: a = xi + yj + zk in 3D; |a| = √(x² + y² + z²)
Unit vector: â = a/|a| — magnitude 1
Parallel: b = λa · Collinear: AB = λBC through a shared point
Position vectors: AB = b − a; midpoint ½(a + b); ratio m : n ⇒ a + m/(m+n) (b − a)
Distance in 3D: √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
In context: forces and velocities add as vectors; equilibrium ⇒ each component sums to 0
Topic 10 is short but heavily examined — and it is the bridge into the Mechanics paper. Press Finish to see your score.
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Mini-lesson complete!
⭐⭐⭐
You have worked through Vectors for Edexcel A-level Mathematics (9MA0). 🎉