• A vector has both magnitude (size) and direction — unlike a scalar, which is just a number.
• Vectors can be written as column vectors (x, y) or using unit vectors i and j as xi + yj.
• The magnitude (length) of a vector is found using Pythagoras' theorem: |v| = √(x² + y²).
• A unit vector has magnitude 1 — you find one by dividing a vector by its own magnitude.
• Position vectors describe where a point is relative to the origin O. Displacement vectors describe the journey between two points.
• To add or subtract vectors, add or subtract their matching components (top with top, bottom with bottom).
• Multiplying a vector by a scalar changes its size (and flips its direction if the scalar is negative) but keeps it parallel to the original.
• Vectors are used to solve geometry problems: proving lines are parallel, finding midpoints, proving points are collinear, and working with ratios on a line.
• Vectors can represent velocity too — and position at time t is found using r = r₀ + vt.

Think of a vector as a set of travel directions, not a place. "Walk 3 steps east and 7 steps south" is a vector — it tells you a magnitude (how far) and a direction (which way). Compare that to a scalar, which is just a plain number with no direction attached — like "it's 20°C outside" or "the bag weighs 5 kg."

A classic example from the chapter: two objects moving at 7 m/s and −7 m/s are travelling at the same speed (7 m/s each) but in opposite directions. Speed doesn't care about direction — velocity does. That's the difference between a scalar and a vector in a nutshell.

Vectors don't just appear in "arrow on a grid" questions — you've already met them when translating graphs (shifting a curve left/right and up/down is literally applying a vector), and they show up constantly in mechanics to model forces, displacement, velocity, and acceleration.

There are two main notations you need to be fluent switching between:

If you know the start point (A) and end point (B), you can write the vector as AB with an arrow drawn above it, showing it goes specifically from A to B. Note that AB and BA are not the same — they point in opposite directions, so BA = -AB.

Vectors can also be written using a single bold lowercase letter, like a or b. Since you can't write in bold in an exam, you should underline it instead: a.

The magnitude tells you the vector's length — literally how far the "journey" travels in a straight line. Because the x and y components form the two shorter sides of a right-angled triangle, you find the magnitude using Pythagoras' theorem.

One detail examiners like to test: if a component is negative (like −6j), you can ignore the minus sign when squaring it, since (−6)² = 36 either way. If the question asks for an exact value, leave your answer as a simplified surd rather than rounding it.

A unit vector is any vector with a magnitude of exactly 1. The vectors i and j you've been using this whole time are themselves unit vectors — i points 1 unit in the positive x-direction, and j points 1 unit in the positive y-direction. They're perpendicular to each other, which is why they work so well as building blocks for any 2D vector.

To find the unit vector in the same direction as a given vector, simply divide that vector by its own magnitude. This shrinks (or stretches) it down to a length of exactly 1, while keeping it pointing the same way.

A position vector is a special type of vector that always starts at the origin O and ends at a specific point. The position vector of point A is written a = OA. Because it always starts at the same fixed place (the origin), a position vector effectively tells you the exact coordinates of a point — if a = 3i − 2j, then point A sits at coordinates (3, −2).

This is different from a displacement vector like AB, which just describes the journey between two points and doesn't care where the origin is.

Here's the key relationship that connects the two: if you know the position vectors of A and B (call them a and b), you can find the displacement vector between them by "going via the origin" — travel backwards from A to O, then forwards from O to B:

This single idea is one of the most useful tools in the whole chapter — you'll use it constantly for distance, midpoint, and collinearity problems.

Adding vectors describes the total journey when you follow one path, then another — you combine the two "trips" into a single overall trip called the resultant vector. To add or subtract vectors numerically, you simply add or subtract the matching components — top with top, bottom with bottom (or i-terms with i-terms, j-terms with j-terms).

Geometrically, if you draw vector a, then draw vector b starting from where a ended ("nose to tail"), the resultant a + b is the single arrow going directly from the start of a to the end of b — the shortcut route.

Subtracting a vector is exactly the same as adding its negative. So a − b = a + (−b), where −b is simply b with both components flipped in sign (same length, opposite direction).

One neat fact: adding PQ and QP together always gives the zero vector, written as a bold 0 — makes sense, since going from P to Q and then straight back to P leaves you exactly where you started.

Two vectors are parallel if and only if one is a scalar multiple of the other — meaning every component has been multiplied by the same constant number, k.

Multiplying a vector by a positive scalar stretches or shrinks it (changes its magnitude) but keeps it pointing the same way. Multiplying by a negative scalar also flips the direction — the arrow now points the opposite way.

For example, a, 2a, ½a, and −3a are all parallel to each other — they all sit on the same line of direction, just at different lengths (and some reversed).

Building on position vectors: if OA = a and OB = b, then AB = b − a. This "go via the origin" trick applies to any three points, not just when one of them is the origin — you can always find a missing vector by adding up known vectors along any valid path between the same two points.

Most exam problems on vectors fall into one of these categories:

• Proving two vectors/lines are parallel — show one is a scalar multiple of the other (a = kb).
• Finding a midpoint — using position vectors.
• Proving three points are collinear (all lie on the same straight line).
• Geometric shape problems — often parallelograms or rhombuses, frequently involving points that split a line in a given ratio.

If A has position vector a and B has position vector b, the midpoint M of AB has a beautifully simple position vector — just the average of the two.

Why this works: AB = b − a (as we found earlier). The midpoint is halfway along that journey, so AM = ½AB = ½(b − a). Starting from the origin: OM = OA + AM = a + ½(b − a), which simplifies to ½(a + b).

Collinear points are points that all sit on exactly the same straight line. There are two standard methods to prove three points A, B, C are collinear:

Method 1: Show that AB = k·AC for some constant k. Since the vectors are scalar multiples, they point in the same direction (parallel) — and because they both start from the same point A, that means B and C must lie on the same line through A. Collinear confirmed.

Method 2: Show that AB = k·BC and that point B lies on both vectors AB and BC (i.e., B is a shared endpoint). This works the same way but pivots around the middle point.

Which method to use just depends on which vectors the question gives you easy access to.

These questions usually give you a grid or shape made of parallelograms, define two "base" vectors (often a and b), and ask you to describe other vectors in the diagram in terms of them. The trick is recognising that many other vectors in the picture are equal to, parallel to, or a scalar multiple of your two known vectors.

An important distinction: a specific vector like AB refers only to the arrow starting exactly at A and ending exactly at B. A general vector like a refers to any vector of the same length and direction anywhere on the diagram — so if RS happens to be the same length and direction as AB, then RS = a too, even though R and S are different points entirely.

To find any unknown vector between two points on the grid, just pick any valid path between them (going along the grid lines) and add up the known vectors along that path — exactly like the "journey via the origin" idea, just generalised to any route.

Everything you've learned about "journey" vectors applies directly to velocity — a vector describing how fast something moves and in which direction. A velocity vector like (3, 7) m/s means: moving at 3 m/s in the positive x-direction and 7 m/s in the positive y-direction, simultaneously.

Speed is different from velocity. Speed is a scalar (just a number — "how fast," no direction). Velocity is a vector (speed and direction). Since speed is the magnitude of velocity, you find it exactly the same way you find the magnitude of any vector — Pythagoras' theorem.

If an object starts at position vector r₀ and moves with constant velocity v, then after t seconds its new position is:

For example, a particle starting at (2, −3) moving with constant velocity (4i + 7j) m/s will, after 10 seconds, be at:

Sometimes an object is influenced by more than one velocity at once — like a boat pushed by its engine and a water current. The overall movement is the resultant velocity, found simply by adding the individual velocity vectors together.

Collision problems are a favourite exam pattern: two objects move with different (constant) velocities from different starting points. They collide at time t exactly when their position vectors become equal — meaning both the i-components and j-components must match simultaneously. This gives you two equations (or one equation you can solve, then check with the other component) to find t.

• I can explain the difference between a scalar and a vector
• I can convert between column vector notation and i, j notation
• I can find the magnitude of a vector using Pythagoras' theorem
• I can find a unit vector in the direction of a given vector
• I understand the difference between a position vector and a displacement vector
• I can find the vector AB using AB = b − a
• I can add, subtract, and scalar-multiply vectors in both notations
• I can prove two vectors are parallel using a = kb
• I can find the midpoint of two points using position vectors
• I can prove three points are collinear (and explain why parallel + shared point is required)
• I can solve geometric shape problems by finding paths using known vectors
• I can find the resultant of two or more velocity vectors
• I can distinguish speed (scalar) from velocity (vector)
• I can use r = r₀ + vt to find a particle's position at a given time
• I can solve collision problems by equating two position vector expressions