To learn Vectors quickly and correctly, start by thinking about movement with both size and direction. A vector can describe a journey five units east, a movement three units up and two units left, or the displacement between two points. Unlike an ordinary number, a vector needs direction as well as magnitude. These eduKateSG Mathematics Quick Notes teach how to read coordinate components, add and subtract vectors, find vector lengths and check your result with a simple arrow sketch.
- Read the components: (x, y) means horizontal displacement x and vertical displacement y.
- Keep direction: positive x is right, negative x is left; positive y is up, negative y is down.
- Add vectors: add corresponding components.
- Subtract vectors: subtract corresponding components.
- Scale a vector: multiply each component by the same scalar.
- Find magnitude: use the Pythagorean relationship √(x² + y²).
- Check: sketch an arrow and test the final displacement.
First, read a vector as a movement rather than merely a coordinate point. Let v = (3, 4) represent a displacement of three units right and four units up. On a square grid, draw an arrow that begins anywhere and ends three squares right and four squares higher. The arrow can move to a new starting point without changing the vector, provided its length and direction stay the same. A point (3, 4) represents a location, while a displacement vector (3, 4) describes a change of position. The same two numbers can be used in both settings, but the meaning depends on the question.
Second, add movements component by component. Suppose your first displacement is u = (4, 1), meaning four right and one up. Your next displacement is v = (−2, 3), meaning two left and three up. The total is u + v = (4 + (−2), 1 + 3) = (2, 4). You finish two units right and four units above your starting position. This is the head-to-tail rule: place the second arrow’s tail at the first arrow’s head, then draw the resultant from the original start to the final end. Addition is not about adding the lengths alone.
Third, use subtraction and scaling to describe the reverse or a repeated movement. With u = (4, 1) and v = (−2, 3), subtracting gives u − v = (6, −2). The negative of v is −v = (2, −3), which points in the opposite direction but has the same magnitude. Doubling u gives 2u = (8, 2), an arrow twice as long pointing the same way. Multiplying by a negative scalar reverses the direction as well as changing the length. These rules work because each component describes an independent signed displacement.
Worked example: find the vector between two points
Let A = (−2, 1) and B = (4, 9). The displacement from A to B is final position minus initial position: AB = B − A = (4 − (−2), 9 − 1) = (6, 8). The arrow goes six units right and eight units up. Reversing the direction gives BA = (−6, −8). Notice that AB and BA are not the same vector: their magnitudes match, but they point in opposite directions. Label the start and end carefully before subtracting coordinates.
Worked example: magnitude of a vector
Take v = (6, 8). Its horizontal and vertical components form the perpendicular sides of a right triangle. The length, or magnitude, of v is |v| = √(6² + 8²) = √(36 + 64) = √100 = 10 units. This is a direct application of Pythagoras’ theorem. The magnitude of (−6, −8) is also 10 because squaring removes the component signs. A magnitude can never be negative, even though individual vector components can be.
Worked example: translate a point
A point P = (−1, 6) is translated by displacement vector (3, −4). Add the vector’s components to the point’s coordinates: new x = −1 + 3 = 2; new y = 6 − 4 = 2. Thus the image point is P′ = (2, 2). You can check by drawing an arrow from (−1, 6) to (2, 2): it moves three to the right and four down. This is how vectors connect to coordinate geometry and transformations.
Worked example: several journeys in one
A traveller moves 4 units east and 3 units north, then 1 unit west and 7 units south. The first displacement is (4, 3) and the second is (−1, −7). Add them to find the total displacement: (4 − 1, 3 − 7) = (3, −4). The traveller finishes three units east and four units south of the start. The magnitude of the displacement is √(3² + (−4)²) = 5 units. This is different from the total distance travelled along the two stages, which is √(4² + 3²) + √(1² + 7²) = 5 + √50 units if both stages are straight.
Vector addition, explained by a triangle
If a journey goes from point A to B and then from B to C, its overall displacement is AB + BC = AC. Draw the arrows so the second starts where the first ends. This geometric relationship is the triangle rule for vectors. It also explains why switching from AB to BA changes the direction: AB + BA = (0, 0), the zero vector. You have travelled from A to B and then back to A, so the total displacement is zero, even though some distance was travelled.
How to tell whether two vectors are parallel
Two non-zero vectors are parallel when one is a scalar multiple of the other. For example, (2, 3) and (4, 6) point in the same direction because (4, 6) = 2(2, 3). The vectors (2, 3) and (−4, −6) lie on parallel directions but point oppositely because (−4, −6) = −2(2, 3). If two components have inconsistent scaling, such as (2, 3) and (4, 5), they are not parallel. Compare both components instead of one.
How to find a unit vector
A unit vector has magnitude 1. To obtain a unit vector in the direction of a non-zero vector v, divide each component by its magnitude: v/|v|. For v = (3, 4), its length is 5, so the unit vector is (3/5, 4/5). Check its magnitude: √((3/5)² + (4/5)²) = √(9/25 + 16/25) = √1 = 1. This idea helps separate a movement’s direction from its length.
A quick vector calculation guide
- u + v: add x-components, then y-components.
- u − v: subtract x-components, then y-components.
- ku: multiply both components of u by k.
- Vector AB: subtract coordinates A from coordinates B.
- Magnitude: |(x, y)| = √(x² + y²).
- Translate point P: add the displacement vector to P’s coordinates.
- Parallel non-zero vectors: one is a scalar multiple of the other.
Common mistakes that spoil vector answers
- Confusing a point with a displacement: points are locations; vectors are directed changes.
- Subtracting B from A when finding AB: end minus start gives the correct direction.
- Adding vector magnitudes instead of components: use coordinate addition for the resultant.
- Dropping a negative component: left and down displacements are negative in the standard coordinate convention.
- Claiming a negative magnitude: lengths are non-negative.
- Calling two vectors equal because their lengths match: equal vectors must also have the same direction.
- Confusing distance travelled with displacement: a return journey can have zero displacement but positive distance.
Mini-test: answer from the vector rules
- Let u = (3, −2) and v = (−1, 5). Find u + v.
- For the same vectors, find u − v.
- Find the magnitude of (−5, 12).
- Find the vector from A = (2, −1) to B = (7, 3).
- Translate P = (4, 2) by displacement (−6, 3).
Answers and explanations
1. u + v = (3 − 1, −2 + 5) = (2, 3). 2. u − v = (3 − (−1), −2 − 5) = (4, −7). 3. √(25 + 144) = √169 = 13. 4. B − A = (7 − 2, 3 − (−1)) = (5, 4). 5. The translated point is (4 − 6, 2 + 3) = (−2, 5). Sketch each movement to check the signs.
Your ten-minute vector learning routine
- Minutes 1–2: draw an arrow for a vector with one negative component.
- Minutes 3–4: add and subtract two vectors component by component.
- Minutes 5–6: find a displacement between two labelled points.
- Minutes 7–8: calculate a magnitude using Pythagoras’ theorem.
- Minutes 9–10: draw a translation and explain the difference between distance and displacement.
The strongest vector habit is to draw or describe the direction first, then calculate the components. If every step describes a clear movement, vector arithmetic becomes a reliable language for geometry rather than a collection of signs.
Continue with related Mathematics Quick Notes
Build your next skill with Coordinate Graphs · Pythagoras’ Theorem · Trigonometric Ratios · Mathematics Word Problems. Return to a fresh practice question tomorrow without copying today’s working.
For the wider learning pathway, read Secondary 1 Mathematics Tutor Clementi | Small Groups Tutorials. The original reference remains unchanged.