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O-Level Mathematics · Secondary 3 and 4

Vectors

A vector has both size and direction, unlike a scalar such as mass or time, which has size only. In O-Level Mathematics you work with vectors in two dimensions: column vectors, magnitudes, addition, subtraction, multiplication by a scalar and position vectors. You then use them to solve geometry problems, such as showing that lines are parallel or that points are collinear, and finding ratios of lengths. Vectors build on coordinate geometry and similar triangles, and they are studied further at A-Level.

What the syllabus expects

  • Use vector notation, including column vectors and the notation AB for the vector from A to B
  • Find the magnitude of a column vector (x, y) using √(x² + y²)
  • Add and subtract vectors and multiply a vector by a scalar
  • Use position vectors, including AB = OB − OA
  • Express vectors in terms of two given vectors, using ratios along lines
  • Use vectors to show that lines are parallel and that points are collinear, and to find ratios of lengths

Notation and column vectors

A vector can be written as a column vector, with the x-movement on top and the y-movement below. On this page column vectors are written in brackets as (x, y), with x on top. So (−6, 8) means 6 units left and 8 units up. The vector from point A to point B is written AB with an arrow above it, and a single-letter vector such as a is printed in bold or underlined when handwritten. The magnitude, or length, of (x, y) is √(x² + y²), so the magnitude of (−6, 8) is √(36 + 64) = 10. Two vectors are equal if they have the same magnitude and the same direction, wherever they are drawn. The vector −a has the same length as a but the opposite direction, so BA = −AB.

Adding, subtracting and scaling

To add column vectors, add the top numbers and add the bottom numbers: (2, 5) + (3, −1) = (5, 4). Subtraction works the same way. Geometrically, a + b means travel along a and then along b, which is the triangle law. Multiplying by a scalar k multiplies both components by k: 3(2, −1) = (6, −3). If k is positive the direction is unchanged, and if k is negative the direction reverses. This gives the key test for parallel vectors: two vectors are parallel if one is a scalar multiple of the other. For example, (4, −6) = 2(2, −3), so these vectors are parallel and the first is twice as long. In algebraic form, 6a − 3b = 3(2a − b), so 6a − 3b is parallel to 2a − b and three times as long.

Position vectors

The position vector of a point A is the vector OA from the origin O to A. If A is (2, 1), its position vector is (2, 1). The vector between two points is found by going back to the origin and out again: AB = AO + OB = −OA + OB, which is usually written as AB = OB − OA, or end minus start. For A(2, 1) and B(5, 7), AB = (5, 7) − (2, 1) = (3, 6), and its magnitude is √(9 + 36) = √45 = 6.71 units to 3 significant figures. To find a point from a vector, add: if BC = (−4, 2) and B is (5, 7), then OC = OB + BC = (1, 9), so C is the point (1, 9).

Vector geometry with ratios

In geometry questions you are usually given two base vectors, often OA = a and OB = b, and you express other vectors as journeys along known lines. If P lies on AB with AP : PB = 1 : 3, then AP is 1/4 of AB, so AP = (1/4)(b − a). Then OP = OA + AP = a + (1/4)(b − a) = (3/4)a + (1/4)b. To show that three points X, Y and Z are collinear, show that XZ is a scalar multiple of XY and state that the two vectors share the point X, so the points lie on one straight line. Parallel vectors alone only prove that lines are parallel. Ratios of lengths come from the scalar: if XZ = 3XY, then XY : YZ = 1 : 2. Area ratios can then follow from similar triangles or from triangles with the same height.

Worked example

In triangle OAB, OA = a and OB = b. M is the midpoint of OB, and P lies on AB such that AP : PB = 2 : 1. (a) Express AB, OP and MP in terms of a and b. (b) The point C is such that OC = 2a + b. Show that MP is parallel to OC and find the ratio MP : OC.

  1. AB = OB − OA = b − a.
  2. AP = (2/3)AB = (2/3)(b − a), because AP is 2 parts out of 3 of AB.
  3. OP = OA + AP = a + (2/3)b − (2/3)a = (1/3)a + (2/3)b.
  4. OM = (1/2)b, since M is the midpoint of OB.
  5. MP = OP − OM = (1/3)a + (2/3)b − (1/2)b = (1/3)a + (1/6)b.
  6. Factorise: MP = (1/6)(2a + b) = (1/6)OC. Since MP is a scalar multiple of OC, MP is parallel to OC.
  7. The scalar 1/6 gives the ratio of lengths, so MP : OC = 1 : 6.

Answer: (a) AB = b − a, OP = (1/3)a + (2/3)b, MP = (1/3)a + (1/6)b. (b) MP = (1/6)OC, so MP is parallel to OC and MP : OC = 1 : 6.

Common mistakes

  • Subtracting the wrong way round. AB = OB − OA, end minus start. Writing OA − OB gives BA, which points the opposite way.
  • Using the wrong fraction from a ratio. If AP : PB = 2 : 1, AP is 2/3 of AB, not 1/2 or 2. Add the parts of the ratio to find the denominator.
  • Treating a vector like a length. The magnitude of a + b is generally not the magnitude of a plus the magnitude of b. Find the column vector first, then its magnitude.
  • Claiming collinearity from parallel vectors alone. Two parallel vectors could lie on separate parallel lines. You must also state a common point.
  • Dropping the vector notation. In handwritten answers, underline single-letter vectors and put arrows over AB, so that a vector is not confused with a length or a number.
  • Expanding brackets carelessly. a + (2/3)(b − a) is (1/3)a + (2/3)b. Forgetting to multiply −a by 2/3 gives the wrong coefficient for a.

Exam tips

  • Draw arrows on the diagram for every vector you find. A route that follows an arrow forwards gives a plus sign, and going against an arrow gives a minus sign.
  • When asked to show that two vectors are parallel, factorise to make the scalar multiple visible, such as (1/3)a + (1/6)b = (1/6)(2a + b), and then state the conclusion in words.
  • Keep fractions exact. Decimals such as 0.333a make it hard to see scalar multiples and ratios.

Practice questions

1. Given p = (3, −4) and q = (1, 2), find the magnitude of p and the column vector 2p − q.

Show answer

The magnitude of p is √(3² + (−4)²) = √25 = 5. 2p − q = (6, −8) − (1, 2) = (5, −10).

2. OA = (3, 2) and OB = (7, −1). Find AB and its magnitude, and the position vector of C, where AC = 2AB.

Show answer

AB = (7 − 3, −1 − 2) = (4, −3), with magnitude √(16 + 9) = 5. AC = 2(4, −3) = (8, −6), so OC = OA + AC = (3, 2) + (8, −6) = (11, −4).

3. OP = 2a and OQ = 3b. R lies on PQ with PR : RQ = 1 : 2, and S is the point with OS = 4a + 3b. Express OR in terms of a and b, and show that O, R and S are collinear.

Show answer

PQ = 3b − 2a, so PR = (1/3)(3b − 2a) = b − (2/3)a. OR = OP + PR = 2a + b − (2/3)a = (4/3)a + b. Then OS = 4a + 3b = 3((4/3)a + b) = 3OR. OS is a scalar multiple of OR and both start at O, so O, R and S are collinear, with OR : RS = 1 : 2.

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