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Cambridge IGCSE0606

Vectors in two dimensions

Mathematics Additional 0606 Chapter Notes

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Vectors in two dimensions
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In this chapter

  1. 1Vector notation
  2. 2Magnitude, sums and equal vectors
  3. 3Position vectors and displacement
  4. 4Unit vectors
  5. 5Ratios, parallel vectors and collinear points
  6. 6Two routes to one vector
  7. 7Velocity, position and collision
  8. 8Composing and resolving velocities

1. Vector notation

A vector is written as a column vector beginpmatrixa bendpmatrix, as ai + bj, as overrightarrowAB, or as a single bold letter such as p. In handwriting, underline the letter, underlinep, so that it is not mistaken for a scalar.

The column vector and the i, j form give the same vector, so choose one and keep to it. Never put i or j inside a column vector. A question may say that i is a unit vector due east and j is a unit vector due north: read that line before you start.

overrightarrowAB is the vector from A to B. Going the other way reverses the sign.

beginpmatrixa bendpmatrix = ai + bj

overrightarrowBA = -overrightarrowAB

Examiner insight

“Candidates should be aware of correct vector notation. The use of i and j within a column vector was sometimes seen in this question.” “Few candidates used the convention of underlining vectors to make them distinct from the scalars.”

Worked example 12 marks

The vector p = 4i - 7j. Write p as a column vector. Write -p in the form ai + bj.

  1. 1

    p = beginpmatrix4 -7endpmatrix [1]

  2. 2

    -p = -4i + 7j [1]

2. Magnitude, sums and equal vectors

Add, subtract and multiply by a scalar component by component. A scalar multiplies both components.

The magnitude of a vector is its length, found with Pythagoras' theorem. It is a scalar, so it is never added to another magnitude to find the magnitude of a sum. Add the vectors first, then find the magnitude of the result.

Two vectors are equal only if both components are equal. To equate like vectors, compare the i parts and compare the j parts. This gives two equations to solve together.

|xi + yj| = √(x² + y²)

kbeginpmatrixa bendpmatrix = beginpmatrixka kbendpmatrix

Examiner insight

“A few candidates simply summed 39 and 34 and stated 73 as the magnitude, misunderstanding what was needed.” “Many candidates gave final answers that were inexact decimals. This was not condoned as an exact answer had been requested.”

Worked example 16 marks

a = 2i - 3j and b = -i + 4j.

(a) Find the exact value of |2a + b|.

(b) Find the values of p and q such that pa + qb = 7i - 18j.

  1. 1

    (a) 2a + b = (4 - 1)i + (-6 + 4)j = 3i - 2j [1]

  2. 2

    |2a + b| = √(3² + (-2)²) = √13 [1]

  3. 3

    (b) i parts: 2p - q = 7 [1]

  4. 4

    j parts: -3p + 4q = -18 [1]

  5. 5

    from the first, q = 2p - 7, so -3p + 8p - 28 = -18 and 5p = 10 [1]

  6. 6

    p = 2, q = -3 [1]

3. Position vectors and displacement

The position vector of a point A is overrightarrowOA, the vector from the origin O to A. Its components are the coordinates of A.

To go from A to B, go back to O and then out to B. So overrightarrowAB is the position vector of B minus the position vector of A: end minus start.

To reach a point on a line, follow a vector route such as overrightarrowOC = overrightarrowOA + overrightarrowAC. Mark each vector on a diagram with an arrow in its direction.

overrightarrowAB = overrightarrowOB - overrightarrowOA

overrightarrowOC = overrightarrowOA + overrightarrowAC

Examiner insight

“Some candidates summed the position vectors instead of subtracting them.” “Candidates who did not annotate the diagram in any way, sometimes stated vectors with incorrect directions.”

Worked example 16 marks

The position vectors of points A and B relative to an origin O are overrightarrowOA = beginpmatrix4 -2endpmatrix and overrightarrowOB = beginpmatrix-4 6endpmatrix.

The point C lies on AB and is such that AC : CB = 3 : 1.

The point D lies on OA such that overrightarrowOD = λoverrightarrowOA and overrightarrowCD = beginpmatrix5 -5.5endpmatrix.

(a) Find overrightarrowOD.

(b) Hence find the value of λ.

  1. 1

    (a) overrightarrowAB = overrightarrowOB - overrightarrowOA = beginpmatrix-8 8endpmatrix [1]

  2. 2

    AC : CB = 3 : 1, so overrightarrowAC = 3/4overrightarrowAB = beginpmatrix-6 6endpmatrix [1]

  3. 3

    overrightarrowOC = overrightarrowOA + overrightarrowAC = beginpmatrix-2 4endpmatrix [1]

  4. 4

    overrightarrowOD = overrightarrowOC + overrightarrowCD [1]

  5. 5

    = beginpmatrix3 -1.5endpmatrix [1]

  6. 6

    (b) beginpmatrix3 -1.5endpmatrix = λbeginpmatrix4 -2endpmatrix, so 3 = 4λ and λ = 3/4 [1]

4. Unit vectors

A unit vector has magnitude 1. The unit vector in the same direction as a is a divided by its magnitude.

A vector equals its magnitude multiplied by the unit vector in its direction. So to build a vector of a given length in a given direction, find the unit vector and multiply by that length.

Divide to get a unit vector. Multiplying by the magnitude makes the vector longer, not shorter.

unit vector = a/(|a|)

vector = magnitude × unit vector

Examiner insight

“Although many correct responses were seen, it was evident that many candidates were unaware of, or did not make use of, the fact that a vector is equal to the product of its magnitude and a unit direction vector.” “Those few who were incorrect sometimes multiplied by the magnitude, rather than dividing by it, not fully understanding the meaning of the term unit vector.”

Worked example 14 marks

(a) Find the unit vector in the direction of 12i - 5j.

(b) Hence find the vector of magnitude 39 in this direction.

  1. 1

    (a) magnitude = √(12² + (-5)²) = √(169) = 13 [1]

  2. 2

    unit vector = 1/13(12i - 5j) [1]

  3. 3

    (b) vector = 39 × 1/13(12i - 5j) [1]

  4. 4

    = 36i - 15j [1]

5. Ratios, parallel vectors and collinear points

Read a ratio as parts of the whole. If AC : CB = 1 : 3, the line AB has 4 parts and C is one part along it. So overrightarrowAC = 1/4overrightarrowAB. Check it against overrightarrowAC = 1/3overrightarrowCB.

If one vector is a scalar multiple of another, the two vectors are parallel. If they also share a point, the three points are collinear.

To prove lines parallel, write each as a vector in terms of a and b. Then show that one is a multiple of the other.

AC : CB = m : n ⇒ overrightarrowAC = m/(m+n)overrightarrowAB

Examiner insight

“Those candidates who offered an incorrect vector commonly made a direction error or misinterpreted the ratio as AP : AB = 1 : 3.” “Some candidates would improve their answers by stating that the parallel vectors have a common point in addition to being multiples of each other.”

Worked example 15 marks

The parallelogram OABC is such that overrightarrowOA = a and overrightarrowOB = b.

The point D lies on OB such that OD : OB = 1 : 3. The point E lies on AC such that AE : AC = 2 : 3.

Show that overrightarrowDE is parallel to overrightarrowAB.

  1. 1

    overrightarrowAB = b - a [1]

  2. 2

    overrightarrowOD = 1/3b. OABC is a parallelogram, so overrightarrowOC = b - a and overrightarrowAC = b - 2a [1]

  3. 3

    overrightarrowOE = a + 2/3(b - 2a) = 2/3b - 1/3a [1]

  4. 4

    overrightarrowDE = overrightarrowOE - overrightarrowOD = 2/3b - 1/3a - 1/3b = 1/3(b -a) [1]

  5. 5

    overrightarrowDE = 1/3overrightarrowAB, a multiple of overrightarrowAB, so overrightarrowDE is parallel to overrightarrowAB [1]

6. Two routes to one vector

Some questions give a scalar for each of two routes to the same point. Write the vector in terms of a and b by the first route. Write it again by the second route, with its own scalar.

Then equate like vectors: the coefficients of a match and the coefficients of b match. This gives two equations in scalars only, which you solve together.

Vectors cannot be divided. Never divide one vector by another to find a scalar.

pa + qb = ra + sb ⇒ p = r, q = s

Examiner insight

“A few candidates attempted to form proportional relationships. These were not valid as division of vectors is not appropriate.” “However, it was clear that some did not recognise that equating scalars was necessary here.”

Worked example 18 marks

A triangle OAB is such that overrightarrowOA = a and overrightarrowOB = b.

The point X lies on OA such that overrightarrowOA = 3overrightarrowOX.

The point Y lies on XB such that overrightarrowXB = 5overrightarrowXY.

The point Z lies on OB such that overrightarrowOB = moverrightarrowOZ, where m is a scalar.

overrightarrowOX = noverrightarrowZY, where n is a scalar.

Use a vector method to find the values of m and n.

  1. 1

    overrightarrowOX = 1/3a, so overrightarrowXB = b - 1/3a [1]

  2. 2

    overrightarrowXY = 1/5(b - 1/3a) [1]

  3. 3

    overrightarrowOZ = 1/mb [1]

  4. 4

    overrightarrowZY = overrightarrowZO + overrightarrowOX + overrightarrowXY = -1/mb + 1/3a + 1/5(b - 1/3a) [2]

  5. 5

    overrightarrowOX = noverrightarrowZY gives 1/3a = n[4/15a + (1/5 - 1/m)b] [1]

  6. 6

    b parts: 0 = 1/5 - 1/m, so m = 5 [1]

  7. 7

    a parts: 1/3 = 4/15n, so n = 5/4 [1]

7. Velocity, position and collision

A particle starts at position vector r₀ and moves with constant velocity v. At time t its position vector is r₀ + tv. Multiply the velocity by t first, then add the start.

The speed is the magnitude of the velocity. To find a velocity from a speed and a direction vector, find the unit vector and multiply it by the speed.

Two particles collide if they are in the same place at the same time. Equate their position vectors and compare components. If both equations give the same t, they collide. If the values of t differ, they do not.

r = r₀ + tv

v = speed × d/(|d|)

Examiner insight

“Some candidates equated the position vectors of A and B at time t and then formed and solved one equation only. These candidates usually stated that the particles collided as a positive value of t existed.” “The most common error was to omit t or to misplace it.”

Worked example 14 marks

In this question, time is in seconds and distances are in metres. At time t = 0, particle A starts at the origin and moves with velocity beginpmatrix2 1endpmatrix. At the same time, particle B starts from the point with position vector beginpmatrix-3 4endpmatrix and moves with velocity beginpmatrix4 -2endpmatrix.

Determine whether or not the particles collide.

  1. 1

    A: r_A = tbeginpmatrix2 1endpmatrix. B: r_B = beginpmatrix-3 4endpmatrix + tbeginpmatrix4 -2endpmatrix [1]

  2. 2

    collision needs r_A = r_B [1]

  3. 3

    i parts: 2t = -3 + 4t. j parts: t = 4 - 2t [1]

  4. 4

    first gives t = 3/2, second gives t = 4/3. The values differ, so the particles do not collide [1]

8. Composing and resolving velocities

To compose velocities, add them as vectors. The sum is the resultant. Its magnitude is the resultant speed.

To resolve a velocity, split it into i and j components with trigonometry. With i east and j north, a speed v on a bearing θ has an east component vsinθ and a north component vcosθ.

Draw a vector triangle or a sketch first. Use calculation, not a scale drawing, when the question asks you to calculate.

v = vsinθ i + vcosθ j

Examiner insight

“It was very clear at the start of this question that j was the unit vector due north. Many candidates did not take note of this and used i as the unit vector due north.” “Candidates were asked to calculate the values and scale drawings are not accepted in this case.”

Worked example 18 marks

In this question, i is a unit vector due east and j is a unit vector due north. Distances are measured in kilometres and time is measured in hours.

Point P has position vector 2i + 5j relative to an origin O. At 13 00, boat A sails from point P with velocity vector 8i + 8j.

(a) Find the position vector of A at the end of 2 hours of sailing.

(b) Write down the position vector of A at the end of t hours of sailing.

(c) Point Q has position vector 20i + 8j relative to the origin O. At 15 00, boat B sails from point Q on a bearing of 300° with a constant speed of 5√3. Find the position vector of B at 17 00.

(d) Find the distance between the two boats at 17 00.

  1. 1

    (a) (2i + 5j) + 2(8i + 8j) = 18i + 21j [1]

  2. 2

    (b) (2i + 5j) + t(8i + 8j) [1]

  3. 3

    (c) B sails for 2 hours. East: 5√3sin300° = -7.5. North: 5√3cos300° = (5√3)/2 [1]

  4. 4

    overrightarrowOB = 20i + 8j + 2(-7.5i + (5√3)/2j) [1]

  5. 5

    = 5i + (8 + 5√3)j [1]

  6. 6

    (d) A has sailed 4 hours: 2i + 5j + 4(8i + 8j) = 34i + 37j [1]

  7. 7

    A minus B: 29i + (29 - 5√3)j [1]

  8. 8

    distance = √(29² + (29 - 5√3)²) = 35.4 km [1]

Recap

  • A vector is a column vector, ai+bj or overrightarrowAB. Never put i or j inside a column vector.
  • Magnitude: √(x²+y²). Add the vectors first, then find the magnitude.
  • Equal vectors have equal components: equate like vectors to get two equations.
  • Position vector of A is overrightarrowOA. overrightarrowAB = overrightarrowOB - overrightarrowOA: end minus start.
  • Unit vector = vector divided by its magnitude. Vector = magnitude × unit vector.
  • Ratio AC : CB = m : n means overrightarrowAC = m/(m+n)overrightarrowAB. Scalar multiples are parallel.
  • Two routes to one point: equate coefficients of a and of b. Never divide by a vector.
  • Position at time t: r₀ + tv. Collision needs the same t from both equations.
  • Resolve a speed v on bearing θ: east vsinθ, north vcosθ.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Write 3i - 5j as a column vector. Write beginpmatrix-4 0endpmatrix in the form ai + bj.2 marks
  2. Do not use a calculator. Find the magnitude of 5i - 12j.2 marks
  3. p = beginpmatrix2 -3endpmatrix and q = beginpmatrix-1 4endpmatrix. Find 3p - 2q.3 marks
  4. Do not use a calculator. Given that (p + 2q)i + (3p - q)j = 7i + 7j, find p and q.4 marks
  5. The points A and B have position vectors beginpmatrix3 -2endpmatrix and beginpmatrix-5 4endpmatrix relative to the origin O. Find overrightarrowAB and the position vector of the midpoint M of AB.3 marks
  6. Do not use a calculator. Find the unit vector in the direction of beginpmatrix-8 15endpmatrix.2 marks
  7. In triangle OAB, overrightarrowOA = a and overrightarrowOB = b. The point P lies on AB such that AP : PB = 2 : 3. Find overrightarrowOP in terms of a and b.3 marks
  8. Do not use a calculator. In this question, i is a unit vector due east and j is a unit vector due north. A boat is steered with speed 8 km / h on a bearing of 120°. A current adds a velocity of (-√3i + 7j) km / h. Find the resultant velocity of the boat in the form ai + bj and find its speed.4 marks
  9. A triangular sail OAB has overrightarrowOA = a and overrightarrowOB = b. A rope is fixed at the point P on OA such that OP : PA = 2 : 1. A seam runs from O to the point Q on AB such that AQ : QB = 1 : 2. The rope runs in a straight line from B to P and crosses the seam OQ at the point X. Use a vector method to find overrightarrowOX in terms of a and b. Hence find the ratio OX : XQ.7 marks
  10. Do not use a calculator. In this question, i is a unit vector due east and j is a unit vector due north. Distances are in kilometres and time is in hours. At 12 00 a ferry is at the point with position vector 2i + 16j and moves with constant velocity 12i - 5j. At 12 30 a patrol boat leaves the point with position vector 2i - 12j and moves with constant speed 20 in the direction of 4i + 3j. Determine whether or not the patrol boat meets the ferry. If it does, find the time and the position vector of the point where they meet.8 marks

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