13 Vectors in two dimensions | IGCSE Additional Maths 0606

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🎯What you need to be able to do

Read and write vectors in every form: \( \begin{pmatrix} a \\ b \end{pmatrix} \), \( \overrightarrow{AB} \), \( \mathbf{p} \), \( a\mathbf{i} + b\mathbf{j} \).
Use position vectors, find magnitudes and unit vectors, and add, subtract and scale vectors.
Solve geometry problems by writing one vector in two ways and equating coefficients.
Compose and resolve velocities, and use \( \mathbf{r} = \mathbf{r}_{0} + t\mathbf{v} \) for position, including when two objects collide.

📚The mathematics

Notation and the basics

A vector has size and direction. \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors along the \(x\)- and \(y\)-axes (in context, often east and north), so \( 3\mathbf{i} - 4\mathbf{j} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} \). The syllabus expects correct notation: underline or bold a vector when you write it by hand, and put an arrow over \( \overrightarrow{AB} \).

\( \left|a\mathbf{i} + b\mathbf{j}\right| = \sqrt{a^{2} + b^{2}} \)

unit vector in the direction of \( \mathbf{a} \): \( \dfrac{\mathbf{a}}{|\mathbf{a}|} \)

\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \)

The position vector of \(A\) is \( \overrightarrow{OA} \), its vector from the origin. The equation \( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} \) is the one to reach for first in almost every question.

Ratio points and parallel vectors

If \(P\) divides \(AB\) with \( AP : PB = m : n \), then \( \overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB} \), so

\[ \overrightarrow{OP} = \mathbf{a} + \frac{m}{m + n}(\mathbf{b} - \mathbf{a}) \]

Two vectors are parallel when one is a scalar multiple of the other. If \( \mathbf{a} \) and \( \mathbf{b} \) are not parallel and \( \lambda\mathbf{a} + \mu\mathbf{b} = p\mathbf{a} + q\mathbf{b} \), then \( \lambda = p \) and \( \mu = q \) — this is “equating like vectors”, and it is how intersection problems are solved: write the same vector along two different routes, and match the coefficients.

Triangle OAB with OA = a and OB = b. P lies on AB one third of the way from A, and Q is the midpoint of OB. The lines OP and AQ cross at X, which is three quarters of the way along OP and halfway along AQ. — Worked example 1: \(X\) is reached two ways, \( \lambda\overrightarrow{OP} \) and \( \mathbf{a} + \mu\overrightarrow{AQ} \). Matching coefficients gives \( \lambda = \tfrac{3}{4} \), \( \mu = \tfrac{1}{2} \).

Velocity, position and collisions

Speed is the magnitude of the velocity. To resolve a velocity of speed \(v\) on a bearing \( \theta \) into components: \( v\sin\theta\,\mathbf{i} + v\cos\theta\,\mathbf{j} \) (east and north). To compose velocities, such as a boat’s velocity through the water plus the current, add the vectors.

An object that starts at \( \mathbf{r}_{0} \) and moves with constant velocity \( \mathbf{v} \) is at

\[ \mathbf{r} = \mathbf{r}_{0} + t\mathbf{v} \]

after time \(t\). Two objects collide only if they are at the same place at the same time: set the \( \mathbf{i} \) components equal and the \( \mathbf{j} \) components equal, and check that both give the same \(t\). Paths that cross at different times do not collide.

Two ships' paths on a grid with i east and j north. Ship A starts at (2, 3) and moves with velocity 4i + 2j; ship B starts at (12, minus 2) and moves with velocity minus i + 4.5j. Their positions at t = 0, 1 and 2 hours are marked, and both reach (10, 7) at t = 2, where they collide. — Worked example 2: both \( \mathbf{i} \) and \( \mathbf{j} \) components agree at \( t = 2 \), so the ships collide at 14:00.

✏️Worked example 1 (no calculator)

\( \overrightarrow{OA} = \mathbf{a} \) and \( \overrightarrow{OB} = \mathbf{b} \). The point \(P\) lies on \(AB\) with \( AP : PB = 1 : 2 \), and \(Q\) is the midpoint of \(OB\). The lines \(OP\) and \(AQ\) meet at \(X\), where \( \overrightarrow{OX} = \lambda\overrightarrow{OP} \) and \( \overrightarrow{AX} = \mu\overrightarrow{AQ} \). Use a vector method to find \( \lambda \) and \( \mu \). [6]

\( \overrightarrow{OP} = \mathbf{a} + \tfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \tfrac{2}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b} \) and \( \overrightarrow{AQ} = -\mathbf{a} + \tfrac{1}{2}\mathbf{b} \). Two routes to \(X\):

\[ \overrightarrow{OX} = \tfrac{2}{3}\lambda\,\mathbf{a} + \tfrac{1}{3}\lambda\,\mathbf{b} \]

\[ \overrightarrow{OX} = \mathbf{a} + \mu\left(-\mathbf{a} + \tfrac{1}{2}\mathbf{b}\right) = (1 - \mu)\mathbf{a} + \tfrac{1}{2}\mu\,\mathbf{b} \]

\( \mathbf{a} \) and \( \mathbf{b} \) are not parallel, so equate coefficients: \( \tfrac{2}{3}\lambda = 1 - \mu \) and \( \tfrac{1}{3}\lambda = \tfrac{1}{2}\mu \). The second gives \( \mu = \tfrac{2}{3}\lambda \); substituting, \( \tfrac{4}{3}\lambda = 1 \), so \( \lambda = \tfrac{3}{4} \) and \( \mu = \tfrac{1}{2} \).

Check it. Both routes give \( \overrightarrow{OX} = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{4}\mathbf{b} \) ✓. So \(X\) is the midpoint of \(AQ\), which the diagram agrees with.

Going the wrong way along a side. \( \overrightarrow{AQ} \) starts at \(A\), so it is \( \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \tfrac{1}{2}\mathbf{b} \), not \( \mathbf{a} + \tfrac{1}{2}\mathbf{b} \). Trace the route on the diagram and write one vector per step.

✏️Worked example 2 (calculator)

In this question \( \mathbf{i} \) is a unit vector due east, \( \mathbf{j} \) due north, and distances are in kilometres. At 12:00 ship \(A\) is at \( 2\mathbf{i} + 3\mathbf{j} \) and moves with velocity \( (4\mathbf{i} + 2\mathbf{j}) \) km h⁻¹; ship \(B\) is at \( 12\mathbf{i} - 2\mathbf{j} \) and moves with velocity \( (-\mathbf{i} + 4.5\mathbf{j}) \) km h⁻¹. (a) Find the speed of \(A\) and the bearing on which it travels. [3] (b) Show that the ships collide, and find when and where. [4] (c) Find the distance between the ships at 13:00. [2]

(a) Speed \( = \sqrt{4^{2} + 2^{2}} = \sqrt{20} = 4.47 \) km h⁻¹. Bearing: the angle from north is \( \tan^{-1}\tfrac{4}{2} = 63.4^{\circ} \), so \( 063.4^{\circ} \).

(b) After \(t\) hours, \( \mathbf{r}_{A} = (2 + 4t)\mathbf{i} + (3 + 2t)\mathbf{j} \) and \( \mathbf{r}_{B} = (12 - t)\mathbf{i} + (-2 + 4.5t)\mathbf{j} \). Equal \( \mathbf{i} \) components: \( 2 + 4t = 12 - t \Rightarrow t = 2 \). Equal \( \mathbf{j} \) components: \( 3 + 2t = -2 + 4.5t \Rightarrow t = 2 \). The same time, so they collide at 14:00, at \( 10\mathbf{i} + 7\mathbf{j} \).

(c) At \( t = 1 \): \( \mathbf{r}_{A} = 6\mathbf{i} + 5\mathbf{j} \), \( \mathbf{r}_{B} = 11\mathbf{i} + 2.5\mathbf{j} \). \( \overrightarrow{AB} = 5\mathbf{i} - 2.5\mathbf{j} \), so the distance is \( \sqrt{31.25} = 5.59 \) km.

Check it. Halfway to the collision, the separation should be half the starting separation (both move at constant velocity): at 12:00, \( \sqrt{10^{2} + 5^{2}} = 11.18 \), and half of that is 5.59 ✓.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) \( \mathbf{a} = 3\mathbf{i} - 4\mathbf{j} \). Find \( |\mathbf{a}| \) and the unit vector in the direction of \( \mathbf{a} \). [2]

\( |\mathbf{a}| = \sqrt{9 + 16} = 5 \); unit vector \( \tfrac{1}{5}(3\mathbf{i} - 4\mathbf{j}) = 0.6\mathbf{i} - 0.8\mathbf{j} \).

2. (No calculator.) \( \mathbf{p} = \begin{pmatrix} 2 \\ k \end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix} k \\ 8 \end{pmatrix} \). Find the values of \(k\) for which \( \mathbf{p} \) and \( \mathbf{q} \) are parallel. [3]

Parallel means \( \mathbf{q} = s\mathbf{p} \): \( k = 2s \) and \( 8 = ks \). So \( 8 = \tfrac{k^{2}}{2} \), \( k^{2} = 16 \), \( k = \pm 4 \).

3. (No calculator.) \( \overrightarrow{OA} = 2\mathbf{i} + 5\mathbf{j} \) and \( \overrightarrow{OB} = 8\mathbf{i} - 3\mathbf{j} \). The point \(C\) lies on \(AB\) with \( AC : CB = 3 : 1 \). Find \( \overrightarrow{OC} \). [3]

\( \overrightarrow{AB} = 6\mathbf{i} - 8\mathbf{j} \). \( \overrightarrow{OC} = \overrightarrow{OA} + \tfrac{3}{4}\overrightarrow{AB} = 2\mathbf{i} + 5\mathbf{j} + 4.5\mathbf{i} - 6\mathbf{j} = 6.5\mathbf{i} - \mathbf{j} \).

Points A(2, 5) and B(8, minus 3) with their position vectors from O. C(6.5, minus 1) lies on AB, three parts of the way from A and one part from B, and the vector OC is drawn. — \(C\) is three quarters of the way from \(A\) to \(B\).

4. (No calculator.) \( \mathbf{a} \) and \( \mathbf{b} \) are not parallel, and \( (\lambda + 2\mu)\mathbf{a} + (3\lambda - \mu)\mathbf{b} = 5\mathbf{a} + \mathbf{b} \). Find \( \lambda \) and \( \mu \). [3]

\( \lambda + 2\mu = 5 \) and \( 3\lambda - \mu = 1 \). From the second, \( \mu = 3\lambda - 1 \); then \( 7\lambda - 2 = 5 \), so \( \lambda = 1 \), \( \mu = 2 \).

5. (Calculator.) A boat travels at 12 km h⁻¹ on a bearing of \( 150^{\circ} \). Write its velocity in the form \( p\mathbf{i} + q\mathbf{j} \), where \( \mathbf{i} \) is east and \( \mathbf{j} \) is north. [2]

\( 12\sin 150^{\circ}\,\mathbf{i} + 12\cos 150^{\circ}\,\mathbf{j} = 6\mathbf{i} - 10.4\mathbf{j} \) (the \( \mathbf{j} \) component is \( -6\sqrt{3} \)).

6. (Calculator.) A plane flies with velocity \( (200\mathbf{i} + 150\mathbf{j}) \) km h⁻¹ relative to the air, and the wind has velocity \( (-30\mathbf{i} + 40\mathbf{j}) \) km h⁻¹. Find the plane’s resultant speed and its bearing. [4]

Resultant \( 170\mathbf{i} + 190\mathbf{j} \). Speed \( \sqrt{170^{2} + 190^{2}} = \sqrt{65000} = 255 \) km h⁻¹. Bearing \( \tan^{-1}\tfrac{170}{190} = 41.8^{\circ} \), i.e. \( 041.8^{\circ} \).

A vector triangle: the air velocity 200i + 150j from the origin, followed by the wind velocity minus 30i + 40j, and the resultant 170i + 190j closing the triangle. The bearing of the resultant, 41.8 degrees from north, is marked. — Adding the wind to the air velocity gives the resultant; its bearing is measured from north.

7. (Calculator.) A particle starts at \( \mathbf{i} - 2\mathbf{j} \) and moves with constant velocity \( (3\mathbf{i} + 4\mathbf{j}) \) m s⁻¹. Find its speed, and the time at which it is 13 m from the origin. [5]

Speed \( = 5 \) m s⁻¹. Position \( (1 + 3t)\mathbf{i} + (-2 + 4t)\mathbf{j} \). \( (1 + 3t)^{2} + (4t - 2)^{2} = 169 \Rightarrow 25t^{2} - 10t - 164 = 0 \Rightarrow t = \dfrac{10 + \sqrt{16500}}{50} = 2.77 \) s (the other root is negative).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

GeoGebra — drag the vertices of a triangle and watch a ratio point’s position vector update
Desmos — plot \( (x_{0} + at,\ y_{0} + bt) \) for two objects with a slider on \(t\) to test a collision