Add Maths · 13 Vectors in two dimensions

13 Vectors in two dimensions

Before you start

Pythagoras' theorem, and simplifying surds such as $\sqrt{20} = 2 \sqrt{5}$ .
Right-angled triangle trigonometry, and the exact values $\sin 30^{\circ} = \frac{1}{2}$ , $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ , $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ , $\cos 60^{\circ} = \frac{1}{2}$ .
Bearings: measured clockwise from north and written with three figures, such as $060^{\circ}$ .
Solving a pair of linear simultaneous equations.
Coordinates of points, and distance = speed × time.

Syllabus objectives covered (4)

Understand and use vector notation.
Know and use position vectors and unit vectors.
Find the magnitude of a vector; add and subtract vectors and multiply vectors by scalars.
Compose and resolve velocities.

Part 1 of 8

What a vector is, and the ways it is written

A vector has a size and a direction. A scalar has a size only. A journey of ' $3$ units right and $2$ units down' is a vector; a distance of $5$ m is a scalar. In the same way, velocity is a vector and speed is a scalar.
A column vector writes the two parts one above the other. The top number is the step in the $x$ -direction and the bottom number is the step in the $y$ -direction. The signs carry the direction: negative means left or down.
$(\binom{3}{- 2}) means 3 right and 2 down$
$𝐢$ is a vector of length $1$ in the positive $x$ -direction, and $𝐣$ is a vector of length $1$ in the positive $y$ -direction. Any vector in two dimensions is some number of $𝐢$ steps plus some number of $𝐣$ steps, so the same vector can be written two ways.
$(\binom{3}{- 2}) = 3 𝐢 - 2 𝐣$
graph diagram — not drawn yet
Square grid with x from 0 to 5 and y from −2 to 3, no scale numbers needed. A bold arrow from the point (1, 2) to the point (4, 0) labelled 3i − 2j. A dashed arrow from (1, 2) to (4, 2) labelled 3i, and a dashed arrow from (4, 2) down to (4, 0) labelled −2j, so that the two dashed arrows and the bold arrow form a right-angled triangle.
The vector $3 𝐢 - 2 𝐣$ is $3$ steps of $𝐢$ followed by $2$ steps of $- 𝐣$ . The two parts are the shorter sides of a right-angled triangle.
A vector can also be named by a single letter. Printed, it is bold: $𝐚$ . You cannot write bold by hand, so underline it: $a$ . A plain $a$ means a number, so a missing underline changes what you have written. The syllabus expects correct vector notation, so underline every vector letter in your working.
$\vec{A B}$ means the displacement from the point $A$ to the point $B$ . The order of the letters is the direction: $\vec{B A}$ is the same length pointing the other way.
$\vec{B A} = - \vec{A B}$
Two vectors are equal when they have the same size and the same direction, wherever they are drawn. Moving an arrow across the page without turning it does not change the vector. That is why, in a parallelogram $O A B C$ , $\vec{A B} = \vec{O C}$ : the two sides are parallel, the same length and point the same way.

Part 2 of 8

Magnitude: the length of the arrow

The magnitude of a vector is its length. It is written with straight bars: $| 𝐚 |$ or $| \vec{A B} |$ . The magnitude is a scalar, so it is never negative.
The two components are the shorter sides of a right-angled triangle, and the vector is the hypotenuse. So the magnitude comes from Pythagoras.
$| x 𝐢 + y 𝐣 | = \sqrt{x^{2} + y^{2}}$
For $5 𝐢 - 12 𝐣$ : square each component, sign and all. ${(- 12)}^{2} = 144$ , so $| 5 𝐢 - 12 𝐣 | = \sqrt{25 + 144} = \sqrt{169} = 13$ . The minus sign disappears in the squaring; it never turns the plus inside the root into a minus.
On the non-calculator paper, leave a magnitude as a simplified surd. $| 2 𝐢 + 4 𝐣 | = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$ . Write $2 \sqrt{5}$ , not $4.47$ .

Part 3 of 8

Adding, subtracting and multiplying by a scalar

To add vectors, add the $𝐢$ parts and add the $𝐣$ parts separately. This works because each part is a step in its own direction, and steps in the same direction simply add.
$(2 𝐢 + 5 𝐣) + (3 𝐢 - 7 𝐣) = 5 𝐢 - 2 𝐣$
On a diagram, adding means following one arrow and then the next, nose to tail. The sum, called the resultant, goes straight from the start of the first arrow to the end of the last. With letters, the inner letters link up and drop out.
$\vec{A B} + \vec{B C} = \vec{A C}$
Subtracting $𝐛$ means adding $- 𝐛$ , which is $𝐛$ reversed. In components, subtract part by part: $(4 𝐢 + 𝐣) - (𝐢 - 2 𝐣) = 3 𝐢 + 3 𝐣$ . Watch the double negative: $1 - (- 2) = 3$ .
Multiplying by a scalar $k$ multiplies every component by $k$ . If $k > 0$ the direction stays the same; if $k < 0$ it reverses. Either way the length is multiplied by $| k |$ .
$3 (2 𝐢 - 𝐣) = 6 𝐢 - 3 𝐣, - \frac{1}{2} (4 𝐢 + 6 𝐣) = - 2 𝐢 - 3 𝐣$
That gives the test for parallel vectors. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other. $6 𝐢 - 9 𝐣 = 3 (2 𝐢 - 3 𝐣)$ , so these two are parallel. In practice, check that the components are in the same ratio: $\frac{6}{2} = \frac{- 9}{- 3} = 3$ .

Part 4 of 8

Position vectors, and the vector from A to B

The position vector of a point $A$ is $\vec{O A}$ , the displacement from the origin $O$ to $A$ . It is written $𝐚$ . A point with coordinates $(2, 5)$ has position vector $2 𝐢 + 5 𝐣$ : the same two numbers, now read as steps from $O$ .
To get from $A$ to $B$ , go back from $A$ to $O$ , which is $- 𝐚$ , then out from $O$ to $B$ , which is $+ 𝐛$ . So the vector from $A$ to $B$ is 'end minus start'.
$\vec{A B} = \vec{A O} + \vec{O B} = - 𝐚 + 𝐛 = 𝐛 - 𝐚$
graph diagram — not drawn yet
Axes with x from −1 to 9 and y from −4 to 6, origin labelled O. Point A at (2, 5) and point B at (8, −3), each marked with a dot and labelled. An arrow from O to A labelled a, an arrow from O to B labelled b, and a bold arrow from A to B labelled b − a.
The route $A \to O \to B$ is $- 𝐚 + 𝐛$ , so $\vec{A B} = 𝐛 - 𝐚$ .
The trap is to write $𝐚 - 𝐛$ . That is $\vec{B A}$ : the right length, but pointing the wrong way, so every later step that uses it is wrong.
The distance between two points is the magnitude of the vector between them: $A B = | 𝐛 - 𝐚 |$ .
A point part of the way along $A B$ is found the same way you would walk to it: start at $A$ and go the right fraction of $\vec{A B}$ . If $A P : P B = m : n$ , then $P$ is $\frac{m}{m + n}$ of the way from $A$ to $B$ .
$\vec{O P} = 𝐚 + \frac{m}{m + n} (𝐛 - 𝐚)$
For the midpoint, $m = n$ , so the fraction is $\frac{1}{2}$ and $\vec{O M} = \frac{1}{2} (𝐚 + 𝐛)$ . Do not learn the general formula as a separate fact; ask 'what fraction of the way along?' and build it each time.
Three points $A$ , $B$ and $C$ lie on one straight line (they are collinear) when $\vec{A C} = k \vec{A B}$ for some scalar $k$ . The vectors are parallel and they share the point $A$ , so they lie on the same line, not on two parallel lines.

Part 5 of 8

Unit vectors and vectors of a given length

A unit vector has magnitude $1$ . $𝐢$ and $𝐣$ are unit vectors.
To get the unit vector in the same direction as $𝐚$ , divide $𝐚$ by its own magnitude. Dividing by a positive number keeps the direction and shrinks the length to exactly $1$ .
$unit vector in the direction of 𝐚 = \frac{𝐚}{| 𝐚 |}$
For $𝐚 = 3 𝐢 - 4 𝐣$ : $| 𝐚 | = \sqrt{9 + 16} = 5$ , so the unit vector is $\frac{1}{5} (3 𝐢 - 4 𝐣) = \frac{3}{5} 𝐢 - \frac{4}{5} 𝐣$ . Check it: $\sqrt{\frac{9}{25} + \frac{16}{25}} = 1$ .
A vector of magnitude $k$ in the direction of $𝐚$ is $k$ times the unit vector. The vector of magnitude $15$ in the direction of $3 𝐢 - 4 𝐣$ is $15 \times \frac{1}{5} (3 𝐢 - 4 𝐣) = 9 𝐢 - 12 𝐣$ .
$vector of magnitude k along 𝐚 = k \frac{𝐚}{| 𝐚 |}$
This is exactly how a speed and a direction become a velocity: velocity = speed × unit vector in the direction of motion. You will use it again in the velocity sections.

Part 6 of 8

Equating like vectors

Two vectors written in $𝐢$ and $𝐣$ are equal only when their $𝐢$ parts are equal and their $𝐣$ parts are equal. So one vector equation gives you two ordinary equations.
If $(p + 2) 𝐢 + 3 𝐣 = 5 𝐢 + (q - 1) 𝐣$ , then $p + 2 = 5$ and $3 = q - 1$ , so $p = 3$ and $q = 4$ .
Use this to find unknown scalars. To find $λ$ and $μ$ with $λ (𝐢 + 2 𝐣) + μ (3 𝐢 - 𝐣) = 11 𝐢 + 𝐣$ , collect the $𝐢$ parts and the $𝐣$ parts.
$𝐢 : λ + 3 μ = 11, 𝐣 : 2 λ - μ = 1$
From the second, $μ = 2 λ - 1$ . Substitute: $λ + 6 λ - 3 = 11$ , so $λ = 2$ and $μ = 3$ . Check: $2 (𝐢 + 2 𝐣) + 3 (3 𝐢 - 𝐣) = 11 𝐢 + 𝐣$ .
The same idea works with letters. If $𝐚$ and $𝐛$ are non-zero and not parallel, then $p 𝐚 + q 𝐛 = r 𝐚 + s 𝐛$ means $p = r$ and $q = s$ . The reason is the same as for $𝐢$ and $𝐣$ : neither vector can be made out of the other, so the amount of each must match.
In vector geometry this is the key move. When two lines cross at a point $X$ , write $\vec{O X}$ twice: once by travelling along the first line, with an unknown scalar such as $λ$ , and once along the second line, with a second unknown such as $μ$ . Equate the coefficients of $𝐚$ and of $𝐛$ , and solve the two equations. The second worked example does this in full.

Part 7 of 8

Velocity as a vector: where is it at time t?

A velocity vector is the displacement in each unit of time. A velocity of $(3 𝐢 + 4 𝐣)$ m/s means that every second the particle moves $3$ m in the $𝐢$ direction and $4$ m in the $𝐣$ direction. Its speed is the magnitude: $\sqrt{9 + 16} = 5$ m/s.
Suppose a particle starts at the point with position vector $𝐫_{0}$ when $t = 0$ , and moves with constant velocity $𝐯$ . After $t$ seconds it has made $t$ lots of the displacement $𝐯$ , so its position vector is
$𝐫 = 𝐫_{0} + t 𝐯$
With $𝐫_{0} = 2 𝐢 - 𝐣$ and $𝐯 = 3 𝐢 + 4 𝐣$ , the position at time $t$ is $𝐫 = (2 + 3 t) 𝐢 + (- 1 + 4 t) 𝐣$ . At $t = 2$ that is $8 𝐢 + 7 𝐣$ . Collect the $𝐢$ terms and the $𝐣$ terms in brackets like this; you will need each part on its own.
Turned round, the same rule gives a velocity from two positions: velocity = change in position ÷ time taken. A particle that moves from $𝐢 + 2 𝐣$ to $7 𝐢 - 6 𝐣$ in $2$ s has velocity $\frac{1}{2} (6 𝐢 - 8 𝐣) = (3 𝐢 - 4 𝐣)$ m/s, so its speed is $5$ m/s.
Two particles collide when they are at the same position at the same time. Write both position vectors with the same $t$ . Equate the $𝐢$ parts and solve for $t$ . Then put that $t$ into both $𝐣$ parts. If they match, the particles collide at that time; if they do not, the particles are never in the same place at the same moment, even if their paths cross.
The distance between two particles at time $t$ is the magnitude of the vector between them, $| 𝐫_{B} - 𝐫_{A} |$ , worked out with that value of $t$ .

Part 8 of 8

Composing and resolving velocities

To compose velocities is to add them. A boat moves through the water, and the water itself moves; a plane moves through the air, and the wind moves the air. The actual velocity, the resultant, is the vector sum: the boat's velocity relative to the water plus the velocity of the current.
A boat is steered straight across a river at $4$ m/s, and the current flows along the river at $3$ m/s. The two velocities are at right angles, so the resultant speed is $\sqrt{4^{2} + 3^{2}} = 5$ m/s. The boat moves at $\tan^{- 1} \frac{4}{3} = {53.1}^{\circ}$ to the bank, carried downstream as it crosses.
graph diagram — not drawn yet
Two horizontal parallel lines for the river banks, with a small arrow along the river labelled 'current'. From a point on the lower bank, an arrow straight up towards the far bank labelled 4 m/s (boat relative to water); from its tip, an arrow to the right labelled 3 m/s (current); and a bold arrow from the start point to the end of the 3 m/s arrow labelled 5 m/s (resultant). Mark the right angle between the 4 m/s and 3 m/s arrows and the angle 53.1° between the resultant and the bank.
Nose to tail: the boat's own velocity, then the current. The resultant is the third side of the triangle and is the way the boat actually moves.
To resolve a velocity is to split it into components. If a velocity of size $V$ makes an angle $θ$ with the positive $x$ -direction, its components are the adjacent and opposite sides of a right-angled triangle.
$𝐯 = V \cos θ 𝐢 + V \sin θ 𝐣$
For $20$ m/s at $30^{\circ}$ above the positive $x$ -direction: $20 \cos 30^{\circ} = 10 \sqrt{3}$ and $20 \sin 30^{\circ} = 10$ , so $𝐯 = (10 \sqrt{3} 𝐢 + 10 𝐣)$ m/s. Use the exact values on the non-calculator paper.
With bearings, the angle is measured clockwise from north. When $𝐢$ is east and $𝐣$ is north, the north side is now the adjacent side, so cosine goes with north and sine with east. Do not learn this as a rule; sketch the angle from north and read off the triangle each time.
$V on a bearing of θ : 𝐯 = V \sin θ 𝐢 + V \cos θ 𝐣$
Wind is described by where it comes from. 'A wind of $70$ km/h from the west' blows towards the east, so its velocity is $+ 70 𝐢$ km/h, not $- 70 𝐢$ . A current is described by where it flows to, so 'a current of $3$ km/h due south' is $- 3 𝐣$ km/h.
Going back the other way: the speed is the magnitude of the resultant, and the direction comes from $\tan$ of the two components. Sketch the components first, so that you can tell which quadrant the direction is in and turn the angle into a bearing.
When the two velocities are not at right angles, resolve each into $𝐢$ and $𝐣$ components and add. You can instead draw the velocity triangle and use the sine and cosine rules, which are on the List of formulas.

Worked examples

Example 1. Relative to an origin $O$ , the points $A$ and $B$ have position vectors $2 𝐢 + 5 𝐣$ and $8 𝐢 - 3 𝐣$ . Find (a) $\vec{A B}$ , (b) $| \vec{A B} |$ , (c) the unit vector in the direction of $\vec{A B}$ , (d) the position vector of the point $P$ on $A B$ such that $A P : P B = 1 : 3$ .

Show the working

(a) End minus start: $\vec{A B} = 𝐛 - 𝐚 = (8 - 2) 𝐢 + (- 3 - 5) 𝐣$ .
$\vec{A B} = 6 𝐢 - 8 𝐣$
(b) Pythagoras on the components, squaring the $- 8$ as a whole.
$| \vec{A B} | = \sqrt{6^{2} + {(- 8)}^{2}} = \sqrt{100} = 10$
(c) Divide the vector by its magnitude.
$\frac{1}{10} (6 𝐢 - 8 𝐣) = \frac{3}{5} 𝐢 - \frac{4}{5} 𝐣$
Check that it has length $1$ : $\frac{9}{25} + \frac{16}{25} = 1$ , and $\sqrt{1} = 1$ .
(d) $A P : P B = 1 : 3$ splits $A B$ into $4$ equal parts, and $P$ is after the first one, so $P$ is $\frac{1}{4}$ of the way from $A$ to $B$ . Start at $A$ and add a quarter of $\vec{A B}$ .
$\vec{O P} = (2 𝐢 + 5 𝐣) + \frac{1}{4} (6 𝐢 - 8 𝐣) = (2 + \frac{3}{2}) 𝐢 + (5 - 2) 𝐣$
So $\vec{O P} = \frac{7}{2} 𝐢 + 3 𝐣$ . Check the ratio: $\vec{A P} = \frac{3}{2} 𝐢 - 2 𝐣$ and $\vec{P B} = \frac{9}{2} 𝐢 - 6 𝐣$ , which is exactly $3$ times $\vec{A P}$ .

Answer: (a) $6 𝐢 - 8 𝐣$ (b) $10$ (c) $\frac{3}{5} 𝐢 - \frac{4}{5} 𝐣$ (d) $\frac{7}{2} 𝐢 + 3 𝐣$

Example 2. In the triangle $O A B$ , $\vec{O A} = 𝐚$ and $\vec{O B} = 𝐛$ , where $𝐚$ and $𝐛$ are not parallel. $P$ is the midpoint of $O A$ . $Q$ is the point on $A B$ such that $A Q : Q B = 1 : 2$ . The lines $O Q$ and $B P$ meet at $X$ . Find $\vec{O X}$ in terms of $𝐚$ and $𝐛$ .

Show the working

Sketch the triangle and mark $P$ , $Q$ and $X$ before writing any vectors. A sketch shows you which routes exist.
graph diagram — not drawn yet
No axes. Triangle with O at bottom left, A at bottom right and B at the top, roughly equal sides. Arrow from O to A labelled a and arrow from O to B labelled b. P marked as the midpoint of OA. Q marked on AB one third of the way from A to B. Straight segments OQ and BP drawn, crossing at a point labelled X.
$X$ lies on both $O Q$ and $B P$ , so $\vec{O X}$ can be reached along either line.
Find the vectors along each line first. $\vec{A B} = 𝐛 - 𝐚$ , and $Q$ is $\frac{1}{3}$ of the way from $A$ to $B$ .
$\vec{O Q} = 𝐚 + \frac{1}{3} (𝐛 - 𝐚) = \frac{2}{3} 𝐚 + \frac{1}{3} 𝐛$
$\vec{O P} = \frac{1}{2} 𝐚$ , so going from $B$ to $P$ is end minus start.
$\vec{B P} = \frac{1}{2} 𝐚 - 𝐛$
Route 1: $X$ is on $O Q$ , so $\vec{O X}$ is some multiple $λ$ of $\vec{O Q}$ .
$\vec{O X} = λ (\frac{2}{3} 𝐚 + \frac{1}{3} 𝐛) = \frac{2}{3} λ 𝐚 + \frac{1}{3} λ 𝐛$
Route 2: go to $B$ , then some multiple $μ$ of the way along $B P$ . Use a different letter, because the fraction along $B P$ need not equal the fraction along $O Q$ .
$\vec{O X} = 𝐛 + μ (\frac{1}{2} 𝐚 - 𝐛) = \frac{1}{2} μ 𝐚 + (1 - μ) 𝐛$
Both are the same vector, and $𝐚$ and $𝐛$ are not parallel, so equate the coefficients of $𝐚$ and of $𝐛$ .
$𝐚 : \frac{2}{3} λ = \frac{1}{2} μ, 𝐛 : \frac{1}{3} λ = 1 - μ$
From the first, $μ = \frac{4}{3} λ$ . Substitute into the second: $\frac{1}{3} λ = 1 - \frac{4}{3} λ$ , so $\frac{5}{3} λ = 1$ and $λ = \frac{3}{5}$ . Then $μ = \frac{4}{5}$ .
Put $λ = \frac{3}{5}$ into route 1: $\vec{O X} = \frac{2}{5} 𝐚 + \frac{1}{5} 𝐛$ . Check with route 2: $\frac{1}{2} \times \frac{4}{5} = \frac{2}{5}$ and $1 - \frac{4}{5} = \frac{1}{5}$ . Both routes agree.

Answer: $\vec{O X} = \frac{2}{5} 𝐚 + \frac{1}{5} 𝐛$ (with $λ = \frac{3}{5}$ and $μ = \frac{4}{5}$ ).

Example 3. Positions are in metres and time $t$ is in seconds. At $t = 0$ , particle $A$ is at the point with position vector $(𝐢 + 12 𝐣)$ m and moves with constant velocity $(4 𝐢 - 2 𝐣)$ m/s. At the same moment, particle $B$ is at $13 𝐢$ m and moves with constant velocity $(𝐢 + 𝐣)$ m/s. Show that the particles collide, and find the time and the position vector of the collision. Find also the speed of $A$ .

Show the working

Write each position at time $t$ using $𝐫 = 𝐫_{0} + t 𝐯$ , with the $𝐢$ and $𝐣$ parts collected.
$𝐫_{A} = (1 + 4 t) 𝐢 + (12 - 2 t) 𝐣, 𝐫_{B} = (13 + t) 𝐢 + t 𝐣$
Equate the $𝐢$ parts: $1 + 4 t = 13 + t$ , so $3 t = 12$ and $t = 4$ .
This is not yet a collision. Put $t = 4$ into both $𝐣$ parts: $12 - 2 (4) = 4$ and $4$ . They are equal, so at $t = 4$ both particles are at the same point, and they collide.
Find the point from either particle: $𝐫_{A} = (1 + 16) 𝐢 + (12 - 8) 𝐣 = 17 𝐢 + 4 𝐣$ . Check with $B$ : $(13 + 4) 𝐢 + 4 𝐣$ , the same.
The speed of $A$ is the magnitude of its velocity.
$| 4 𝐢 - 2 𝐣 | = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}$

Answer: They collide at $t = 4$ s at the point with position vector $(17 𝐢 + 4 𝐣)$ m. The speed of $A$ is $2 \sqrt{5}$ m/s ( $4.47$ m/s to 3 s.f.).

Example 4. In this question $𝐢$ is a unit vector due east and $𝐣$ is a unit vector due north. A boat is steered so that its velocity relative to the water is $8$ km/h on a bearing of $060^{\circ}$ . The water flows with a current of $3$ km/h due south. Find the resultant velocity of the boat in terms of $𝐢$ and $𝐣$ , its speed, and the bearing on which it actually travels.

Show the working

Resolve the boat's own velocity. The bearing is measured from north, so north is the adjacent side: north component $8 \cos 60^{\circ} = 4$ , east component $8 \sin 60^{\circ} = 4 \sqrt{3}$ .
$𝐯_{boat} = 4 \sqrt{3} 𝐢 + 4 𝐣$
The current flows due south, which is the negative $𝐣$ direction: $𝐯_{current} = - 3 𝐣$ .
Compose: add the two velocities.
$𝐯 = 4 \sqrt{3} 𝐢 + (4 - 3) 𝐣 = 4 \sqrt{3} 𝐢 + 𝐣$
The speed is the magnitude.
$| 𝐯 | = \sqrt{{(4 \sqrt{3})}^{2} + 1^{2}} = \sqrt{48 + 1} = 7$
For the bearing, sketch the resultant: $4 \sqrt{3}$ east and $1$ north, so it points just north of east. The angle from north has east as the opposite side and north as the adjacent side.
$θ = \tan^{- 1} \frac{4 \sqrt{3}}{1} = {81.8}^{\circ} (1 d.p.)$
The angle is clockwise from north and less than $90^{\circ}$ , so it is the bearing as it stands: ${081.8}^{\circ}$ . The current has turned the boat ${21.8}^{\circ}$ away from where it is steered.

Answer: $𝐯 = (4 \sqrt{3} 𝐢 + 𝐣)$ km/h, speed $7$ km/h, bearing ${081.8}^{\circ}$ .

Where people go wrong

Mistake: Writing $\vec{A B} = 𝐚 - 𝐛$ .

Instead: $\vec{A B} = 𝐛 - 𝐚$ : end minus start.

The route from $A$ to $B$ is back to $O$ ( $- 𝐚$ ) and out to $B$ ( $+ 𝐛$ ). $𝐚 - 𝐛$ is $\vec{B A}$ , which points the opposite way, so every vector built from it is wrong.

Mistake: Finding a magnitude by adding the components, or by putting a minus inside the root: $| 5 𝐢 - 12 𝐣 | = \sqrt{25 - 144}$ .

Instead: Square each component as a whole: ${(- 12)}^{2} = 144$ , so $| 5 𝐢 - 12 𝐣 | = \sqrt{25 + 144} = 13$ .

The components are the two shorter sides of a right-angled triangle, and the lengths of sides are positive. Pythagoras adds their squares.

Mistake: Finding $t$ from the $𝐢$ parts of two positions and stating that the particles collide.

Instead: Put that $t$ into the $𝐣$ parts as well. They collide only if the $𝐣$ parts are also equal at that $t$ .

Equal $𝐢$ parts only mean the particles have the same east–west position at that moment. A collision needs the same position in both directions at the same time.

Mistake: Using the same unknown for both routes in a vector geometry question, e.g. $\vec{O X} = λ \vec{O Q}$ and $\vec{B X} = λ \vec{B P}$ .

Instead: Use a different scalar for each line, such as $λ$ and $μ$ , and find both by equating coefficients.

$X$ can be a different fraction of the way along each line. In the second worked example it is $\frac{3}{5}$ of the way along $O Q$ but $\frac{4}{5}$ of the way along $B P$ .

Mistake: Taking 'a wind from the west' as a velocity of $- 70 𝐢$ , or using $V \cos θ$ for the east component of a bearing.

Instead: A wind from the west blows towards the east: $+ 70 𝐢$ . For a bearing $θ$ with $𝐢$ east and $𝐣$ north, the east part is $V \sin θ$ and the north part is $V \cos θ$ .

Wind is named by where it comes from. Bearings start from north, not from east, so the north side of the triangle is the adjacent side.

Check yourself

Write $- 5 𝐢 + 2 𝐣$ as a column vector, and write $(\binom{0}{- 3})$ in terms of $𝐢$ and $𝐣$ .
Answer
$(\binom{- 5}{2})$ , and $- 3 𝐣$ (there is no $𝐢$ part).
$𝐚 = 4 𝐢 + 𝐣$ and $𝐛 = 𝐢 - 2 𝐣$ . Find $𝐚 + 2 𝐛$ and its magnitude, giving the magnitude as a simplified surd.
Answer
$𝐚 + 2 𝐛 = 6 𝐢 - 3 𝐣$ . Its magnitude is $\sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5}$ .
Find the unit vector in the direction of $7 𝐢 - 24 𝐣$ .
Answer
The magnitude is $\sqrt{49 + 576} = 25$ , so the unit vector is $\frac{1}{25} (7 𝐢 - 24 𝐣) = \frac{7}{25} 𝐢 - \frac{24}{25} 𝐣$ .
$P$ and $Q$ have position vectors $- 2 𝐢 + 5 𝐣$ and $4 𝐢 - 3 𝐣$ . Find $\vec{P Q}$ , the distance $P Q$ , and the position vector of the midpoint $M$ of $P Q$ .
Answer
$\vec{P Q} = 𝐪 - 𝐩 = 6 𝐢 - 8 𝐣$ , so $P Q = \sqrt{36 + 64} = 10$ . $\vec{O M} = \frac{1}{2} (𝐩 + 𝐪) = 𝐢 + 𝐣$ .
Find the values of $p$ and $q$ such that $p (2 𝐢 + 𝐣) + q (𝐢 - 3 𝐣) = 7 𝐢 - 7 𝐣$ .
Answer
$𝐢$ : $2 p + q = 7$ ; $𝐣$ : $p - 3 q = - 7$ . From the second, $p = 3 q - 7$ , so $2 (3 q - 7) + q = 7$ , giving $q = 3$ and $p = 2$ .
Given that $(k - 1) 𝐢 + 6 𝐣$ is parallel to $2 𝐢 + 3 𝐣$ , find $k$ .
Answer
Parallel means the components are in the same ratio: $\frac{k - 1}{2} = \frac{6}{3} = 2$ , so $k - 1 = 4$ and $k = 5$ . The vector is $4 𝐢 + 6 𝐣 = 2 (2 𝐢 + 3 𝐣)$ .
A plane has velocity $240$ km/h due north in still air. A wind blows at $70$ km/h from the west. With $𝐢$ east and $𝐣$ north, find the resultant velocity, the plane's speed, and the bearing on which it travels.
Answer
Wind from the west blows east, so $𝐯 = (70 𝐢 + 240 𝐣)$ km/h. Speed $= \sqrt{70^{2} + 240^{2}} = 250$ km/h. Bearing: $\tan^{- 1} \frac{70}{240} = {16.3}^{\circ}$ , so ${016.3}^{\circ}$ .
A particle starts at the point with position vector $(3 𝐢 - 2 𝐣)$ m and moves with constant velocity $(- 𝐢 + 4 𝐣)$ m/s. Find its position vector after $3$ s, and its distance from $O$ at that time.
Answer
$𝐫 = (3 𝐢 - 2 𝐣) + 3 (- 𝐢 + 4 𝐣) = 10 𝐣$ m, so it is $10$ m from $O$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Relative to an origin $O$ , the points $A$ and $B$ have position vectors $3 𝐢 + 7 𝐣$ and $8 𝐢 - 5 𝐣$ .

(a) Write down $\vec{A B}$ in terms of $𝐢$ and $𝐣$ . 1 mark No calculator

Mark scheme

B1 $5 𝐢 - 12 𝐣$

(b) Find the vector of magnitude $39$ in the same direction as $\vec{A B}$ . 2 marks No calculator

Mark scheme

M1 $\frac{39}{\sqrt{5^{2} + 12^{2}}}$ multiplied by their $\vec{A B}$ , e.g. $3 (5 𝐢 - 12 𝐣)$ ; FT their $\vec{A B}$
A1 $15 𝐢 - 36 𝐣$ or $(\binom{15}{- 36})$ cao

(c) The point $C$ has position vector $λ 𝐢 - 17 𝐣$ . Given that $A$ , $B$ and $C$ lie on a straight line, find the value of $λ$ . 2 marks No calculator

Mark scheme

M1 $\vec{A C} = (λ - 3) 𝐢 - 24 𝐣$ set equal to $k (5 𝐢 - 12 𝐣)$ oe, with components equated (giving $k = 2$ ); or any equivalent method, e.g. $\vec{B C} = k \vec{A B}$ or equal component ratios
A1 $λ = 13$

Question 2. $O A B C$ is a parallelogram, with $\vec{O A} = 𝐚$ and $\vec{O C} = 𝐜$ . $M$ is the midpoint of $A B$ . The lines $O M$ and $A C$ meet at the point $X$ . Sketch the parallelogram before you start.

(a) Find $\vec{O M}$ in terms of $𝐚$ and $𝐜$ . 2 marks No calculator

Mark scheme

M1 $\vec{O A} + \frac{1}{2} \vec{A B}$ with $\vec{A B} = 𝐜$ soi
A1 $𝐚 + \frac{1}{2} 𝐜$

(b) Given that $\vec{O X} = λ \vec{O M}$ and $\vec{A X} = μ \vec{A C}$ , find the values of $λ$ and $μ$ . 5 marks No calculator

Mark scheme

B1 $\vec{O X} = λ 𝐚 + \frac{1}{2} λ 𝐜$ ; FT their $\vec{O M}$
B1 $\vec{O X} = 𝐚 + μ (𝐜 - 𝐚)$ oe
M1 equates coefficients of $𝐚$ and of $𝐜$ to obtain two equations, e.g. $λ = 1 - μ$ and $\frac{1}{2} λ = μ$
M1 dep on the previous M1: solves their pair of equations
A1 $λ = \frac{2}{3}$ and $μ = \frac{1}{3}$

(c) Write down the ratio $A X : X C$ . 1 mark No calculator

Mark scheme

B1 $1 : 2$ ; FT their $μ$

Question 3. In this question, positions are in metres and time $t$ is in seconds. At $t = 0$ , particle $A$ is at the point with position vector $(- 4 𝐢 + 9 𝐣)$ m and moves with constant velocity $(3 𝐢 - 𝐣)$ m/s. At $t = 0$ , particle $B$ is at the point with position vector $(14 𝐢 - 9 𝐣)$ m and moves with constant speed $10$ m/s in the direction of the vector $- 3 𝐢 + 4 𝐣$ .

(a) Find the velocity of $B$ . 2 marks

Mark scheme

M1 $10 \times \frac{- 3 𝐢 + 4 𝐣}{\sqrt{3^{2} + 4^{2}}}$
A1 $(- 6 𝐢 + 8 𝐣)$ m/s

(b) Show that $A$ and $B$ collide, and find the position vector of the point where they collide. 5 marks

Mark scheme

B1 $𝐫_{A} = (- 4 + 3 t) 𝐢 + (9 - t) 𝐣$
B1 $𝐫_{B} = (14 - 6 t) 𝐢 + (- 9 + 8 t) 𝐣$ ; FT their velocity of $B$
M1 equates one pair of components and solves for $t$ , e.g. $- 4 + 3 t = 14 - 6 t$ giving $t = 2$
A1 $t = 2$ substituted into the other components of both particles, both giving $7$ , with a conclusion that they collide
A1 $(2 𝐢 + 7 𝐣)$ m

How this is examined

Finding where two lines in a vector diagram cross, by writing one vector two ways: a multi-part question on a triangle or parallelogram described with $𝐚$ and $𝐛$ : first express some vectors, then 'given that $\vec{O X} = λ \vec{O M}$ and $\vec{A X} = μ \vec{A C}$ , find $λ$ and $μ$ ' Each route written as a sum of known vectors with its own scalar (for example $\vec{O X} = 𝐚 + μ (𝐜 - 𝐚)$ ), then the coefficients of $𝐚$ and of $𝐛$ equated to give two equations (M), those solved (M) and both values stated (A). The first vectors in the question, such as $\vec{A B} = 𝐛 - 𝐚$ , feed every later mark, so check their direction.
Position at time t from a start point and a velocity, and testing for a collision: a context with $𝐢$ and $𝐣$ defined, two particles or ships, and a sequence: position at time $t$ , then 'Show that they collide' or the distance between them at a given time $𝐫 = 𝐫_{0} + t 𝐯$ written for each body with the same $t$ ; the $𝐢$ parts equated and solved for $t$ ; that $t$ substituted into both $𝐣$ parts and shown to give the same value; the meeting point stated as a position vector.
Adding two velocities to get a resultant speed and direction: a boat in a current or a plane in a wind, with speeds and bearings given, asking for the resultant speed and the bearing of travel Both velocities resolved into components (or a correct velocity triangle), added, the speed found as the magnitude, and the angle found with $\tan$ and turned into a bearing; to 1 d.p. for angles on the calculator paper.
Turning a speed and a direction into a velocity vector: 'moves with speed $10$ m/s in the direction of $- 3 𝐢 + 4 𝐣$ ', or the vector of a given magnitude parallel to $\vec{A B}$ The direction vector divided by its own magnitude to make a unit vector (M), then multiplied by the speed or the required length (A). For example $10 \times \frac{1}{5} (- 3 𝐢 + 4 𝐣) = - 6 𝐢 + 8 𝐣$ .
The vector between two points, its length, and the collinear test: opening parts such as 'Find $\vec{A B}$ ' and 'Find $| \vec{A B} |$ ', and a 'given that $A$ , $B$ and $C$ lie on a straight line, find …' $𝐛 - 𝐚$ (not $𝐚 - 𝐛$ ); Pythagoras on the components with surds left exact on Paper 1; for collinear points, $\vec{A C} = k \vec{A B}$ written down and the components equated.

Words to know

Vector: A quantity with both a size and a direction, such as a displacement or a velocity.
Scalar: A quantity with a size only, such as a distance, a speed or a number that multiplies a vector.
Magnitude: The length of a vector, $| x 𝐢 + y 𝐣 | = \sqrt{x^{2} + y^{2}}$ . It is never negative.
$𝐢$ and $𝐣$: Unit vectors in the positive $x$ - and $y$ -directions. In a bearings context a question may define $𝐢$ as east and $𝐣$ as north.
Position vector: The vector $\vec{O A}$ from the origin to a point $A$ , written $𝐚$ .
Unit vector: A vector of magnitude $1$ . The unit vector in the direction of $𝐚$ is $\frac{𝐚}{| 𝐚 |}$ .
Parallel vectors: Non-zero vectors where one is a scalar multiple of the other, $𝐛 = k 𝐚$ .
Collinear: Lying on one straight line. $A$ , $B$ and $C$ are collinear when $\vec{A C} = k \vec{A B}$ .
Resultant: The single vector equal to the sum of two or more vectors; for velocities, the actual velocity of the moving body.
Compose (velocities): Add two or more velocities to find the resultant velocity.
Resolve (a velocity): Split a velocity into components in two perpendicular directions, such as $V \cos θ 𝐢 + V \sin θ 𝐣$ .
Speed: The magnitude of a velocity.

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