Vectors in Two Dimensions — Revision Notes

The question	Where it is answered	Syllabus
How do I write a vector down, and how long is it?	12.1 Notation, 12.1 Magnitude	7.1, 7.2, 7.5
What is a + b? What is a − b?	12.2 The triangle law, 12.2 Subtraction	7.2
What does 3a mean, and are these two vectors parallel?	12.3 Scalar multiples, 12.3 Parallel vectors	7.7
Where is the point, and where does it go if I translate it?	12.4 Position vectors, 12.4 Translation	7.3, 7.4
Express this vector in terms of a and b.	12.5 In terms of two others	7.6
Show that these lines are parallel / these points are in a straight line / find this ratio of areas.	12.6 Geometric problems	7.8

Scalars and vectors

Every quantity you have met so far in this course has been a single number with a unit attached. Some quantities need more than that.

A scalar quantity has only a magnitude (a size).
A vector quantity has both a magnitude and a direction. A non-zero vector can be represented by a directed line segment — an arrow whose length is the magnitude and whose direction is the direction.

Fig. 12.1

The initial point is where the arrow starts and the terminal point is where it ends; the arrowhead sits at the terminal point and nowhere else. This one vector can be written

\vec{P Q}

or a.

Scalar (size only)	Vector (size and direction)
distance — "he walked 5 km"	displacement — "he ended up 5 km north of where he started"
speed — "80 km/h"	velocity — "80 km/h on a bearing of 070°"
mass, time, area, temperature	force, acceleration, a translation on a grid

A vector is not tied to one spot. Two arrows drawn in different places are the same vector if they have the same length and point the same way, which is why the arrow above can be called a without saying where it starts.

The one exception is a position vector, which is deliberately anchored to the origin — see 12.4.

The notation — five ways of writing, five different meanings

Outcome 7.1 is not "be able to do vectors": it is a list of notations, and it is examined directly. Here is the whole list, with what each form means and the mistake that each one invites.

Written	Said	What it is	The mark-losing slip
$\vec{A B}$	"vector `AB`"	The vector from `A` to `B`. The arrow above always points right, whichever way the vector itself points; the order of the letters carries the direction.	Leaving the arrow off. `AB` with no arrow is the length of a line segment — a number, not a vector.
a	"vector `a`"	A vector given a single name. In print it is bold and upright; the scalar `a` is thin and italic. They are different objects.	You cannot write bold by hand. Underline it: write a. An un-underlined `a` in your working is a number.
$(\begin{matrix} x \\ y \end{matrix})$	"the column vector `x`, `y`"	The same vector as components: `x` across (right positive) on top, `y` up (up positive) underneath.	Writing it sideways as (`x`, `y`). That is a point. A column vector is written in a column, in round brackets, with no comma.
$\| \vec{A B} \|$	"the magnitude of $\vec{A B}$ "	A length: how long that arrow is, in units. It is a scalar and it is never negative.	Answering a "find $\| \vec{A B} \|$ " question with a column vector. The bars mean "how long", not "which way".
$\| a \|$	"the magnitude of `a`"	Same idea for a vector with a single name: the length of a.	Confusing it with the absolute value of a number. Same bars, different object — the textbook flags this itself.

\vec{B A} = - \vec{A B}

Reversing the letters reverses the vector: same length, opposite direction. So

\vec{A B} + \vec{B A} = 0

, and the magnitudes are equal:

| \vec{A B} | = | \vec{B A} |

.

Column vectors on a grid

Put the arrow on squared paper and it turns into two numbers: how far across and how far up. That is outcome 7.2 — representing a vector as a directed line segment — and outcome 7.1's column notation, meeting each other.

The column vector

(\begin{matrix} x \\ y \end{matrix})

means "x units in the positive x-direction, then y units in the positive y-direction". A negative x means left; a negative y means down. The top number is the x-component and the bottom number is the y-component.

Fig. 12.2

Left:

(\begin{matrix} 4 \\ 3 \end{matrix})

— 4 right and 3 up, drawn here from (1, 1), but it would be the same vector drawn from anywhere. Right:

(\begin{matrix} 3 \\ −4 \end{matrix})

— 3 right and 4 down, so the bottom number is negative. Count the squares, not the centimetres.

Read a vector off a diagram from tail to head: across first (right +, left −), then up (up +, down −). To draw one, reverse that, and put the arrowhead at the finishing end — the arrowhead is what carries the direction.

Magnitude — the one formula in the chapter

The components are the two short sides of a right-angled triangle and the vector itself is the hypotenuse, so the length comes straight out of Pythagoras' Theorem. That is outcome 7.5.

The magnitude of the column vector

(\begin{matrix} x \\ y \end{matrix})

is

| (\begin{matrix} x \\ y \end{matrix}) | = \sqrt{x^{2} + y^{2}}

and it is measured in units.

The signs do not survive. Both components are squared, so

(\begin{matrix} 3 \\ 4 \end{matrix})

,

(\begin{matrix} −3 \\ 4 \end{matrix})

,

(\begin{matrix} 3 \\ −4 \end{matrix})

and

(\begin{matrix} −3 \\ −4 \end{matrix})

all have a magnitude of 5 units. Four different vectors; one length.

Surds are not on this syllabus. When the square root is not exact there are two answers: leave it as

\sqrt{41}

, or evaluate and round once, to 3 s.f. —

6.40

.

A magnitude is never negative. An answer of −13 units comes from rooting a bracket instead of squaring it first: square both components, add, then take the root.

More detail

Simplifying a surd is not wanted here: $\sqrt{20}$ stays as it stands and is never rewritten as $2 \sqrt{5}$ . The textbook's own phrase "in surd form" means nothing more than "leave the square root sign in", so a root left standing is already a finished answer.

Equal, negative, parallel — and the zero vector

Four words that the "show that" questions in 12.6 are built out of. Each one is a statement about magnitude, direction, or both.

Equal vectors. a = b if and only if they have the same magnitude and the same direction. In column form,

(\begin{matrix} p \\ q \end{matrix}) = (\begin{matrix} r \\ s \end{matrix})

if and only if p = r and q = s — so one vector equation is really two ordinary equations.

Negative vectors. If a and c have the same magnitude but opposite directions, then c is the negative vector of a: c = −a (and a = −c).

Parallel vectors. Two vectors are parallel if they have the same or opposite directions. Their magnitudes may be equal or different — parallel says nothing about length. (The test for it is in 12.3.)

The zero vector.

0 = (\begin{matrix} 0 \\ 0 \end{matrix})

is the zero vector: magnitude 0, no direction. It is written bold 0 (underlined by hand), because the scalar 0 and the vector 0 are different objects. Any round trip gives it:

\vec{P Q} + \vec{Q P} = 0

.

Fig. 12.3

Left: a = b — same length, same direction, drawn in different places, and still the same vector. Middle: a and −a — same length, opposite direction. Right: p and q are parallel — here q is twice as long as p and points the other way, and parallel does not mind either.

"Opposite directions" still counts as parallel. Only "different directions" — genuinely at an angle — makes two vectors non-parallel.

Walkthroughs — Notation & Directed Line Segments

Step through each one; the why beside every line is the part a marker is actually reading.

Walkthrough 1 — from words to a column vector Basic

The vector a takes you 6 units to the left and 8 units up. Write a as a column vector and find

| a |

.

Across: 6 to the left, so the x-component is −6. Up: 8, so the y-component is 8.

Fig. 12.4

6 left and 8 up, drawn here starting from (7, −1). The dashed legs are the two components and the arrow is the hypotenuse.

The components are read in a fixed order, across first and up second, and the direction words become the signs: left is negative x, down would be negative y. Drawing the arrow on a grid, from anywhere, checks both signs in seconds.
a = $(\begin{matrix} −6 \\ 8 \end{matrix})$

The answer is written as a column in round brackets, with no comma. Written sideways as (−6, 8) it would be the coordinates of a point, which is a different thing, and the notation mark in outcome 7.1 is exactly this.
$| a | = \sqrt{{(−6)}^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100}$

Magnitude means length, so it is Pythagoras' Theorem on the two components. Both get squared, which is what makes the minus sign on the 6 irrelevant: a step to the left is exactly as long as a step to the right.
= 10 units

100 is a perfect square, so this length is exact: no rounding, no decimal point, no "(3 s.f.)". Write units — a magnitude is a measurement even when the axes carry no scale.

Walkthrough 2 — exact, and not exact Basic

(a) Given that a =

(\begin{matrix} 9 \\ −12 \end{matrix})

, find

| a |

. (b) Given that b =

(\begin{matrix} −7 \\ 4 \end{matrix})

, find

| b |

, correct to 3 significant figures.

(a) $| a | = \sqrt{9^{2} + {(−12)}^{2}}$

Straight into the magnitude formula, each component in its own brackets so that (−12)² cannot quietly become −144. Quoting the substitution line is the working the marker is looking for.
= $\sqrt{81 + 144} = \sqrt{225} = 15$

81 + 144 = 225, and 225 is a perfect square, so this magnitude comes out exactly. Nothing to round.
(b) $| b | = \sqrt{{(−7)}^{2} + 4^{2}} = \sqrt{49 + 16} = \sqrt{65}$

The same formula on the second vector. 49 + 16 = 65, which is not a perfect square — so this answer will not be exact, and the question has already warned you by asking for 3 significant figures.
= 8.0622…

Copy several figures off the calculator before doing anything else. Rounding here and then using the rounded value later is rounding twice, which is how a correct method still ends up with a wrong last digit.
(a) 15 units (b) 8.06 units (to 3 s.f.)

The first three significant figures of 8.0622… are 8, 0 and 6; the next figure is 2, so nothing rounds up. Label the inexact answer "(to 3 s.f.)" and leave the exact one unlabelled, since 15 units is exact already.

Walkthrough 3 — one vector equation, two ordinary ones Intermediate

It is given that u =

(\begin{matrix} 4 h - k \\ 9 \end{matrix})

and v =

(\begin{matrix} 11 \\ h + k \end{matrix})

, and that u = v. Find the value of h and of k.

$(\begin{matrix} 4 h - k \\ 9 \end{matrix}) = (\begin{matrix} 11 \\ h + k \end{matrix})$

Two vectors are equal only when they have the same magnitude and the same direction, and in column form that means both components match. So the single vector statement u = v is written out as one equation of columns before anything is unpacked.
4h − k = 11 ——(1)
h + k = 9 ——(2)

Top with top, bottom with bottom — that is the whole content of "equal column vectors". Numbering the two equations costs a second and makes the elimination line below readable.
(1) + (2): 5h = 20, so h = 4

The k terms are −k and +k, so adding the equations kills k outright. Choosing the operation that eliminates rather than substituting first is what keeps this to one line.
In (2): 4 + k = 9, so k = 5

Substituting back into the simpler of the two equations gives the second unknown with no algebra worth the name.
h = 4 and k = 5

Check both components before you stop: 4(4) − 5 = 11 ✓ and 4 + 5 = 9 ✓. Both must work — a pair that fits only the top row means the two vectors are not equal at all.

Check yourself

Write as a column vector: the vector that takes you 9 units right and 5 units down.
Answer
$(\begin{matrix} 9 \\ −5 \end{matrix})$ . Right is positive on top; down is negative underneath.
Find $| c |$ where c = $(\begin{matrix} −5 \\ 12 \end{matrix})$ .
Answer
$| c | = \sqrt{{(−5)}^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$ 13 units, exactly.
Find $| d |$ where d = $(\begin{matrix} 6 \\ −4 \end{matrix})$ , correct to 3 significant figures.
Answer
$| d | = \sqrt{6^{2} + {(−4)}^{2}} = \sqrt{36 + 16} = \sqrt{52}$ = 7.2111… = 7.21 units (to 3 s.f.). Do not rewrite $\sqrt{52}$ as $2 \sqrt{13}$ — surds are not on this syllabus.
$\vec{Q P} = - \vec{P Q}$ and $| \vec{Q P} | = | \vec{P Q} |$ . Explain why.
Answer
$\vec{P Q}$ is the vector from P to Q; $\vec{Q P}$ is the same arrow travelled backwards, so it has the same length but the opposite direction, which is exactly what − $\vec{P Q}$ means. Because the two magnitudes are lengths of the same segment, they are equal.
$(\begin{matrix} 2 m \\ t + 3 \end{matrix})$ and $(\begin{matrix} −8 \\ 1 \end{matrix})$ are equal vectors. Find m and t.
Answer
Top: 2m = −8, so m = −4. Bottom: t + 3 = 1, so t = −2. Two components, two equations.
A student writes " $\vec{A B}$ = 5". What is wrong with it?
Answer
$\vec{A B}$ is a vector and 5 is a number, so the two sides are different kinds of object and the statement cannot be true. What the student meant was $| \vec{A B} |$ = 5 units — the bars turn the vector into its length. This is the notation slip outcome 7.1 is written to catch.

The Triangle Law — "get there via `B`"

Walk from A to B, then from B to C, and you end up where the single journey from A to C would have put you. That sentence is the whole of vector addition.

Triangle Law of Vector Addition. If the ending point of the first vector is the starting point of the second, then

\vec{A B} + \vec{B C} = \vec{A C}

The vector

\vec{A C}

is called the resultant vector, and in diagrams it is drawn with a double arrowhead so you can see at a glance which one is the answer.

Fig. 12.5

The middle letter is the one that cancels:

\vec{A B} + \vec{B C} = \vec{A C}

. Read the letters left to right and the B at the end of the first meets the B at the start of the second.

The two vectors meet nose to tail:

\vec{A B} + \vec{B C}

works because the first stops at B and the second starts there.

More detail

A pair that does not chain can be repaired. $\vec{A B} + \vec{C B}$ is not a triangle-law sum, because the second vector starts at C, not at B. Reversing one of them, $\vec{C B} = - \vec{B C}$ , puts the two back nose to tail.

Longer chains, and the round trip

Nothing stops at two. Any chain of vectors where each one starts where the last one finished collapses to a single vector from the very first point to the very last.

\vec{A B} + \vec{B C} + \vec{C D} = \vec{A D}

— and in general, in a chain of any length, every letter cancels except the first and the last.

Fig. 12.6

Three hops or one — you finish in the same place, so the two are the same vector. Only the A at the start and the D at the end survive.

A round trip is the zero vector. When the chain closes, the first and last letters match and nothing is left:

\vec{P Q} + \vec{Q R} + \vec{R P} = \vec{P P} = 0

Spot the closed loop.

Vector addition is commutative: a + b = b + a, so a chain may be reordered until the letters meet.

\vec{Q R} + \vec{P Q}

does not chain, but

\vec{P Q} + \vec{Q R} = \vec{P R}

does.

The Parallelogram Law — when both vectors start together

Diagrams do not always hand you a nose-to-tail pair. Very often the two vectors you are given both start at the same point — and then there is a second picture of the same fact.

Parallelogram Law of Vector Addition. If a and b both start at the same point, complete the parallelogram; then a + b is the diagonal from that point. In the parallelogram OACB with

\vec{O A} = a

and

\vec{O B} = b

,

\vec{O C} = a + b

Fig. 12.7

The same sum, drawn the other way round. Note which diagonal it is:

\vec{O C}

, from the corner the two vectors share. The other diagonal,

\vec{A B}

, is a difference — see below.

The two laws are the same law. Use whichever arrangement the diagram hands you; the textbook prints both side by side and either earns the marks.

Why this works

The parallelogram picture contains the triangle picture: $\vec{O C} = \vec{O A} + \vec{A C}$ is just $\vec{O C} = a + b$ with $\vec{A C}$ replaced by the equal vector $\vec{O B}$ , the opposite side of the parallelogram.

Adding in column form — component by component

Once the vectors are columns, addition and subtraction stop being geometry and become arithmetic done twice.

(\begin{matrix} p \\ q \end{matrix}) + (\begin{matrix} r \\ s \end{matrix}) = (\begin{matrix} p + r \\ q + s \end{matrix})

(\begin{matrix} p \\ q \end{matrix}) - (\begin{matrix} r \\ s \end{matrix}) = (\begin{matrix} p - r \\ q - s \end{matrix})

Fig. 12.8

The triangle law with real numbers:

(\begin{matrix} 5 \\ 2 \end{matrix}) + (\begin{matrix} −2 \\ 4 \end{matrix}) = (\begin{matrix} 3 \\ 6 \end{matrix})

. Go 5 right and 2 up, then 2 left and 4 up; you are 3 right and 6 up from where you began.

Tops with tops, bottoms with bottoms. Keeping the two columns vertically aligned as you write is what keeps the pairing right.

Subtraction — and the one line worth more than any other

There is no new law to learn. Subtracting a vector is adding its negative, and the negative is the same arrow turned round.

a − b = a + (−b) — go along a, then backwards along b.

Triangle Law of Vector Subtraction. If a and b start at the same point, then a − b is the vector drawn from the ending point of b to the ending point of a. In short: end minus start.

Fig. 12.9

Both vectors leave O. Left:

\vec{B A} = a - b

— the arrow runs from the end of b to the end of a, and the letters of the answer read end then start backwards. Right: the same segment the other way is

\vec{A B} = b - a

.

The single most useful line in the topic. When two vectors share a starting point — in position-vector questions, the origin —

\vec{A B} = \vec{O B} - \vec{O A}

Read it as end minus start.

Which way round?

a - b

and

b - a

are opposites. Say the arrow out loud, "from B to A", and write the ending letter's vector first:

\vec{B A}

ends at A, so a comes first.

More detail

In words, "end minus start" is the letter you finish at, minus the letter you started at, both measured from the shared point. Every question in 12.5 and 12.6 runs on this one line, so it is worth writing out in letters before any numbers go in.

Magnitudes do not add

Adding two vectors and adding their two lengths are different operations, and they give different answers. For two non-zero, non-parallel vectors a and b,

| a + b | \neq | a | + | b |

| a - b | \neq | a | - | b |

Find the resultant first, then take its magnitude. To get

| a + b |

, add the columns and only then apply the magnitude formula.

Why this works

It is the reason a shortcut is shorter than going round two sides of a triangle: $| a + b |$ is the direct route across the triangle and $| a | + | b |$ is the way round two of its sides. The two agree only when a and b point the same way, which is why the statement above names non-parallel vectors.

Walkthroughs — Adding & Subtracting Vectors

Step through each one; the why beside every line is the part a marker is actually reading.

Walkthrough 4 — simplifying a chain Basic

Simplify each of the following. (a)

\vec{P Q} + \vec{Q R}

(b)

\vec{M N} + \vec{N P} + \vec{P Q}

(c)

\vec{D E} + \vec{E F} + \vec{F D}

(d)

\vec{K L} - \vec{M L}

(a) $\vec{P Q} + \vec{Q R} = \vec{P R}$

The first vector ends at Q and the second starts at Q, so they are nose to tail and the Triangle Law applies directly. The repeated middle letter disappears; the outer two survive.
(b) $\vec{M N} + \vec{N P} + \vec{P Q} = \vec{M P} + \vec{P Q} = \vec{M Q}$

Three hops, each starting where the last finished, so the same cancelling happens twice: the Ns go, then the Ps. What is left is the first letter and the last.
(c) $\vec{D E} + \vec{E F} + \vec{F D} = \vec{D F} + \vec{F D} = \vec{D D} = 0$

This chain comes back to D, so the two surviving letters are the same letter — a journey that ends where it began. That displacement is zero, and it is written as the bold zero vector, not the number 0.
(d) $\vec{K L} - \vec{M L} = \vec{K L} + \vec{L M} = \vec{K M}$

The two do not chain as written, because both end at L. Turning $- \vec{M L}$ into $\vec{L M}$ reverses its direction and makes it start where the first one finishes.
(a) $\vec{P R}$ (b) $\vec{M Q}$ (c) $0$ (d) $\vec{K M}$

Every answer is a single vector (or the zero vector) — that is what "simplify" was asking for. Check each one by reading the outer two letters of the original chain; they should be the two letters of your answer, in that order.

Walkthrough 5 — adding columns, and a warning about magnitudes Basic

It is given that a =

(\begin{matrix} 4 \\ −3 \end{matrix})

and b =

(\begin{matrix} −9 \\ 7 \end{matrix})

. (i) Express a + b as a column vector. (ii) Find

| a |

,

| b |

and

| a + b |

, giving each non-exact answer correct to 3 significant figures. (iii) Is

| a + b |

equal to

| a | + | b |

?

$a + b = (\begin{matrix} 4 \\ −3 \end{matrix}) + (\begin{matrix} −9 \\ 7 \end{matrix}) = (\begin{matrix} 4 + (−9) \\ (−3) + 7 \end{matrix})$

Tops with tops and bottoms with bottoms: 4 + (−9) and (−3) + 7. Writing the un-simplified column first keeps the two additions separate, which is where sign slips otherwise happen.
a + b = $(\begin{matrix} −5 \\ 4 \end{matrix})$

4 − 9 = −5 and −3 + 7 = 4. The answer is a column vector, exactly like the two it came from — adding vectors gives a vector, never a number.
$| a | = \sqrt{4^{2} + {(−3)}^{2}} = \sqrt{25} = 5$

Magnitude of the first vector, straight from the formula. 16 + 9 = 25 is a perfect square, so this one is exact.
$| b | = \sqrt{{(−9)}^{2} + 7^{2}} = \sqrt{130} = 11.4017…$

The same formula on b. 81 + 49 = 130 is not a perfect square, so the value is carried to several figures now and rounded only in the final line.
$| a + b | = \sqrt{{(−5)}^{2} + 4^{2}} = \sqrt{41} = 6.4031…$

The magnitude of the resultant is found from the resultant's own components, −5 and 4 — the column worked out two steps ago. There is no shortcut from $| a |$ and $| b |$ to this number.
$| a | = 5$ , $| b | = 11.4$ , $| a + b | = 6.40$ (to 3 s.f.)
No: $| a | + | b |$ = 5 + 11.4 = 16.4, which is not 6.40.

The two are equal only when the vectors point the same way; here 5 units one way and 11.4 units in quite another leave you 6.40 units from the start. Round once, at the end, and label the accuracy.

Walkthrough 6 — end minus start Intermediate

In the diagram,

\vec{O A} = a

and

\vec{O B} = b

. (i) Express

\vec{A B}

and

\vec{B A}

in terms of a and b. (ii) Given that a =

(\begin{matrix} 7 \\ −2 \end{matrix})

and b =

(\begin{matrix} −3 \\ 5 \end{matrix})

, express

\vec{B A}

as a column vector.

Fig. 12.10

The two given vectors both leave O; the side AB is drawn without an arrowhead because its direction is what the question is asking you to decide.

$\vec{A B} = \vec{O B} - \vec{O A} = b - a$

Both given vectors start at O, so this is the Triangle Law of Vector Subtraction. The arrow wanted runs from A to B, so B is the end and A is the start: end minus start.
$\vec{B A} = - \vec{A B} = - (b - a) = a - b$

The same segment travelled the other way, so it is the negative of the line above. Writing it as −(b − a) first and then expanding is the safe order — the sign of every term flips, not just the first.
$\vec{B A} = (\begin{matrix} 7 \\ −2 \end{matrix}) - (\begin{matrix} −3 \\ 5 \end{matrix}) = (\begin{matrix} 7 - (−3) \\ −2 - 5 \end{matrix})$

Now the letters are replaced by their columns, in the order the expression demands: a first because $\vec{B A}$ = a − b. Substituting into the wrong one of the two answers is the commonest way to lose this mark.
$\vec{B A} = (\begin{matrix} 10 \\ −7 \end{matrix})$

7 − (−3) = 10 and −2 − 5 = −7. A useful check: $\vec{B A}$ should be the exact negative of $\vec{A B}$ = $(\begin{matrix} −10 \\ 7 \end{matrix})$ , and it is.

Check yourself

Simplify $\vec{X Y} + \vec{Y Z}$ .
Answer
The two Ys meet, so the answer is $\vec{X Z}$ .
Simplify $\vec{A B} + \vec{B C} + \vec{C A}$ .
Answer
The chain closes back on A, so it is a round trip: $0$ , the zero vector — bold, not the number 0.
Simplify $\vec{P R} - \vec{Q R}$ .
Answer
$\vec{P R} - \vec{Q R} = \vec{P R} + \vec{R Q} = \vec{P Q}$ The subtraction is turned into an addition by reversing the second vector, and then the Rs meet.
Given p = $(\begin{matrix} −2 \\ 5 \end{matrix})$ and q = $(\begin{matrix} 6 \\ −1 \end{matrix})$ , find p + q, p − q, and $| p - q |$ .
Answer
p + q = $(\begin{matrix} 4 \\ 4 \end{matrix})$ ; p − q = $(\begin{matrix} −8 \\ 6 \end{matrix})$ ; $| p - q | = \sqrt{64 + 36} = \sqrt{100}$ = 10 units, exactly.
Using the same p and q, find $| p + q |$ correct to 3 s.f., and compare it with $| p | + | q |$ .
Answer
$| p + q | = \sqrt{4^{2} + 4^{2}} = \sqrt{32}$ = 5.6568… = 5.66 units (3 s.f.). Meanwhile $| p | + | q |$ = $\sqrt{29} + \sqrt{37}$ = 5.3851… + 6.0827… = 11.5 units. The two are not equal, and there is no rule that says they should be.
In a diagram, $\vec{O M} = m$ and $\vec{O N} = n$ . Write down $\vec{M N}$ and $\vec{N M}$ .
Answer
Both start at O, so it is end minus start: $\vec{M N} = n - m$ and $\vec{N M} = m - n$ . Each is the negative of the other.

Multiplying a vector by a number

a + a is two steps of a in the same direction, so it is twice as long and points the same way; write it 2a. That is outcome 7.7.

For a real number k (a scalar) and a vector a, the vector ka is parallel to a, is

| k |

times as long, and points

the same way as a when k is positive;
the opposite way when k is negative.

In column form, the scalar multiplies both components:

k (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} k x \\ k y \end{matrix})

and the magnitude scales by the size of the scalar:

| k a | = | k | | a |

Fig. 12.11

a, 3a and −2a. All three lie along parallel lines: 3a is three times as long as a and points the same way, and −2a is twice as long and points the opposite way.

$| k a | = | k | | a |$ , not $| k a | = k | a |$ . The magnitude scales by the size of the scalar, because a length cannot be negative.

Fractions are scalars too.

\frac{1}{2} a

is half as long as a and points the same way;

- \frac{1}{3} a

is a third as long and points backwards. That is how ratios get into the algebra; see 12.5.

More detail

A number makes the modulus visible. If $| a |$ = 7 units, then $| −3 a |$ = 3 × 7 = 21 units, not −21: the sign of the scalar turns the vector round, while the length only grows or shrinks.

Parallel vectors — the test, and the inference

Everything about parallel vectors comes out of one statement, and that statement is the engine of every "show that" question in 12.6.

If a and b are two non-zero parallel vectors, then

a = k b

for some scalar (real number) k ≠ 0. And the converse is the useful direction: if you can write one vector as a number times the other, they are parallel.

More detail

The careful version. If a = kb for some real number k, there are three possibilities:

a and b are parallel;
a = b = 0 (the zero vector is a special case — it has no direction to be parallel to);
k = 0, which again makes a the zero vector.

In an exam the vectors are non-zero, so the first line is the one in play — and saying "non-zero" in a proof covers the other two.

Testing two column vectors

Given two columns, are they parallel? Assume they are, find the k that the top row forces, and then see whether the bottom row agrees.

To test whether

(\begin{matrix} p \\ q \end{matrix})

and

(\begin{matrix} r \\ s \end{matrix})

are parallel, solve

(\begin{matrix} p \\ q \end{matrix}) = k (\begin{matrix} r \\ s \end{matrix})

which gives p = kr and q = ks. If the two values of k are the same, the vectors are parallel; if they differ, they are not.

Write the conclusion in words. A pair of equal ks is the working; "∴ the two vectors are parallel" is the answer, and that sentence is the line a "show that" is marked on.

Walkthroughs — Scalar Multiples & Parallel Vectors

Step through each one; the why beside every line is the part a marker is actually reading.

Walkthrough 7 — a combination of two vectors Basic

It is given that a =

(\begin{matrix} 5 \\ −2 \end{matrix})

and b =

(\begin{matrix} −3 \\ 4 \end{matrix})

. (i) Express 3a − 2b as a column vector. (ii) Find

| 3 a - 2 b |

, correct to 3 significant figures.

$3 a - 2 b = 3 (\begin{matrix} 5 \\ −2 \end{matrix}) - 2 (\begin{matrix} −3 \\ 4 \end{matrix})$

Substitute the two columns first and leave the scalars outside the brackets. Doing the multiplying and the subtracting in one leap is where a sign gets dropped; this line costs nothing and makes the next one mechanical.
$= (\begin{matrix} 15 \\ −6 \end{matrix}) - (\begin{matrix} −6 \\ 8 \end{matrix})$

Each scalar multiplies both components of its own vector: 3 × 5 and 3 × (−2); 2 × (−3) and 2 × 4. The minus sign in front of the second column is still waiting — it has not been used yet.
$= (\begin{matrix} 15 - (−6) \\ −6 - 8 \end{matrix}) = (\begin{matrix} 21 \\ −14 \end{matrix})$

Now subtract, top from top and bottom from bottom: 15 − (−6) = 21 and −6 − 8 = −14. The double minus on the top row is exactly what the bracketed −6 was there to protect.
$| 3 a - 2 b | = \sqrt{21^{2} + {(−14)}^{2}} = \sqrt{441 + 196} = \sqrt{637} = 25.238…$

The magnitude is taken from the answer to part (i), not from a and b separately — magnitudes do not survive addition. 441 + 196 = 637, which is not a perfect square.
= 25.2 units (to 3 s.f.)

25.238… rounds to 25.2: the fourth significant figure is 3, so the third stays as it is. State the accuracy, and the units.

Walkthrough 8 — which pairs are parallel? Intermediate

State, with reasons, which of the following pairs of vectors are parallel. (a)

(\begin{matrix} 6 \\ −10 \end{matrix})

and

(\begin{matrix} −9 \\ 15 \end{matrix})

(b)

(\begin{matrix} 12 \\ 8 \end{matrix})

and

(\begin{matrix} 9 \\ 5 \end{matrix})

(c)

(\begin{matrix} −4 \\ 7 \end{matrix})

and

(\begin{matrix} 8 \\ −14 \end{matrix})

(a) $(\begin{matrix} 6 \\ −10 \end{matrix}) = k (\begin{matrix} −9 \\ 15 \end{matrix})$

Assume they are parallel and see whether a single k can do the job. Setting the equation up before touching the numbers is what turns "do they look parallel?" into something a marker can award.
Top: 6 = −9k, so $k = - \frac{2}{3}$ . Bottom: −10 = 15k, so $k = - \frac{2}{3}$ .
The two values agree, so the vectors are parallel.

Each row gives its own value of k, and they came out the same, so one scalar really does turn the second vector into the first. Notice the scalar is negative: parallel, but pointing in opposite directions — which still counts.
(b) Top: 12 = 9k, so $k = \frac{4}{3}$ . Bottom: 8 = 5k, so $k = \frac{8}{5}$ .
$\frac{4}{3} \neq \frac{8}{5}$ , so the vectors are not parallel.

No single number can stretch $(\begin{matrix} 9 \\ 5 \end{matrix})$ into $(\begin{matrix} 12 \\ 8 \end{matrix})$ , because the top row wants one scale factor and the bottom row wants a different one. Both rows must agree; one out of two is a fail.
(c) Top: −4 = 8k, so $k = - \frac{1}{2}$ . Bottom: 7 = −14k, so $k = - \frac{1}{2}$ .
The two values agree, so the vectors are parallel.

Again one scalar serves both rows, so the second vector is $- \frac{1}{2}$ times the first, and they lie along parallel lines.
(a) parallel (b) not parallel (c) parallel

The question said "with reasons", so the values of k and the sentence that compares them are part of the answer, not scratch work. A bare "yes, no, yes" scores nothing.

Walkthrough 9 — making two vectors parallel Intermediate

(a) Given that

(\begin{matrix} 10 \\ −4 \end{matrix})

and

(\begin{matrix} −15 \\ t \end{matrix})

are parallel vectors, find the value of t. (b) It is given that u =

(\begin{matrix} −8 \\ 6 \end{matrix})

, and that v is parallel to u, points in the same direction, and has

| v |

= 25 units. Find v.

(a) $(\begin{matrix} 10 \\ −4 \end{matrix}) = k (\begin{matrix} −15 \\ t \end{matrix})$

Parallel means one is a scalar multiple of the other, so the equation is written down first with k unknown. Two unknowns, k and t, and two rows to find them from.
Top: 10 = −15k, so $k = - \frac{2}{3}$

The top row is the one with no t in it, so it pins k down on its own. Starting with the row that holds a single unknown keeps this to one line.
Bottom: $−4 = - \frac{2}{3} t$ , so t = 6

With k known, the bottom row becomes an ordinary equation in t. Check it: $- \frac{2}{3} (6) = −4$ ✓.
(b) $| u | = \sqrt{{(−8)}^{2} + 6^{2}} = \sqrt{100} = 10$

Before v can be scaled to length 25, the length of u has to be known. 64 + 36 = 100, so this one is exact and the scale factor will be a clean number.
$v = \frac{25}{10} u = 2.5 (\begin{matrix} −8 \\ 6 \end{matrix}) = (\begin{matrix} −20 \\ 15 \end{matrix})$

v is parallel to u, so v = ku; at 25 units against u's 10, k = 2.5. It is positive because the question said the same direction; the opposite direction would give k = −2.5.
(a) t = 6 (b) v = $(\begin{matrix} −20 \\ 15 \end{matrix})$

Check the magnitude you were given: $| v | = \sqrt{400 + 225} = \sqrt{625}$ = 25 ✓. Without the phrase "same direction", $(\begin{matrix} 20 \\ −15 \end{matrix})$ would be a second answer, which is why the question says it.

Check yourself

a = $(\begin{matrix} −2 \\ 7 \end{matrix})$ . Write 4a as a column vector.
Answer
$(\begin{matrix} −8 \\ 28 \end{matrix})$ — the 4 multiplies both rows.
$| a |$ = 6 units. Write down $| - 3 a |$ .
Answer
$| - 3 a | = | −3 | | a | = 3 \times 6$ 18 units. The minus sign reverses the direction but cannot shorten anything.
Are $(\begin{matrix} 14 \\ −21 \end{matrix})$ and $(\begin{matrix} −6 \\ 9 \end{matrix})$ parallel?
Answer
Top: 14 = −6k, so $k = - \frac{7}{3}$ . Bottom: −21 = 9k, so $k = - \frac{7}{3}$ . The two agree, so yes — parallel, in opposite directions.
$(\begin{matrix} 5 \\ −2 \end{matrix})$ and $(\begin{matrix} −20 \\ w \end{matrix})$ are parallel. Find w.
Answer
Top: −20 = 5k, so k = −4. Bottom: w = −4 × (−2) = 8.
a = $(\begin{matrix} 12 \\ −5 \end{matrix})$ . Find the vector parallel to a, in the same direction, with magnitude 39 units.
Answer
$| a | = \sqrt{12^{2} + {(−5)}^{2}} = \sqrt{169} = 13$ so the scale factor is 39 ÷ 13 = 3, giving 3a = $(\begin{matrix} 36 \\ −15 \end{matrix})$ . (Check: $\sqrt{1296 + 225} = \sqrt{1521} = 39$ .)
Why is "a = kb" not by itself a proof that a and b are parallel?
Answer
Because k might be 0, or one of the vectors might be the zero vector, and the zero vector has no direction to be parallel to. The full statement needs "a and b are non-zero and k ≠ 0" — which in an exam question is always true, but is worth saying in a proof.

Free vectors and position vectors

Up to now a vector has been allowed to float: the same arrow drawn anywhere on the page is the same vector. That freedom is exactly what has to be given up to describe where a point is.

A free vector has no fixed starting point — a and b in 12.1 were free vectors.
The position vector of a point P has a fixed starting point, the origin O. It is written

\vec{O P}

, and for the point P(x, y),

\vec{O P} = (\begin{matrix} x \\ y \end{matrix})

A point's position vector and its coordinates carry the same two numbers, one written down a column and the other across in brackets. Written with a single letter, the position vector of P is p: small bold letter for the capital point.

Why this works

The match is not a coincidence: $\vec{O P}$ is the vector that takes you from the origin to P, which is precisely what the coordinates of P describe. The same convention runs through the chapter: the position vector of Q is q, of R is r, and so on.

$\vec{P Q} = \vec{O Q} - \vec{O P}$ — the move everything else rests on

Two points, two position vectors, both starting at the same place. That is the Triangle Law of Vector Subtraction from 12.2, and it gives the line you will use more than any other in this chapter.

\vec{P Q} = \vec{O Q} - \vec{O P} = q - p

— end minus start, both measured from the origin.

Fig. 12.12

Both position vectors leave the origin. The vector from P to Q is what is left when you undo p and then do q:

(\begin{matrix} 8 \\ 2 \end{matrix}) - (\begin{matrix} 3 \\ 6 \end{matrix}) = (\begin{matrix} 5 \\ −4 \end{matrix})

— 5 right and 4 down, which is exactly what the arrow does.

The order decides the sign.

\vec{P Q}

starts at P, so P's position vector is the one subtracted. Say it out loud, "from P to Q", and write the arrival first.

More detail

The other route gives the same answer.

\vec{P Q} = \vec{P O} + \vec{O Q} = - \vec{O P} + \vec{O Q} = \vec{O Q} - \vec{O P}

Some diagrams make the addition version easier to see; the subtraction version is quicker to write. Both are worth full marks.

Translation by a vector

A vector is a set of instructions for a move, so the same object does the job of the translations you met in Secondary 2. That is outcome 7.3.

The vector

\vec{A B}

can be read as the translation that moves the point A to the point B; used this way it is called a translation vector. In column form it says how far right and how far up the whole object moves, and

\vec{O B} = \vec{O A} + \vec{A B}

Fig. 12.13

A(−4, 2) translated by

(\begin{matrix} 7 \\ −5 \end{matrix})

lands on B(3, −3): −4 + 7 = 3 and 2 + (−5) = −3. Every point of a translated shape moves by the same vector, which is why the shape does not turn or change size.

Both directions of the same rule. To find where a point lands, add the translation vector to its position vector; to find where it came from, subtract it. One line of arithmetic either way.

Past-paper questions

G3 N2019 · Paper 1 · Q13 3 marks

The point A is translated to B (5, −2) by the vector $(\begin{matrix} −3 \\ 7 \end{matrix})$ .

Find the coordinates of point A. [1]
Find the magnitude of $(\begin{matrix} −3 \\ 7 \end{matrix})$ . [2]

Worked solution

(a) A translation adds the vector to the starting point, so $\vec{O B} = \vec{O A} + (\begin{matrix} −3 \\ 7 \end{matrix})$ and therefore $\vec{O A} = \vec{O B} - (\begin{matrix} −3 \\ 7 \end{matrix})$ .

$\vec{O A} = (\begin{matrix} 5 \\ −2 \end{matrix}) - (\begin{matrix} −3 \\ 7 \end{matrix}) = (\begin{matrix} 8 \\ −9 \end{matrix})$ so A is the point (8, −9).

(b) ${| (\begin{matrix} −3 \\ 7 \end{matrix}) |}^{2} = {(−3)}^{2} + 7^{2} = 9 + 49 = 58$

$| (\begin{matrix} −3 \\ 7 \end{matrix}) | = \sqrt{58} = 7.6157…$ $=$ 7.62 units (to 3 s.f.).

Why this works. Two of the three marks are for the magnitude, and the first turns on direction: the arrow runs from A to B, so B is the finish and the translation vector is subtracted from it. Check by translating (8, −9) forwards to (5, −2). Part (b) wants a length, so the answer is one number in units.

Going back to coordinates

A vector answer often has to be turned back into a point before the question is finished, and the notation changes when it does: $\vec{O R} = (\begin{matrix} 7 \\ 9 \end{matrix})$ means the coordinates of R are (7, 9). Column vector in, coordinates out — round brackets, comma, across then up.

Where a question asks for the coordinates, an answer left as a column vector is one step short of the mark.

Not every capital letter is a position vector.

\vec{O A}

is one, because it starts at the origin;

\vec{A B}

is a free vector joining two named points. Only vectors from O become single bold letters a, b, p…

Past-paper questions

G3 N2015 · Paper 1 · Q8 2 marks

A is the point (−1, 4) and $\vec{B A} = (\begin{matrix} −3 \\ 6 \end{matrix})$ .

Find the coordinates of point B. [1]
Calculate $| \vec{A B} |$ . [1]

Worked solution

(a) The arrow runs from B to A, so $\vec{B A} = \vec{O A} - \vec{O B}$ and $\vec{O B} = \vec{O A} - \vec{B A}$ .

$\vec{O B} = (\begin{matrix} −1 \\ 4 \end{matrix}) - (\begin{matrix} −3 \\ 6 \end{matrix}) = (\begin{matrix} 2 \\ −2 \end{matrix})$ so the coordinates of B are (2, −2).

(b) $\vec{A B} = - \vec{B A} = (\begin{matrix} 3 \\ −6 \end{matrix})$ , and reversing an arrow does not change its length, so

${| \vec{A B} |}^{2} = 3^{2} + {(−6)}^{2} = 9 + 36 = 45$

$| \vec{A B} | = \sqrt{45} = 6.7082…$ $=$ 6.71 units (to 3 s.f.).

Why this works. Both marks are about reading the letters. Part (a) gives $\vec{B A}$ rather than $\vec{A B}$ , so B is the start, subtracted from A's position vector. The bars in (b) ask "how long": one number in units, from either $\vec{A B}$ or $\vec{B A}$ , since squaring removes the signs; $\sqrt{45}$ is not exact, so leave $\sqrt{45}$ or round once.

G3 N2018 · Paper 1 · Q18 5 marks

$\vec{P Q} = (\begin{matrix} a \\ 3 a \end{matrix})$ .
$| \vec{P Q} | = \frac{5 \sqrt{10}}{2}$
Find the two possible values of a. [2]
A line joins the two points A (−5, 1) and B (43, 38).
1. Find $\vec{A B}$ . [1]
2. Use vectors to show whether or not the point C (11, 13) lies on this line. [2]

Worked solution

(a) ${| \vec{P Q} |}^{2} = a^{2} + {(3 a)}^{2} = a^{2} + 9 a^{2} = 10 a^{2}$

Squaring the given magnitude as well: ${(\frac{5 \sqrt{10}}{2})}^{2} = \frac{25 \times 10}{4} = \frac{250}{4}$ , so $10 a^{2} = \frac{250}{4}$ .

$a^{2} = \frac{250}{40} = \frac{25}{4}$ a = $\frac{5}{2}$ or a = $- \frac{5}{2}$ (that is, 2.5 or −2.5).

(b)(i) $\vec{A B} = \vec{O B} - \vec{O A} = (\begin{matrix} 43 \\ 38 \end{matrix}) - (\begin{matrix} −5 \\ 1 \end{matrix}) =$ $(\begin{matrix} 48 \\ 37 \end{matrix})$

(ii) $\vec{A C} = (\begin{matrix} 11 \\ 13 \end{matrix}) - (\begin{matrix} −5 \\ 1 \end{matrix}) = (\begin{matrix} 16 \\ 12 \end{matrix})$

If C were on the line then $\vec{A C}$ would be a multiple of $\vec{A B}$ , because the two arrows would share the point A. Matching the x-components needs $48 \div 16 = 3$ , but $3 \times (\begin{matrix} 16 \\ 12 \end{matrix}) = (\begin{matrix} 48 \\ 36 \end{matrix})$ , and 36 $\neq$ 37.

$\vec{A C}$ is not a scalar multiple of $\vec{A B}$ , so C does not lie on the line.

Why this works. Part (a) runs the magnitude formula backwards: $\sqrt{a^{2} + 9 a^{2}} = \sqrt{10 a^{2}}$ , and a squared unknown has two answers. Part (b)(ii) says "use vectors", so build the second vector from A and test whether one column is a multiple of the other. Scaling has to fit both rows: the top wants $3$ , the bottom does not follow. Say the conclusion.

G3 N2020 · Paper 2 · Q8 9 marks

ABCD is a parallelogram.
A is the point (−7, −1) and B is the point (−4, 5).
$\vec{B C} = (\begin{matrix} 8 \\ −2 \end{matrix})$ .

Find the length of the line AB. [2]
Find the equation of the line CD. [4]
X is the point where the diagonals of the parallelogram intersect.
1. Find $\vec{X C}$ . [2]
2. Find the position vector of X. [1]

Worked solution

(a) $\vec{A B} = (\begin{matrix} −4 \\ 5 \end{matrix}) - (\begin{matrix} −7 \\ −1 \end{matrix}) = (\begin{matrix} 3 \\ 6 \end{matrix})$

${| \vec{A B} |}^{2} = 3^{2} + 6^{2} = 9 + 36 = 45$

$| \vec{A B} | = \sqrt{45} = 6.7082…$ $=$ 6.71 units (to 3 s.f.).

(b) Going round the parallelogram, $\vec{O C} = \vec{O B} + \vec{B C}$ :

$\vec{O C} = (\begin{matrix} −4 \\ 5 \end{matrix}) + (\begin{matrix} 8 \\ −2 \end{matrix}) = (\begin{matrix} 4 \\ 3 \end{matrix})$ so C is (4, 3).

$\vec{A D} = \vec{B C}$ (opposite sides), so $\vec{O D} = (\begin{matrix} −7 \\ −1 \end{matrix}) + (\begin{matrix} 8 \\ −2 \end{matrix}) = (\begin{matrix} 1 \\ −3 \end{matrix})$ and D is (1, −3).

$gradient of C D = \frac{3 - (−3)}{4 - 1} = \frac{6}{3} = 2$

Substituting C (4, 3) into y = 2x + c gives 3 = 8 + c, so c = −5: y = 2x − 5.

(c)(i) The diagonals of a parallelogram bisect each other, so X is the midpoint of AC and $\vec{X C} = \frac{1}{2} \vec{A C}$ .

$\vec{A C} = (\begin{matrix} 4 \\ 3 \end{matrix}) - (\begin{matrix} −7 \\ −1 \end{matrix}) = (\begin{matrix} 11 \\ 4 \end{matrix})$

$\vec{X C} = \frac{1}{2} (\begin{matrix} 11 \\ 4 \end{matrix}) =$ $(\begin{matrix} 5.5 \\ 2 \end{matrix})$

(ii) $\vec{O X} = \vec{O C} - \vec{X C} = (\begin{matrix} 4 \\ 3 \end{matrix}) - (\begin{matrix} 5.5 \\ 2 \end{matrix}) =$ $(\begin{matrix} −1.5 \\ 1 \end{matrix})$

Why this works. Nine marks, mostly coordinate work that this chapter feeds. Get C and D down first from Triangle-Law lines, using $\vec{A D}$ = $\vec{B C}$ — opposite sides of a parallelogram are equal vectors, not merely equal lengths. In (c) "the diagonals bisect each other" is the whole of the working, and (ii) asks for a position vector, so that answer stays a column.

Walkthroughs — Position Vectors & Translations

Step through each one; the why beside every line is the part a marker is actually reading.

Walkthrough 10 — position vectors and the vector between two points Basic

The points A and B have coordinates (−3, 8) and (5, 2) respectively. (i) Write down the position vectors of A and of B. (ii) Express

\vec{A B}

as a column vector. (iii) Find

| \vec{A B} |

.

$\vec{O A} = (\begin{matrix} −3 \\ 8 \end{matrix}), \vec{O B} = (\begin{matrix} 5 \\ 2 \end{matrix})$

A position vector is measured from the origin, so its two components are the coordinates — written down a column instead of across. Nothing is being calculated here; the notation is being changed, and outcome 7.1 pays for exactly that.
$\vec{A B} = \vec{O B} - \vec{O A} = (\begin{matrix} 5 \\ 2 \end{matrix}) - (\begin{matrix} −3 \\ 8 \end{matrix})$

The vector wanted runs from A to B, so B is the end and A is the start: end minus start. Quoting the rule before substituting is the line that earns the method mark.
$= (\begin{matrix} 5 - (−3) \\ 2 - 8 \end{matrix}) = (\begin{matrix} 8 \\ −6 \end{matrix})$

Top from top and bottom from bottom: 5 − (−3) = 8 and 2 − 8 = −6. The bracket round the −3 is what stops the first subtraction quietly becoming 5 − 3.
$| \vec{A B} | = \sqrt{8^{2} + {(−6)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$

The magnitude comes from the components of $\vec{A B}$ itself. 64 + 36 = 100, a perfect square, so the length is exact.
(i) $\vec{O A} = (\begin{matrix} −3 \\ 8 \end{matrix})$ , $\vec{O B} = (\begin{matrix} 5 \\ 2 \end{matrix})$ (ii) $\vec{A B}$ = $(\begin{matrix} 8 \\ −6 \end{matrix})$ (iii) 10 units

Three kinds of answer, and each is written in its own notation: two position vectors as columns, one free vector as a column, and one magnitude as a number with units. Marks are given for the form as well as the value.

Walkthrough 11 — translating a point, forwards and backwards Basic

(a) The point C(−1, 7) is translated by the vector

\vec{C D} = (\begin{matrix} 6 \\ −11 \end{matrix})

to the point D. Find the coordinates of D. (b) A point E is translated by the same vector to the point F(4, −2). Find the coordinates of E.

$\vec{O D} = \vec{O C} + \vec{C D} = (\begin{matrix} −1 \\ 7 \end{matrix}) + (\begin{matrix} 6 \\ −11 \end{matrix})$

A translation is a journey from C to D, so the Triangle Law applies with the origin as the starting corner: go to C, then travel the translation vector. Writing the rule in letters first sets up part (b), the same rule rearranged.
$= (\begin{matrix} −1 + 6 \\ 7 + (−11) \end{matrix}) = (\begin{matrix} 5 \\ −4 \end{matrix})$

−1 + 6 = 5 and 7 + (−11) = −4. The negative bottom component moved the point down, which is what a translation vector with a negative y-component always does.
The coordinates of D are (5, −4).

The question asked for coordinates, so the column vector is turned back into a point: round brackets, comma, across then up. Leaving it as a column would be answering a different question.
$\vec{O E} = \vec{O F} - \vec{E F} = (\begin{matrix} 4 \\ −2 \end{matrix}) - (\begin{matrix} 6 \\ −11 \end{matrix})$

This time the finish is known and the start is not, so the same rule is rearranged: subtract the translation vector instead of adding it. Reading it as "undo the move" is the check that the sign is the right way round.
$= (\begin{matrix} 4 - 6 \\ −2 - (−11) \end{matrix}) = (\begin{matrix} −2 \\ 9 \end{matrix})$
The coordinates of E are (−2, 9).

4 − 6 = −2 and −2 − (−11) = 9. Check by going forwards: (−2) + 6 = 4 and 9 + (−11) = −2, which is F ✓.

Walkthrough 12 — the missing vertex of a parallelogram Intermediate

The points P, Q and S have coordinates (2, −1), (7, 3) and (4, 6) respectively. Find the coordinates of R if PQRS is a parallelogram.

Since PQRS is a parallelogram, $\vec{S R} = \vec{P Q}$

Opposite sides of a parallelogram are equal in length and parallel, so as vectors they are equal when travelled the same way. Round the vertices PQRS, $\vec{S R}$ matches $\vec{P Q}$ . $\vec{R S}$ and $\vec{P Q}$ point opposite ways.
$\vec{O R} - \vec{O S} = \vec{O Q} - \vec{O P}$

Both sides are rewritten with the "end minus start" rule so that everything is in terms of position vectors, which are the things the coordinates actually give. The only unknown left is $\vec{O R}$ .
$\vec{O R} = \vec{O Q} - \vec{O P} + \vec{O S} = (\begin{matrix} 7 \\ 3 \end{matrix}) - (\begin{matrix} 2 \\ −1 \end{matrix}) + (\begin{matrix} 4 \\ 6 \end{matrix})$

Making $\vec{O R}$ the subject turns the geometric statement into a single addition. Substituting the numbers only now, and not earlier, keeps the algebra in one place.
$= (\begin{matrix} 5 \\ 4 \end{matrix}) + (\begin{matrix} 4 \\ 6 \end{matrix}) = (\begin{matrix} 9 \\ 10 \end{matrix})$

7 − 2 = 5 and 3 − (−1) = 4, giving $\vec{P Q}$ = $(\begin{matrix} 5 \\ 4 \end{matrix})$ ; adding that to $\vec{O S}$ gives 4 + 5 and 6 + 4.
The coordinates of R are (9, 10).

Check the other pair of sides: $\vec{Q R}$ = $(\begin{matrix} 2 \\ 7 \end{matrix})$ and $\vec{P S}$ = $(\begin{matrix} 2 \\ 7 \end{matrix})$ ✓. Both pairs equal confirms the parallelogram, and the answer is given as coordinates because that is what was asked.

Check yourself

The point A is (6, −4). Write down $\vec{O A}$ .
Answer
$(\begin{matrix} 6 \\ −4 \end{matrix})$ — the coordinates, written down a column.
P is (−2, 3) and Q is (5, −1). Find $\vec{P Q}$ as a column vector, and $| \vec{P Q} |$ correct to 3 s.f.
Answer
$\vec{P Q} = (\begin{matrix} 5 \\ −1 \end{matrix}) - (\begin{matrix} −2 \\ 3 \end{matrix}) = (\begin{matrix} 7 \\ −4 \end{matrix})$ and $| \vec{P Q} | = \sqrt{49 + 16} = \sqrt{65}$ = 8.0622… = 8.06 units (3 s.f.).
$\vec{O M} = (\begin{matrix} −3 \\ −5 \end{matrix})$ and $\vec{M N} = (\begin{matrix} 8 \\ 2 \end{matrix})$ . Find the coordinates of N.
Answer
$\vec{O N} = \vec{O M} + \vec{M N} = (\begin{matrix} −3 \\ −5 \end{matrix}) + (\begin{matrix} 8 \\ 2 \end{matrix}) = (\begin{matrix} 5 \\ −3 \end{matrix})$ so N is (5, −3).
The point A(4, −7) is translated by $(\begin{matrix} −9 \\ 5 \end{matrix})$ . Find the coordinates of its image B.
Answer
4 + (−9) = −5 and −7 + 5 = −2, so B is (−5, −2).
K(1, 2), L(6, 4) and M(8, 9) are three vertices of a parallelogram KLMN. Find the coordinates of N.
Answer
$\vec{N M} = \vec{K L} = (\begin{matrix} 5 \\ 2 \end{matrix})$ so $\vec{O N} = \vec{O M} - (\begin{matrix} 5 \\ 2 \end{matrix}) = (\begin{matrix} 8 \\ 9 \end{matrix}) - (\begin{matrix} 5 \\ 2 \end{matrix}) = (\begin{matrix} 3 \\ 7 \end{matrix})$ and N is (3, 7). Check the other pair: $\vec{K N}$ = $\vec{L M}$ = $(\begin{matrix} 2 \\ 5 \end{matrix})$ ✓.
Why can $\vec{O A}$ be replaced by the single letter a, while $\vec{A B}$ cannot?
Answer
Because $\vec{O A}$ starts at the origin and so is a position vector: naming it a tells you where A is. $\vec{A B}$ is a free vector joining two named points; calling it a as well would make the letter mean two different things in the same solution.

Every vector in the plane, out of two others

This is outcome 7.6 — the hardest of the eight, and the one that pays for most of the marks in a vectors question. The idea is small; the notation is what makes it look hard.

Any vector

\vec{A B}

can be expressed uniquely in terms of two other non-zero, non-parallel, coplanar vectors u and v:

\vec{A B} = m u + n v

where m and n are real numbers.

"Uniquely" is a promise you can use. There is exactly one pair m, n that works, so two different answers cannot both be right, and two expressions that are equal match term for term.

More detail

Each of the three conditions rules out a way it could fail. Non-zero: the zero vector points nowhere and can build nothing. Non-parallel: if u and v lay along the same line, every combination of them would lie along that line too, and most vectors would be out of reach. Coplanar is the syllabus's own word (7.6 says "two coplanar vectors") and means in the same plane — on a flat page it costs nothing, which is why any two vectors here are automatically coplanar.

"Uniquely" is worth using as a check. If you and a friend reach 3u + 2v and 2u + 3v by different routes, one of you is wrong, because the answers cannot both be right. Two expressions that are equal, such as 3u + 2v and 2v + 3u, match term for term.

Past-paper questions

G3 N2023 · Paper 2 · Q6(a) 4 marks

$p = (\begin{matrix} −4 \\ 5 \end{matrix})$ and $q = (\begin{matrix} 2 \\ −3 \end{matrix})$ .

X is the point (7, 3) and Y is the point (3, 4).
$\vec{X Y} = m p + n q$ , where m and n are constants.
Find the value of m and the value of n. [4]

Worked solution

$\vec{X Y} = \vec{O Y} - \vec{O X} = (\begin{matrix} 3 \\ 4 \end{matrix}) - (\begin{matrix} 7 \\ 3 \end{matrix}) = (\begin{matrix} −4 \\ 1 \end{matrix})$

$m (\begin{matrix} −4 \\ 5 \end{matrix}) + n (\begin{matrix} 2 \\ −3 \end{matrix}) = (\begin{matrix} −4 \\ 1 \end{matrix})$

Two column vectors are equal only when both rows match, which gives a pair of simultaneous equations:

−4m + 2n = −4 …(1)
5m − 3n = 1 …(2)

Scale each equation so the n terms cancel when they are added:

(1) × 3: −12m + 6n = −12
(2) × 2: 10m − 6n = 2

Adding: −2m = −10, so m = 5.

Substituting into (1): −20 + 2n = −4, so n = 8.

Check: $5 (\begin{matrix} −4 \\ 5 \end{matrix}) + 8 (\begin{matrix} 2 \\ −3 \end{matrix}) = (\begin{matrix} −20 \\ 25 \end{matrix}) + (\begin{matrix} 16 \\ −24 \end{matrix}) = (\begin{matrix} −4 \\ 1 \end{matrix})$ — the required $\vec{X Y}$ .

Why this works. Outcome 7.6 in column form: p and q are not parallel, so exactly one pair (m, n) works and the four marks are for finding it. The first mark is $\vec{X Y}$ , finish minus start, since reversing it flips both signs. "Equal column vectors" then splits one vector equation into two ordinary ones, and substituting back checks the answer.

The method: find a route made only of us and vs

To express $\vec{A B}$ in terms of u and v, travel from A to B by a detour whose every leg is a multiple of u or of v, and add the legs up with the Triangle Law.

The routing recipe.

Mark on the diagram every vector you already know in terms of u and v — including the ones you get free from equal vectors (opposite sides of a parallelogram) and from fractions of a known side.
Find a chain of those known vectors that starts at the right letter and finishes at the right letter.
Add them with the Triangle Law and collect the u terms and the v terms.

Fig. 12.14

Two detours, one answer. Left: go 2u to C, then 3v to B. Right: go 3v to D first, then 2u. Either way

\vec{A B} = 2 u + 3 v

, because vector addition is commutative — and "uniquely" says there is no third answer waiting.

Move 1 — go the wrong way, then reverse it.

\vec{B A} = - \vec{A B}

.

Move 2 — if everything is measured from one point, subtract.

\vec{A B} = \vec{O B} - \vec{O A}

. With an origin in the diagram this is usually the fastest route there is.

Points that cut a line in a ratio

Most of the difficulty in a 7.6 question is not the vectors: it is turning "P is the point on AB such that AP : PB = 2 : 3" into something algebraic.

If P lies on AB with AP : PB = m : n, then AB is m + n parts and AP is m of them, so

\vec{A P} = \frac{m}{m + n} \vec{A B}

In particular, if M is the midpoint of AB, then

\vec{A M} = \frac{1}{2} \vec{A B}

.

Draw the parts on a line before you write anything. │—2—│——3——│ with A, P, B underneath, and the fraction

\frac{2}{5}

reads itself off. That is how the textbook does it.

A ratio of lengths does not fix a direction. "SU = ¼SR" is true whichever way round the arrows are drawn, so read the diagram before attaching the sign.

More detail

"SU = ¼SR" is a statement about two lengths, so it does not automatically follow that $\vec{S U} = \frac{1}{4} \vec{S R}$ — it could just as well give $\vec{S U} = - \frac{1}{4} \vec{R S}$ . Looking at which way each arrow actually points, and only then attaching the sign, is the step the harder questions turn on.

Past-paper questions

G3 N2019 · Paper 2 · Q7 9 marks

P is the point (4, −3), Q is the point (−5, 1).
1. Write down the column vector $\vec{P Q}$ . [1]
2. R has coordinates (h, 3) and $\vec{P R} = k \vec{P Q}$ .
  Find the value of h and the value of k. [2]
OAC is a triangle and B is a point on OC.
$\vec{O A} = a$ , $\vec{O B} = b$ and OB : BC = 2 : 3.
X is the point on AB such that AX : XB = 1 : 2.
1. Express $\vec{A C}$ in terms of a and b, as simply as possible. [2]
2. Express $\vec{X B}$ in terms of a and b, as simply as possible. [2]
3. Y is the point on OC such that AXYC is a trapezium.
  Find, in terms of a and b, $\vec{X Y}$ . [2]

Worked solution

(a)(i) $\vec{P Q} = \vec{O Q} - \vec{O P} = (\begin{matrix} −5 \\ 1 \end{matrix}) - (\begin{matrix} 4 \\ −3 \end{matrix}) =$ $(\begin{matrix} −9 \\ 4 \end{matrix})$

(ii) $\vec{P R} = (\begin{matrix} h - 4 \\ 6 \end{matrix})$ and this must equal $k (\begin{matrix} −9 \\ 4 \end{matrix}) = (\begin{matrix} - 9 k \\ 4 k \end{matrix})$ .

Bottom row: 4k = 6, so k = 1.5.
Top row: h − 4 = −9(1.5) = −13.5, so h = −9.5.

(b)(i) OB : BC = 2 : 3 makes OC five of those parts, so $\vec{O C} = \frac{5}{2} \vec{O B} = \frac{5}{2} b$ .

$\vec{A C} = \vec{A O} + \vec{O C} =$ $\frac{5}{2} b - a$

(ii) $\vec{A B} = \vec{A O} + \vec{O B} = b - a$ , and AX : XB = 1 : 2 makes XB two of the three parts of AB:

$\vec{X B} = \frac{2}{3} \vec{A B} = \frac{2}{3} (b - a) =$ $\frac{2}{3} b - \frac{2}{3} a$

(iii) In the trapezium AXYC the sides AX and YC lie along AB and OC, which are not parallel, so the parallel pair must be XY and CA. Hence $\vec{X Y} = λ \vec{A C}$ for some number λ.

$\vec{O X} = \vec{O A} + \frac{1}{3} \vec{A B} = a + \frac{1}{3} (b - a) = \frac{2}{3} a + \frac{1}{3} b$

Y is on OC, so $\vec{O Y} = t b$ for some number t, and

$\vec{X Y} = t b - \frac{2}{3} a - \frac{1}{3} b = - \frac{2}{3} a + (t - \frac{1}{3}) b$

Comparing with $λ (\frac{5}{2} b - a)$ : the a terms give $λ = \frac{2}{3}$ , and then the b terms give $t - \frac{1}{3} = \frac{2}{3} \times \frac{5}{2} = \frac{5}{3}$ , so $t = 2$ .

$\vec{X Y} = \frac{2}{3} \vec{A C} = \frac{2}{3} (\frac{5}{2} b - a) =$ $\frac{5}{3} b - \frac{2}{3} a$

Check: $\vec{O Y} = 2 b$ while $\vec{O C} = \frac{5}{2} b$ , so Y really is between O and C.

Why this works. Every part turns a ratio into a fraction of a vector. Read the parts: OB : BC = 2 : 3 makes OC five parts, so $\vec{O C}$ = $\frac{5}{2} b$ ; AX : XB = 1 : 2 makes XB two thirds of AB. In (iii), "trapezium" names the pair: write $\vec{X Y}$ = λ $\vec{A C}$ and compare coefficients.

G3 N2021 · Paper 2 · Q8 12 marks

The position vector of point R is $(\begin{matrix} −4 \\ 5 \end{matrix})$ .

The position vector of point S is $(\begin{matrix} 3 \\ −1 \end{matrix})$ .
1. 1. Find the vector that represents the translation from R to S. [1]
  2. Find the magnitude of $\vec{R S}$ . [2]
2. The position vector of T is $(\begin{matrix} k \\ 7 \end{matrix})$ and $| \vec{S T} | = 10$ .
  Find the two possible values of k. [3]
ABCD is a parallelogram.
F is the point on AB such that AF : AB = 1 : 2.
G is the point on BC such that BG : GC = 3 : 1.
$\vec{A D} = p$ and $\vec{A G} = q$ .
1. Express $\vec{C D}$ , as simply as possible, in terms of p and q. [2]
2. H is the point such that $\vec{D H} = \vec{G F}$ .
  Express $\vec{A H}$ , as simply as possible, in terms of p and q. [4]

Worked solution

(a)(i)(a) $\vec{R S} = \vec{O S} - \vec{O R} = (\begin{matrix} 3 \\ −1 \end{matrix}) - (\begin{matrix} −4 \\ 5 \end{matrix}) =$ $(\begin{matrix} 7 \\ −6 \end{matrix})$

(a)(i)(b) ${| \vec{R S} |}^{2} = 7^{2} + {(−6)}^{2} = 49 + 36 = 85$

$| \vec{R S} | = \sqrt{85} = 9.2195…$ $=$ 9.22 units (to 3 s.f.).

(a)(ii) $\vec{S T} = (\begin{matrix} k \\ 7 \end{matrix}) - (\begin{matrix} 3 \\ −1 \end{matrix}) = (\begin{matrix} k - 3 \\ 8 \end{matrix})$

${(k - 3)}^{2} + 8^{2} = 10^{2}$

${(k - 3)}^{2} = 100 - 64 = 36$ , so k − 3 = 6 or k − 3 = −6:

k = 9 or k = −3

(b)(i) $\vec{B C} = \vec{A D} = p$ (opposite sides of the parallelogram), and BG : GC = 3 : 1 makes $\vec{B G} = \frac{3}{4} \vec{B C} = \frac{3}{4} p$ , so

$\vec{A B} = \vec{A G} + \vec{G B} = q - \frac{3}{4} p$

In the parallelogram $\vec{D C} = \vec{A B}$ , so $\vec{C D}$ is its reverse:

$\vec{C D} = - \vec{A B} = - (q - \frac{3}{4} p) =$ $\frac{3}{4} p - q$

(ii) $\vec{A F} = \frac{1}{2} \vec{A B} = \frac{1}{2} (q - \frac{3}{4} p)$ , so

$\vec{G F} = \vec{G A} + \vec{A F} = - q + \frac{1}{2} q - \frac{3}{8} p = - \frac{3}{8} p - \frac{1}{2} q$

$\vec{A H} = \vec{A D} + \vec{D H} = \vec{A D} + \vec{G F} = p - \frac{3}{8} p - \frac{1}{2} q =$ $\frac{5}{8} p - \frac{1}{2} q$

Why this works. q is $\vec{A G}$ , to a point inside the parallelogram, so $\vec{A B}$ is not given; building it, $\vec{A B}$ = q − $\frac{3}{4} p$ , earns most marks. Then follow the letters: $\vec{C D}$ runs C to D, and $\vec{D H}$ = $\vec{G F}$ reaches H by $\vec{G F}$ , not $\vec{F G}$ .

G3 N2022 · Paper 2 · Q8(b) 5 marks

OCD is a triangle.
A is a point on OC and B is a point on OD.
$\vec{O A} = a$ and $\vec{O B} = b$ .
OA = $\frac{1}{3}$ OC and OB : BD = 3 : 2.
X is a point on OC such that BX is parallel to DC.
Find $\vec{X D}$ .
Give your answer as simply as possible in terms of a and b. [5]

Worked solution

$\vec{O C} = 3 a$ , and OB : BD = 3 : 2 makes OD five of those parts, so $\vec{O D} = \frac{5}{3} b$ .

Because BX is parallel to DC, triangles OBX and ODC have equal corresponding angles, so they are similar and

$\frac{O X}{O C} = \frac{O B}{O D} = \frac{3}{5}$

$\vec{O X} = \frac{3}{5} \vec{O C} = \frac{3}{5} \times 3 a = \frac{9}{5} a$

$\vec{X D} = \vec{X O} + \vec{O D} = - \frac{9}{5} a + \frac{5}{3} b$

$\vec{X D} = \frac{5}{3} b - \frac{9}{5} a$

Check without similarity: put $\vec{O X} = t \vec{O C} = 3 t a$ . Then $\vec{B X} = 3 t a - b$ and $\vec{D C} = 3 a - \frac{5}{3} b$ ; these are parallel only when $3 t = \frac{3}{5} \times 3$ , that is $t = \frac{3}{5}$ , giving the same $\vec{O X}$ .

Why this works. Every intermediate statement is paid for here. Read the two ratios carefully: OA = $\frac{1}{3}$ OC makes OC three as, while OB : BD = 3 : 2 makes OD five thirds of b, not $\frac{3}{5}$ . BX parallel to DC makes X cut OC in the same 3 : 5, and the first leg from X is minus $\vec{O X}$ .

Walkthroughs — Vectors in Terms of Two Others

Step through each one; the why beside every line is the part a marker is actually reading.

Walkthrough 13 — every side and diagonal of a parallelogram Basic

The diagram shows a parallelogram OACB in which

\vec{O A} = 3 a

and

\vec{O B} = 4 b

. Express the following in terms of a and/or b. (i)

\vec{B C}

(ii)

\vec{A C}

(iii)

\vec{O C}

(iv)

\vec{C O}

(v)

\vec{A B}

(vi)

\vec{B A}

Fig. 12.15

OACB: the letters go round the parallelogram, so OA and BC are one pair of opposite sides and OB and AC are the other. The dashed lines are the two diagonals.

$\vec{B C} = \vec{O A} = 3 a$ (equal vectors)

BC and OA are opposite sides of the parallelogram, so they are parallel and the same length; and read in the order given, both arrows point the same way (left to right). Same length, same direction, so they are equal vectors.
$\vec{A C} = \vec{O B} = 4 b$ (equal vectors)

The other pair of opposite sides, and again both arrows run the same way (upwards on the diagram). Reversing the order would give $\vec{C A}$ = −4b, so the letters come from the question rather than the picture.
$\vec{O C} = \vec{O A} + \vec{O B} = 3 a + 4 b$ (Parallelogram Law)

OC is the diagonal from the corner where the two vectors meet, so it is their sum. The Triangle Law agrees: $\vec{O C} = \vec{O A} + \vec{A C} = 3 a + 4 b$ .
$\vec{C O} = - \vec{O C} = - (3 a + 4 b) = - 3 a - 4 b$

The same diagonal travelled backwards, so it is the negative of the line above. Expanding the bracket flips the sign of both terms — writing −3a + 4b here is the classic half-done negative.
$\vec{A B} = \vec{O B} - \vec{O A} = 4 b - 3 a$ (Triangle Law of Subtraction)

This is the other diagonal, and both a and b start at O, so it is a difference. The arrow ends at B, so B's vector comes first: end minus start.
$\vec{B A} = - \vec{A B} = 3 a - 4 b$
(i) 3a (ii) 4b (iii) 3a + 4b (iv) −3a − 4b (v) 4b − 3a (vi) 3a − 4b

$\vec{B A}$ is the reverse of $\vec{A B}$ , so it is again a sign change. Notice the shape of the answer: the two diagonals are the sum and the difference, each with a negative twin.

Walkthrough 14 — reading a vector off a grid, two ways Basic

The diagram shows two vectors u and v and the vector

\vec{A B}

. Express

\vec{A B}

in terms of u and v, in two different ways.

Fig. 12.16

The two given vectors are drawn in the strip on the left; the same

\vec{A B}

is shown with each of the two detours.

From the reference arrows, u is 1 right and 2 up, and v is 2 right and 0 up.

Before any route can be built, the two building blocks have to be read off the grid in squares. Counting them once, carefully, is what makes every later leg checkable.
Draw a line from A parallel to u, and a line back from B parallel to v. They meet at C(3, 5).

The detour is being constructed, not guessed: one leg parallel to each of the two given vectors, so that every leg is bound to be a multiple of u or of v. Where the two lines cross is the corner of the journey.
$\vec{A C} = 2 u, \vec{C B} = 3 v$

AC is 2 right and 4 up, which is twice 1 right and 2 up; CB is 6 right and 0 up, which is three times 2 right. Each leg is counted in squares and then divided by the block it is made of.
$\vec{A B} = \vec{A C} + \vec{C B} = 2 u + 3 v$

The Triangle Law on the detour: the C at the end of the first leg is the C at the start of the second, so the two legs add to the single vector from A to B.
Going the other way round, via D(7, 1): $\vec{A B} = \vec{A D} + \vec{D B} = 3 v + 2 u$ so $\vec{A B} = 2 u + 3 v$ , by either route.

Same two legs, taken in the other order, so the answer has to agree — vector addition is commutative. And it must agree, because 7.6 says the pair of numbers 2 and 3 is unique: there is no third expression to find.

Walkthrough 15 — the midpoint of a side Intermediate

In the triangle OAB,

\vec{O A} = 4 a

and

\vec{O B} = 6 b

. M is the midpoint of AB. Express, in terms of a and b: (i)

\vec{A B}

(ii)

\vec{A M}

(iii)

\vec{O M}

Fig. 12.17

The tick marks say AM = MB, which is the whole content of “M is the midpoint”.

$\vec{A B} = \vec{O B} - \vec{O A} = 6 b - 4 a$

Both given vectors start at O, so the side AB is a difference: end minus start. This first line is the one everything else in the question is built on, so it is worth writing the rule out in letters before substituting.
$\vec{A M} = \frac{1}{2} \vec{A B} = \frac{1}{2} (6 b - 4 a) = 3 b - 2 a$

M is the midpoint, so AM is half of AB — and in the same direction, since M lies between A and B. Halving multiplies both terms by ½.
$\vec{O M} = \vec{O A} + \vec{A M} = 4 a + 3 b - 2 a$

To reach M from O, go to A first and then along AM — a two-leg journey whose legs are both already known. This is the routing recipe in one line.
$\vec{O M} = 2 a + 3 b$

Collecting the a terms: 4a − 2a = 2a. Like terms in a vector expression collect exactly like terms in ordinary algebra, because a and b are non-parallel and so cannot be turned into each other.
(i) $6 b - 4 a$ (ii) $3 b - 2 a$ (iii) $2 a + 3 b$

The midpoint of AB is the "average" of A and B: $\frac{1}{2} (\vec{O A} + \vec{O B}) = \frac{1}{2} (4 a + 6 b) = 2 a + 3 b$ A result worth remembering.

Walkthrough 16 — a trapezium, and what the answer tells you Intermediate

ABCD is a trapezium in which AD is parallel to BC and BC = 2AD. It is given that

\vec{A B} = u

and

\vec{A D} = v

. P is the midpoint of AB and Q is the midpoint of DC. Express, in terms of u and v: (i)

\vec{A C}

(ii)

\vec{D C}

(iii)

\vec{A Q}

(iv)

\vec{P Q}

(v) What does your answer to (iv) tell you about PQ and AD?

Fig. 12.18

AD and BC are the parallel sides, with BC twice the length of AD. The single ticks mark the midpoint P and the double ticks the midpoint Q.

$\vec{B C} = 2 \vec{A D} = 2 v$

BC is parallel to AD and twice as long, so it is 2 times AD. The diagram shows both arrows pointing the same way, so the scalar is positive — a ratio of lengths alone would not settle that.
$\vec{A C} = \vec{A B} + \vec{B C} = u + 2 v$

A route from A to C made of two known legs: along the bottom to B, then up the side to C. Both legs are now in terms of u and v, which is all the question wants.
$\vec{D C} = \vec{A C} - \vec{A D} = u + 2 v - v = u + v$

D and C are both reachable from A, so the shared start makes this a difference: end minus start. The v terms collect to leave one of each letter.
$\vec{A Q} = \vec{A D} + \frac{1}{2} \vec{D C} = v + \frac{1}{2} (u + v) = \frac{1}{2} u + \frac{3}{2} v$

Go up the left side to D, then half way along DC, because Q is its midpoint. The expression for $\vec{D C}$ from the line above is what makes that half-leg usable.
$\vec{P Q} = \vec{A Q} - \vec{A P} = \frac{1}{2} u + \frac{3}{2} v - \frac{1}{2} u = \frac{3}{2} v$

P is the midpoint of AB, so $\vec{A P}$ = ½u, and the difference of the two routes from A is the vector from P to Q. The u terms cancel exactly — which is the interesting part.
$\vec{P Q}$ = $\frac{3}{2} \vec{A D}$ , so PQ is parallel to AD (and therefore to BC), and PQ = 1½AD.

A vector that is a scalar multiple of another is parallel to it — the inference from 12.3 and what part (v) asks for. The scalar is positive, so the two point the same way; its size is the ratio of their lengths.

Walkthrough 17 — going through the position vectors Advanced

In the diagram,

\vec{O U} = 9 u

and

\vec{O V} = 12 v

. A is the point on OU such that

\vec{O A} = \frac{1}{3} \vec{O U}

, B is the point on OV such that

\vec{O B} = \frac{1}{4} \vec{O V}

, and X is the point on AV such that

\vec{A X} = \frac{1}{3} \vec{A V}

. (i) Express

\vec{U V}

and

\vec{A B}

in terms of u and v. (ii) Find the position vector of X relative to O. (iii) Hence express

\vec{X B}

in terms of u and v, as simply as possible.

Fig. 12.19

A lies a third of the way along OU, B a quarter of the way along OV, and X a third of the way along the dashed line AV.

$\vec{U V} = \vec{O V} - \vec{O U} = 12 v - 9 u = 3 (4 v - 3 u)$

Both U and V are given from O, so the side UV is end minus start. Taking the common factor 3 out is not required, but it is what "as simply as possible" would want later.
$\vec{O A} = \frac{1}{3} (9 u) = 3 u, \vec{O B} = \frac{1}{4} (12 v) = 3 v$

A fraction of a known vector is a scalar multiple of it, so each of these is one multiplication. Both A and B are now position vectors measured from O, which is what makes every later step a subtraction.
$\vec{A B} = \vec{O B} - \vec{O A} = 3 v - 3 u = 3 (v - u)$

End minus start again, this time on the two points just found. That finishes part (i).
$\vec{A X} = \frac{1}{3} \vec{A V} = \frac{1}{3} (\vec{O V} - \vec{O A}) = \frac{1}{3} (12 v - 3 u) = 4 v - u$

AX is a third of AV, and AV itself is end minus start from A to V. The whole leg has to be found before a third of it means anything.
$\vec{O X} = \vec{O A} + \vec{A X} = 3 u + 4 v - u = 2 u + 4 v$

To reach X from the origin, go to A and then along AX — the routing recipe, with both legs already known. Collecting: 3u − u = 2u.
$\vec{X B} = \vec{O B} - \vec{O X} = 3 v - (2 u + 4 v) = - 2 u - v = - (2 u + v)$

One last end-minus-start, from X to B, using the position vector just found. "As simply as possible" asks for the factor to be taken out at the end, and the minus sign is real: $\vec{X B}$ points back towards O.

Check yourself

OACB is a parallelogram with $\vec{O A} = a$ and $\vec{O B} = b$ . Write down $\vec{O C}$ and $\vec{A B}$ .
Answer
$\vec{O C} = a + b$ (the diagonal from the shared corner is the sum) and $\vec{A B} = b - a$ (the other diagonal is a difference — end minus start).
In a triangle OXY, $\vec{O X} = 2 x$ and $\vec{O Y} = 5 y$ . N is the midpoint of XY. Express $\vec{X Y}$ and $\vec{O N}$ in terms of x and y.
Answer
$\vec{X Y} = 5 y - 2 x$ $\vec{O N} = \vec{O X} + \frac{1}{2} \vec{X Y} = 2 x + \frac{5}{2} y - x = x + \frac{5}{2} y$ (the midpoint is always the average of the two ends).
In a triangle OAB, $\vec{O A} = a$ and $\vec{O B} = b$ , and P lies on AB with $\vec{A P} = \frac{1}{4} \vec{A B}$ . Express $\vec{O P}$ in terms of a and b.
Answer
$\vec{O P} = a + \frac{1}{4} (b - a) = \frac{3}{4} a + \frac{1}{4} b = \frac{1}{4} (3 a + b)$
$\vec{A B} = 3 u - 2 v$ and $\vec{B C} = u + 5 v$ . Express $\vec{A C}$ in terms of u and v.
Answer
$\vec{A C} = \vec{A B} + \vec{B C} = 3 u - 2 v + u + 5 v = 4 u + 3 v$
Why must the two vectors in outcome 7.6 be non-parallel?
Answer
If v were parallel to u then v = ku, so every combination mu + nv would collapse to (m + nk)u — a multiple of u alone. Every vector you could build would lie along one line, and any vector pointing off that line would be unreachable.
$\vec{A B} = m u + n v$ and also $\vec{A B} = 7 u - 4 v$ . What can you say about m and n?
Answer
m = 7 and n = −4. Outcome 7.6 says the expression is unique, so the two forms must match term for term — there is no second pair of numbers that also works.

Two inferences — and outcome 7.8 is made of them

By the end of a "show that" question you will have some vector expression such as $\vec{P R} = 2 \vec{A B}$ . The whole point of the question is what you are then allowed to say — and there are only two things, both about scalar multiples.

Inference 1 — parallel and in what ratio. If

\vec{A B} = k \vec{C D}

(with k a non-zero number and neither vector the zero vector), then

AB is parallel to CD;
AB = $| k |$ × CD — the lengths are in the ratio $| k |$ : 1;
they point the same way if k > 0 and opposite ways if k < 0.

Inference 2 — collinear. If

\vec{A B} = k \vec{C D}

and the two vectors share a point, then that point cannot be a corner: the three points lie on one straight line. So to prove that X, Y and Z are collinear, show

\vec{X Y} = k \vec{Y Z}

— two vectors that both use Y.

The shared point is not optional.

\vec{A B} = k \vec{C D}

on its own says only that the two segments are parallel; the letter they have in common is what makes three points collinear.

Finish in words. The algebra is the working; the sentence is the answer. "∴

\vec{P R} = 2 \vec{A B}

, so PR is parallel to AB and PR = 2AB. (shown)"

More detail

Two parallel segments drawn in different places are not collinear at all, so the shared letter is what closes the argument — a proof that leaves it unsaid is an incomplete proof. And the sentence in words is the last line the "show that" is asking for: the algebra on its own stops one step short.

Past-paper questions

G3 N2016 · Paper 1 · Q25 6 marks

OABC is a quadrilateral.
$\vec{O A} = 5 a$ , $\vec{O C} = 5 c$ , $\vec{C B} = 10 a + 5 c$ and $\vec{A Q} = 4 a + 8 c$ .
AP : PC = 4 : 1.

Write each of the following in terms of a and c.
Give your answers in their simplest form.
1. $\vec{O P}$ [2]
2. $\vec{P Q}$ [2]
1. Explain why PQ is parallel to CB. [1]
2. Find the ratio CB : PQ. [1]

Worked solution

(a)(i) $\vec{A C} = \vec{A O} + \vec{O C} = - 5 a + 5 c$

AP : PC = 4 : 1 makes AP four of the five parts of AC:

$\vec{A P} = \frac{4}{5} \vec{A C} = \frac{4}{5} (- 5 a + 5 c) = - 4 a + 4 c$

$\vec{O P} = \vec{O A} + \vec{A P} = 5 a - 4 a + 4 c =$ $a + 4 c$

(ii) $\vec{P Q} = \vec{P A} + \vec{A Q} = (4 a - 4 c) + (4 a + 8 c) =$ $8 a + 4 c$

(b)(i) $\vec{P Q} = 8 a + 4 c = 4 (2 a + c)$

$\vec{C B} = 10 a + 5 c = 5 (2 a + c)$

So $\vec{C B} = \frac{5}{4} \vec{P Q}$ . One vector is a scalar multiple of the other, so PQ is parallel to CB.

(ii) Since $\vec{C B} = \frac{5}{4} \vec{P Q}$ , the lengths are in the same proportion: CB : PQ = 5 : 4.

Why this works. This is Inference 1: the multiple gives parallel, and the scalar gives the ratio of lengths. Factorising over one bracket, $4 (2 a + c)$ and $5 (2 a + c)$ , makes the common direction visible. In (a)(ii) $\vec{P Q}$ starts at P, opening with $\vec{P A}$ , which is minus $\vec{A P}$ .

Midpoints: the shape hiding inside every quadrilateral

Join the midpoints of the four sides of any quadrilateral, however lopsided, and you always get a parallelogram. Vectors prove it in six lines; without them it is genuinely hard.

Fig. 12.20

The outer quadrilateral KLMN is deliberately irregular. P, Q, R and S are the midpoints of its four sides, and the inner shape PQRS comes out a parallelogram every time. The dashed diagonal KM is the link between the two halves of the proof.

The engine is one line, used twice: in a triangle, the segment joining the midpoints of two sides is half the third side and parallel to it. If P and Q are the midpoints of LK and LM, then $\vec{K M} = 2 \vec{P Q}$ .

Apply that in the triangle on each side of the diagonal, and both halves give the same vector.

Collinearity, and the ratio that comes free with it

Once you have shown $\vec{C M} = k \vec{C N}$ , you have not only proved the three points are in a line — you have also found how the line is divided, at no extra cost.

Fig. 12.21

A parallelogram OABC, with M the midpoint of AB and N two thirds of the way along the diagonal OB. The amber line shows what the algebra proves: C, N and M really are in a straight line.

From the scalar to the ratio. If $\vec{C M} = \frac{3}{2} \vec{C N}$ then, since both start at C, N lies on CM with CN : CM = 2 : 3, and therefore CN : NM = 2 : 1.

More detail

The two ratios are different questions. "CN to the whole" and "CN to the rest" give different answers — here 2 : 3 against 2 : 1 — and the question says which one it wants, so it is worth reading that line twice before writing the ratio down.

Past-paper questions

G3 N2020 · Paper 1 · Q24 7 marks

OABC is a quadrilateral.
$\vec{O A} = a$ , $\vec{O C} = c$ and $\vec{C P} = m (a - c)$ .

Write $\vec{O P}$ in terms of a, c and m. [1]
OB→=4(1−m)c+4ma.
1. Show that O, P and B lie on a straight line. [2]
2. $\vec{C B} = k a$ .
  Find the value of m and the value of k. [3]
3. Write down the ratio CP : CA. [1]

Worked solution

(a) $\vec{O P} = \vec{O C} + \vec{C P} = c + m (a - c) =$ $m a + (1 - m) c$

(b)(i) $\vec{O B} = 4 m a + 4 (1 - m) c = 4 [m a + (1 - m) c] = 4 \vec{O P}$

$\vec{O B}$ is a scalar multiple of $\vec{O P}$ , so OB is parallel to OP; and the two lines share the point O. Two parallel lines through a common point are the same line, so O, P and B lie on a straight line (shown).

(ii) $\vec{C B} = \vec{O B} - \vec{O C} = 4 m a + 4 (1 - m) c - c$

$\vec{C B} = 4 m a + (3 - 4 m) c$

$\vec{C B} = k a$ has no c in it, and a and c are not parallel, so the c coefficient must be zero:

3 − 4m = 0 ⇒ m = $\frac{3}{4}$ , and then k = 4m = 3.

(iii) $\vec{C A} = \vec{C O} + \vec{O A} = a - c$ , and $\vec{C P} = \frac{3}{4} (a - c) = \frac{3}{4} \vec{C A}$ , so CP : CA = 3 : 4.

Why this works. Part (b)(i) is the whole collinearity argument, and both halves are needed: the multiple gives "parallel", and the shared point O makes it "one straight line". Since a and c are not parallel, the equation splits per letter, so $\vec{C B}$ = ka says the c coefficient is zero. (iii) then follows: $\vec{C P}$ is $\frac{3}{4}$ of $\vec{C A}$ .

G3 N2023 · Paper 1 · Q27 5 marks

OMN is a triangle.
P is the point on MN such that MP : PN = 3 : 2.
$\vec{O M} = a$ and $\vec{O N} = b$ .

Show that $\vec{O P} = \frac{1}{5} (2 a + 3 b)$ . [2]
Q is the point such that $\vec{M Q} = \frac{1}{5} (a + 9 b)$ .
Explain why O, P and Q lie on a straight line. [3]

Worked solution

(a) $\vec{M N} = \vec{M O} + \vec{O N} = b - a$ , and MP : PN = 3 : 2 makes MP three of the five parts of MN.

$\vec{O P} = \vec{O M} + \frac{3}{5} \vec{M N} = a + \frac{3}{5} (b - a)$

$= \frac{2}{5} a + \frac{3}{5} b =$ $\frac{1}{5} (2 a + 3 b)$ (shown)

(b) $\vec{O Q} = \vec{O M} + \vec{M Q} = a + \frac{1}{5} (a + 9 b)$

$= \frac{6}{5} a + \frac{9}{5} b = \frac{3}{5} (2 a + 3 b)$

$\vec{O Q} = 3 \times \frac{1}{5} (2 a + 3 b) = 3 \vec{O P}$

$\vec{O Q}$ is a scalar multiple of $\vec{O P}$ , so OQ is parallel to OP; they also share the point O. Hence O, P and Q lie on a straight line (and OP : PQ = 1 : 2).

Why this works. In (a), 3 : 2 makes MP three fifths of MN, the printed $\frac{1}{5} (2 a + 3 b)$ . Then: $\vec{O Q}$ in a and b; $\vec{O Q}$ = 3 $\vec{O P}$ ; and a conclusion naming the common point. From $\vec{M Q}$ : $\vec{O Q}$ = $\vec{O M}$ + $\vec{M Q}$ .

Ratios of areas ⓘ Where this comes from

The last part of a long vectors question very often asks for a ratio of areas, and with no formula printed and no heights given there are exactly two rules that can answer it. Deciding which one applies is the whole skill.

Rule 1 — similar triangles. For two similar triangles with areas

A_{2}

and

A_{1}

and corresponding lengths

l_{2}

and

l_{1}

,

\frac{A_{2}}{A_{1}} = {(\frac{l_{2}}{l_{1}})}^{2}

Rule 2 — triangles with the same height. For two triangles with the same height and bases

b_{2}

and

b_{1}

,

\frac{A_{2}}{A_{1}} = \frac{b_{2}}{b_{1}}

Fig. 12.22

Left: similar triangles — BC is parallel to DE, so △ABC and △ADE are similar and their areas are in the ratio of the squares of corresponding lengths. Right: the same height — △PQR and △PRS stand on the same line with the same apex, so their areas are in the ratio of their bases.

Which rule? Ask what the two triangles have in common.

Two angles equal → similar → ratio of squares.
Bases along one straight line, same apex → same height → ratio of bases.

The vectors did the hard part already. Vector algebra produces a ratio of lengths, and one of the two rules above turns it into a ratio of areas.

More detail

The equal angles in Rule 1 usually come from a pair of parallel sides, giving corresponding or alternate angles; Rule 2 needs only a common apex over bases on one line. Choosing between squaring and not squaring is what this last part is testing.

A ratio like $\frac{S U}{R U}$ comes straight out of an earlier part, where you found $\vec{S U} = 2 a, \vec{R U} = - 6 a$ . That is how the two halves of a 7.8 question fit together: the vector algebra produces a ratio of lengths, and one of the two rules above turns it into a ratio of areas.

Past-paper questions

G3 N2015 · Paper 2 · Q2 8 marks

OCD is a triangle where A and M are points on OC and B is a point on OD.
$\vec{O A} = a$ , $\vec{O B} = b$ , $\vec{M B} = b - 2 a$ , and $\vec{M D} = 3 b - 2 a$ .
AM = MC.

Express, as simply as possible, in terms of a and/or b,
1. $\vec{A B}$ [1]
2. $\vec{B D}$ [1]
Show that $\vec{C D} = 3 b - 3 a$ . [1]
Show that triangles OAB and OCD are similar.
Give a reason for each statement you make. [3]
Find the ratio of the area of triangle OAB to the area of quadrilateral ABDC. [2]

Worked solution

(a)(i) $\vec{A B} = \vec{A O} + \vec{O B} = - a + b =$ $b - a$

(ii) $\vec{B D} = \vec{B M} + \vec{M D} = - (b - 2 a) + (3 b - 2 a) =$ $2 b$

(b) First locate M and C: $\vec{O M} = \vec{O B} + \vec{B M} = b + 2 a - b = 2 a$ .

So $\vec{A M} = 2 a - a = a$ , and since AM = MC, $\vec{M C} = a$ and $\vec{O C} = 3 a$ .

$\vec{O D} = \vec{O B} + \vec{B D} = b + 2 b = 3 b$ , so

$\vec{C D} = \vec{C O} + \vec{O D} = - 3 a + 3 b =$ $3 b - 3 a$ (shown)

(c) $\vec{O C} = 3 \vec{O A}$ , so OC = 3 × OA.
$\vec{O D} = 3 \vec{O B}$ , so OD = 3 × OB.
$\vec{C D} = 3 (b - a) = 3 \vec{A B}$ , so CD = 3 × AB.

All three pairs of corresponding sides are in the same ratio 1 : 3, so triangles OAB and OCD are similar (SSS similarity). Equivalently, AB is parallel to CD (one is a scalar multiple of the other), so the corresponding angles $∠ O A B$ and $∠ O C D$ are equal and the angle at O is common.

(d) For similar triangles the areas are in the ratio of the squares of corresponding sides:

$\frac{area O A B}{area O C D} = {(\frac{1}{3})}^{2} = \frac{1}{9}$

The quadrilateral ABDC is what is left of triangle OCD when triangle OAB is removed, so it is 9 − 1 = 8 of those parts:

area OAB : area ABDC = 1 : (9 − 1) = 1 : 8

Why this works. One fact is never stated: $\vec{O M}$ = 2a, got by reversing the given $\vec{M B}$ ; from there AM = MC makes OC = 3a. Part (c) wants "a reason for each statement", so three sentences, about lengths or angles. Part (d) asks for the quadrilateral, not the triangle, so subtract: 1 : 9 answers a different question.

G3 N2017 · Paper 2 · Q6 9 marks

ABCD is a parallelogram.
$\vec{A B} = 2 p$ and $\vec{B C} = 3 q$ .
R is the point on AD such that AR : RD = 2 : 1.

Express $\vec{R C}$ , as simply as possible, in terms of p and q. [2]
S is a point on AD and T is a point on CD such that RC→=2ST→.
1. Show that triangles RCD and STD are similar.
  Give a reason for each statement you make. [3]
2. Find the ratio area triangle STD : area parallelogram ABCD. [2]
3. Q is a point on BC.
  Triangle QAB is congruent to triangle RCD.
  Find $\vec{Q R}$ , as simply as possible, in terms of p and q. [2]

Worked solution

(a) $\vec{A D} = \vec{B C} = 3 q$ (opposite sides of the parallelogram), and AR : RD = 2 : 1 makes $\vec{R D} = \frac{1}{3} \vec{A D} = q$ .

$\vec{R C} = \vec{R D} + \vec{D C} = q + 2 p =$ $2 p + q$

(b)(i) $\vec{R C} = 2 \vec{S T}$ , so $\vec{S T}$ is a scalar multiple of $\vec{R C}$ and ST is parallel to RC.

$∠ D S T = ∠ D R C$ (corresponding angles, ST parallel to RC)
$∠ D T S = ∠ D C R$ (corresponding angles, ST parallel to RC)
$∠ S D T = ∠ R D C$ (same angle)

Three pairs of corresponding angles are equal, so triangles STD and RCD are similar (AA).

(b)(ii) The similar triangles have corresponding sides in the ratio ST : RC = 1 : 2, so their areas are in the ratio 1² : 2² = 1 : 4.

Triangle RCD stands on base RD with apex C, and the parallelogram stands on AD with the same perpendicular height h:

$\frac{area R C D}{area A B C D} = \frac{\frac{1}{2} \times R D \times h}{A D \times h} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

$area S T D = \frac{1}{4} \times \frac{1}{6} = \frac{1}{24} of the parallelogram$

area STD : area ABCD = 1 : 24

(b)(iii) In the congruent triangles, QB corresponds to RD, so $\vec{Q B} = - q$ — Q lies one third of the way along BC, with $\vec{B Q} = q$ . (Check: $\vec{Q A} = - 2 p - q = \vec{R C}$ reversed, and $\vec{A B} = 2 p = - \vec{C D}$ .)

$\vec{Q R} = \vec{Q B} + \vec{B A} + \vec{A R} = - q - 2 p + 2 q =$ $q - 2 p$

Why this works. Mostly geometry that vectors unlocked. In (a), AR : RD = 2 : 1 makes RD a third of AD. Part (ii) needs two rules: similar triangles compare by squares, while RCD and the parallelogram compare by base and height, giving the $\frac{1}{2}$ . In (iii) matching QAB to RCD pins Q, and $\vec{Q R}$ is a route.

G3 N2024 · Paper 1 · Q27 7 marks

OAB is a triangle.
C is the point on BA such that BC : CA = 2 : 1.
$\vec{O A} = 3 a$ and $\vec{O B} = 6 b$ .

Show that the position vector of C is given by $\vec{O C} = 2 a + 2 b$ . [2]
P is the midpoint of OC and Q is a point on OB such that APQ is a straight line.
AQ = mAP and OQ = nOB where m and n are numbers.
Find the ratio OQ : QB. [4]
The area of triangle OBC = 25 cm².
Find the area of triangle OAC. [1]

Worked solution

(a) $\vec{B A} = \vec{O A} - \vec{O B} = 3 a - 6 b$

BC : CA = 2 : 1 makes BC two of the three parts of BA, so

$\vec{O C} = \vec{O B} + \frac{2}{3} \vec{B A} = 6 b + \frac{2}{3} (3 a - 6 b)$

$= 6 b + 2 a - 4 b =$ $2 a + 2 b$ (shown)

(b) $\vec{O P} = \frac{1}{2} \vec{O C} = a + b$ , so

$\vec{A P} = \vec{O P} - \vec{O A} = a + b - 3 a = b - 2 a$

$\vec{O Q} = \vec{O A} + m \vec{A P} = 3 a + m (b - 2 a) = (3 - 2 m) a + m b$

But $\vec{O Q} = n \vec{O B} = 6 n b$ , which has no a in it. Comparing coefficients (a and b are not parallel):

3 − 2m = 0 ⇒ m = $\frac{3}{2}$ , and 6n = m ⇒ n = $\frac{1}{4}$ .

So $\vec{O Q} = \frac{1}{4} \vec{O B}$ and QB is the other three quarters: OQ : QB = 1 : 3.

(c) Triangles OBC and OAC stand on BC and CA, which lie on the same line BA, and both have apex O — so they have the same perpendicular height and their areas are in the ratio of their bases, 2 : 1.

$area O A C = \frac{1}{2} \times 25 = 12.5$ area = 12.5 cm²

Why this works. Two unknowns and one vector equation is enough, because an equation in two non-parallel directions is really two. Write $\vec{O Q}$ twice, once along APQ and once along OB, then compare a and b coefficients; the a coefficient vanishes because $\vec{O Q}$ is a multiple of b. n = $\frac{1}{4}$ is not the answer: OQ : QB is 1 : 3.

Walkthroughs — Geometric Problems

Step through each one; the why beside every line is the part a marker is actually reading.

Walkthrough 18 — joining the midpoints Intermediate

KLMN is a quadrilateral, and P, Q, R and S are the midpoints of KL, LM, MN and NK respectively. Show that (i) KM is parallel to PQ, and KM = 2PQ, (ii) PQRS is a parallelogram.

Fig. 12.23

The midpoints joined in order, with the diagonal KM dashed.

Let $\vec{L P} = p$ and $\vec{L Q} = q$ .
$\vec{P Q} = \vec{L Q} - \vec{L P} = q - p$

Naming two vectors that both start at L turns the picture into algebra, and it makes $\vec{P Q}$ a plain difference — end minus start. Everything in part (i) will now be written in p and q.
$\vec{L K} = 2 \vec{L P} = 2 p, \vec{L M} = 2 \vec{L Q} = 2 q$

P is the midpoint of KL, so LK is twice LP; likewise for LM. Both are written from L so that they can be subtracted in the next line.
$\vec{K M} = \vec{L M} - \vec{L K} = 2 q - 2 p = 2 (q - p)$

The diagonal is end minus start from K to M, with both position vectors measured from L. Taking the 2 out is what makes the comparison with $\vec{P Q}$ visible.
$\vec{K M} = 2 \vec{P Q}$
Since $\vec{K M} = 2 \vec{P Q}$ , KM is parallel to PQ and KM = 2PQ. (shown)

A vector equal to a scalar multiple of another is parallel to it, and the scalar gives the ratio of the lengths — Inference 1. The sentence in words is the part the "show that" is marked on.
In the same way, in triangle NKM with S and R the midpoints of NK and NM, $\vec{K M} = 2 \vec{S R}$

The diagonal KM splits the quadrilateral into two triangles, and the argument just used works word for word in the second one. Nothing new is needed — only the observation that S and R are midpoints of the two sides of that triangle.
$2 \vec{P Q} = 2 \vec{S R}, so \vec{P Q} = \vec{S R}$
So PQ is parallel to SR and PQ = SR: one pair of opposite sides of PQRS is equal and parallel, so PQRS is a parallelogram. (shown)

Two expressions both equal to $\vec{K M}$ must be equal to each other, and equal vectors means same length and same direction — which is exactly the condition for a parallelogram. Notice that nothing anywhere assumed the outer quadrilateral was any particular shape.

Walkthrough 19 — proving three points are in a straight line Advanced

OABC is a parallelogram in which

\vec{O A} = a

and

\vec{O C} = c

. M is the midpoint of AB, and N is the point on OB such that

\vec{O N} = \frac{2}{3} \vec{O B}

. (i) Express

\vec{O B}

,

\vec{C M}

and

\vec{C N}

in terms of a and c. (ii) Show that C, N and M lie on a straight line. (iii) Find CN : NM.

Fig. 12.24

OABC in order round the parallelogram, so AB is the side opposite OC. M is the midpoint of AB; N sits two thirds of the way along the diagonal OB.

$\vec{O B} = \vec{O A} + \vec{O C} = a + c$ (Parallelogram Law)

OB is the diagonal from the corner where a and c meet, so it is their sum. This is the vector both of the interesting points are measured along, so it is worth having first.
$\vec{O M} = \vec{O A} + \frac{1}{2} \vec{A B} = a + \frac{1}{2} c$

To reach M, go along OA and then half of AB — and AB is the side opposite OC, so it is the equal vector c. Spotting that equal pair is what keeps this to one line.
$\vec{C M} = \vec{O M} - \vec{O C} = a + \frac{1}{2} c - c = a - \frac{1}{2} c = \frac{1}{2} (2 a - c)$

End minus start from C to M, using the position vector just found. The half is factored out at once, because the comparison in part (ii) is going to be with another multiple of the same bracket.
$\vec{C N} = \vec{O N} - \vec{O C} = \frac{2}{3} (a + c) - c = \frac{2}{3} a - \frac{1}{3} c = \frac{1}{3} (2 a - c)$

Same move from C to N, with $\vec{O N} = \frac{2}{3} (a + c)$ . Collecting the c terms: ⅔c − c = −⅓c, and taking ⅓ out leaves the same bracket as $\vec{C M}$ .
$\vec{C M} = \frac{3}{2} \vec{C N}$
$\vec{C M}$ and $\vec{C N}$ are parallel and both pass through C, so C, N and M are collinear. (shown)

Dividing one expression by the other gives a pure number, so one vector is a scalar multiple of the other: they are parallel. They also share the point C, and parallel plus a shared point is a straight line.
$\vec{N M} = \vec{C M} - \vec{C N} = \frac{3}{2} \vec{C N} - \vec{C N} = \frac{1}{2} \vec{C N}$
CN : NM = 2 : 1

$\vec{C M} = \frac{3}{2} \vec{C N}$ says CM is 1½ times CN, so what is left over, NM, is half of CN. The question asked for CN to NM, which is 2 : 1.

Walkthrough 20 — from vectors to a ratio of areas Advanced

In the diagram, SPQR is a parallelogram in which

\vec{P Q} = 8 a

and

\vec{P S} = 6 b

. U is the point on SR such that SU = ¼SR, and the lines PS and QU, when produced, meet at T. (i) Express

\vec{P R}

,

\vec{S U}

and

\vec{R U}

in terms of a and/or b. (ii) Show that △TSU and △TPQ are similar, and hence find TS : SP. (iii) Calculate the value of (a)

\frac{area of △ T S U}{area of △ Q R U}

, (b)

\frac{area of △ T S U}{area of △ P S U}

.

Fig. 12.25

SPQR is the parallelogram; U is a quarter of the way along SR from S. The lines PS and QU are both produced upwards and meet at T.

$\vec{P R} = \vec{P Q} + \vec{Q R} = \vec{P Q} + \vec{P S} = 8 a + 6 b$ (Parallelogram Law)

PR is the diagonal from P, the corner where the two given vectors meet, so it is their sum — using $\vec{Q R} = \vec{P S}$ , the opposite side of the parallelogram.
$\vec{S U} = \frac{1}{4} \vec{S R} = \frac{1}{4} \vec{P Q} = \frac{1}{4} (8 a) = 2 a$

SR is the side opposite PQ, and the diagram shows both arrows running the same way, so they are equal vectors and $\vec{S U}$ is a genuine quarter of 8a. Arrows drawn the other way would flip the sign.
$\vec{R U} = - \frac{3}{4} \vec{S R} = - \frac{3}{4} (8 a) = - 6 a$

U is a quarter of the way from S, so UR is the remaining three quarters, and $\vec{R U}$ runs back towards U, which is where the minus sign comes from.
Since $\vec{S U} = 2 a, \vec{P Q} = 8 a$ , SU is parallel to PQ, so ∡TSU = ∡TPQ (corresponding angles) and ∡T is common. So △TSU and △TPQ are similar (AA).
$\frac{T S}{T P} = \frac{S U}{P Q} = \frac{2}{8} = \frac{1}{4}$

Two pairs of equal angles is the AA similarity test, and the parallel sides are what produce the first pair. Corresponding sides of similar triangles are in the same ratio, and SU and PQ are the pair whose lengths are already known.
$\frac{T S}{S P} = \frac{1}{3}, i.e. T S : S P = 1 : 3$

S lies between T and P on the line TSP, so SP is what is left of TP after TS is taken away. Three quarters of TP against one quarter gives the ratio.
(a) △TSU and △QRU are similar (∡TUS = ∡QUR, vertically opposite; ∡TSU = ∡QRU, alternate angles, since PT is parallel to QR), and $\frac{S U}{R U} = \frac{2}{6} = \frac{1}{3}$ $\frac{area of △ T S U}{area of △ Q R U} = {(\frac{1}{3})}^{2} = \frac{1}{9}$

Similar triangles, so their areas are in the ratio of the squares of corresponding sides — and the ratio of the sides comes straight out of part (i), where SU = 2a and UR = 6a were found.
(b) △TSU and △PSU have the same height from U to the line TSP, so $\frac{area of △ T S U}{area of △ P S U} = \frac{T S}{S P} = \frac{1}{3}$ (a) $\frac{1}{9}$ (b) $\frac{1}{3}$

These two share an apex and have their bases on one straight line, so it is the same height rule — ratio of bases, not of squares. Two area questions, two different rules; choosing between them is the mark.

Check yourself

$\vec{P Q} = 2 u + 6 v$ and $\vec{R S} = 3 u + 9 v$ . Are PQ and RS parallel?
Answer
$\vec{R S} = 3 u + 9 v = \frac{3}{2} (2 u + 6 v) = \frac{3}{2} \vec{P Q}$ so yes — one is a scalar multiple of the other, and the scalar is positive, so they point the same way.
$\vec{X Y} = 4 a - 2 b$ and $\vec{Y Z} = 6 a - 3 b$ . Show that X, Y and Z are collinear, and find XY : YZ.
Answer
$\vec{X Y} = 4 a - 2 b = \frac{2}{3} (6 a - 3 b) = \frac{2}{3} \vec{Y Z}$ so $\vec{X Y}$ is parallel to $\vec{Y Z}$ ; and they share the point Y, so X, Y and Z are collinear. The scalar ⅔ makes XY : YZ = 2 : 3.
In triangle OAB, P lies on AB with AP : PB = 3 : 5. Find area △OAP : area △OPB.
Answer
Both triangles have the same apex O and bases along the same line AB, so they have the same height and the areas are in the ratio of the bases: 3 : 5. No squaring.
Two similar triangles have corresponding sides in the ratio 2 : 5. What is the ratio of their areas?
Answer
4 : 25 — similar triangles, so the areas are in the ratio of the squares: 2² : 5².
A student writes: " $\vec{A B} = 3 \vec{C D}$ , so A, B, C and D are collinear." What is wrong?
Answer
The scalar multiple proves only that AB is parallel to CD. Collinearity needs the two segments to share a point, and these four letters are all different, so nothing of the kind follows — two parallel segments can sit anywhere on the page.
Show that if M is the midpoint of AB, then $\vec{O M} = \frac{1}{2} (\vec{O A} + \vec{O B})$ .
Answer
$\vec{O M} = \vec{O A} + \frac{1}{2} \vec{A B} = \vec{O A} + \frac{1}{2} (\vec{O B} - \vec{O A}) = \frac{1}{2} \vec{O A} + \frac{1}{2} \vec{O B}$ — the position vector of a midpoint is the average of the two end position vectors.

The notation, in one table

Nothing here is printed on the paper. Learn this table first — two of the eight outcomes are about nothing else.

Written	Means	Kind of object
$\vec{A B}$	the vector from `A` to `B`	vector
a (underlined by hand)	a vector given one name	vector
$(\begin{matrix} x \\ y \end{matrix})$	`x` across, `y` up	vector
$\| \vec{A B} \|$ or $\| a \|$	the length of that arrow, in units	scalar — never negative
$0$	the zero vector: no length, no direction	vector
$\vec{O P}$	the position vector of `P`, measured from the origin	vector

The three laws

\vec{A B} + \vec{B C} = \vec{A C}

\vec{O C} = a + b

\vec{A B} = \vec{O B} - \vec{O A} = b - a

1. Triangle Law of Vector Addition — nose to tail. Every letter cancels except the first and the last; a closed loop gives 0.

2. Parallelogram Law — both from the same corner. With

\vec{O A} = a

and

\vec{O B} = b

in the parallelogram OACB, the sum is the diagonal from the corner the two share.

3. Triangle Law of Vector Subtraction — end minus start. This is the single most useful line in the chapter. Everything in 12.5 and 12.6 starts here.

The arithmetic

(\begin{matrix} p \\ q \end{matrix}) \pm (\begin{matrix} r \\ s \end{matrix}) = (\begin{matrix} p \pm r \\ q \pm s \end{matrix})

k (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} k x \\ k y \end{matrix})

| (\begin{matrix} x \\ y \end{matrix}) | = \sqrt{x^{2} + y^{2}}

| k a | = | k | | a |, | a + b | \neq | a | + | b |

The two inferences that answer "show that"

Parallel.

\vec{A B} = k \vec{C D}

(with k ≠ 0 and neither vector zero) ⇒ AB is parallel to CD, and AB =

| k |

× CD.
Collinear. The same equation plus a shared point ⇒ the three points lie on one straight line. To prove X, Y, Z collinear, show

\vec{X Y} = k \vec{Y Z}

.
Ratio of areas. Similar triangles → ratio of the squares of corresponding lengths. Triangles with the same height → ratio of the bases.

Which tool for which question

The question asks for…	Reach for	Where the marks go
a column vector from a description or a grid	across first (right +), then up (up +)	the column form itself — not (`x`, `y`)
a magnitude	$\sqrt{x^{2} + y^{2}}$	the substitution line; brackets round negatives; units; 3 s.f. only if it is not exact
an unknown in two equal vectors	top = top and bottom = bottom	both equations written down, and both checked
simplify a chain of vectors	the Triangle Law; reverse any vector that will not chain	a single vector as the answer — or bold 0 for a round trip
whether two vectors are parallel	$a = k b$ , solved from both rows	the two values of `k`, and the sentence comparing them
a vector in terms of a and b	a route of known legs; equal vectors and fractions of sides for the legs	every leg named; like terms collected; the bracket taken out if it says "as simply as possible"
the vector between two points	$\vec{A B} = \vec{O B} - \vec{O A}$	the order — end minus start, never the other way
the coordinates of a point	the position vector, then rewritten across	coordinates, not a column vector
show that two lines are parallel	write one vector as `k` times the other	the concluding sentence in words
show that three points are collinear	the same, plus naming the shared point	the shared point — the commonest omission
a ratio of areas	similar → squares; same height → bases	saying why the triangles are similar, or which height is shared

The habits that earn the marks

1. Put the arrow above every

\vec{A B}

and underline every hand-written a — the notation is two of the eight outcomes.
2. Never write a vector where a magnitude is wanted, or a magnitude where a vector is wanted.
3. Mark every known vector on the diagram before starting; equal vectors and fractions of sides are free.
4. Say the arrow out loud — "from A to B" — then write end minus start.
5. A ratio of lengths does not fix a direction: check the arrows on the diagram before you attach a sign.
6. Expand a bracket after a minus sign in full: −(a + b) = −a − b.
7. Collect like terms; a and b are non-parallel and cannot merge.
8. "As simply as possible" means take the common factor out at the end.
9. Finish a "show that" with a sentence, and name the shared point in any collinearity proof.
10. Round once, at the end, to 3 s.f. — and only if the answer is not exact.

Syllabus at a glance

The eight G3 Mathematics (K310) outcomes this chapter covers — the largest single block in the syllabus — quoted from the Secondary Three/Four section, and, as in chapters 5 and 6, the reason the G2 panel beside it is empty.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
7.1	use of notations: $(\begin{matrix} x \\ y \end{matrix})$ , $\vec{A B}$ , a, $\| \vec{A B} \|$ , and $\| a \|$ ✕ Not in G2	12.1 The notation, 12.1 Column vectors, 12.4 Position vectors
7.2	representing a vector as a directed line segment ✕ Not in G2	12.1 Scalars and vectors, 12.1 Column vectors on a grid
7.3	translation by a vector ✕ Not in G2	12.4 Translation by a vector
7.4	position vectors ✕ Not in G2	12.4 Free and position vectors, 12.4 $\vec{P Q} = \vec{O Q} - \vec{O P}$
7.5	magnitude of a vector $(\begin{matrix} x \\ y \end{matrix})$ as $\sqrt{x^{2} + y^{2}}$ ✕ Not in G2	12.1 Magnitude
7.6	use of sum and difference of two vectors to express given vectors in terms of two coplanar vectors ✕ Not in G2	12.5 In terms of two others, 12.5 The routing method, 12.5 Ratios
7.7	multiplication of a vector by a scalar ✕ Not in G2	12.3 Scalar multiples, 12.3 Parallel vectors
7.8	geometric problems involving the use of vectors ✕ Not in G2	12.6 Two inferences, 12.6 Collinearity, 12.6 Ratios of areas

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus, including 7.1's punctuation (the syllabus writes "…, $| \vec{A B} |$ , and $| a |$ ", with the comma before "and"; the Secondary Five version of the same row drops that one comma, which is the only difference between the two lists anywhere in G7). Several things this chapter uses are not named in any of the eight: the Triangle Law and the Parallelogram Law themselves, which are the textbook's names for what 7.6 calls "sum and difference"; free vectors, which are authored only as the contrast that makes 7.4 mean something; and the ratio of areas used in 12.6, which arrives through 7.8 "geometric problems" and through the congruence-and-similarity outcomes rather than through G7. Three-dimensional vectors, unit vectors and the dot product are in neither list and are not taught here — the block is called "Vectors in two dimensions".

🧭 Vectors in Two Dimensions — Revision Notes

The whole chapter on one page

Scalars and vectors

The notation — five ways of writing, five different meanings

Column vectors on a grid

Magnitude — the one formula in the chapter

Equal, negative, parallel — and the zero vector

Walkthroughs — Notation & Directed Line Segments

Check yourself

The Triangle Law — "get there via B"

Longer chains, and the round trip

The Parallelogram Law — when both vectors start together

Adding in column form — component by component

Subtraction — and the one line worth more than any other

Magnitudes do not add

Walkthroughs — Adding & Subtracting Vectors

Check yourself

Multiplying a vector by a number

Parallel vectors — the test, and the inference

Testing two column vectors

Walkthroughs — Scalar Multiples & Parallel Vectors

Check yourself

Free vectors and position vectors

PQ→=OQ→−OP→ — the move everything else rests on

Translation by a vector

Past-paper questions

Going back to coordinates

Past-paper questions

Walkthroughs — Position Vectors & Translations

Check yourself

Every vector in the plane, out of two others

Past-paper questions

The method: find a route made only of us and vs

Points that cut a line in a ratio

Past-paper questions

Walkthroughs — Vectors in Terms of Two Others

Check yourself

Two inferences — and outcome 7.8 is made of them

Past-paper questions

Midpoints: the shape hiding inside every quadrilateral

Collinearity, and the ratio that comes free with it

Past-paper questions

Ratios of areas ⓘ Where this comes from

Past-paper questions

Walkthroughs — Geometric Problems

Check yourself

The notation, in one table

The three laws

The arithmetic

The two inferences that answer "show that"

Which tool for which question

The habits that earn the marks

Syllabus at a glance

The Triangle Law — "get there via `B`"

$\vec{P Q} = \vec{O Q} - \vec{O P}$ — the move everything else rests on