Ext Maths · 7 Transformations and vectors

7.4 Vector geometry

Before you start

Adding, subtracting and scalar-multiplying vectors, and that $k 𝐚$ is parallel to $𝐚$ (7.2).
Expanding brackets and collecting like terms, including fractions: $𝐚 + \frac{1}{4} (𝐛 - 𝐚) = \frac{3}{4} 𝐚 + \frac{1}{4} 𝐛$ .
Turning a ratio into a fraction of the whole: in the ratio $2 : 3$ the first part is $\frac{2}{5}$ of the whole.
Properties of a parallelogram and a trapezium, and what makes two triangles similar (4.4).

Syllabus objectives covered (4)

Represent vectors by directed line segments.
Use position vectors.
Use the sum and difference of two or more vectors to express given vectors in terms of two coplanar vectors.
Use vectors to reason and to solve geometric problems.

Part 1 of 6

Vectors as arrows: directed line segments

In this subtopic vectors are mostly letters rather than numbers. $𝐚$ stands for one fixed movement, a fixed length in a fixed direction, drawn as a line segment with an arrow on it. That is a directed line segment.
Any segment with the same length and the same direction represents the same vector, wherever it is drawn. In a parallelogram $O A B C$ , if $\vec{O A} = 𝐚$ then the opposite side $\vec{C B}$ is also $𝐚$ .
Direction is everything. Travelling along a segment against its arrow means a minus sign: if $\vec{O A} = 𝐚$ then $\vec{A O} = - 𝐚$ .
Scalar multiples keep the line but change the length: $2 𝐚$ is parallel to $𝐚$ and twice as long; $- \frac{1}{2} 𝐚$ is parallel, half as long, and points the other way.
When you write these by hand, underline the letters ( $a$ ) so it is clear they are vectors, not numbers.

Part 2 of 6

Position vectors

Choose a fixed point $O$ , the origin. The position vector of a point $A$ is $\vec{O A}$ , the journey from $O$ to $A$ . It is usually named with the matching small bold letter, $𝐚$ .
On a coordinate grid with $O$ at $(0, 0)$ , the point $A (4, 1)$ has position vector $(\binom{4}{1})$ . Same numbers, but the column vector is a movement from $O$ , not a place.
The most useful fact in this subtopic comes from going from $A$ to $B$ **via $O$ **: first back along $𝐚$ , then out along $𝐛$ .
$\vec{A B} = \vec{A O} + \vec{O B} = - 𝐚 + 𝐛 = 𝐛 - 𝐚$
Remember it as end minus start. $𝐚 - 𝐛$ is $\vec{B A}$ , the journey the wrong way round.
With numbers: $A (4, 1)$ and $B (6, - 3)$ give $\vec{A B} = (\binom{6}{- 3}) - (\binom{4}{1}) = (\binom{2}{- 4})$ .
It works in reverse too: the position vector of $B$ is $𝐚 + \vec{A B}$ , so if you know where $A$ is and how to get from $A$ to $B$ , you know where $B$ is.

Part 3 of 6

Expressing a vector along a route

Questions give you two vectors, such as $𝐚$ and $𝐜$ , that lie in the same flat plane and are not parallel: two coplanar vectors. Every other vector in the diagram can be written as some amount of $𝐚$ plus some amount of $𝐜$ .
The method: find a route from the start point to the end point that uses only segments you already know. Write each leg (with a minus for any leg taken against its arrow), then add them up.
Take parallelogram $O A B C$ with $\vec{O A} = 𝐚$ and $\vec{O C} = 𝐜$ . Opposite sides of a parallelogram are equal and parallel, so $\vec{A B} = 𝐜$ and $\vec{C B} = 𝐚$ .
Each route adds up legs you know. $\vec{A C}$ goes against $𝐚$ (minus) then along $𝐜$ (plus), so it is $𝐜 - 𝐚$ .
$\vec{O B}$ : go $O \to A \to B$ , which is $𝐚 + 𝐜$ .
$\vec{A C}$ : go $A \to O \to C$ . The first leg is against $𝐚$ , so it is $- 𝐚$ ; the second is $𝐜$ . So $\vec{A C} = - 𝐚 + 𝐜 = 𝐜 - 𝐚$ .
Any route gives the same answer, which is a free check. $A \to B \to C$ is $𝐜$ then $- 𝐚$ (going from $B$ to $C$ is against $\vec{C B} = 𝐚$ ), again $𝐜 - 𝐚$ . If two routes disagree, a sign is wrong somewhere.
Midpoints are halves. $M$ is the midpoint of $A B$ , so $\vec{A M} = \frac{1}{2} \vec{A B} = \frac{1}{2} 𝐜$ , and $\vec{O M} = 𝐚 + \frac{1}{2} 𝐜$ .
'Simplest form' means brackets expanded and like terms collected, so that $𝐚$ and $𝐜$ each appear once, e.g. $\frac{3}{2} 𝐚 - \frac{1}{2} 𝐜$ . Taking out a common factor, $\frac{1}{2} (3 𝐚 - 𝐜)$ , is also simplified, and it is often the form that shows a parallel line.

Part 4 of 6

A point that divides a line in a ratio

If $P$ lies on $A B$ with $A P : P B = 2 : 3$ , split $A B$ into $2 + 3 = 5$ equal parts. $A P$ is 2 of those 5 parts.
$A P : P B = 2 : 3$ means 5 parts in total, so $A P$ is $\frac{2}{5}$ of $A B$ , not $\frac{2}{3}$ .
In general:
$A P : P B = m : n \Rightarrow \vec{A P} = \frac{m}{m + n} \vec{A B}$
The trap is writing $\frac{2}{3}$ . A ratio compares one part with the other part; a fraction compares a part with the whole. Add the ratio numbers to get the whole.
Then build the position vector of $P$ in triangle $O A B$ with $\vec{O A} = 𝐚$ and $\vec{O B} = 𝐛$ . Go $O \to A \to P$ , using $\vec{A B} = 𝐛 - 𝐚$ :
$\vec{O P} = 𝐚 + \frac{2}{5} (𝐛 - 𝐚) = 𝐚 + \frac{2}{5} 𝐛 - \frac{2}{5} 𝐚 = \frac{3}{5} 𝐚 + \frac{2}{5} 𝐛$
Two quick checks for any point on $A B$ : the two fractions add up to 1 ( $\frac{3}{5} + \frac{2}{5} = 1$ ), and the bigger fraction belongs to the end the point is nearer. $P$ is nearer $A$ , so it has more $𝐚$ .
Read the order of the letters carefully. $O P : P A = 1 : 2$ puts $P$ on $O A$ , one third of the way from $O$ , so $\vec{O P} = \frac{1}{3} 𝐚$ .

Part 5 of 6

Showing that lines are parallel

Two vectors are parallel exactly when one is a scalar multiple of the other. So to show $P Q$ is parallel to $R S$ , you must show $\vec{P Q} = k \vec{R S}$ for some number $k$ .
Write both vectors in terms of $𝐚$ and $𝐛$ in simplest form, then take out a common factor to expose the multiple. For example, $\vec{P Q} = 6 𝐚 - 9 𝐛 = 3 (2 𝐚 - 3 𝐛)$ , and if $\vec{R S} = 2 𝐚 - 3 𝐛$ then $\vec{P Q} = 3 \vec{R S}$ .
Now say it in words. The algebra alone does not show anything until you state the conclusion. The wording that earns the mark is:
$\vec{P Q} = 3 \vec{R S}, so P Q is parallel to R S .$
The scalar also gives the lengths: $P Q$ is 3 times as long as $R S$ . A negative scalar still means parallel, just pointing the opposite way.
Not a multiple means not parallel. $4 𝐚 + 6 𝐛$ and $2 𝐚 + 5 𝐛$ are not parallel: to get from $2 𝐚$ to $4 𝐚$ you multiply by 2, but $5 𝐛 \times 2 = 10 𝐛$ , not $6 𝐛$ .
Because $𝐚$ and $𝐛$ are not parallel, the only way $h 𝐚 + k 𝐛$ can equal $3 𝐚 - 2 𝐛$ is $h = 3$ and $k = - 2$ . Matching the amounts of $𝐚$ and of $𝐛$ like this (comparing coefficients) is how you find an unknown number in a vector question.
The link to similarity: if $D$ is on $A B$ and $E$ is on $A C$ with $\vec{D E} = \frac{1}{2} \vec{B C}$ , then $D E$ is parallel to $B C$ and half as long. The parallel lines make the corresponding angles equal, so triangle $A D E$ is similar to triangle $A B C$ with lengths in the ratio $1 : 2$ .

Part 6 of 6

Showing that three points are collinear

Collinear means lying on one straight line.
Parallel on its own is not enough. Opposite sides of a parallelogram are parallel, but they are on different lines. You need two facts: the two vectors are parallel, AND they share a common point.
So choose two vectors that meet at one of the three points: $\vec{P Q}$ and $\vec{P R}$ (both start at $P$ ), or $\vec{P Q}$ and $\vec{Q R}$ (they meet at $Q$ ). Then show one is a multiple of the other.
The wording that earns the final mark names both facts:
$\vec{P R} = 3 \vec{P Q}, so P Q and P R are parallel. They share the point P, so P, Q, R are collinear.$
The multiple also tells you where the middle point is. $\vec{P R} = 3 \vec{P Q}$ means $P Q$ is one third of $P R$ , so $P Q : Q R = 1 : 2$ .
The situation in worked example 3: $M$ , $N$ and $C$ lie on one straight line, with $\vec{M C} = 3 \vec{M N}$ .

Worked examples

Example 1. $O A B C$ is a parallelogram with $\vec{O A} = 𝐚$ and $\vec{O C} = 𝐜$ . $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$ . (a) Find $\vec{A C}$ in terms of $𝐚$ and $𝐜$ . (b) Find $\vec{M N}$ in terms of $𝐚$ and $𝐜$ . (c) Show that $M N$ is parallel to $A C$ .

Show the working

Label the diagram first. Opposite sides are equal vectors: $\vec{A B} = 𝐜$ and $\vec{C B} = 𝐚$ , so $\vec{B C} = - 𝐚$ .
(a) Route $A \to O \to C$ : against $𝐚$ , then along $𝐜$ .
$\vec{A C} = - 𝐚 + 𝐜 = 𝐜 - 𝐚$
(b) Route $M \to B \to N$ . $M$ is halfway along $A B$ , so $\vec{M B} = \frac{1}{2} 𝐜$ . $N$ is halfway along $B C$ , so $\vec{B N} = \frac{1}{2} (- 𝐚)$ .
$\vec{M N} = \frac{1}{2} 𝐜 - \frac{1}{2} 𝐚$
(c) Take out the factor $\frac{1}{2}$ so the comparison with (a) is visible.
$\vec{M N} = \frac{1}{2} (𝐜 - 𝐚) = \frac{1}{2} \vec{A C}$
State the conclusion in words: $\vec{M N}$ is a scalar multiple of $\vec{A C}$ , so the lines are parallel. The scalar $\frac{1}{2}$ also tells you $M N$ is half the length of $A C$ .

Answer: (a) $𝐜 - 𝐚$ (b) $\frac{1}{2} 𝐜 - \frac{1}{2} 𝐚$ (c) $\vec{M N} = \frac{1}{2} \vec{A C}$ , so $M N$ is parallel to $A C$ .

Example 2. In triangle $O A B$ , $\vec{O A} = 𝐚$ and $\vec{O B} = 𝐛$ . $P$ lies on $A B$ with $A P : P B = 1 : 3$ . Find $\vec{O P}$ in terms of $𝐚$ and $𝐛$ , in its simplest form.

Show the working

Find the side $P$ lies on, end minus start.
$\vec{A B} = 𝐛 - 𝐚$
The ratio $1 : 3$ has $1 + 3 = 4$ parts, and $A P$ is 1 of them.
$\vec{A P} = \frac{1}{4} (𝐛 - 𝐚)$
Route $O \to A \to P$ .
$\vec{O P} = 𝐚 + \frac{1}{4} (𝐛 - 𝐚)$
Expand and collect: $𝐚 - \frac{1}{4} 𝐚 = \frac{3}{4} 𝐚$ .
$\vec{O P} = 𝐚 + \frac{1}{4} 𝐛 - \frac{1}{4} 𝐚 = \frac{3}{4} 𝐚 + \frac{1}{4} 𝐛$
Check: $\frac{3}{4} + \frac{1}{4} = 1$ , and the larger fraction is on $𝐚$ , which fits because $P$ is much nearer $A$ .

Answer: $\vec{O P} = \frac{3}{4} 𝐚 + \frac{1}{4} 𝐛$

Example 3. In triangle $O A B$ , $\vec{O A} = 𝐚$ and $\vec{O B} = 𝐛$ . $M$ is the midpoint of $O B$ . $N$ lies on $A B$ with $A N : N B = 1 : 2$ . The line $O A$ is extended to $C$ so that $A C = O A$ . Show that $M$ , $N$ and $C$ are collinear.

Show the working

Write the position vector of each point from $O$ . $\vec{O M} = \frac{1}{2} 𝐛$ . $C$ is twice as far along the line as $A$ , so $\vec{O C} = 2 𝐚$ .
For $N$ : $A N$ is $\frac{1}{3}$ of $A B$ .
$\vec{O N} = 𝐚 + \frac{1}{3} (𝐛 - 𝐚) = \frac{2}{3} 𝐚 + \frac{1}{3} 𝐛$
Pick two vectors that share a point. Both $\vec{M N}$ and $\vec{M C}$ start at $M$ . Each is end minus start.
$\vec{M N} = \frac{2}{3} 𝐚 + \frac{1}{3} 𝐛 - \frac{1}{2} 𝐛 = \frac{2}{3} 𝐚 - \frac{1}{6} 𝐛$
In the same way, $\vec{M C}$ is $\vec{O C}$ minus $\vec{O M}$ .
$\vec{M C} = 2 𝐚 - \frac{1}{2} 𝐛$
Take out factors to expose a common bracket: $\vec{M N} = \frac{1}{6} (4 𝐚 - 𝐛)$ and $\vec{M C} = \frac{1}{2} (4 𝐚 - 𝐛)$ . Since $\frac{1}{2} \div \frac{1}{6} = 3$ :
$\vec{M C} = 3 \vec{M N}$
Conclude with both facts: $\vec{M C} = 3 \vec{M N}$ , so $M N$ and $M C$ are parallel; they share the point $M$ , so $M$ , $N$ and $C$ are collinear.
Bonus information from the scalar: $M N$ is one third of $M C$ , so $M N : N C = 1 : 2$ .

Answer: $\vec{M C} = 3 \vec{M N}$ , so $M N$ is parallel to $M C$ , and they share the point $M$ . Therefore $M$ , $N$ and $C$ are collinear.

Example 4. Points $A$ and $B$ have position vectors $(\binom{- 1}{4})$ and $(\binom{8}{- 2})$ . $P$ lies on $A B$ with $A P : P B = 2 : 1$ . Find the position vector of $P$ .

Show the working

End minus start gives the vector along the line.
$\vec{A B} = (\binom{8}{- 2}) - (\binom{- 1}{4}) = (\binom{9}{- 6})$
$2 + 1 = 3$ parts, and $A P$ is 2 of them.
$\vec{A P} = \frac{2}{3} (\binom{9}{- 6}) = (\binom{6}{- 4})$
Start at $O$ , go to $A$ , then along $\vec{A P}$ .
$\vec{O P} = (\binom{- 1}{4}) + (\binom{6}{- 4}) = (\binom{5}{0})$
Check: $\vec{P B} = (\binom{8}{- 2}) - (\binom{5}{0}) = (\binom{3}{- 2})$ , which is half of $\vec{A P}$ , matching the ratio $2 : 1$ .

Answer: The position vector of $P$ is $(\binom{5}{0})$ , so $P$ is the point $(5, 0)$ .

Where people go wrong

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

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

The route from $A$ to $B$ via $O$ is $- 𝐚 + 𝐛$ . $𝐚 - 𝐛$ is the journey from $B$ to $A$ .

Mistake: Turning $A P : P B = 2 : 3$ into $\vec{A P} = \frac{2}{3} \vec{A B}$ .

Instead: $\vec{A P} = \frac{2}{5} \vec{A B}$ .

A ratio compares part with part. The whole line has $2 + 3 = 5$ parts.

Mistake: Travelling against an arrow without a minus sign.

Instead: Every leg taken against its arrow is negative: if $\vec{O A} = 𝐚$ then $\vec{A O} = - 𝐚$ .

A vector includes its direction. Check by working out the same vector along a second route.

Mistake: Claiming three points are collinear because two vectors are parallel.

Instead: Show the vectors are parallel and state that they share a common point.

Parallel lines can be separate lines. Only a shared point forces them onto the same line.

Mistake: Finishing a 'show that' with algebra and no conclusion.

Instead: End with the sentence, e.g. ' $\vec{M N} = \frac{1}{2} \vec{A C}$ , so $M N$ is parallel to $A C$ '.

A 'Show (that)' asks for structured evidence leading to the result, so the result has to be stated.

Check yourself

$A$ and $B$ have position vectors $(\binom{2}{5})$ and $(\binom{- 1}{7})$ . Find $\vec{A B}$ .
Answer
$(\binom{- 1}{7}) - (\binom{2}{5}) = (\binom{- 3}{2})$ .
$\vec{A B} = 3 𝐩 - 𝐪$ . Write down $\vec{B A}$ .
Answer
$\vec{B A} = - (3 𝐩 - 𝐪) = 𝐪 - 3 𝐩$ .
$X$ lies on $P Q$ with $P X : X Q = 3 : 2$ . Write $\vec{P X}$ as a fraction of $\vec{P Q}$ .
Answer
$\vec{P X} = \frac{3}{5} \vec{P Q}$ .
In triangle $O A B$ , $\vec{O A} = 𝐚$ , $\vec{O B} = 𝐛$ and $M$ is the midpoint of $A B$ . Find $\vec{O M}$ in simplest form.
Answer
$𝐚 + \frac{1}{2} (𝐛 - 𝐚) = \frac{1}{2} 𝐚 + \frac{1}{2} 𝐛$ .
Are $4 𝐚 - 6 𝐛$ and $- 2 𝐚 + 3 𝐛$ parallel? Explain.
Answer
Yes. $4 𝐚 - 6 𝐛 = - 2 (- 2 𝐚 + 3 𝐛)$ , a scalar multiple. It is twice as long and points the opposite way.
What two things must you show to prove that three points are collinear?
Answer
That two vectors joining the points are parallel (one is a multiple of the other), and that those two vectors share a common point.
$\vec{P Q} = (k - 2) 𝐚 + 3 𝐛$ is parallel to $2 𝐚 + 6 𝐛$ , where $𝐚$ and $𝐛$ are not parallel. Find $k$ .
Answer
$3 𝐛$ is half of $6 𝐛$ , so $\vec{P Q} = \frac{1}{2} (2 𝐚 + 6 𝐛) = 𝐚 + 3 𝐛$ . So $k - 2 = 1$ and $k = 3$ .

Exam questions

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

Question 1. $O P Q R$ is a trapezium with $\vec{O P} = 𝐩$ , $\vec{O R} = 𝐫$ and $\vec{R Q} = 2 𝐩$ . $M$ is the midpoint of $P Q$ .

(a) Find $\vec{P Q}$ , in terms of $𝐩$ and $𝐫$ , in its simplest form. 2 marks No calculator

Mark scheme

M1 correct route, e.g. $- 𝐩 + 𝐫 + 2 𝐩$
A1 $𝐩 + 𝐫$

(b) Find $\vec{O M}$ , in terms of $𝐩$ and $𝐫$ , in its simplest form. 2 marks No calculator

Mark scheme

M1 $𝐩 + \frac{1}{2}$ (their $\vec{P Q}$ ) oe
A1 $\frac{3}{2} 𝐩 + \frac{1}{2} 𝐫$ oe

(c) The line $R Q$ is extended to the point $S$ so that $\vec{Q S} = 𝐩$ . Show that $O$ , $M$ and $S$ are collinear. 3 marks No calculator

Mark scheme

B1 $\vec{O S} = 3 𝐩 + 𝐫$
M1 $\vec{O S} = 2 \vec{O M}$ oe
A1 states $O M$ and $O S$ are parallel and share the point $O$ , so collinear

(d) Write down the ratio $O M : M S$ . 1 mark No calculator

Mark scheme

B1 $1 : 1$

Question 2. In triangle $O A B$ , $\vec{O A} = 𝐚$ and $\vec{O B} = 𝐛$ . $P$ is the point on $O A$ with $O P : P A = 1 : 2$ . $Q$ is the point on $A B$ with $A Q : Q B = 2 : 1$ .

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

Mark scheme

B1 $𝐛 - 𝐚$ oe

(b) Find the position vector $\vec{O Q}$ , in terms of $𝐚$ and $𝐛$ , in its simplest form. 2 marks No calculator

Mark scheme

M1 $𝐚 + \frac{2}{3} (𝐛 - 𝐚)$ oe
A1 $\frac{1}{3} 𝐚 + \frac{2}{3} 𝐛$

(c) Show that $P Q$ is parallel to $O B$ . 3 marks No calculator

Mark scheme

M1 correct route for $\vec{P Q}$ , e.g. $- \frac{1}{3} 𝐚 + \vec{O Q}$ or $\frac{2}{3} 𝐚 + \frac{2}{3} (𝐛 - 𝐚)$
A1 $\vec{P Q} = \frac{2}{3} 𝐛$
A1 $\vec{P Q} = \frac{2}{3} \vec{O B}$ , so $P Q$ is parallel to $O B$ , stated

(d) Write down the ratio $P Q : O B$ . 1 mark No calculator

Mark scheme

B1 $2 : 3$

(e) Explain why triangle $A P Q$ is similar to triangle $A O B$ . 2 marks No calculator

Mark scheme

B1 angle $A$ is common to both triangles
B1 $P Q$ is parallel to $O B$ , so angle $A P Q$ = angle $A O B$ (corresponding angles) oe

How this is examined

Writing a vector along a route in terms of the two given vectors: A chain of 1- to 2-mark 'Find ... in terms of $𝐚$ and $𝐛$ ' parts, usually with 'in its simplest form', each part using the answer before A correct route written out, e.g. $\vec{O M} = 𝐚 + \frac{1}{2} \vec{A B}$ , earns method credit even if a later sign slips; the final answer with like terms collected earns the accuracy mark.
Showing that three points are collinear: A 2- or 3-mark 'Show that $P$ , $Q$ and $R$ are collinear', near the end of a vector question Two vectors that share a point, each in simplest form; one written as a multiple of the other, e.g. $\vec{P R} = 3 \vec{P Q}$ ; and the sentence naming both facts: 'parallel and share the point $P$ , so collinear'.
Using a ratio to place a point on a line: A 2-mark 'Find $\vec{O P}$ ' where $P$ divides a side in a given ratio The fraction of the whole side, $\frac{m}{m + n}$ , applied to the side written as end minus start, e.g. $𝐚 + \frac{2}{5} (𝐛 - 𝐚)$ ; then simplified to $\frac{3}{5} 𝐚 + \frac{2}{5} 𝐛$ .
Showing that two lines are parallel: A 2- or 3-mark 'Show that $M N$ is parallel to $A C$ ' Both vectors in simplest form, a factor taken out to show the multiple, e.g. $\vec{M N} = \frac{1}{2} \vec{A C}$ , and the words 'so $M N$ is parallel to $A C$ '.
Ratios of lengths and similar triangles from the scalar: A 1- or 2-mark 'Write down the ratio' or 'Explain why the triangles are similar' after a parallel result Reading the scalar as a length ratio ( $\vec{P Q} = \frac{2}{3} \vec{O B}$ gives $P Q : O B = 2 : 3$ ); for similarity, the parallel lines giving equal corresponding angles and the shared angle.

Words to know

Directed line segment: A line segment with an arrow showing its direction. It represents a vector: its length is the magnitude, its arrow the direction.
Position vector: The vector from the origin $O$ to a point. The position vector of $A$ is $\vec{O A}$ , often written $𝐚$ .
Coplanar vectors: Vectors lying in the same plane. Two non-parallel coplanar vectors can be combined to make any other vector in that plane.
Parallel vectors: Vectors where one is a scalar multiple of the other, $𝐮 = k 𝐯$ .
Collinear: Lying on the same straight line. Shown by two vectors being parallel and sharing a common point.
Simplest form: Brackets expanded and like terms collected, so each given vector appears once, e.g. $\frac{3}{4} 𝐚 + \frac{1}{4} 𝐛$ .

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