Vectors can represent both positions and movements. Positions use coordinates (x,y) while movements use vectors with the change in x and y written vertically. Vectors can be added by adding the x and y components. Vectors can also be scaled by multiplying each component by a scalar value. Exam questions often involve finding a vector in part (a) and then using it to find another vector in part (b).

1. GCSE: Vectors
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Last modified: 20th March 2017

2. 𝑥
𝑦
1
2
3
4
5
6
-1
-2
1 2 3 4 5 6 7
2,3
4
1
Coordinates represent a position.
We write 𝑥, 𝑦 to indicate the 𝑥
position and 𝑦 position.
Vectors represent a movement.
We write
𝑥
𝑦 to indicate the
change in 𝑥 and change in 𝑦.
?
?
?
?
Note we put the numbers
vertically. Do NOT write
4
1
;
there is no line and vectors are
very different to fractions!
You may have seen vectors
briefly before if you’ve done
transformations; we can use
them to describe the
movement of a shape in a
translation.
Coordinates vs Vectors

3. 𝒂
𝒃
𝒄
𝒅
𝒆
𝒇
𝒂 =
1
3
𝒃 =
5
1
𝒄 =
2
−3
𝒅 =
−4
2
𝒆 =
−4
−4
𝒇 =
−1
1
𝒈 =
0
−3
𝒈
? ?
? ?
? ?
?
Vectors to represent movements

4. You’re used to variables representing numbers in maths. They can also
represent vectors!
𝒂
𝒃
𝑋
𝑌 What can you say about how we
use variables for vertices (points) vs
variables for vectors?
We use capital letters for vertices
and lower case letters for vectors.
There’s 3 ways in which can
represent the vector from point 𝑋
to 𝑍:
1. 𝒂 (in bold)
2. 𝑎 (with an ‘underbar’)
3. 𝑋𝑍
𝑍 ?
?
?
?
Writing Vectors

5. 𝒂
𝒃
𝑋
𝑌
𝑍
𝑋𝑍 = 𝒂 =
2
5
𝑍𝑌 = 𝒃 =
5
3
𝑋𝑌 =
7
8
?
?
?
What do you notice about the numbers in
7
8
when compared to
2
5
and
5
3
?
We’ve simply added the 𝑥 values and 𝑦
values to describe the combined movement.
i.e.
𝒂 + 𝒃 =
2
5
+
5
3
=
7
8
𝑋𝑍 + 𝑍𝑌 = 𝑋𝑌
?
Bro Important Note: The point is that we can use
any route to get from the start to finish, and the
vector will always be the same.
• Route 1: We go from 𝑋 to 𝑌 via 𝑍.
𝑋𝑍 =
2
5
+
5
3
=
7
8
• Route 2: Use the direct line from 𝑋 to 𝑌:
7
8
Adding Vectors

6. 𝒂
Scaling Vectors
We can ‘scale’ a vector by
multiplying it by a normal number,
aptly known as a scalar.
If 𝒂 =
4
3
, then
2𝒂 = 2
4
3
=
8
6
What is the same about 𝒂 and 2𝒂 and
what is different?
2𝒂
Bro Note: Note that
vector letters are bold
but scalars are not.
Same:
Same direction / Parallel
Different:
The length of the vector, known as
the magnitude, is longer.
?
?
?

7. More on Adding/Subtracting Vectors
𝐴
𝑋
𝒂
𝒃
2𝒄
If 𝑂𝐴 = 𝒂, 𝐴𝐵 = 𝒃 and 𝑋𝐵 = 2𝒄,
then find the following in terms
of 𝑎, 𝑏 and 𝑐:
𝑂𝐵 = 𝒂 + 𝒃
𝑂𝑌 = 𝒂 + 2𝒃
𝐴𝑋 = 𝒃 − 2𝒄
𝑋𝑂 = 2𝒄 − 𝒃 − 𝒂
𝑌𝑋 = −𝒃 − 2𝒄
Note: Since −
𝑥
𝑦 =
−𝑥
−𝑦 ,
subtracting a vector goes in
the opposite direction.
?
?
?
𝒃
𝐵
𝑌
𝑂 𝒃
𝒂
Bro Side Note: Since 𝑏 + 𝑎
would end up at the same
finish point, we can see
𝒃 + 𝒂 = 𝒂 + 𝒃 (i.e. vector
addition, like normal
addition, is ‘commutative’)
?
?

8. a. 𝐵𝐴 = −𝑎
b. 𝐴𝐶 = 𝑎 + 𝑏
c. 𝐷𝐵 = 2𝑎 − 𝑏
d. 𝐴𝐷 = −𝑎 + 𝑏
a. 𝑀𝐾 = −𝑎 − 𝑏
b. 𝑁𝐿 = 3𝑎 − 𝑏
c. 𝑁𝐾 = 2𝑎 − 𝑏
d. 𝐾𝑁 = −2𝑎 + 𝑏
a. 𝑍𝑋 = 𝑎 + 𝑏
b. 𝑌𝑊 = 𝑎 − 2𝑏
c. 𝑋𝑌 = −𝑎 + 𝑏
d. 𝑋𝑍 = −𝑎 − 𝑏
a. 𝐴𝐵 = −𝑎 + 𝑏
b. 𝐹𝑂 = −𝑎 + 𝑏
c. 𝐴𝑂 = −2𝑎 + 𝑏
d. 𝐹𝐷 = −3𝑎 + 2𝑏
2
3
4
5
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Exercise 1 (on provided sheet)
1 State the value of each vector.
𝒂
𝒃
𝒄
𝒅
𝒆 𝒇
𝒂 =
3
1
𝒃 =
3
−2
𝒄 =
−3
0
𝒅 =
0
5
𝒆 =
−1
4
𝒇 =
−3
−4
? ? ?
? ? ?

9. Starter
Expand and simplify the following expressions:
2 𝒂 + 𝒃 + 𝒃 = 2𝒂 + 3𝒃
1
2
𝒂 − 𝒃 + 𝒃 =
1
2
𝒂 +
1
2
𝒃
𝒂 +
1
2
𝒂 + 𝒃 =
3
2
𝒂 +
1
2
𝒃
𝒂 − 2 𝒂 − 2𝒃 = −𝒂 + 4𝒃
1
3
𝒂 + 2𝒃 − 𝒃 =
1
3
𝒂 −
1
3
𝒃
1
2
3
4
5
?
?
?
?
?

10. The ‘Two Parter’ exam question
Many exams questions follow a two-part format:
a) Find a relatively easy vector using skills from Lesson 1.
b) Find a harder vector that uses a fraction of your vector from part (a).
Edexcel June 2013 1H Q27
𝑆𝑄 = −𝒃 + 𝒂
For (b), there’s two possible paths
to get from 𝑁 to 𝑅: via 𝑆 or via 𝑄.
But which is best?
In (a) we found 𝑺 to 𝑸 rather than
𝑸 to 𝑺, so it makes sense to go in
this direction so that we can use
our result in (a).
a
b 𝑁𝑅 =
2
5
𝑆𝑄 + 𝒃
=
2
5
−𝒃 + 𝒂 + 𝒃
=
2
5
𝒂 +
3
5
𝒃
Bro Tip: This ratio
wasn’t in the
original diagram. I
like to add the ratio
as a visual aid.
?
?
Bro Workings Tip: While you’re
welcome to start your working with
the second line, I recommend the
first line so that your chosen route
is clearer.
𝑄𝑅 is also 𝑏
because it is exactly
the same
movement as 𝑃𝑆.
?

11. Test Your Understanding
𝐴𝐵 = 𝐴𝑂 + 𝑂𝐵
= −𝒂 + 𝒃
?
𝑂𝑃 = 𝒂 +
3
4
𝐴𝐵
= 𝒂 +
3
4
−𝒂 + 𝒃
=
1
4
𝒂 +
3
4
𝒃
?
You MUST
expand and
simplify.
Edexcel June 2012

12. Further Test Your Understanding
𝒂
𝒃
𝑂 𝐴
𝐵
𝑋
𝑂𝑋: 𝑋𝐵 = 1: 3
𝐴𝑋 = −𝒂 +
1
4
𝑂𝐵
= −𝒂 +
1
4
𝒂 + 𝒃
= −𝒂 +
1
4
𝒂 +
1
4
𝒃
= −
3
4
𝒂 +
1
4
𝒃
First Step?
?
𝒂
𝒃
𝑂 𝐴
𝐵
𝑌
𝐴𝑌: 𝑌𝐵 = 2: 3
𝑂𝑌 = 𝒂 +
2
5
𝐴𝐵
= 𝒂 +
2
5
−𝒂 + 𝒃
= 𝒂 −
2
5
𝒂 +
2
5
𝒃
=
3
5
𝒂 +
2
5
𝒃
First Step?
?
A B

13. a. 𝐴𝐵 = −𝒂 + 𝒃
b. 𝐴𝑋 = −
1
5
𝒂 +
1
5
𝒃
c. 𝑂𝑋 =
4
5
𝒂 +
1
5
𝒃
d. 𝐵𝑋 =
4
5
𝒂 −
4
5
𝒃
a. 𝐴𝑌 = −
1
3
𝒂 +
1
3
𝒃
b. 𝑂𝑌 =
2
3
𝒂 +
1
3
𝒃
c. 𝑌𝑂 = −
2
3
𝒂 −
1
3
𝒃
1
2
?
?
?
?
?
?
?
Exercise 2 (on provided sheet)
𝑋 is a point such
that 𝐴𝑋: 𝑋𝐵 = 1: 4
𝑌 is a point such
that 𝑌𝐵 = 2𝐴𝑌
3
[June 2009 2H Q23]
a) Find 𝐴𝐵 in terms of 𝒂 and 𝒃.
𝐴𝐵 = −𝒂 + 𝒃 𝑜𝑟 𝒃 − 𝒂
b) 𝑃 is on 𝐴𝐵 such that 𝐴𝑃: 𝑃𝐵 = 3: 2.
Show that 𝑂𝑃 =
1
5
2𝒂 + 3𝒃
?
?

14. Exercise 2 (on provided sheet)
5
a. 𝑂𝑅 =
3
5
𝒂 +
2
5
𝒃
b. 𝐵𝑆 =
1
4
𝒂
c. 𝑂𝑆 =
1
4
𝒂 + 𝒃
d. 𝑅𝑆 = −
7
20
𝒂 +
3
5
𝒃
6
a. 𝐷𝐶 = −𝒛 + 𝒚
b. 𝐷𝑀 = −
1
2
𝒛 +
1
2
𝒚
c. 𝐴𝑀 = 𝐴𝐷 + 𝐷𝑀 =
1
2
𝒛 +
1
2
𝒚
d. 𝐵𝑀 = 𝐵𝐴 + 𝐴𝑀 = −𝒙 +
1
2
𝒛 +
1
2
𝒚
e. 𝐵𝑃 =
2
3
𝐵𝑀 = −
2
3
𝒙 +
1
3
𝒛 +
1
3
𝒚
f. 𝐴𝑃 = 𝐴𝐵 + 𝐵𝑃 =
1
3
𝒙 +
1
3
𝒛 +
1
3
𝒚
?
?
?
?
?
?
?
?
?
?
𝑂𝐴𝐶𝐵 is a parallelogram. 𝑅
is a point such that
𝐴𝑅: 𝑅𝐵 = 2: 3. 𝑆 is a point
such that 𝐵𝑆: 𝑆𝐶 = 1: 3.
𝑀 is the midpoint of
𝐶𝐷, 𝐵𝑃: 𝑃𝑀 = 2: 1
4 [Nov 2010 1H Q27] 𝑀 is the
midpoint of 𝑂𝑃.
a) Express 𝑂𝑀 in terms of 𝒂 and 𝒃.
𝑂𝑀 =
1
2
𝒂 + 𝒃 𝑜𝑟
1
2
𝒂 +
1
2
𝒃
b) Express 𝑇𝑀 in terms of 𝒂 and 𝒃
giving your answer in its simplest
form.
𝑇𝑀 = −𝒂 +
1
2
𝑂𝑀
= −𝒂 +
1
2
𝒂 + 𝒃
=
1
2
𝒃 −
1
2
𝒂
?
?

15. a. 𝐴𝐵 = −𝒂 + 𝒃
b. 𝐵𝐶 = 𝒂
c. 𝑀𝐵 = −
1
2
𝒂 +
1
2
𝒃
d. 𝑀𝑋 = −
1
6
𝒂 +
1
2
𝒃
e. 𝑋𝐴 =
2
3
𝒂 − 𝒃
f. 𝐶𝑀 = −
1
2
𝒂 −
1
2
𝒃
g. 𝑋𝑂 = −
1
3
𝒂 − 𝒃
7
?
8
9
?
?
?
?
?
?
a. 𝑂𝐵 = 𝒂 + 2𝒃
b. 𝐵𝐶 = 𝒂 − 𝒃
c. 𝐴𝑀 =
1
2
𝒂 + 𝒃
d. 𝑂𝑀 =
3
2
𝒂 + 𝒃
?
?
?
?
a. 𝑂𝐶 = −𝒂 + 𝒃
b. 𝑋𝐶 = −
3
4
𝒂 +
1
2
𝒃
c. 𝐴𝑋 = −
5
4
𝒂 +
1
2
𝒃
?
?
?
Exercise 2 (on provided sheet)

16. ?
Recap

17. Types of vector ‘proof’ questions
“Prove that … is
a straight line.”
“Show that … is
parallel to …”
“Prove these two
vectors are equal.”

18. ?
Showing vectors are equal
There is nothing
mysterious about this
kind of ‘prove’ question.
Just find any vectors
involved, in this case,
𝑂𝑋 and
2
5
𝑂𝑌. Hopefully
they should be equal!
Edexcel March 2013
2
5
𝑂𝑌 =
2
5
6𝑏 + 5𝑎 − 𝑏
= 2𝑎 + 2𝑏
𝑂𝑋 = 3𝑎 +
1
3
−3𝑎 + 6𝑏
= 2𝑎 + 2𝑏
∴ 𝑂𝑋 =
2
5
𝑂𝑌
With proof questions
you should restate the
thing you are trying to
prove, as a ‘conclusion’.
?

19. What do you notice?
𝒂 2𝒂 𝒂 + 2𝒃
2𝒂 + 4𝒃
𝒂
−
3
2
𝒂
! Vectors are parallel
if one is a multiple of
another.
?

20. Parallel or not parallel?
Vector 1 Vector 2 Parallel?
(in general)
𝑎 + 𝑏 𝑎 + 2𝑏
𝑎 + 𝑏 3𝑎 + 3𝑏
𝑎 + 2𝑏 −3𝑎 − 6𝑏
𝑎 − 𝑏 −𝑎 + 𝑏
 
No Yes


No Yes


No Yes


No Yes
For ones which are parallel, show it diagrammatically.

21. How to show two vectors are parallel
𝒂
𝑴
𝑶 𝑩
𝑿
𝒃
𝑨 𝑪
𝑋 is a point on 𝐴𝐵 such that 𝐴𝑋: 𝑋𝐵 = 3: 1. 𝑀 is the midpoint of 𝐵𝐶.
Show that 𝑋𝑀 is parallel to 𝑂𝐶.
𝑂𝐶 = 𝑎 + 𝑏
𝑋𝑀 =
1
4
−𝑎 + 𝑏 +
1
2
𝑎 =
1
4
𝑎 +
1
4
𝑏
=
1
4
𝑎 + 𝑏
𝑋𝑀 is a multiple of 𝑂𝐶 ∴ parallel.
The key is to factor out a scalar such
that we see the same vector.
For any proof question always find the vectors
involved first, in this case 𝑋𝑀 and 𝑂𝐶.
The magic words here are
“is a multiple of”.
?

22. Test Your Understanding
𝑃
𝐵
𝑂
𝐴
2𝒂
3𝒃
a) Find 𝐴𝐵 in terms of 𝒂 and 𝒃.
−2𝒂 + 3𝒃
b) 𝑃 is the point on 𝐴𝐵 such that
𝐴𝑃: 𝑃𝐵 = 2: 3.
Show that 𝑂𝑃 is parallel to the vector
𝒂 + 𝒃.
?
Edexcel June 2011 Q26
Bro Exam Note: Notice that the mark
scheme didn’t specifically require “is a
multiple of” here (but write it anyway!), but
DID explicitly factorise out the
6
5
?

23. Proving three points form a straight line
Points A, B and C form a straight line if:
𝑨𝑩 and 𝑩𝑪 are parallel (and B is a common point).
Alternatively, we could show 𝑨𝑩 and 𝑨𝑪 are parallel. This tends to be easier.
A
B
C
?
A
B
C

24. 𝐴𝑁 = 2𝒃, 𝑁𝑃 = 𝒃
𝐵 is the midpoint of 𝐴𝐶. 𝑀 is the midpoint of 𝑃𝐵.
a) Find 𝑃𝐵 in terms of 𝒂 and 𝒃.
b) Show that 𝑁𝑀𝐶 is a straight line.
Straight Line Example
−3𝒃 + 𝒂
𝑁𝑀 = 𝑏 +
1
2
𝑎 − 3𝑏
= 𝑏 +
1
2
𝑎 −
3
2
𝑏
=
1
2
𝑎 −
1
2
𝑏
=
𝟏
𝟐
(𝒂 − 𝒃)
𝑁𝐶 = −2𝑏 + 2𝑎
= 𝟐 𝒂 − 𝒃
𝑵𝑪 is a multiple of 𝑵𝑴 ∴ 𝑵𝑪 is
parallel to 𝑵𝑴.
𝑵 is a common point.
∴ 𝑵𝑴𝑪 is a straight line.
b
a
?
?

25. November 2013 1H Q24
𝑂𝐴 = 𝑎 and 𝑂𝐵 = 𝑏
𝐷 is the point such that 𝐴𝐶 = 𝐶𝐷
The point 𝑁 divides 𝐴𝐵 in the ratio 2: 1.
(a) Write an expression for 𝑂𝑁 in terms of 𝒂 and 𝒃.
𝑶𝑵 = 𝒃 +
𝟏
𝟑
−𝒃 + 𝒂 =
𝟏
𝟑
𝒂 +
𝟐
𝟑
𝒃
(b) Prove that 𝑂𝑁𝐷 is a straight line.
𝑶𝑫 = 𝒂 + 𝟐𝒃
𝑶𝑵 =
𝟏
𝟑
(𝒂 + 𝟐𝒃)
𝑶𝑵 is a multiple of 𝑶𝑫 and 𝑶 is a common point.
∴ 𝑶𝑵𝑫 is a straight line.
?
?
Test Your Understanding

26. Exercise 3
𝑇 is the point on 𝐴𝐵 such that
𝐴𝑇: 𝑇𝐵 = 5: 1. Show that 𝑂𝑇 is
parallel to the vector 𝒂 + 2𝒃.
𝑂𝑇 = 2𝒃 +
1
6
−2𝒃 + 5𝒂
= 2𝒃 −
1
3
𝒃 +
5
6
𝒂
=
5
6
𝒂 +
5
3
𝒃
=
5
6
𝒂 + 2𝒃
1
𝑀 is the midpoint of 𝑂𝐴. 𝑁 is the
midpoint of 𝑂𝐵. Prove that 𝐴𝐵 is
parallel to 𝑀𝑁.
𝐴𝐵 = −2𝒎 + 2𝒏
= 2 −𝒎 + 𝒏
𝑀𝑁 = −𝒎 + 𝒏
𝐴𝐵 is a multiple of 𝑀𝑁 ∴ parallel.
2
?
?

27. Exercise 3
3
𝐴𝐶𝐸𝐹 is a parallelogram. 𝐵 is the
midpoint of 𝐴𝐶. 𝑀 is the midpoint of 𝐵𝐸.
Show that 𝐴𝑀𝐷 is a straight line.
𝐷𝑀 = −𝒃 +
1
2
3𝒃 + 𝒂
= −𝒃 +
3
2
𝒃 +
1
2
𝒂
=
1
2
𝒂 +
1
2
𝒃 =
1
2
𝒂 + 𝒃
𝐷𝐴 = 2𝒃 + 2𝒂 = 2 𝒂 + 𝒃
𝐷𝐴 is a multiple of 𝐷𝑀 and 𝐷 is a
common point, so 𝐴𝑀𝐷 is a straight line.
4
𝐶𝐷 = 𝒂, 𝐷𝐸 = 𝒃 and 𝐹𝐶 = 𝒂 − 𝒃
i) Express 𝐶𝐸 in terms of 𝒂 and 𝒃.
𝒂 + 𝒃
ii) Prove that 𝐹𝐸 is parallel to 𝐶𝐷.
𝐹𝐸 = 𝑎 − 𝑏 + 𝑎 + 𝑏 = 2𝒂 which
is a multiple of 𝐶𝐷
iii) 𝑋 is the point on 𝐹𝑀 such that such that
𝐹𝑋: 𝑋𝑀 = 4: 1. Prove that 𝐶, 𝑋 and 𝐸 lie on the
same straight line.
𝐶𝑋 = −𝒂 + 𝒃 +
4
5
2𝒂 −
1
2
𝒃 =
3
5
𝒂 +
3
5
𝒃
=
3
5
𝒂 + 𝒃
𝐶𝐸 = 𝒂 + 𝒃
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28. Exercise 3
5
𝑂𝐴𝐵𝐶 is a parallelogram. 𝑃 is the point on
𝐴𝐶 such that 𝐴𝑃 =
2
3
𝐴𝐶.
i) Find the vector 𝑂𝑃. Give your answer in
terms of 𝒂 and 𝒄.
𝑂𝑃 = 6𝒂 +
2
3
−6𝒂 + 6𝒄
= 2𝒂 + 4𝒄 = 2 𝒂 + 2𝒄
ii) Given that the midpoint of 𝐶𝐵 is 𝑀,
prove that 𝑂𝑃𝑀 is a straight line.
𝑂𝑀 = 3𝒂 + 6𝒄 = 3 𝒂 + 2𝒄
𝑂𝑀 is a multiple of 𝑂𝑃 and 𝑂 is a common
point, therefore 𝑂𝑃𝑀 is a straight line.
6
𝑂𝐴 = 3𝒂 and 𝐴𝑄 = 𝒂 and 𝑂𝐵 = 𝒃 and
𝐵𝐶 =
1
2
𝒃. 𝑀is the midpoint of 𝑄𝐵. Prove
that 𝐴𝑀𝐶 is a straight line.
𝐴𝑀 = 𝒂 +
1
2
−4𝒂 + 𝒃 = −𝒂 +
1
2
𝒃
𝐴𝐶 = −3𝒂 +
3
2
𝒃 = 3 −𝒂 +
1
2
𝒃
𝐴𝐶 is a multiple of 𝐴𝑀 and 𝐴 is a
common point, therefore 𝐴𝑀𝐶 is a
straight line.
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