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Level 1
Level 2
Level 3
Level 4
Level 5
Column vectors
16 questions
Lesson
Not started
Mark as done
a
=
(
3
1
)
\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}
a
=
(
3
1
)
and
b
=
(
2
5
)
\mathbf{b} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}
b
=
(
2
5
)
. Work out
a
+
b
\mathbf{a} + \mathbf{b}
a
+
b
.
●
●●●●
Level 1
1 mark
Start
→
Mark as done
a
=
(
4
−
2
)
\mathbf{a} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}
a
=
(
4
−
2
)
. Work out
3
a
3\mathbf{a}
3
a
.
●
●●●●
Level 1
1 mark
Start
→
Mark as done
p
=
(
6
2
)
\mathbf{p} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}
p
=
(
6
2
)
and
q
=
(
1
4
)
\mathbf{q} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}
q
=
(
1
4
)
. Work out
p
−
q
\mathbf{p} - \mathbf{q}
p
−
q
.
●
●●●●
Level 1
1 mark
Start
→
Mark as done
a
=
(
5
−
2
)
\mathbf{a} = \binom{5}{-2}
a
=
(
−
2
5
)
and
b
=
(
−
3
4
)
\mathbf{b} = \binom{-3}{4}
b
=
(
4
−
3
)
.
(a)
Work out
a
+
b
\mathbf{a} + \mathbf{b}
a
+
b
as a column vector.
(b)
Work out
a
−
b
\mathbf{a} - \mathbf{b}
a
−
b
as a column vector.
(c)
Work out
3
b
3\mathbf{b}
3
b
as a column vector.
●●
●●●
Level 2
3 marks
Start
→
Mark as done
a
=
(
2
3
)
\mathbf{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}
a
=
(
2
3
)
and
b
=
(
−
1
4
)
\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}
b
=
(
−
1
4
)
(a)
Work out
a
−
b
\mathbf{a} - \mathbf{b}
a
−
b
(1 mark)
(b)
Work out
3
a
+
2
b
3\mathbf{a} + 2\mathbf{b}
3
a
+
2
b
(1 mark)
(c)
P
P
P
is the point
(
1
,
2
)
(1, 2)
(
1
,
2
)
and
Q
Q
Q
is the point
(
4
,
8
)
(4, 8)
(
4
,
8
)
. Write
P
Q
→
\overrightarrow{PQ}
P
Q
as a column vector. (1 mark)
●●
●●●
Level 2
3 marks
Start
→
Mark as done
a
=
(
3
−
2
)
\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}
a
=
(
3
−
2
)
and
b
=
(
−
1
4
)
\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}
b
=
(
−
1
4
)
(a)
Work out
2
a
+
b
2\mathbf{a} + \mathbf{b}
2
a
+
b
as a column vector. (1 mark)
(b)
Work out the magnitude of
a
\mathbf{a}
a
. Give your answer correct to 3 significant figures. (2 marks)
●●
●●●
Level 2
3 marks
Start
→
Mark as done
a
=
(
3
1
)
\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}
a
=
(
3
1
)
and
b
=
(
−
2
4
)
\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}
b
=
(
−
2
4
)
(a)
Work out
2
a
−
b
2\mathbf{a} - \mathbf{b}
2
a
−
b
[1]
(b)
Write down the exact magnitude of
a
\mathbf{a}
a
. [1]
(c)
P
P
P
is
(
−
1
,
2
)
(-1, 2)
(
−
1
,
2
)
and
Q
Q
Q
is
(
3
,
−
4
)
(3, -4)
(
3
,
−
4
)
. Write
P
Q
→
\overrightarrow{PQ}
P
Q
as a column vector. [1]
(d)
Show that
(
6
2
)
\begin{pmatrix} 6 \\ 2 \end{pmatrix}
(
6
2
)
is parallel to
a
\mathbf{a}
a
. [1]
●●
●●●
Level 2
4 marks
Start
→
Mark as done
A
A
A
is the point
(
2
,
7
)
(2, 7)
(
2
,
7
)
and
B
B
B
is the point
(
9
,
3
)
(9, 3)
(
9
,
3
)
.
(a)
Write
A
B
→
\overrightarrow{AB}
A
B
as a column vector.
(b)
Write down
B
A
→
\overrightarrow{BA}
B
A
as a column vector.
(c)
Work out
∣
A
B
→
∣
|\overrightarrow{AB}|
∣
A
B
∣
, giving your answer correct to
1
1
1
decimal place.
●●●
●●
Level 3
5 marks
Start
→
Mark as done
A drone flies from
P
P
P
to
Q
Q
Q
, and then from
Q
Q
Q
to
R
R
R
.
P
Q
R
Figure 1
(not accurately drawn)
P
Q
→
=
(
4
3
)
\overrightarrow{PQ} = \binom{4}{3}
P
Q
=
(
3
4
)
and
Q
R
→
=
(
−
6
1
)
\overrightarrow{QR} = \binom{-6}{1}
QR
=
(
1
−
6
)
.
(a)
Write
P
R
→
\overrightarrow{PR}
P
R
as a column vector.
(b)
P
P
P
is the point
(
1
,
2
)
(1, 2)
(
1
,
2
)
. Write down the coordinates of
R
R
R
.
(c)
Write down
R
P
→
\overrightarrow{RP}
R
P
as a column vector.
●●●
●●
Level 3
5 marks
Start
→
Mark as done
a
=
(
6
−
8
)
\mathbf{a} = \binom{6}{-8}
a
=
(
−
8
6
)
and
b
=
(
−
9
12
)
\mathbf{b} = \binom{-9}{12}
b
=
(
12
−
9
)
.
(a)
Work out
∣
a
∣
|\mathbf{a}|
∣
a
∣
.
(b)
Show that
b
\mathbf{b}
b
is parallel to
a
\mathbf{a}
a
.
(c)
Hence write down the value of
∣
b
∣
|\mathbf{b}|
∣
b
∣
.
●●●●
●
Level 4
5 marks
Start
→
Mark as done
a
=
(
4
−
1
)
\mathbf{a} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}
a
=
(
4
−
1
)
\qquad
b
=
(
−
1
3
)
\mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}
b
=
(
−
1
3
)
The vector
(
t
+
2
3
t
)
\begin{pmatrix} t + 2 \\ 3t \end{pmatrix}
(
t
+
2
3
t
)
is parallel to the vector
a
+
2
b
\mathbf{a} + 2\mathbf{b}
a
+
2
b
Find the value of
t
t
t
.
(3 marks)
●●●●
●
Level 4
3 marks
Start
→
Mark as done
A
B
→
=
(
6
−
4
)
\overrightarrow{AB} = \binom{6}{-4}
A
B
=
(
−
4
6
)
and
B
B
B
is the point
(
4
,
−
1
)
(4, -1)
(
4
,
−
1
)
.
(a)
Find the coordinates of
A
A
A
.
(b)
M
M
M
is the midpoint of
A
B
AB
A
B
. Find the coordinates of
M
M
M
.
●●●●
●
Level 4
4 marks
Start
→
Mark as done
a
=
(
3
−
1
)
\mathbf{a} = \binom{3}{-1}
a
=
(
−
1
3
)
and
b
=
(
−
2
5
)
\mathbf{b} = \binom{-2}{5}
b
=
(
5
−
2
)
.
(a)
Work out
2
a
−
3
b
2\mathbf{a} - 3\mathbf{b}
2
a
−
3
b
as a column vector.
(b)
The vector
(
k
20
)
\binom{k}{20}
(
20
k
)
is parallel to
b
\mathbf{b}
b
. Find the value of
k
k
k
.
●●●●
●
Level 4
5 marks
Start
→
Mark as done
Given that the vector
(
2
x
5
)
+
(
y
−
3
)
\begin{pmatrix} 2x \\ 5 \end{pmatrix} + \begin{pmatrix} y \\ -3 \end{pmatrix}
(
2
x
5
)
+
(
y
−
3
)
is parallel to the vector
(
x
2
)
+
(
3
y
1
)
\begin{pmatrix} x \\ 2 \end{pmatrix} + \begin{pmatrix} 3y \\ 1 \end{pmatrix}
(
x
2
)
+
(
3
y
1
)
find an expression for
y
y
y
in terms of
x
x
x
.
(3 marks)
●●●●●
Level 5
3 marks
Start
→
Mark as done
v
=
(
t
−
4
)
\mathbf{v} = \binom{t}{-4}
v
=
(
−
4
t
)
, where
t
t
t
is a positive number, and
∣
v
∣
=
5
|\mathbf{v}| = 5
∣
v
∣
=
5
.
(a)
Find the value of
t
t
t
.
(b)
Write down the value of
∣
−
5
v
∣
|-5\mathbf{v}|
∣
−
5
v
∣
.
●●●●●
Level 5
4 marks
Start
→
Mark as done
A
A
A
is the point
(
−
3
,
5
)
(-3, 5)
(
−
3
,
5
)
and
B
B
B
is the point
(
7
,
−
1
)
(7, -1)
(
7
,
−
1
)
.
(a)
Write
A
B
→
\overrightarrow{AB}
A
B
as a column vector.
(b)
C
C
C
is the point such that
B
C
→
=
1
2
A
B
→
\overrightarrow{BC} = \tfrac12\overrightarrow{AB}
B
C
=
2
1
A
B
. Find the coordinates of
C
C
C
.
(c)
Work out
∣
A
C
→
∣
|\overrightarrow{AC}|
∣
A
C
∣
, giving your answer correct to
1
1
1
decimal place.
●●●●●
Level 5
7 marks
Start
→
Vectors on a shape
13 questions
Lesson
Not started
Mark as done
P
Q
R
S
PQRS
P
QR
S
is a parallelogram.
P
Q
→
=
a
\overrightarrow{PQ} = \mathbf{a}
P
Q
=
a
and
P
S
→
=
b
\overrightarrow{PS} = \mathbf{b}
P
S
=
b
.
a
b
P
Q
R
S
The diagram is a sketch and is not accurately drawn.
(a)
Write down, in terms of
a
\mathbf{a}
a
and/or
b
\mathbf{b}
b
, the vector
S
R
→
\overrightarrow{SR}
S
R
.
(b)
Give a reason for your answer to part (a).
(c)
Find
Q
S
→
\overrightarrow{QS}
QS
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
.
●●
●●●
Level 2
4 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
M
M
M
is the midpoint of
A
B
AB
A
B
.
(a)
Write
A
B
→
\overrightarrow{AB}
A
B
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. (1 mark)
(b)
Write
O
M
→
\overrightarrow{OM}
O
M
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form. (2 marks)
●●●
●●
Level 3
3 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
M
M
M
is the midpoint of
A
B
AB
A
B
.
a
b
O
A
B
M
The diagram is a sketch and is not accurately drawn.
(a)
Find
A
B
→
\overrightarrow{AB}
A
B
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
.
(b)
Find
O
M
→
\overrightarrow{OM}
O
M
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●
●●
Level 3
4 marks
Start
→
Mark as done
A
B
C
D
ABCD
A
B
C
D
is a trapezium.
A
B
→
=
a
\overrightarrow{AB} = \mathbf{a}
A
B
=
a
and
B
C
→
=
b
\overrightarrow{BC} = \mathbf{b}
B
C
=
b
.
A
D
AD
A
D
is parallel to
B
C
BC
B
C
, and
A
D
=
3
×
B
C
AD = 3 \times BC
A
D
=
3
×
B
C
.
a
b
A
B
C
D
The diagram is a sketch and is not accurately drawn.
(a)
Write down
A
D
→
\overrightarrow{AD}
A
D
in terms of
b
\mathbf{b}
b
.
(b)
Find
D
C
→
\overrightarrow{DC}
D
C
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●
●●
Level 3
4 marks
Start
→
Mark as done
A
B
C
D
ABCD
A
B
C
D
is a quadrilateral.
A
B
→
=
a
\overrightarrow{AB} = \mathbf{a}
A
B
=
a
,
B
C
→
=
b
\overrightarrow{BC} = \mathbf{b}
B
C
=
b
and
C
D
→
=
2
b
−
a
\overrightarrow{CD} = 2\mathbf{b} - \mathbf{a}
C
D
=
2
b
−
a
.
a
b
2b − a
A
B
C
D
The diagram is a sketch and is not accurately drawn.
(a)
Find
A
D
→
\overrightarrow{AD}
A
D
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
(b)
Find
D
B
→
\overrightarrow{DB}
D
B
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●
●●
Level 3
4 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
P
P
P
is the point on
A
B
AB
A
B
such that
A
P
:
P
B
=
2
:
1
AP : PB = 2 : 1
A
P
:
P
B
=
2
:
1
.
M
M
M
is the midpoint of
O
B
OB
O
B
.
(a)
Find
O
P
→
\overrightarrow{OP}
O
P
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form. [3]
(b)
Find
P
M
→
\overrightarrow{PM}
P
M
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form. [2]
●●●●
●
Level 4
5 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
The point
P
P
P
lies on
A
B
AB
A
B
so that
A
P
:
P
B
=
1
:
4
AP:PB = 1:4
A
P
:
P
B
=
1
:
4
.
a
b
O
A
B
P
The diagram is a sketch and is not accurately drawn.
Find
O
P
→
\overrightarrow{OP}
O
P
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●●
●
Level 4
4 marks
Start
→
Mark as done
O
A
B
C
OABC
O
A
B
C
is a trapezium.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
C
→
=
b
\overrightarrow{OC} = \mathbf{b}
O
C
=
b
.
C
B
CB
C
B
is parallel to
O
A
OA
O
A
, and
C
B
=
3
×
O
A
CB = 3 \times OA
C
B
=
3
×
O
A
.
M
M
M
is the midpoint of
A
B
AB
A
B
.
a
b
O
A
B
C
M
The diagram is a sketch and is not accurately drawn.
(a)
Find
O
B
→
\overrightarrow{OB}
O
B
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
.
(b)
Find
O
M
→
\overrightarrow{OM}
O
M
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●●
●
Level 4
5 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
P
P
P
is the point on
A
B
AB
A
B
such that
A
P
:
P
B
=
1
:
2
AP : PB = 1 : 2
A
P
:
P
B
=
1
:
2
.
a
b
O
A
B
P
Diagram NOT accurately drawn
(a)
Write
A
B
→
\overrightarrow{AB}
A
B
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. (1 mark)
(b)
Show that
O
P
→
=
2
3
a
+
1
3
b
\overrightarrow{OP} = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}
O
P
=
3
2
a
+
3
1
b
(3 marks)
●●●●●
Level 5
4 marks
Start
→
Mark as done
O
A
B
C
OABC
O
A
B
C
is a quadrilateral.
C
B
T
CBT
C
B
T
is a straight line.
6a
4b
3a
O
A
B
C
T
M
N
M
M
M
is the point on
O
A
OA
O
A
such that
O
M
:
M
A
=
1
:
2
OM : MA = 1 : 2
O
M
:
M
A
=
1
:
2
N
N
N
is the midpoint of
A
B
AB
A
B
.
O
A
→
=
6
a
\overrightarrow{OA} = 6\mathbf{a}
O
A
=
6
a
\qquad
O
C
→
=
4
b
\overrightarrow{OC} = 4\mathbf{b}
O
C
=
4
b
\qquad
C
B
→
=
3
a
\overrightarrow{CB} = 3\mathbf{a}
C
B
=
3
a
B
T
→
=
k
C
B
→
\overrightarrow{BT} = k\overrightarrow{CB}
B
T
=
k
C
B
where
k
k
k
is a scalar.
Given that
M
N
T
MNT
M
N
T
is a straight line, find the value of
k
k
k
.
You must show all your working.
(5 marks)
●●●●●
Level 5
5 marks
Start
→
Mark as done
O
P
Q
R
OPQR
O
P
QR
is a quadrilateral.
a
b
kb
O
P
Q
R
X
Y
X
X
X
is the point on
O
P
OP
O
P
such that
O
X
:
X
P
=
1
:
3
OX : XP = 1 : 3
O
X
:
X
P
=
1
:
3
Y
Y
Y
is the point on
R
Q
RQ
R
Q
such that
R
Y
:
Y
Q
=
2
:
1
RY : YQ = 2 : 1
R
Y
:
Y
Q
=
2
:
1
O
P
→
=
a
\overrightarrow{OP} = \mathbf{a}
O
P
=
a
\qquad
O
R
→
=
b
\overrightarrow{OR} = \mathbf{b}
O
R
=
b
\qquad
P
Q
→
=
k
b
\overrightarrow{PQ} = k\mathbf{b}
P
Q
=
k
b
where
k
k
k
is a positive integer.
(a)
Find
X
Y
→
\overrightarrow{XY}
X
Y
in terms of
k
k
k
,
a
\mathbf{a}
a
and
b
\mathbf{b}
b
.
Give your answer in its simplest form.
(4 marks)
(b)
Is
X
Y
XY
X
Y
parallel to
O
P
OP
O
P
?
Give a reason for your answer.
(1 mark)
●●●●●
Level 5
5 marks
Start
→
Mark as done
O
A
B
C
OABC
O
A
B
C
is a parallelogram.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
C
→
=
b
\overrightarrow{OC} = \mathbf{b}
O
C
=
b
.
D
D
D
is the point on
A
B
AB
A
B
such that
A
D
:
D
B
=
3
:
1
AD:DB = 3:1
A
D
:
D
B
=
3
:
1
.
E
E
E
is the midpoint of
O
C
OC
O
C
.
a
b
O
A
B
C
D
E
The diagram is a sketch and is not accurately drawn.
(a)
Find
O
D
→
\overrightarrow{OD}
O
D
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
.
(b)
Find
E
D
→
\overrightarrow{ED}
E
D
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●●●
Level 5
6 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
M
M
M
is the midpoint of
O
A
OA
O
A
.
N
N
N
is the point on
A
B
AB
A
B
such that
B
N
:
N
A
=
1
:
3
BN:NA = 1:3
B
N
:
N
A
=
1
:
3
.
a
b
O
A
B
M
N
The diagram is a sketch and is not accurately drawn.
(a)
Find
O
N
→
\overrightarrow{ON}
O
N
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
(b)
Find
M
N
→
\overrightarrow{MN}
M
N
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
, giving your answer in its simplest form.
●●●●●
Level 5
6 marks
Start
→
Vector proofs
12 questions
Lesson
Not started
Mark as done
P
P
P
,
Q
Q
Q
,
R
R
R
and
S
S
S
are four points.
P
Q
→
=
2
a
+
6
b
\overrightarrow{PQ} = 2\mathbf{a} + 6\mathbf{b}
P
Q
=
2
a
+
6
b
and
R
S
→
=
3
a
+
9
b
\overrightarrow{RS} = 3\mathbf{a} + 9\mathbf{b}
R
S
=
3
a
+
9
b
.
(a)
Show that
P
Q
PQ
P
Q
is parallel to
R
S
RS
R
S
.
(b)
Freya says: "This working also proves that
P
P
P
,
Q
Q
Q
,
R
R
R
and
S
S
S
all lie on the same straight line."
Freya is wrong. Explain why.
●●●
●●
Level 3
3 marks
Start
→
Mark as done
A
A
A
,
B
B
B
and
C
C
C
are three points.
A
B
→
=
4
a
−
2
b
\overrightarrow{AB} = 4\mathbf{a} - 2\mathbf{b}
A
B
=
4
a
−
2
b
and
B
C
→
=
10
a
−
5
b
\overrightarrow{BC} = 10\mathbf{a} - 5\mathbf{b}
B
C
=
10
a
−
5
b
.
Prove that
A
A
A
,
B
B
B
and
C
C
C
are collinear.
●●●
●●
Level 3
3 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
2
a
\overrightarrow{OA} = 2\mathbf{a}
O
A
=
2
a
and
O
B
→
=
2
b
\overrightarrow{OB} = 2\mathbf{b}
O
B
=
2
b
.
M
M
M
is the midpoint of
O
A
OA
O
A
and
N
N
N
is the midpoint of
O
B
OB
O
B
.
O
A
B
M
N
2a
2b
Diagram not accurately drawn.
(a)
Find
M
N
→
\overrightarrow{MN}
M
N
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form.
(b)
Prove that
M
N
MN
M
N
is parallel to
A
B
AB
A
B
.
(c)
Hence write down the ratio
M
N
:
A
B
MN : AB
M
N
:
A
B
.
●●●●
●
Level 4
6 marks
Start
→
Mark as done
P
Q
R
S
PQRS
P
QR
S
is a quadrilateral.
P
Q
→
=
5
a
+
2
b
\overrightarrow{PQ} = 5\mathbf{a} + 2\mathbf{b}
P
Q
=
5
a
+
2
b
,
Q
R
→
=
3
a
−
4
b
\quad\overrightarrow{QR} = 3\mathbf{a} - 4\mathbf{b}
QR
=
3
a
−
4
b
,
R
S
→
=
−
5
a
−
2
b
\quad\overrightarrow{RS} = -5\mathbf{a} - 2\mathbf{b}
R
S
=
−
5
a
−
2
b
.
(a)
Find
P
R
→
\overrightarrow{PR}
P
R
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form.
(b)
Prove that
P
Q
R
S
PQRS
P
QR
S
is a parallelogram.
●●●●
●
Level 4
5 marks
Start
→
Mark as done
O
O
O
,
A
A
A
and
B
B
B
are three points, with
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
.
The point
C
C
C
is such that
O
C
→
=
1
5
a
+
4
5
b
\overrightarrow{OC} = \tfrac{1}{5}\mathbf{a} + \tfrac{4}{5}\mathbf{b}
O
C
=
5
1
a
+
5
4
b
.
(a)
Show that
C
C
C
lies on the line
A
B
AB
A
B
.
(b)
Write down the ratio
A
C
:
C
B
AC : CB
A
C
:
C
B
.
●●●●
●
Level 4
4 marks
Start
→
Mark as done
A
B
C
D
ABCD
A
B
C
D
is a quadrilateral.
A
B
→
=
4
a
+
2
b
\overrightarrow{AB} = 4\mathbf{a} + 2\mathbf{b}
A
B
=
4
a
+
2
b
,
B
C
→
=
a
−
3
b
\quad\overrightarrow{BC} = \mathbf{a} - 3\mathbf{b}
B
C
=
a
−
3
b
,
C
D
→
=
−
2
a
−
8
b
\quad\overrightarrow{CD} = -2\mathbf{a} - 8\mathbf{b}
C
D
=
−
2
a
−
8
b
.
(a)
Find
A
D
→
\overrightarrow{AD}
A
D
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form.
(b)
Show that
A
D
AD
A
D
is parallel to
B
C
BC
B
C
.
(c)
Aisha says that
A
B
C
D
ABCD
A
B
C
D
is a parallelogram. Aisha is wrong. Explain why.
●●●●
●
Level 4
5 marks
Start
→
Mark as done
O
A
C
B
OACB
O
A
C
B
is a parallelogram.
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
B
→
=
b
\overrightarrow{OB} = \mathbf{b}
O
B
=
b
M
M
M
is the midpoint of
A
C
AC
A
C
and
N
N
N
is the midpoint of
B
C
BC
B
C
.
a
b
O
A
C
B
M
N
Diagram NOT accurately drawn
Prove that
M
N
MN
M
N
is parallel to
A
B
AB
A
B
. (4 marks)
●●●●●
Level 5
4 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle.
O
A
→
=
2
a
\overrightarrow{OA} = 2\mathbf{a}
O
A
=
2
a
and
O
B
→
=
2
b
\overrightarrow{OB} = 2\mathbf{b}
O
B
=
2
b
.
P
P
P
is the point on
A
B
AB
A
B
such that
A
P
:
P
B
=
1
:
3
AP : PB = 1 : 3
A
P
:
P
B
=
1
:
3
.
Q
Q
Q
is the point on
O
B
OB
O
B
such that
O
Q
:
Q
B
=
1
:
3
OQ : QB = 1 : 3
O
Q
:
QB
=
1
:
3
.
Prove that
P
Q
PQ
P
Q
is parallel to
O
A
OA
O
A
. [5]
●●●●●
Level 5
5 marks
Start
→
Mark as done
O
A
B
C
OABC
O
A
B
C
is a quadrilateral.
C
B
T
CBT
C
B
T
is a straight line.
6a
4b
3a
O
A
B
C
T
M
N
M
M
M
is the point on
O
A
OA
O
A
such that
O
M
:
M
A
=
1
:
2
OM : MA = 1 : 2
O
M
:
M
A
=
1
:
2
N
N
N
is the midpoint of
A
B
AB
A
B
.
O
A
→
=
6
a
\overrightarrow{OA} = 6\mathbf{a}
O
A
=
6
a
\qquad
O
C
→
=
4
b
\overrightarrow{OC} = 4\mathbf{b}
O
C
=
4
b
\qquad
C
B
→
=
3
a
\overrightarrow{CB} = 3\mathbf{a}
C
B
=
3
a
B
T
→
=
k
C
B
→
\overrightarrow{BT} = k\overrightarrow{CB}
B
T
=
k
C
B
where
k
k
k
is a scalar.
Given that
M
N
T
MNT
M
N
T
is a straight line, find the value of
k
k
k
.
You must show all your working.
(5 marks)
●●●●●
Level 5
5 marks
Start
→
Mark as done
O
P
Q
R
OPQR
O
P
QR
is a quadrilateral.
a
b
kb
O
P
Q
R
X
Y
X
X
X
is the point on
O
P
OP
O
P
such that
O
X
:
X
P
=
1
:
3
OX : XP = 1 : 3
O
X
:
X
P
=
1
:
3
Y
Y
Y
is the point on
R
Q
RQ
R
Q
such that
R
Y
:
Y
Q
=
2
:
1
RY : YQ = 2 : 1
R
Y
:
Y
Q
=
2
:
1
O
P
→
=
a
\overrightarrow{OP} = \mathbf{a}
O
P
=
a
\qquad
O
R
→
=
b
\overrightarrow{OR} = \mathbf{b}
O
R
=
b
\qquad
P
Q
→
=
k
b
\overrightarrow{PQ} = k\mathbf{b}
P
Q
=
k
b
where
k
k
k
is a positive integer.
(a)
Find
X
Y
→
\overrightarrow{XY}
X
Y
in terms of
k
k
k
,
a
\mathbf{a}
a
and
b
\mathbf{b}
b
.
Give your answer in its simplest form.
(4 marks)
(b)
Is
X
Y
XY
X
Y
parallel to
O
P
OP
O
P
?
Give a reason for your answer.
(1 mark)
●●●●●
Level 5
5 marks
Start
→
Mark as done
O
A
B
C
OABC
O
A
B
C
is a parallelogram, with
O
A
→
=
a
\overrightarrow{OA} = \mathbf{a}
O
A
=
a
and
O
C
→
=
b
\overrightarrow{OC} = \mathbf{b}
O
C
=
b
.
M
M
M
is the midpoint of
A
B
AB
A
B
.
N
N
N
is the point on the diagonal
O
B
OB
O
B
such that
O
N
→
=
2
3
O
B
→
\overrightarrow{ON} = \tfrac{2}{3}\overrightarrow{OB}
O
N
=
3
2
O
B
.
a
b
O
A
B
C
M
N
Diagram not accurately drawn.
(a)
Find
C
N
→
\overrightarrow{CN}
C
N
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form.
(b)
Prove that
C
C
C
,
N
N
N
and
M
M
M
lie on the same straight line.
●●●●●
Level 5
7 marks
Start
→
Mark as done
O
A
B
OAB
O
A
B
is a triangle, with
O
A
→
=
6
a
\overrightarrow{OA} = 6\mathbf{a}
O
A
=
6
a
and
O
B
→
=
6
b
\overrightarrow{OB} = 6\mathbf{b}
O
B
=
6
b
.
M
M
M
is the midpoint of
O
A
OA
O
A
.
N
N
N
is the point on
A
B
AB
A
B
such that
A
N
:
N
B
=
2
:
1
AN : NB = 2 : 1
A
N
:
N
B
=
2
:
1
.
P
P
P
is the point on
O
B
OB
O
B
extended such that
O
P
→
=
p
b
\overrightarrow{OP} = p\,\mathbf{b}
O
P
=
p
b
, where
p
p
p
is a number.
(a)
Find
M
N
→
\overrightarrow{MN}
M
N
in terms of
a
\mathbf{a}
a
and
b
\mathbf{b}
b
. Give your answer in its simplest form.
(b)
M
M
M
,
N
N
N
and
P
P
P
lie on the same straight line.
Work out the value of
p
p
p
.
●●●●●
Level 5
6 marks
Start
→