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© T Madas
12
AB BC CD ( )+ + = + + +a b a buur uur uur
2 3+i j
© T Madas
A vector is a line with a start and a finish.It therefore has:
1. line of action2. a direction3. a given size (magnitude)
A B ABuuur
= a
A B BAuur
-= a
© T Madas
a
ABuuur
a
we write vectors in the following ways:By writing the starting point and the finishing point in capitals with an arrow over them
With a lower case letter which:is printed in boldor underlined when handwrittenIn component form if the vector is drawn on a grid:45
© T Madas
E
F
A
B
C
D
G
H
Let AB = auuur
CD =uuur
2a
EF =uur
12a
HG =uuur
-2a
© T Madas
45
A
B
-5 4
C
D
AB =CD =
© T Madas
© T Madas
A
B C
D
Let AB = auuur
Let AD =buuur
a a
b
b
ABCD is a parallelogram
DC =uuur
a
BC =uuur
b
© T Madas
A
B C
D
Let AB = auuur
Let AD =buuur
a a
b
b
ABCD is a parallelogram
DC =uuur
a
BC =uuur
b
Now adding vectors
AC =uuur
ADuuur
DC+ =uuur
+ab = +a b
AB=uuur
BC+ =uuur
+ba
© T Madas
A
B C
D
Let AB = auuur
Let AD =buuur
a a
b
b
ABCD is a parallelogram
DC =uuur
a
BC =uuur
b
Now adding vectors
BD =uuur
BA=uur
AD+ =uuur
+b-a = -b a
BCuuur
CD+ =uuur
- ab
© T Madas
© T Madas
A
B
C
M
ABC is a triangle
N
with M the midpoint of ABand N the
midpoint of BC.Let AB = auuur
and AC =buuur
AM =uuur
12a
MB =uuur
BC =uuur
BAuur
AC+ =uuur
+b-a = -b a
12a
12a
12a
b
© T Madas
A
B
C
M
ABC is a triangle
N
with M the midpoint of ABand N the
midpoint of BC.Let AB = auuur
and AC =buuur
AM =uuur
12a
MB =uuur
BN =uuur
NC =uuur
BC =uuur
BAuur
AC+ =uuur
+b-a = -b a
12a
1 12 2
-b a1 12 2
-b a
12a
12a
1 12 2
-b a
1 12 2
-b a
b
© T Madas
A
B
C
M
ABC is a triangle
N
with M the midpoint of ABand N the
midpoint of BC.Let AB = auuur
and AC =buuur
AM =uuur
12a
MB =uuur
BN =uuur
NC =uuur
MN =uuur
MBuuur
BN+ =uuur
1 12 2
-+ b a12a
12a
1 12 2
-b a1 12 2
-b a
12a
1 12 2
-b a
1 12 2
-b a
12
= b
12b
b
12a
What is the relationship between AC and MN ?
© T Madas
12
- b
A
B
C
M
ABC is a triangle
N
with M the midpoint of ABand N the
midpoint of BC.Let AB = auuur
and AC =buuur
AM =uuur
12a
MB =uuur
BN =uuur
NC =uuur
NA =uur
NMuuur
MA+ =uuur
12
- b 12
- a
12a
1 12 2
-b a1 12 2
-b a
12a
1 12 2
-b a
1 12 2
-b a
12
- ( )=
12b
b
12a
b+a 12
- ( )= +a b
NA =uur
NBuuur
BA+ =uur
- a12
+ a 12
-= b 12
- a12bNA =
uurNCuuur
CA+ =uur
- b12
- a 12
-= b 12
- a
etc etc
etc etc
© T Madas
© T Madas
ABCDEF is a regular hexagon.
, and .AB BC CD= = =a b cuuur uuur uuur
A
B
C D
M is the midpoint of CE.
E
F
M
a
b
c
Write and simplify expressions in terms of a, b and c for :
(a) ,CEuuur
(b) MDuuur
and (c) MBuuur
a
b
c
CE =uuur
CDuuur
DE+ =uuur
c- a
CM =uuur
ME =uuur
12c 1
2- a
1 12 2
-c aMD =uuur
MEuuur
ED+uuur
12
= c 12
- a +a
12
= c 12
+ asolution
© T Madas
ABCDEF is a regular hexagon.
, and .AB BC CD= = =a b cuuur uuur uuur
A
B
C D
M is the midpoint of CE.
E
F
M
a
b
c
Write and simplify expressions in terms of a, b and c for :
(a) ,CEuuur
(b) MDuuur
and (c) MBuuur
a
b
c
1 12 2
-c a
MB =uuur
MCuuur
CB+uur
12
-= c 12
+ a - b
12
- c12
= a - b
12
( )= a- c 2- b
12
( 2 )= - -a b csolution
© T Madas
© T Madas
MA
B C
D
N
ABCD is a parallelogram.
and .AB AD= =a buuur uuur
M is the midpoint of AD
Write and simplify expressions in terms of a and b for :
(a) ,BDuuur
(b) BNuuur
and (d) NCuuur
N is a point of BD so that BN : ND = 2 : 1.
(c) MNuuur
a
12b 1
2b
a
b
2 23 3
-b a
solution
BD =uuur
BAuur
AD+ =uuur
-a+b = -b a
BN =uuur
23
BD =uuur
23
- a23
b
© T Madas
MA
B C
D
N
ABCD is a parallelogram.
and .AB AD= =a buuur uuur
M is the midpoint of AD
Write and simplify expressions in terms of a and b for :
(a) ,BDuuur
(b) BNuuur
and (d) NCuuur
N is a point of BD so that BN : ND = 2 : 1.
(c) MNuuur
a
12b 1
2b
a
b
MN =uuur
MAuuur
BN+ =uuur
12
- b +a
2 23 3
-b a
AB+uuur 2 2
3 3é ù+ -ê úë û
b a
13
= a 16
+ b
solution
© T Madas
MA
B C
D
N
ABCD is a parallelogram.
and .AB AD= =a buuur uuur
M is the midpoint of AD
Write and simplify expressions in terms of a and b for :
(a) ,BDuuur
(b) BNuuur
and (d) NCuuur
N is a point of BD so that BN : ND = 2 : 1.
(c) MNuuur
a
12b 1
2b
a
b
NC =uuur
NBuuur
23
- b 23
+ a
2 23 3
-b a
BC+ =uuur
+b 23
= a 13
+ b
(e) Using your answers from parts (c) and (d), show that M, N and C lie on a straight line
solution
© T Madas
MA
B C
D
N
ABCD is a parallelogram.
and .AB AD= =a buuur uuur
M is the midpoint of AD
Write and simplify expressions in terms of a and b for :
(a) ,BDuuur
(b) BNuuur
and (d) NCuuur
N is a point of BD so that BN : ND = 2 : 1.
(c) MNuuur
a
12b 1
2b
a
b
2 13 3
NC = +a buuur2 2
3 3-b a
(e) Using your answers from parts (c) and (d), show that M, N and C lie on a straight line
1 13 6
MN = +a buuur
2 13 3
NC = +a buuur
13
( )= 2a +b
1 13 6
MN = +a buuur
16
( )= 2a +b What is theratio MN : NC ?
solution
© T Madas
© T Madas
6 and 6 .AB AD= =a buuur uuur
ABCD is a parallelogram. M is the midpoint of AB
(a) Find the vector in terms of a and b (b) Prove that MNC is a straight line
13
BN BD=uuur uuur
N is a point of BD so that .
A
B C
D
M
NCuuur
3a
3a
6b
6a
6b
BD =uuur
BCuuur
BD+ =uuur
6b 6- a
BN =uuur
13
BD =uuur
2b 2- a
2 2-b aN
© T Madas
13
BN BD=uuur uuur
N is a point of BD so that . 6 and 6 .AB AD= =a buuur uuur
ABCD is a parallelogram. M is the midpoint of AB
(a) Find the vector in terms of a and b (b) Prove that MNC is a straight line
A
B C
D
M
NCuuur
3a
3a
6b
6a
6b
2 2-b a
NC =uuur
NBuuur
- (2 2 )-b aBC+ =uuur
6+ b-2= b 2+ a 6+ b2= a 4+ b
2 4+a bN
© T Madas
2+a b
13
BN BD=uuur uuur
N is a point of BD so that . 6 and 6 .AB AD= =a buuur uuur
ABCD is a parallelogram. M is the midpoint of AB
(a) Find the vector in terms of a and b (b) Prove that MNC is a straight line
A
B C
D
M
N
NCuuur
3a
3a
6b
6a
6b
2 2-b a2 4+a b
MN =uuur
MBuuur
(2 2 )+ -b aBN+ =uuur
3a2- a3= a 2+ b
= a 2+ b
© T Madas
2+a b
13
BN BD=uuur uuur
N is a point of BD so that . 6 and 6 .AB AD= =a buuur uuur
ABCD is a parallelogram. M is the midpoint of AB
(a) Find the vector in terms of a and b (b) Prove that MNC is a straight line
A
B C
D
M
N
NCuuur
3a
3a
6b
6a
6b
2 2-b a2 4+a b
2MN = +a buuur
2 4NC = +a buuur
MC =uuur
3 6+a b
, and have the same directionMN NC MCuuur uuur uuur
, and are collinearM N C\
© T Madas
© T Madas
ABCD is a quadrilateral and M, N , P and Q are the midpoints of AB, BC, CD and DA respectively.
AM = a, BN = b and CP = c.
Find in terms of a, b and c:
a) AD
b) AQ
c) MQ
d) NP
e) Deduce a geometric fact about the quadrilateral MNPQ
A
B
C
D
M
N
P
Q
a
b
c
a b
c
AD =AB + BC + CD= 2a+ 2b + 2c= 2(a + b + c)
AQ =a + b + c
a + b + c a + b + c
© T Madas
ABCD is a quadrilateral and M, N , P and Q are the midpoints of AB, BC, CD and DA respectively.
AM = a, BN = b and CP = c.
Find in terms of a, b and c:
a) AD
b) AQ
c) MQ
d) NP
e) Deduce a geometric fact about the quadrilateral MNPQ
A
B
C
D
M
N
P
Q
a
b
c
a b
c
MQ =MA+ AQ= -a+ (a + b + c)= b + c
a + b + c a + b + c
b + c
NP = NC + CP = b+ c
b + c
© T Madas
ABCD is a quadrilateral and M, N , P and Q are the midpoints of AB, BC, CD and DA respectively.
AM = a, BN = b and CP = c.
Find in terms of a, b and c:
a) AD
b) AQ
c) MQ
d) NP
e) Deduce a geometric fact about the quadrilateral MNPQ
A
B
C
D
M
N
P
Q
a
b
c
a b
c
a + b + c a + b + c
b + c
b + c
A quadrilateral with a pair of sides equal and parallel is a parallelogram
or can show that MN = QP = a + b
Hence MNPQ is a parallelogram.
© T Madas
