Notes 4.
1 VECTOR
4.1 .1 Vector and scalar quantities
A quantity that only has a size, known as magnitude, is called a
scalar quantity, such as area, volume, mass, temperature and
pressure.
A quantity that has a magnitude and a direction is called a vector
quantity, such as force, displacement, velocity, acceleration,
momentum and impulse.
EXAMPLE
Answers:
a) A scalar quantity ( magnitude only )
b) A vector quantity ( magnitude and direction )
c) A vector quantity ( magnitude and direction )
d) A scalar quantity ( magnitude only )
4.1.2 Magnitude and direction of vectors
The symbol || is used to denote the magnitude of a vector.
For example, |PQ| and |u| denote the magnitude of vectors PQ
and u respectively.
A vector with magnitude of zero and no direction is known as zero
vector. It is denoted as 0.
A negative sign ( - ) is used to denote a vector in the opposite
direction. For example the negative vector for a is a.
4.1.3 Determining whether two vectors are equal
Two vectors are equal if and only if they have the same magnitude
and the same direction.
For negative vectors, PQ = - QP and PQ = QP
Determine whether each of the following quantities is a
scalar or a vector quantity.
a) 40 ms
-1
b) 10E towards the east
c) 20 ms
-2
towards the west
d) 6m
2
a
b
Vector a and b are equal.
4.1.4 Determining whether two vectors are parallel
Two vectors, a and b are parallel if and only if a = mb where m is a
constant meaning
a) If a =mb then vectors a and b are parallel,
b) If vectors a and b are parallel, then a =mb.
If P,Q and R are three collinear points, then vectors PQ and QR are
parallel, meaning that PQ =mQR where m is a constant.
If vectors a and b are not parallel, and ha =kb, then h=k=0
EXAMPLES
Given that PQ = 2v and AB =- -
v, determine whether vectors PQ
and AB are parallel.
Steps Solution
Write down the value of PQ PQ=2v
Then find the value of v
v =
PQ
Write down the value of AB
AB = -
v
= -
PQ )
= -
PQ
Therefore vectors PQ and AB are
parallel.
Given that ( 3 ) p = ( 2 + 7 ) q where and are constants, find
the values of and if vectors p and q are non- zero and non-
parallel.
Steps Solution
Write down the equation ( 3 ) p = ( 2 + 7 ) q
Vectors p and q are non-zero
and non-parallel
3 = 0
= 3 and
2 + 7 = 0
= -
Given that MN = 4u and NP = 7u, show that points M,N and P are
collinear.
Steps Solution
Write down the value of MN MN = 4u
Then find the value of u
u =
MN
Write down the value of NP NP = 7u
= 7 (
MN )
=
MN
Therefore, vectors MN and
NP are parallel.
Since N is a common point, points
M,N and P are collinear.
4.2 ADDITION AND SUBTRACTION OF VECTORS
When two or more vectors are combined and represented by a
single vector, the single vector is known as the resultant vectors.
In the diagram above, PQRS is a trapezium . PQ is parallel to SR.
Given that PQ =
SR, PQ = 5m and |m| = 6, find
a) Vector PQ + SR in terms of m,
b) |PQ = SR|.
Steps Solution
Write down the value of PQ
PQ=
SR
Then find the value of SR
SR =
PQ ( PQ and SR are parallel)
=
( 5m )
= 3m
Find the value of PQ + SR PQ + SR = 5m + 3m
= 8m
Find the value | PQ + SR | |PQ + SR | = |8m|
= 8 |m|
= 8 ( 6 )
= 48 units
The diagram shows a triangle OAB. OA = a , OB = b and M is the
midpoint of OA. State in terms of a and/or b,
(a) MA (b) MB (c) AB
Steps Solution
(a) OM = MA because M is the
midpoint.
OA = a
MA =
OA
=
a
(b) In MOB, express MB in
terms of a and b.
MB = MO + OB
M
-a
O B
b
MB = MO + OB
= - OM + OB
= - -
a + b
(c) In AOB, express AB in
terms of a and b.
AB = AO + OB
AB = - AO + OB
= - a + b
j
5m
S
R
P Q
b
a
O B
A
M
In the diagram above, PQRS is a trapezium. PQ is parallel to SR.
Given that PQ = 2a, PR = 5b and SR = 2PQ.
a) Determine each of the following vectors in terms of a.
i) PR QR
ii) SR
b) Find the vector which is equal to vector 4a 5b.
Steps Solution
a i)
PR QR = PR + ( - QR )
= PR + RQ
= PQ
= 2a
ii) SR = 2PQ ( PQ and SR are parallel)
= 2 ( 2a )
= 4a
b) 4a 5b = 4a + ( - 5b )
= SR + ( - PR )
= SR + RP
= SP
In the diagram above, PTR and QSR are straight lines. Points S and T
lie on QR and PR respectively such that S is the midpoint of QR and
3PT = PR. Given That PQ = 4a and PR = 6b, express each of the
following vectors in terms of a and b.
a) QR b) RS c) PS d) QT
Steps Solution
a) In PQR, find QR
QR = QP + PR
= - PQ + PR
= - 4a + 6b
b) Find RS
S is the midpoint of QR
RS and RQ are parallel
RS =
RQ
RS =
( - QR )
=
( 4a 6b)
= 2a - 3b
c) In PQS, find PS
PS = PQ + QS
= PQ + SR (QS = SR)
= PQ + ( - RS )
= 4a + ( 3b 2a )
= 2a + 3b
2a
5b
S
P
Q
R
2a
5b
P
R
Q
P
R
Q
S
T
P
R
Q
P
Q
S
d) Given 3PT = PR
PT and PR are parallel
In PQT, find QT
3PT = PR
PT =
PR
=
(6b)
= 2b
QT = QP + PT
= - PQ + PT
= - 4a + 2b
4.3 VECTORS IN A CARTESIAN PLANE
a) In the form x + y or (
)
b) vector units I = (
) and j = (
)
The vector in a Cartesian plane which proceeds from point A( x
1
, y
1
)
to point B( x
2
, y
2
) is given by
= (
)
Given points P ( 5 , 1 ) and Q ( -2 , 4 ), express vector PQ in the form
x + y and in the form of (
)
Steps Solution
Use the formula
PQ = (
)
PQ = (-2 5 )I + ( 4 1 )j
= -7i + 3j
PQ = -7i + 3j
= (
)
Given that u = (
) and v = (
), find the value of p such that 2pu - v
is parallel to the x-axis.
Steps Solution
Write down the value of 2pu v
2pu v = 2p (
) - (
)
= (
) - (
)
= (
)
y= 0 ( parallel to x-axis ) 8p + 4 = 0
P =
4.3.1 UNIT VECTOR S IN A GIVEN PLANE
The unit vector in the direction of a is denoted as
If a = x + y, then the unit vector in the direction of a is given by
=
x + y
EXAMPLE
a) Determine the unit vector in the direction of u if u = 4 - 7.
b) Given that v = (
), find
a ) =
x + y
=
4 -7
=
P Q
T
b) =
)
=
)
=
)
= (
)
EXERCISES
SPM 2003
PAPER 1
Diagram 3 shows a parallelogram ABCD with BED as a straight line.
Given that AB = 6p, AD = 4q and DE = 2EB, express, in terms of p
and q
a) BD
b) EC
PAPER 2
6 Given that AB = (
), OB = (
) and CD = (
), find
a) the coordinates of A,
b) the unit vector in the direction of OA,
c) the value of k, if CD is parallel to AB
SPM 2004
PAPER 1
16. Given that O (0,0), A(-3,4) and B(2,16), find in terms of the unit
vectors, i and j ,
a) AB
b) the unit vector in the direction of AB
17. Given that A(-2,6), B(4,2) and C(m,p), find the value of m and of
p such that,AB + 2BC = 10i 12j.
Diagram 3
D C
B
A
E
PAPER 2
8 Diagram 3 shows OAB. The straight AP intersects the straight line
OQ at R. It is given that OP =
OB, AQ =
AB, OP = 6x and OA = 2y.
a) Express in terms of x and / or y :
i) AP
ii) OQ
b) I) given that AR = hAP, state AR in terms of h,x and y
ii) given that RQ = kOQ, state RQ in terms of k,x and y.
c) Using AR and RQ from b), find the value of h and of k.
SPM 2005
PAPER 1
15 Diagram 4 shows vector OA drawn on a Cartesian plane.
a) Express OA in the form (
).
b) Find the unit vector in the direction of OA.
16. Diagram 5 shows a parallelogram, OPQR drawn on a Cartesian
plane.
It is given that OP = 6i + 4j and PQ = -4i +5j.
Find PR.
PAPER 2
6 In diagram 3, ABCD is a quadrilateral. AED and EFC are straight
lines.
R
Diagram 3
O
B
A
P
Q
8
6
4
2
-2
-4
-6
-8
y
-15 -10 -5 5 10 15 x
A
6
4
2
-2
-4
-6
-8
-10
y
-15 -10 -5 5 10 15 x
Diagram 5
O
R
Q
P
F
Diagram 3
D
A B
C E
It is given that AB = 20x, AE = 8y, DC = 25x 24y, AE =
AD and
EF =
EC.
a) Express in terms of x and/ or y :
i) BD ,
ii) EC.
b) Show that the points B, F and D are collinear.
c) If = 2 and = 3, find
SPM 2006
PAPER 1
13 Diagram 6, shows 2 vectors, OA and AB.
Express
a) OA in the form of (
),
b) AB in the form xi + yj
14 The points P,Q and R are collinear. It is given that PQ = 4a 2b
and QR = 3a + (1 + k)b, where k is a constant. Find
a) The value of k
b) the ratio of PQ :QR
PAPER 2
5 Diagram 1 shows a trapezium ABCD.
It is given that AB = 2y, AD = 6x, AE =
AD and BC =
AD.
a) Express AC, in terms of x and y.
b) Point F lies inside the trapezium ABCD such that 2EF = mAB,
and m is a constant.
i) Express AF, interms of m, x and y.
ii) Hence, if the points A,F and C are collinear, find the value
of m.
Diagram 6
x
y
(4,3)
-5
A
O
B
Diagram 1
A
D
B
C
F
E
SPM 2007
PAPER 1
15 Diagram 3 shows a rectangle OABC and the point D lies on
the straight line OB.
It is given that OD = 3DB. Express OD, in terms of x and y.
16. The following information refers to the vectors a and b.
a= (
) and b = (
)
Find a) the vector 2a b
b) the unit vector in the direction of 2a b
PAPER 2
8 Diagram 3 shows AOB . The point P lies on OA and the point Q
lies on AB. The straight line BP intersects the straight line OQ at
point S.
It is given that
OA : OP = 4 : 1, AB : AQ = 2 : 1, OA = 8x, OB = 6y.
a) Express in terms of x and/or y:
i) BP
ii) OQ
b) Using OS = hOQ and BS = kBP, where h and k are constants,
find the value of h and k.
c) Given that = 2 units, = 3 units and AOB = 90
0.
Find
Diagram3
5y
9x
O
A
B
C
D
Diagram 3
S
B
O
A
P
Q
