Scalars and Vectors/Types of vectors
FURTHER MATHEMATICS SS 1 SECOND TERM
WEEK 4 DATE(S)………….
SUBJECT: FURTHER MATHEMATICS
CLASS: SSS 1
TOPIC: VECTORS 1
CONTENT:
Scalars and Vectors
Forms of vectors.
Parts of vectors
Algebra of vectors
The triangle law, parallelogram law and Resolution of vectors
Place vectors
SUB – TOPIC 1: Scalars and Vectors/Forms of vectors
Scalars and Vectors
There are two sorts of bodily portions
These which have solely magnitude
These which have each magnitude and path
Once we discuss of the temperature of a affected person as exceeding the conventional physique temperature of 98.40F or the mass of a baggage as 34kg, or the density of dry air at STP as 1.293kg.m-3 and many others, we’re referring to some particular magnitude in every case and these portions go by the title scalars.
Definition: A scalar is that amount which possesses solely magnitude
Moreover temperature, mass and density, different examples of scalars are; power, velocity, size and time.
Suppose an engine boat strikes with a velocity of 10knots, 450 East of North, then the displacement of the boat may be discovered at any given time if we all know the place to begin. We’ve got two sorts of displacement particularly;
Localized or free displacement: for instance, a displacement 40km east is identical in Lagos as in Accra.
Line situated displacement; as in forces. When a drive acts on a physique, the physique tends to maneuver within the path of the road of drive.
All portions that behave or have the identical traits as displacement are known as Vectors.
Definition:
A vector amount is that which has magnitude, path and sense. The next are vector portions; drive, momentum, velocity, acceleration, displacement, electrical discipline energy and magnetic induction. Vectors which don’t have any explicit instructions related to them are known as Free Vectors. A vector which is situated alongside a straight line known as a Line situated vector. Pressure is an instance of a line situated vector. When a vector is relevant to a selected level, it’s known as Level situated vector.
In printed work, vectors shall be proven in print to tell apart them from scalars which might be in mild print.
The size or magnitude of a vector a is denoted lal. The magnitude of a vector is usually known as its Modulus or Absolute worth or Numerical worth.
Vector which has a magnitude of zero and no explicit path known as a Zero vector or Null vector and is denoted by 0.
A vector which has an absolute worth of unity known as a Unit vector.
The vector which has the identical absolute worth as a vector a however a path reverse to the vector a known as the Detrimental vector of a and is denoted –a.
We will denote the unit vector within the path of the vector a by â and is such that
a = lal â.
Two vectors which have the identical magnitude and are in the identical path are mentioned to be Equal.
CLASS ACTIVITY:
What’s a scalar amount?
Outline a vector amount?
Enumerate three examples of scalar
Outline the followings. (i) Free vectors (ii) Line situated vector (iii) Pointed situated vector (iv) Modulus (v) Unity (vi) zero vector (vii) Equal
SUB – TOPIC: Algebra of vector/Triangle and parallelogram legal guidelines of vector
ALGEBRA OF VECTORS
Let recognized vector routes as a direct journey from one level to the opposite and likewise as a journey on the edges of a proper – angle triangle. On this half, we want to set up that the third vector route can take the type of travelling by the direct – route. Two vectors or extra may be added if their line segments are positioned finish to finish. The sum of two or extra vectors known as the resultant and it’s represented by the road section from the preliminary level of the primary vector to the ultimate level of the final vector.
u
u + v
v
P
Q
R
There are two methods of transferring from P to R.
One can transfer from P to Q after which transfer from Q to R
One can transfer instantly from P to R
The vector PRis known as the sum or resultant of the vectors PQand QR. We are able to write this as
PR = PQ + QR.
Suppose we denote PQ by u and QR by v then PR is the vector u + v. In different phrases.
PR = u + v
You may see that the above determine is a triangle, therefore the legislation that PR = PQ + QRis known as Triangle legislation of vector addition. The triangle legislation states: when two coplanar vectors u and vare added, their sum is u + v and ifuand vare the edges of a triangle, then their third aspect is u + v.
P
u – v
u
R
-v
Q
O
v + u
v
Within the triangle above, R is the mid – level of PQ
RP = -RQ
OR – RP = OR + (-RP)
= OR + PR
= OR + RQ
= OQ
Suppose
OR = u, RP = v and RQ = -v
Then
OP = u + v
OQ = u – v
A
C
O
B
a + b
a
Given two vectors a and b which aren’t parallel, then it’s doable to attract a parallelogram with sides equal to IaI and IbI respectively. The sum of a and b is the same as the vector illustration of one of many diagonals of the parallelogram. That is the import of the parallelogram legislation of vector addition which states that the resultant of two vectors is represented by the diagonal of the parallelogram whose adjoining sides are the 2 vectors.
From the determine,
OB + OA = OB + BC
= OC
i.e. b + a = OC
Definition: the addition of the vectors a and b which aren’t parallel is a + b and this sum is the vector illustration of one of many diagonals of a parallelogram whose sides are a and b.
Notice that if the vectors are parallel then we can not draw a parallelogram with these parallel sides forming adjoining sides.
If a, b and c are vectors, then below addition the next properties maintain:
a + b is a vector demonstrating closure
a + b = b – a, the commutative legislation
(a + b) + c = a + (b + c), the associative legislation
There’s a distinctive vector 0, with the property that; for another vector a,
a + 0 = 0 + a = a, property of identification
a + (-a) = (-a) + a = 0, property of inverse.
P
R
S
Q
RS = PQ or a
The magnitude of the vector amount is written IPQI, or IaI, or just PQ or a.
Notice that QP would signify a vector amount of the identical magnitude however with reverse sense.
A vector amount may be represented graphically by a line drawn in order that;
The size of the road denotes the magnitude of the amount, in line with some said vector scale.
The director of the road denotes the path through which the vector amount acts. The sense of the path is indicated by an arrowhead.
Instance 1: Discover the sum of the vectors PQ, 2RS, SP and –RQ
Answer: To resolve any such downside, we could or could not draw a diagram. All we’d like do is to rearrange the vectors in such a method that letters comply with the pure order P , Q , R and S.
Subsequently, PQ+2RS+SP – RQ = PQ – RQ + SP + 2RS
= PQ + QR + RS + RS + SP
= PS + RP
= RS
Instance 2: PQRS is a quadrilateral, with M and N because the midpoints of SP and RQ respectively. Present that PQ + SR = 2MN.
Answer: since M is midpoint of SP, then SM = MP. Equally, RN = NQ since N is the midpoint of RQ.
Vectorially, PQ+MP+QN = MN …………………………..(1)
MS +SR + RN = MN…………………………..(2)
Including equations (1) and (2) and substituting – MS for MP and – NR for NQ , we get
2MN = PQ +(- MS)+QN +MS+SR+NQ
2MN = PQ +SR since QN = -NQ
CLASS ACTIVITY:
Discover the sum of the next vectors:
AB, -DC, BC and CE (b) PR, – SR, ST and – QT
PQRS is a quadrilateral, with M and N because the mid – factors of the diagonals PR and QS respectively. Present that PQ + PS +RQ +RS = 4MN
SUB – TOPIC: Decision of vectors/element of vectors
If c is the resultant of vectors a and b, i.e. c = a + b then the vectors a and b, are referred to as the elements within the path of vector c.
B
C
A
Within the diagram above, AC + CB = AB by triangle legislation. The vectors AC and CB are the elements of AB.
As an alternative of utilizing arbitrary axes, it’s common to resolve a vector into elements parallel to the oblong axes that are perpendicular to one another. When such is finished, we select unit vectors with the identical instructions and senses because the constructive OX and OY axes.
Y
r
y
x
R(x,y)
N
O
X
From the diagram above, ON = ; NR =
Let OR = r. Then, OR = ON + NR
=
Subsequently, r =
Example1: If , and discover 2p + q + 3r
Answer: 2p + q + 3r =2 ( + + 3 (
=
=
Instance 2: The vectors are given by:
. Discover
Answer: =
=
=
=
Various methodology: may be written as ;
Equally, may be written as ;
Additionally, may be written as ;
Subsequently, =[mediator_tech]
=
CLASS ACTIVITY
On condition that and, discover the scalar m and n such that
Discover the resultant ; ;
Sub – Subject: place vector
The purpose from which a vector begins known as its origin. The sense of the vector is at all times indicated by arrows declaring from that origin. The place vector of a degree P relative to origin, O, is outlined because the vector OP. If we signify OP by then the vector is the place vector of P.
O
Q
P
p
q
PQ = PO + OQ
=- OP + OQ
= – p + q
= q– p
Instance 1: if OP = OQ = OR = . Discover :(i) QP(ii) RP
Answer: (i)QP = PO – OQ
=
=
(ii) RP = PO – OR
=
=
Instance 2: The place vector of the factors A, B and C are respectively:
Discover the place vector of the purpose P which divides: AB within the ratio 2:3
Answer: AP = AO + OP PB = PO + OB
2 = – + p …………….. (i) 3 = – p + …………… (ii)
Dividing equation (ii) by (i) we get
Making p topic of the formulation and substituting for we get
P = =
=
CLASS ACTIVITY
The place vector of the factors A, B and C are respectively:
Discover the place vector of the purpose P which divides: (i) AC within the ratio 1:4 (ii) BC within the ratio 3:5
if OP = OQ = OR = . Discover : (i) RQ (ii) PQ +RP
PRACTICE EXERCISE
ABCD is a quadrilateral whose sides signify vectors, discover a vector equal to AD+DC+CB
Discover the resultant of the vectors
Discover the sum of the next vectors utilizing place vectors
AB, -DC, BC and CE (b) PR, – SR, ST and – QT
If PQ and RS are two vectors such that PQ = 3RS, which of the next is appropriate?
RS is thrice so long as PQ
P, Q, R, S are collinear
PQ is parallel to RS
PQ divides RS internally.
State : (i) triangle legislation of vector (ii) parallelogram legislation of vector
ASSIGNMENT
If AB + AD = CB + CDprove that A and B are coincident.
Show that if BC = DE+AC then BD is parallel to AE.
Plot on a graph paper, the factors A(1,3), B(3,4), C(5,1). Discover the displacement specified by AB, BC and AC. (Trace: Let O(0,0) be the origin and apply the triangle legislation to triangle OAB)
Given OA = OB=and Okay=3, Discover kAB
Draw on a graph paper the next vectors: OB = (4cm,0300), BC = (3cm,3000). Discover vector OC given O as origin.
KEY WORDS/PHRASES
VECTOR
SCALAR/SCALAR QUANTITY
COMPONENT OF VECTOR
POSITION VECTOR
FREE VECTORS
RESULTANT VECTOR/SUM OF TWO OR MORE VECTORS
