Vector Quantities
Scalar quantities
A quantity that can be fully defined by its magnitude alone is called a scalar quantity. e.g.
mass, time, density, work, power, population density, bank balance etc.
Vector quantities
A quantity that can be fully defined only if its direction is known as well as its magnitude is
called a vector quantity. e.g. displacement, velocity, acceleration, momentum, force, torque, etc.
Vector A is written as A or
Vector representation in diagrams
In a diagram a vector is represented by a straight line with an arrow. The arrow represents the
direction of the vector.
If the diagram is not drawn to scale then the magnitude of the vector is written alongside the line.
If it is a scale diagram (also called a vector diagram), then the length (x) of the line is proportional
to the magnitude (y) of the vector.
y
x
y=kx
where k is a constant whose value is obtained from the given scale.
e.g. :
Let 1.0 cm represent 2.0N
Therefore, the scale constant is given by k = 2 N/cm
If length x = 10 cm, then magnitude Y of vector is given by
y=k.x
= 2.0 N/cm x 10 cm
= 20 N
Vectors are fully specified by magnitude and direction
Vectors having the same magnitude and acting in the same direction are identical. All the vectors
shown in the figure below are interchangeable.
Vectors cannot have negative magnitude
If a vector A is of magnitude 10 units, then the vector -A is also of magnitude +10 units but
its direction is opposite to that of vector A
The unit vector
If a vector of p units acting along the x axis is cut into p equal parts, then each part is a unit
vector along the x axis and is conventionally called i.
Similarly, if a vector of Q units along the y axis is cut into Q equal parts then each part is a
unit vector along the y axis and is conventionally called j.
On a two dimensional surface the x-axis and the y-axis co-ordinates are sufficient to define
uniquely the position of a point. But in 3-dimensional space, a z-axis co-ordinate is also required
to locate a point. Hence a unit vector along the z-axis is defined in a similar manner. Thus, if a
vector of S units along the z axis is cut into S equal parts then each part is a unit vector along the
z axis and is conventionally called k.
The zero vector
If vector A = vector B, then A - B is defined as the null or zero vector and is represented by
the symbol 0 or simply 0. It has zero magnitude and no specific direction. A vector which is not
null is a proper vector.
Orthogonal components of a vector
If the coordinates (x1,y1) of the point P on an x-y plane, are given, then the vector OP is uniquely
defined.
The magnitude of the vector OP is given by :-
The direction ( ) with the positive x-axis is given by :-
OP1 and OP2 have magnitudes x1 and y1 and they are directed along the x-axis and y-axis
respectively. OP1 and OP2 are therefore vectors. These two vectors are the components of OP.
Since OP1 and OP2 are perpendicular to each other, they are called the orthogonal
(perpendicular) components of the vector OP.
The vector OP is called the resultant or the vector sum of OP1and OP2.
Thus, given the orthogonal components of a vector, it is possible to determine the
magnitude and the direction, so that the vector sum OP is uniquely known.
If the x-component is a vector of x1 units, and the y-component a vector of y1 units, then,
the two components are given as:
OP1 = x1i ,
and
OP2 = y1j
Vector OP is written as:
P = x1i + y1j
The diagram below shows a vector P given by the equation:
P = 6i + 8j
Its two orthogonal components 6i and 8j are also shown in the figure below. The magnitude of the
vector is given by
In 3-dimensional space a vector P would be given as
P = x1i + y1j + z1k
Resolving a vector into its orthogonal components
If the coordinates of the point P are not supplied but the magnitude r and the direction of the
vector with respect to the positive x-axis are supplied, then the magnitudes of the orthogonal
components can be calculated as shown below:
OP1 = r cos
= x1
OP2 = r sin
= y1
Vectors have no effect perpendicular to their directions
It should be noted that a vector OP has components at angles of and (90 - ). At an angle
itself its component is OP1 = OP cos . If is 90o or 270o however, then,
to
OP1 = OP cos 90o = 0
OP1 = OP cos 270o = 0
or,
Hence a vector has no effect at right angles to itself.
Numericals
1. Find the magnitude and direction of a force which has an x-component of -40 N and a
y-component of -60 N.
[72N; 56.3o]
2. A child pulls a rope attached to a sledge with a force of 60 N. The rope makes an angle of
40o to the ground. Compute a) effective force trying to move the sledge along the ground b)
the force trying to lift the sledge vertically.
[46.0 N; 38.6 N]
Vector Addition and Subtraction
Vector addition
When vectors act in different directions, they cannot be added like scalars. Only vectors
acting along the same straight line can be added like scalars.
There are several different techniques for adding of vectors. These are merely different
conveniences and must yield the same result if the same vectors are being added.
Parallelogram law
If two vectors P and Q can be represented in magnitude and direction by two adjacent sides of a
parallelogram drawn outwards from a point, then their vector sum R is the diagonal of the same
parallelogram drawn outwards from the same point.
The magnitude and direction of the vector sum R are given respectively by:
The opposide side of the parallelogram being of the same magnitude and having the same
direction, must also be the same vector Q. this vector Q can be resolved into components Q
cos parallel to P, and Q sin perpendicular to P as shown below.
Thus there are three vectors to add now. P and Q cos being in the same direction can be
added like scalars yielding a single vector of (P+Q cos ). In the perpendicular direction there is
one vector Q sin .
But (P+Q cos ) and Q sin are the two orthogonal components of the vector R. They must
therefore add up to vector R.
The magnitude of the vector R is therefore given by:
The angle between the vector sum R and the vector component P is given by:
Triangle law
If the vectors P and Q to be added are represented in a diagram as the two sides of a triangle
taken in order, then the vector sum R is the third side of the same triangle taken in reverse order.
It should be noted that the vector sum R is identical to that calculated from the parallelogram.
Special cases
In the special cases below :
[a]
Vectors P and Q are parallel to each other.
i.e.
= 0,
[b]
The vectors P and Q are antiparallel to each other.
= 180o
[c]
The vectors P and Q are perpendicular to each other.
= 90o,
Resolution Method
If there are several vectors to add together, it becomes tedious to add two at a time using
the formula given to find the magnitude and the direction of each answer.
Instead of this needlessly inefficient process, a pair of axes is arbitrarily chosen. Let P, Q, S
and T be the vectors to be added. Let OX and OY be the chosen axes.
Vectors of the same magnitude and having the same direction are identical to each other
regardless of their points of application. Thus each vector can be moved parallel to itself so that it
is drawn outwards from the origin.
Each given vector is resolved into components in these two directions.
The x-axis components are now added like scalars yielding a result of
RX
=
P + Tcos 4 + (-Qcos 2) + (-Scos 3)
=
P + Tcos 4 - Qcos 2 - Scos 3
[the negative signs indicate directions along the negative x-axis]
The y-axis components are also added like scalars yielding a result of
RY
=
Pcos90o + Qsin 2 + (-Tsin 4) + (-Ssin 3)
=
0 + Qsin 2 - Tsin 4 - Ssin 3
[the negative signs indicate directions along the negative y-axis]
Now the sum of RX and RY can be found from:
This gives the magnitude of the sum R of all the vectors P, Q, S and T.
The direction of R with respect to the x-axis is given by:
Vector Subtraction
If a vector Q is to be subtracted from a vector P, the operation required is the same as adding a
vector -Q to the vector P.
Thus,
P - Q = P + (-Q)
In the case of vectors, as with scalars, only comparable quantities can be added or subtracted.
Numericals
1.
Find the sum of two forces P and Q of 30 N and 48 N making an angle of 60 o with
each other. At what angle should the forces P and Q act for their resultant to have
(a)
maximum magnitude
(b)
minimum magnitude
[68N;37.6o;78N;
18N]
2.
A man walks 5 km due north-east, then 4 km due 30onorth of east and finally 2 km due
60o south of east. What is his final displacement with respect to his starting point?
[8.86 km at 26o north of east ]
3.
Three
forces of 3N, 2N and 1N act
(a)
the total force along the x axis
on
a
body
as
shown.
Calculate:-
(b)
the total force along the y axis
(c)
the resultant force on the body
(d)
the single force that requires to be added to keep the body stationary.
[1.5N; 0.866N;1.732N at 30o; 1.732N at 210o]
4.
A man runs north west at a speed of 0.6 m/s on the deck of a ship which steams 30 o S of
east at 1.2 m/s. An ocean current pushes the ship at a speed of 1.0 m/s towards 30 o east of
north and the wind blowing against the ship pushes it due south at 0.8 m/s. Find the
resultant velocity of the man with respect to the earth.
[1.1 ms-1 at 6.7o south of east]
5.
Two vectors are given by a = 4i + 3j and b = -13i + 7j.
What are the x and y components of :(a)
vector a
(b)
vector b
(c)
vector c = a + b
What is the magnitude and the direction of the vector c. [4; 3; 13; 7; 9; 10; 13.5; 42o]
6.
Given that A = 8i + 6j + k and B = 12i - 5j - k, find the vectors P and Q such that
P = A - B and Q = B - A. Hence find R = P+ Q
Vector Multiplication
A vector may be multiplied by a scalar. The product is then a vector of changed magnitude
but unchanged direction. This process is called scaling a vector.
e.g. mass x acceleration = force
Vector quantity
Scalar multiplier
Product
velocity
mass
momentum
acceleration
mass
force
velocity
time
displacement
acceleration
time
velocity
P
3
3P
A vector may be multiplied by another vector only if the two vectors are either
(a)
parallel, or
(b)
perpendicular.
When parallel vectors such as A and B shown below are multiplied, the answer is a scalar.
This answer is called a scalar product or a dot product. It is expressed as:
A .B
If A is a vector of 2 units magnitude and B is a vector of 3 units magnitude, and they are parallel to
each other, then the answer of this mathematical operation yields a scalar of 6 units. Since it is a
scalar it has no direction.
e.g. the force vector F (= 2N) and the displacement vector s (=3m) in the direction of the force,
when multiplied together yields a product of 6 units (joules) of work which is a scalar quantity.
If the perpendicular vectors A and B shown in the diagram below are multiplied, then the answer is
a vector.
This answer is called a VECTOR PRODUCT or a CROSS PRODUCT. It is expressed as:
AxB
or, as
BxA
If A is a vector of 2 units magnitude and B is a vector of 3 units magnitude, and they are
perpendicular to each other, then the answer to each of these mathematical operations yields a
vector of 6 units. Since it is a vector it has a direction.
The direction of the product vector is perpendicular to each of the vectors being multiplied.
The two vectors being multiplied being directed along the positive x axis and the positive y axis as
shown below, the vector product is directed perpendicular to the x-y plane, i.e. it is parallel to the
z-axis.
Let P = A x B
and Q = B x A
P and Q are both vectors. P and Q are both of the same magnitude. P and Q are both directed
perpendicular to the x-y plane as shown. However P is directed exactly opposite to Q.
If it is required to multiply vectors A and B acting at an angle to each other, it is possible
to multiply A with the parallel component (Bcos ) of B yielding a scalar product of magnitude
A Bcos . It is also possible to multiply A with the perpendicular component (Bsin ) of B yielding a
vector of magnitude ABsin . This vector product would have a direction perpendicular to the
plane of the page.
Numerical
1.
2.
With reference to the diagram given below, find the following:
[a]
P.P
[b]
QxQ
[c]
QxP
[d]
Q.P
[e]
PxQ
[f]
3P - 4Q
Calculate A.N, given:
A = 4i + 3j
N = -6i - 2j
3.
For the vectors shown below, calculate the dot product.
4.
5.
Consider the paralellogram OABC which hosts the vectors Pand Q as shown. OA = 2 cm,
OC = 4 cm. Calculate:
[i]
the area of the parallelogram,
[ii]
the magnitude of the vector P x Q.
Given
A = 2i - 3j
B = 3i + 2j
Calculate the value of A x B.
6.
Find the angle between the lines P = 2i + 6j and Q = 5i - 10j