Vectors and scalar quantities in physics

Vectors and scalar

Representations of vectors

Parallelogram law of vector

Addition of two vectors (compounding two vectors)

Resolution of vectors

The resultant of more than two vectors

Relative motion

Vectors and scalar

In physics physical quantities are divided into two; vectors and scalar quantities

What is Vector quantity?  Vector quantities are physical quantities that have both magnitudes and directions.  Examples of vector quantities are displacement, velocity, force, displacement and many more.

What is scalar quantity?  Scalar quantities are physical quantities that have magnitudes, but no directions. Examples are, distance, speed, mass, temperature, etc, .

Scalar quantities unlike vector quantities can be added and subtracted using ordinary algebra.

Representations of vectors

A vector can be represented by line OA, as show in the diagram below. The magnitude of vector is represented by the length of the line OA. If a vector is refers to be OA, it means that the vector have a magnitude equivalent to the length of OA, and act from O to A


A vector of 40 units can be represented on paper by an equivalent length, for example let say, 5cm, it means that 1cm, represent  8units.

The direction of the vector is represented by angle which it makes with a give line OX, as shown on the diagram, by the arrow.

Parallelogram law of vector  



The parallelogram law of vector, states that if two vectors are represented in magnitudes and directions, by the adjacent sides of a parallelogram, the resultant is represented in magnitude and direction by the diagonal of the parallelogram, drawn from the common points.

Addition of two vectors (compounding two vectors)

Two or more vectors acting on a body in given direction can be added or combined to give a single vector, which produce the same effect. The single vector produced after combining two or more vectors is called the resultant.

Consider two forces of magnitude P=30N and Q=40N, acting on a body O in any of the following directions:

(A).P and Q act in the same direction


In this case the resultant R, of the forces, P and Q ,  is given by the algebraic sum of the two force, that is

R= P + Q = 30 + 40 + 70N along OX

(B).P and Q act in opposite direction


For this number (B), the resultant of the two forces is obtained by subtraction, since they are acting in opposite direction to each other; therefore the resultant is given as

R = Q – P = 40 – 30 = 10N along OX

(C).P and Q act at right angle to each other

In the case where the forces are inclined to each other at an angle, the resultant of the two forces cannot be obtained by simple algebraic subtraction, or addition, but by what is called vector addition.

To get the resultant of the vectors inclined to each other, we can use the parallelogram law of vector, trigonometry ratio, or scalar drawing to obtain the magnitude and directions of the resultant vector.

 So from the Pythagoras’s theorem the solution of (c), is given as;

R2 = P2 + Q2 

R = \sqrt {{{30}^2} + {{40}^2}}  = 50N

The direction of the resultant with respect to a given line OX is the angle which R makes with OQ. It is given by: 

The HTML Clipboard

\tan \alpha  = \frac{{opposite{\rm{ side}}}}{{adjacent{\rm{ side}}}} = \frac{P}{Q}

 = , assuming that P=3 and Q =4, then

\tan \alpha  = \frac{3}{4} = 0.75,{\rm{  so }}\alpha {\rm{ = ta}}{{\rm{n}}^{ - 1}}{\rm{ of 0}}{\rm{.75 = 36}}{\rm{.}}{{\rm{9}}^0}

In alternative method, the resultant R and the direction α of the vector can also be found using scale drawing. The  P and  Q are represented using any suitable scale , for example 1cm to represent 10N, so that P (30N),  is represented by 3cm  and  Q (40N)  is represented by  4cm.  The rectangle is completed as shown below.


Now you, measure, the diagonal and convert it into Newton, using the same scale of 1cm=10N to obtain the resultant R.

The direction of the resultant vector, which is the angle α, can also be measured using a protractor.

If you have done your scale drawing well, you will get approximately the same value like the one you got using Pythagoras theorem.

(D).P and Q act at an acute angle Ѳ =60o

If  p and  Q  are inclined at angle 60o, .  In this case, you can use cosine rule to obtain the resultant that is using

R2 =P2 + Q2 – 2PQcos (180-Ѳ)

And because cos (180-Ѳ) = – cosѲ, then we have

R2 = P2 + Q2 + 2PQcosѲ

Where Ѳ is the angle between P and Q,

And α, can be found using the sine rule

\frac{{\sin \alpha }}{{QA}} = \frac{{\sin {{120}^0}}}{{OA}}

But sine 120o = sin (180o -120o) = sin60o, then we have

\frac{{\sin \alpha }}{{QA}} = \frac{{\sin {{60}^0}}}{{OA}}

You can also use scale drawing to obtain, the resultant and the direction, by using scale of 1cm = 10N, as shown, below. Again we use scale of 1cm=10N. we draw the length 4cm to represent Q ( 40N), using protractor , we draw the angle 60o, then we draw length 3cm to represent P (30N). Then we complete the parallelogram, as shown with dotted lines. The resultant and its direction can now be obtained by measuring the resultant R, with meter and measuring the direction of it which is the angle α between OA, and OQ using protractor


(E).P and Q act at an obtuse angle O-120o

If P and Q are inclined to at 120o, the resultant can also be solved using cosine rule, that is

R2 = P2 + Q2 + 2PQ cos120o, but cos120o = – cos60o, therefore,

R2 = P2 + Q2 – 2PQcos60o =36N

Alternatively we can also use scale drawing to solve or obtain resultant and its direction.

Resolution of vectors

A vector, such as force OA  =10N, acting at angle 30o, with direction of OX, can be broken or resolved into two perpendicular parts. This component or parts can be determined by drawing a rectangle for which the force will be represented by diagonal


Using Pythagoras’s theorem, we can obtain

OX =10cos30o (horizontal component) and

OY=10sin30o (vertical component)

These two components called horizontal and vertical component can also be resolved to produce original result, by using vector addition

The resultant of more than two vectors

 To find the resultant of several forces, f1, f2,f3 f4…… acting on a point O, we first reduce the whole system to two perpendicular forces and then compound the force to obtain the resultant

To reduce the system to two perpendicular forces, we simple find the component of each forces, f1, f2,f3,…along two perpendicular direction OX and OY, and then add to obtain   P  and  Q


Let look at the example below, Find the resultant of the system of forces given in diagram below


Stage A, is the given problem needed to be solved, so from stage A, it was reduced to stage B, and to stage C, where  it is finally reduced to two perpendicular force of 4N and 2N. From stage C, the vectors can now be solved using Pythagoras theorem.

{R^2} = {4^2} + {2^2}

  Where R = resultant

{R^2} = \sqrt {{4^2} + {2^2}}  = 4.5N

 , the direction is given by tan-12=63o with horizontal

Relative motion

When two bodies A and B are in motion are moving in a straight line , the velocity of A in  relative to B is found by adding the velocity of  reversed to the velocity of A.

Example1: A car traveling in a straight road at 100km/h passes a bus going in the same direction at 60km/h. What is the velocity of the car in relative to the velocity of the bus?

Answer: the velocity of the car in relative to the bus is 100 – 60 = 40km/h

Example 2: If a car in motion with velocity of 100km/h and a bus with 60km/h are moving in opposite direction, what will be the velocity of the car in relative to bus?

Answer: The relative velocity of the car to that of the bus is 100 + 60 =160km/h


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