# Basic Concepts Of Vectors Part 2 (F.Sc-Physics-Chapter 2)

**Basic Concepts Of Vectors: (Vi) Multiplication Of A Vector By A Scalar: **The product of a vector A and a number n > 0 is defined to be a new vector nA having the same direction as A but a magnitude n times the magnitude of A as illustrated in Fig. 2.4. if the vector is multiplied by a negative number, then its direction is reversed.

In the event that n represents a scalar quantity, the product nA will correspond to a new physical quantity and the dimensions of the two quantities which were multiplied together. For example, when velocity is multiplied by scalar mass m, the product is a new vector quantity called momentum having the dimensions as those of mass and velocity.

**(Vii) Unit Vector**

A unit vector in a given direction is a vector with magnitude one in that direction. It is used to represent the direction of a vector.

A unit vector in the direction of A is written as Â, which we read as A hat, thus

The direction along x, y and z axes are generally represented by unit vectors î, ĵ and k̂ respectively (fig. 2.5 a). The use of unit vectors is not restricted to Cartesian coordinate system only. Unit vectors may be defined for any direction. Two of the more frequently used unit vectors are the vector r̂ which represents the direction of the vector r (fig. 2.5 b) and the vector n̂ which represents the direction of a normal drawn on a specified surface as shown in Fig 2.5 (c)

Null vector is a vector of zero magnitude and arbitrary direction. For example, the sum of a vector and its negative vector is a null vector.

**A + (-A) = 0** …………….. (2.2)

**(ix) Equal Vectors **

Two vectors A and B are said to be equal if they have the same magnitude and direction, regardless of the position of their initial points.

This means that parallel vectors of the same magnitude are equal to each other.

**(X) Rectangular Components of a Vector**

A component of a vector is its effective value in a given direction. A vector may be considered as the resultant of its component vectors along the specified directions. It is usually convenient to resolve a vector into components along mutually perpendicular directions. Such components are called rectangular components.

Let there be a vector A represented by OP making angle θ with the x-axis. Draw projection OM of vector OP on x-axis and projection ON of vector OP on y-axis as shown in Fig.2.6. projection OM being along x-direction is represented by and projection ON = MP along y-direction is represented by . by head and tail rule

Thus and are the components of vector A. since these are at right angle to each other, hence, they are called rectangular components of A. considering the right angled triangle OMP, the magnitude of or x-component of A is

**(Xi) Determination Of A Vector From Its Rectangular Components**

If the rectangular components of a vector, as shown in Fig. 2.6, are given, we can find out the magnitude of the vector by using Pythagorean theorem.

In the right angled ∆ OMP,

The position vector r is a vector that describes the location of a point with respect to the origin. It is represented by a straight line drawn in such a way that its tail coincides with the origin and the head with point P (a,b) as shown in fig. 2.7(a). the projections of position vector r on the x and y axes are the coordinates a and b and they are the rectangular components of the vector r. hence

in there dimensional space, the position vector of a point P (a,b,c) is shown in fig. 2.7 (b) and is represented by

**Example 2.1:** the positions of two aeroplanes at any instant are represented by two points A (2, 3, 4) and B(5, 6, 7) from an origin O in km as shown in Fig. 2.8.

(i) What are their position vectors?

(ii) Calculate the distance between the two aeroplanes.

**Solution:** (i) A position vector r is given by

**r** = a î + b ĵ + 4 k̂

Thus position vector of first aeroplane A is

**OA** = 2 î + 3 ĵ + 4 k̂

And position vector of the second aeroplane B is

**OB** = 5 î + 6 ĵ + 7 k̂

By head and tail rule

**OA + AB = OB**

Therefore, the distance between two aeroplanes is given by

AB = OB – OA = (5 î + 6 ĵ + 7 k̂) – (2 î +3 ĵ + 4 k̂)

= ( 3 î + 3 ĵ + 3 k̂ )

Magnitude of vector AB is the distance between the position of two aeroplanes which is then: