**Scalar and Vector**

The quantities that can be presented completely by the magnitude only are called scalars. Example : temperature, mass, density, electric potential and so on.

The quantities that require both magnitude as well as direction for their complete significance are called vectors. Example : velocity, electric field intensity, force and so on.

*Vector Algebra*

Assume that A and B are two vector quantities and r and s are two constants.

1. Addition:

A + B = B + A

A + (B + C) = (A + B) + C

2. Subtraction:

A + (- B) = A - B

3. Multiplication:

(r + s) (A + B) = r A + r B + s A + s B

4. Division:

(1/r) (A + B) = A / r + B / r

*Scalar Field and Vector Field*

Scalar field is the field in which each point in space can be represented by scalar quantities. For example : temperature field i.e. T = e ^ (xyz / 20)

Vector field is the field in which each point in space can be represented by the vector quantities. For example : velocity field.

*Dot Product*

- Let us consider A and B be two vectors, then the dot product of these two vectors is represented as:

A . B = |A| * |B| * cosθ

where, θ represents the smaller angle between A and B.

- It is considered that: A . B = B . A

- Example of dot product is:

1. Work done by a force is given by:

W = F . L = F * L * cosθ = ∫ F . dL

- For A = A(x) a(x) + A(y) a(y) + A(z) a(z), where a(x), a(y) and a(z) are the unit vector components of A vector and A(x), A(y) and A(z) are constants. Similarly, for B = B(x) b(x) + B(y) b(y) + B(z) b(z)

A . B = A(x) B(x) + A(y) B(y) + A(z) B(z)

- Dot product of similar unit vectors is 1, while of dissimilar unit vectors is 0.

i.e. a(x) . a(x) = a(y) . a(y) = a(z) . a(z) = 1

a(x) . a(y) = a(y) . a(z) = a(z) . a(x) = 0

*Cross Product*

- For the two vectors A and B, the cross product of these two vectors is given by:

A × B = |A| * |B| * sinθ * n^

- It is perpendicular to both the vectors.

- a(x) × a(y) = a(z); a(y) × a(z) = a(x); a(z) × a(x) = a(y)

- a(y) × a(x) = -a(z); a(z) × a(y) = -a(x); a(x) × a(z) = -a(y)

- a(x) × a(x) = a(y) × a(y) = a(z) × a(z) = 0

- A × B != B × A

A × B = | a(x) a(y) a(z) | | A(x) A(y) A(z) | | B(x) B(y) B(z) |

**Coordinate System**

*Rectangular Coordinate System*

- It consists of three coordinate axes mutually perpendicular to each other.

- The axes are x, y and z axes.

- A point is located by its x, y and z coordinates.

- Any vector in rectangular or cartesian coordinate system can be represented by:

B = B(x) a(x) + B(y) a(y) + B(z) a(z)

where, a(x), a(y) and a(z) are the unit vectors along x, y and z axes respectively.

*Circular Cylindrical Component System*

- It is the 3 dimensional version of polar coordinate system.

- It consists of three planes; circular cylinder (p = constant), planes (z = constant and ϕ = constant)

- The unit vectors a(p), a(z) and a(ϕ) are mutually perpendicular to each other.

- a(p) is normal to cylindrical surface, a(ϕ) is normal to plane ϕ = constant and a(z) is normal to plane z = constant.

- For the two cylinders of radius p and p + dp, two radial planes at angles ϕ and ϕ + dϕ, two horizontal planes at elevation z and z + dϕ, can enclose a small volume.

The sides are of length dp, dz and pdϕ.

So, surface areas are pdϕdp, pdϕdz and dpdz.

The volume is pdpdϕdz.

*Spherical Coordinate System*

- It consists of three mutually perpendicular surfaces (sphere; r = constant, cone; θ = constant, and plane; ϕ = constant)

- a(r), a(θ) and a(ϕ) are unit vectors.

- Differential Elements:

The lengths are dr, rdθ and rsinθdϕ

*Rectangular and Cylindrical Coordinate Conversion*

We know that:

p = √(x^2 + y^2)

ϕ = tan-1 (y / x)

Also,

x = p cosϕ

y = p sinϕ

The conversion table is shown in given table:

a(p) | a(ϕ) | a(z) | |

a(x) | cosϕ | - sinϕ | 0 |

a(y) | sinϕ | cosϕ | 0 |

a(z) | 0 | 0 | 1 |

*Rectangular and Spherical Coordinate Conversion*

We know:

r = √(x^2 + y^2 + z^2)

θ = cos-1 (z / (√(x^2 + y^2 + z^2)))

ϕ = tan-1 (y / x)

Also,

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

The conversion table is shown in given table:

a(r) | a(θ) | a(ϕ) | |

a(x) | sinθ cosϕ | cosθ cosϕ | - sinϕ |

a(y) | sinθ sinϕ | cosθ sinϕ | cosϕ |

a(z) | cosθ | -sinθ | 0 |

**Numerical Problems of Coordinate System**

1. Convert B = y a(x) - x a(y) + z a(z) into cylindrical coordinate system.

We have, B = y a(x) - x a(y) + z a(z)

In cylindrical system, let B = B(p) a(p) + B(ϕ) a(ϕ) + B(z) a(z)

Now, from the rectangular and cylindrical conversion table,

1. B(p) = B.a(p) = [y a(x) - x a(y) + z a(z)] . a(p)

= y a(x) . a(p) - x a(y) . a(p) + z a(z) . a(p)

= y cosϕ - x sinϕ

2. B(ϕ) = B.a(ϕ) = [y a(x) - x a(y) + z a(z)] . a(ϕ)

= y a(x) . a(ϕ) - x a(y) . a(ϕ) + z a(z) . a(ϕ)

= - y sinϕ - x cosϕ

3. B(z) = B.a(z) = z

So,

B = (y cosϕ - x sinϕ) a(p) - (y sinϕ + x cosϕ) a(ϕ) + z a(z)

Substituting value of x and y, we get:

B = (p sinϕ cosϕ - p cosϕ sinϕ) a(p) - (p (cosϕ)^2 + p (sinϕ)^2) a(ϕ) + z a(z)

= -pa(ϕ) + za(z)

2. Convert G = (xz / y) a(x) into spherical coordinate system.

Let in spherical coordinate, G = G(r) a(r) + G(θ) a(θ) + G(ϕ) a(ϕ)

Now,

G(r) = G. a(r) = (xz / y) a(x) . a(r) = xz/y sinθ cosϕ

G(θ) = G. a(θ) = (xz / y) a(x) . a(θ) = xz/y cosθ cosϕ

G(ϕ) = G. a(ϕ) = (xz / y) a(x) . a(ϕ) = - xz/y sinϕ

So,

G = xz/y sinθ cosϕ a(r) + xz/y cosθ cosϕ a(θ) - xz/y sinϕ a(ϕ)

Substituting for the value of x, y and z;

G = ( r cosθ cosϕ / sinϕ) * { sinθ cosϕ a(r) + cosθ cosϕ a(θ) - sinϕ a(ϕ) }

3. Transform to cylindrical coordinates: F = 10 a(x) - 8 a(y) + 6 a(z) at point P(10, -8, 6).

Let us consider in cylindrical coordinates: F = F(p) a(p) + F(ϕ) a(ϕ) + F(z) a(z)

So,

F(p) = [10 a(x) - 8 a(y) + 6 a(z)] . a(p) = 10 a(x) . a(p) - 8 a(y) . a(p) + 6 a(z) . a(p) = 10 cosϕ - 8 sinϕ

F(ϕ) = 10 a(x) . a(ϕ) - 8 a(y) . a(ϕ) + 6 a(z) . a(ϕ) = - 10 sinϕ - 8 cosϕ

F(z) = 6

Hence, F = (10 cosϕ - 8 sinϕ) a(p) - (10 sinϕ + 8 cosϕ) a(ϕ) + 6 a(z)

At point P(10, -8, 6):

p = √(x^2 + y^2) = 12.81

ϕ = tan-1 (y / x) = -38.66 degree

Now,

F = 12.81 a(p) + 6 a(z)

4. At point P(-3, -4, 5), express that vector that extends from P to Q(2, 0, -1) in spherical coordinates.

The vector that extends from P to Q can be expressed as vector PQ given by: PQ = 5 a(x) + 4 a(y) - 6 a(z)

In terms of spherical coordinates:

A(r) = 5 a(x) . a(r) + 4 a(y) . a(r) - 6 a(z) . a(r) = 5 sinθ cosϕ + 4 sinθ sinϕ - 6 cosθ

A(θ) = 5 a(x) . a(θ) + 4 a(y) . a(θ) - 6 a(z) . a(θ) = 5 cosθ cosϕ + 4 cosθ sinϕ + 6 sinθ

A(ϕ) = 5 a(x) . a(ϕ) + 4 a(y) . a(ϕ) - 6 a(z) . a(ϕ) = - 5 sinϕ + 4 cosϕ

Hence,

PQ = (5 sinθ cosϕ + 4 sinθ sinϕ - 6 cosθ) a(r) + (5 cosθ cosϕ + 4 cosθ sinϕ + 6 sinθ) a(θ) + (- 5 sinϕ + 4 cosϕ) a(ϕ)

- by **SURAJ AWAL**

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