Scalers:

Those physical quantities which have magnitude but no direction is called scalars. They follow ordinary algebraic rules for the operation of scalars. Eg. speed, mass, length, time, work done, etc.

Vectors:

Those physical quantities which have both magnitude and direction and follow the rules of vector rules of operation. For Eg. velocity, displacement, force, linear momentum, acceleration, etc.

Vectors are represented by the straight lines with arrow heads. Length of the line gives the magnitude of that vector and arrow heads gives the direction.

Laws of addition of vector:

1) Head tail combination:

a) [latex]\vec{P} + \vec{Q}[/latex]

b) [latex]\vec{Q} + \vec{P}[/latex]

Triangle law of vector addition:

It states that if two vectors acting at a point simultaneously be represented in magnitude and direction by two sides of a triangle taken in the same order then third side of the triangle in the reverse order represents magnitude and direction of the resultant vector of these two vectors.

Suppose two vector P and Q acting on a point simultaneously making an angle [latex]\theta[/latex] shown in figure (a). These two vectors are represented in magnitude and direction by two sides [latex]\vec{AB}[/latex] and [latex]\vec{B}[/latex] of a triangle [latex]\Delta[/latex]ABC in the same order then third side of the triangle in a reversed order i.e. [latex]\vec{AC}[/latex] represents magnitude and direction of the resultant vectors as shown in the figure (b).

In [latex]\Delta[/latex]ABC,

[latex]\vec{AC}=\vec{AB}+\vec{BC}[/latex] 

[latex]\vec{R}=\vec{P}+\vec{Q}[/latex]………………….. (i)

Draw CD ⊥ AB produced.

In [latex]\Delta[/latex]BDC,

Cos[latex]\theta = \frac{BD}{BC}[/latex]

Or, BD = QCos[latex]\theta[/latex]

Sin[latex]\theta = \frac{CD}{BC}[/latex]

Or, CD = QSin[latex]\theta[/latex]

Then,

AD = AB + BD

              = P + QCos[latex]\theta[/latex]

Now, using Pythagoras theorem in [latex]\Delta[/latex]ADC, we get,

AC² = AD² + CD²

R² = (P + QCos[latex]\theta[/latex])² + (QSin[latex]\theta[/latex])²

R² = P² + 2PQCos[latex]\theta[/latex] + Q²Cos²[latex]\theta[/latex] + Q²Sin²[latex]\theta[/latex]

R² = P² + 2PQCos[latex]\theta[/latex] + Q²(Cos²[latex]\theta[/latex] + Sin²[latex]\theta[/latex])

R² = P² + 2PQCos[latex]\theta[/latex] + Q²

R = [latex]\sqrt{P^2+Q^2+2PQCos\theta}[/latex] ………………… (ii)

Let, β be the angle made by [latex]\vec{R}[/latex] with [latex]\vec{P}[/latex], then,

Tanβ = [latex]\frac{CD}{AD}[/latex]

Or, Tanβ = [latex]\frac{QSin\theta}{P+QCos\theta}[/latex] ………………. (iii)

Specific cases:

Case I:

If [latex]\theta[/latex] = 0^o, then,

R = [latex]\sqrt{P^2+Q^2+2PQCos0^o}[/latex]

Or, R = [latex]\sqrt{P^2+Q^2+2PQ}[/latex]

Or, R = P + Q

∴ Tanβ = [latex]\frac{QSin0^o}{P+QCos0^o}[/latex] = 0

∴ β = 0^o

Case II:

If [latex]\theta[/latex] = 90^o, then,

R = [latex]\sqrt{P^2+Q^2+2PQCos90^o}[/latex]

Or, R = [latex]\sqrt{P^2+Q^2}[/latex]

∴ Tanβ = [latex]\frac{QSin90^o}{P+QCos90^o} = \frac{Q}{P}[/latex]

∴ β = Tan^-1 [latex](\frac{Q}{P})[/latex]

Case III:

If [latex]\theta[/latex] = 180^o, then,

R = [latex]\sqrt{P^2+Q^2+2PQCos180^o}[/latex]

              = [latex]\sqrt{P^2+Q^2-2PQ}[/latex]

              = P – Q if P > Q

              = Q – P if Q > P

Tanβ = [latex]\frac{QSin180^o}{P+QCos180^o} = 0[/latex]

∴ β = 0^o.

Parallelogram law of vector addition:

It states that, if two vectors acting at a point simultaneously be represented in magnitude and direction by two adjacent sides of a parallelogram then the diagonal of it passing through the point of intersection of these adjacent sides represents magnitude and direction of the resultant vector of these two vectors.

Suppose, two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are acting at a point O simultaneously making an angle [latex]\theta[/latex] as shown in fig. (a). These two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are represented in magnitude and direction by two adjacent sides [latex]\vec{OP}[/latex] and [latex]\vec{OS}[/latex]. On parallelogram OPQS, then, the diagonal [latex]\vec{OQ}[/latex] represents the magnitude and direction of the resultant vector [latex]\vec{R}[/latex].

So, [latex]\vec{OQ}=\vec{OP}+\vec{PQ}[/latex] 

 [latex]\vec{R}=\vec{A}+\vec{B}[/latex]…….…………… (A)

Now, drop a perpendicular from Q on OM so they meet at M.

From right angled triangle [latex]\Delta[/latex]OQM,

OQ² = OM² + QM²

Or, R² = (OP + PM)² + QM²

Or, R² = OP² + 2.OP.PM + PM² + QM² ……………. (i)

In right angled [latex]\Delta[/latex]PQM,

Sin[latex]\theta = \frac{QM}{PQ} = \frac{QM}{B}[/latex]

Or, QM = BSin[latex]\theta[/latex]

And, Cos[latex]\theta = \frac{PM}{PQ} = \frac{PM}{B}[/latex]

Or, PM = BCos[latex]\theta[/latex]

Then, by substituting the values of OP, QM and PM, we get,

R² = A² + 2ABCos[latex]\theta[/latex] + B²Cos²[latex]\theta[/latex] + B²Sin²[latex]\theta[/latex]

Or, R² = A² + 2ABCos[latex]\theta[/latex] + B² (Cos²[latex]\theta[/latex] + Sin²[latex]\theta[/latex])

Or, R² = A² + 2ABCos[latex]\theta[/latex] + B² [Since, Sin²[latex]\theta[/latex] + Cos²[latex]\theta[/latex] = 1]

Or, R = [latex]\sqrt{A^2+2ABCos\theta + B^2}[/latex]

∴ R = [latex]\sqrt{A^2+B^2+2ABCos\theta}[/latex]

Which is the magnitude of resultant of [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex].

Now, for direction of resultant [latex]\vec{R}[/latex], let us suppose, ‘β’ is the angle made by resultant with the direction of [latex]\vec{A}[/latex]

In [latex]\Delta[/latex]OQM,

tanβ = [latex]\frac{QM}{OM} = \frac{BSin\theta}{OM} = \frac{BSin\theta}{A+BCos\theta}[/latex]

Or, β = tan^-1 [latex](\frac{BSin\theta}{A+BCos\theta})[/latex]

This gives the direction of resultant vector.

Special Cases:

1) When two vectors are acting in the same direction, then [latex]\theta[/latex] = 0^o.

∴ R = [latex]\sqrt{A^2+B^2+2ABCos0^o}[/latex]

= [latex]\sqrt{A^2+B^2+2AB}[/latex] [Since, Cos0^o = 1]

Or, R = [latex]\sqrt{(A+B)^2}[/latex]

Or, R = (A + B)

i.e., resultant is algebraic sum of magnitude of two vectors.

And, β = tan^-1 [latex](\frac{BSin0^o}{A+BCos0^o})[/latex]

= tan^-1 ([latex]\frac{B\times 0}{A+B\times 1}[/latex])

Or, β = tan^-1(0) = 0

i.e., the resultant coincides with given two vectors.

2) When two vectors are acting at right angles, then [latex]\theta[/latex] = 90^o.

∴ R = [latex]\sqrt{A^2+B^2+2ABCos90^o}[/latex]

= [latex]\sqrt{A^2+B^2+0}[/latex] [Since, Cos90^o = 0]

Or, R = [latex]\sqrt{A^2+B^2}[/latex]

And, β = tan^-1 [latex](\frac{B}{A})[/latex]

3) When two vectors are acting in the opposite direction, then, [latex]\theta[/latex] = 180^o.

∴ R = [latex]\sqrt{A^2+B^2+2ABCos180^o}[/latex]

= [latex]\sqrt{A^2+B^2-2AB}[/latex] [Since, Cos180^o = -1]

Or, R = [latex]\sqrt{(A-B)^2}[/latex]

Or, R = A – B

i.e., Resultant is the difference of magnitude of two vectors.

And, β = tan^-1 [latex](\frac{BSin180^o}{A+BCos180^o})[/latex]

Or, β = tan^-1(0) = 0

i.e., resultant is directed towards the bigger one.

Polygonal law of vector addition:

It states that, if many vectors acting at a point simultaneously be represented in magnitude and direction by sides of a polygon taken in same order then, the last side of it taken in the reverse order gives magnitude and direction of the resultant vector of these vectors.

Suppose, a number of vectors P, Q, R, S, T and U are acting at a point simultaneously as shown in figure (a). These vectors are represented in magnitude and direction by sides [latex]\vec{AB}[/latex], [latex]\vec{BC}[/latex], [latex]\vec{CD}[/latex], [latex]\vec{DE}[/latex], [latex]\vec{EF}[/latex] and [latex]\vec{FA}[/latex]. A polygon ABCDEF in the same order, the last side of a polygon [latex]\vec{FA}[/latex]  i.e. [latex]\vec{AF}[/latex] in the reverse order gives magnitude and direction of the resultant vector.

Join AC, AD and AE.

In [latex]\Delta[/latex]ABC,

             [latex]\vec{AC}[/latex] = [latex]\vec{AB}[/latex] + [latex]\vec{BC}[/latex]

= [latex]\vec{P}+\vec{Q}[/latex]

In [latex]\vec{\Delta}[/latex]ACD,

             [latex]\vec{AD}[/latex] = [latex]\vec{AC}[/latex] + [latex]\vec{CD} = \vec{P}+\vec{Q}+\vec{S}[/latex]

In [latex]\Delta[/latex]ADE,

             [latex]\vec{AE}=\vec{AD}+\vec{DE}[/latex]

= [latex]\vec{P}+\vec{Q}+\vec{S}+\vec{T}[/latex]

In [latex]\Delta[/latex]ADE,

[latex]\vec{AF}=\vec{AE}+\vec{EF}[/latex]

  [latex]\vec{R}=\vec{P}+\vec{Q}+\vec{S}+\vec{T}+\vec{U}[/latex]

Resolution of a vector: rectangular components:

The process of splitting of a vector into its parts is known as resolution and split parts are called component. If they are perpendicular to each other then they are called rectangular components.

Let, q be the angle made by [latex]\vec{OP}[/latex] = [latex]\vec{F}[/latex] with x – axis. Draw PN perpendicular to x – axis and PM ⊥ y axis.

Then, ON = F_x and OM = F_y are rectangular components of [latex]\vec{F}[/latex].

In [latex]\Delta[/latex]ONP,

Cos[latex]\theta = \frac{ON}{OP}[/latex]

Or, ON = OPCos[latex]\theta[/latex]

Or, F_x = FCos[latex]\theta[/latex] ……………. (i)

Again,

Sin[latex]\theta[/latex] = [latex]\frac{PN}{OP}[/latex]

Or, PN = OPSin[latex]\theta[/latex]

Or, OM = PN = OPSin[latex]\theta[/latex]

Or, F_y = FSin[latex]\theta[/latex] ………………. (ii)

Squatting and adding eqⁿ. (i) and (ii), we get,

F_x² + F_y² = F²Sin²[latex]\theta[/latex] + F²Cos²[latex]\theta[/latex]

              = F² x 1

F_x² + F_y²= F

Or, F = [latex]\sqrt{F_x^2+F_y^2}[/latex]

Dividing eq^n. (ii) by (i), we get,

[latex]\frac{F_y}{F_x} = Tan\theta[/latex]

Or, Tan[latex]\theta = \frac{F_y}{F_x}[/latex]

Dot or Scalar Product:

Let, [latex]\theta[/latex] be the angle between two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex]. The dot or scalar product of these two vectors is defined as:

[latex]\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|Cos\theta[/latex]

              = ABCos[latex]\theta[/latex]  ………………. (i)

Where, [latex]\vec{i}, \vec{j}[/latex] and [latex]\vec{k}[/latex] are unit vectors along x – axis, y – axis and z – axis respectively.

[latex]\hat{i}.\hat{i}=|\hat{i}||\hat{i}|Cos0^o[/latex]

              = 1 x 1 x 1 = 1

∴ [latex]\hat{i}.\hat{i} = \hat{j}.\hat{j}=\hat{k}.\hat{k} = 1[/latex]

∴ [latex]\hat{i}.\hat{j}=|\hat{i}||\hat{j}|Cos90^o[/latex]

              = 1 x 1 x 0 = 0

∴ [latex]\hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}[/latex] = 0

If [latex]\vec{A}[/latex] = A_x [latex]\hat{i}[/latex] + A_y [latex]\hat{j}[/latex] and

 [latex]\vec{B}[/latex] = B_x [latex]\hat{i}[/latex] + B_y [latex]\hat{j}[/latex], then,

             [latex]\vec{A}.\vec{B} = (A_x\hat{i}+A_y\hat{j}+A_z\hat{k}).(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})[/latex]

= A_xB_x + A_yB_y + A_zB_z ………………….. (ii)

Which is required expression for dot or scalar product.