# Scalars and vectors

Scalars and vectors:

Scalars are the physical quantities that can be represented by their magnitude. Examples of such physical quantities include mass, time, length, energy, temperature etc. All of these require a magnitude to be represented. For instance mass is represented by just expressing its magnitude in respective units, like 5 kg, time is expressed in seconds, energy in joules, temperature in Celsius. There is no sense of direction in these physical quantities.

Vectors are the physical quantities that require magnitude and also direction to be represented. Examples include force, velocity, acceleration, etc. All of these require a magnitude and a direction in which they act to be completely defined. For instance force needs to be defined in terms of magnitude in N and also the direction in which it acts; velocity is expressed in ms-1 and needs a direction in which the body is travelling.,

Unit Vector:

A vector having a magnitude equal to 1 and it is denoted by putting a hat (circumflex) on top of the vector symbol.

Unit vector = ${\rm{\hat u}}$

A unit vector in the x-direction can be denoted by    or  ${\rm{\hat i}}$ similarly, unit vector in y-direction can be denoted by   ${\rm{j}}$  or this

Null vector:

A vector having a zero magnitude and arbitrary direction is called as null vector. For example, the result of the sum of a vector and its negative vector is always a null vector.

A+ (-A) =0

Equal vectors:

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

Negative of a vector:

Vector equal in magnitude with the other vector but in the opposite direction.  Such vector is called as negative of a vector.  Suppose we have two vectors A and B both have the same magnitude but they are opposite in direction so they will be written as:

A = -B

Parallelogram law of vectors:

Parallelogram law of vector states that “If the vectors acting simultaneously at a point both in direction and magnitude represented by the adjacent sides of the parallelogram drawn from the point, then the resultant of the vectors both in magnitude and direction are represented by the diagonal of the parallelogram passing through that point”.

Here,${\rm{\: }}\mathop\to\limits_{\rm{P}}$   and ${\rm{\: \: }}\mathop\to\limits_{\rm{Q}}$    are represented by the sides of parallelogram OBCA and the

$\mathop\to\limits_{\rm{R}}$ is represented by OC such that   ${\rm{\: \: }}\mathop\to\limits_{\rm{R}}$    =${\rm{\: \: \: }}\mathop\to\limits_{{\rm{\: P}}}$  + ${\rm{\: \: }}\mathop\to\limits_{\rm{Q}}$

Let the angle between ${\rm{\: }}\mathop\to\limits_{\rm{P}}$    and ${\rm{\: \: }}\mathop\to\limits_{\rm{Q}}$     is ${\rm{\: \: }}\theta$

Magnitude of$\:\:\mathop\to\limits_{\bf{R}}$

To calculate ${\rm{\: \: }}\mathop\to\limits_{\rm{R}}$    , draw the perpendicular from point C to D as shown in figure.

In $\Delta$OCD

${\rm{O}}{{\rm{C}}^2}{\rm{\: }} = {\rm{\: \: }}{\left( {{\rm{OA}} + {\rm{AD\: }}} \right)^2}{\rm{\: }} + {\rm{C}}{{\rm{D}}^2}$                ………………………i

In Δ ACD,

CD = Q sin θ

From I

${\rm{O}}{{\rm{C}}^2}{\rm{\: }} = {\rm{\: \: }}{\left( {{\rm{P}} + {\rm{Qcos}}\theta {\rm{\: }}} \right)^2}{\rm{\: }} + {\left( {{\rm{Qsin}}\theta {\rm{\: }}} \right)^2}$

${\rm{R\: }} = {\rm{\: }}\sqrt {{\rm{\: }}{{\rm{P}}^2} + {\rm{\: }}2{\rm{PQcos}}\theta {\rm{\: }} + {\rm{\: \: }}{{\rm{Q}}^2}}$

Direction of ${\rm{\vec R}}$ is given by

Tan α $= \frac{{{\rm{CD}}}}{{{\rm{OD}}}}{\rm{\: }}$

= $\frac{{{\rm{Q\: sin}}\theta }}{{\left( {{\rm{P\: }} + {\rm{\: Q\:cos}}\theta {\rm{\: }}} \right){\rm{\: }}0}}{\rm{\: }}$

α = ${\tan ^{ - 1}}{\rm{\: Q}}\frac{{{\rm{sin}}\theta }}{{\left( {{\rm{P\: }} + {\rm{\: Q\:cos}}\theta {\rm{\: }}} \right)}}$

Special cases,

i. when P and Q act on the same direction

θ = 0 °

${\rm{R\: \: }} = {\rm{\: }}\sqrt {{\rm{\: }}{{\rm{P}}^2} + {\rm{\: }}2{\rm{PQcos}}0{\rm{\: }} + {\rm{\: \: }}{{\rm{Q}}^2}}$

R = P+Q

Tan α = 0

ii. When they act in opp direction

${\rm{R\: }} = {\rm{\: }}\sqrt {{\rm{\: }}{{\rm{P}}^2} + {\rm{\: }}2{\rm{PQcos}}180{\rm{\: }} + {\rm{\: \: }}{{\rm{Q}}^2}}$

R = A – B

Tan α   = 0

When P an Q are perpendicular to each other θ = 90

R = ${\rm{\: }}\sqrt {{{\rm{P}}^2} + {\rm{\: }}{{\rm{Q}}^2}}$

Tan α = $\frac{{\rm{Q}}}{{\rm{P}}}$

α   = $\frac{{{{\tan }^1}{\rm{Q}}}}{{\rm{P}}}$

The triangle law of vectors states:

If two vectors such as AB and BC are representing the two sides of a triangle ABC, then the third side AC closing the other side of the triangle in opposite direction represents the sum of two vectors both in magnitude and vectors.

$\overrightarrow {{\rm{AC}}}$=  $\overrightarrow {{\rm{AB}}}$  + $\overrightarrow {{\rm{BC}}}$

 Rectangular components of vector Consider a vector ${\rm{\vec V\: }}$acting at a point making an angle θ with positive X-axis as shown in figure Vector${\rm{\vec V}}$ is represented by a line OA. From point A draw a perpendicular AB on X-axis. Suppose OB and BA represents two vectors. Vector O’A is parallel to X-axis and vector BA is parallel to Y-axis. Magnitude of these vectors are Vx and Vy respectively. The sum of these vectors is equal to vector${\rm{\vec V}}$.Thus Vx and Vy are the rectangular components of vector v⃗ figure: : from the figure  Cosθ= $\frac{{{\rm{O\vec B}}}}{{{\rm{O}}\mathop\to\limits_{\rm{A}}{\rm{\: }}}}$ =$\frac{{{{\rm{V}}_{\rm{x}}}}}{{\rm{V}}}$ ${{\rm{V}}_{\rm{x}}}{\rm{\: }} = {\rm{Vcos\: }}\theta$ Sinθ=$\frac{{{\rm{A\vec B}}}}{{{\rm{O\vec A}}}}$= $\frac{{{{\rm{V}}_{\rm{y}}}}}{{{\rm{V\: }}}}$ ${{\rm{v}}_{\rm{y}}} = {\rm{V\: sin}}\theta$

Tanθ=$\frac{{{{\rm{v}}_{\rm{y}}}}}{{{{\rm{V}}_{\rm{x}}}}}$

The geometrical interpretation of the scalar product of two vectors:

If the product of two vectors is a scalar quantity, then the operation is called scalar product.

Let us consider two vectors$\overrightarrow {{\rm{\: P}}}$and${\rm{\vec Q}}$inclined at an angle θ to each other as shown in figure

Their scalar product is defined as${\rm{\: \vec P}}$${\rm{\: \vec Q}} . = PQ cos θ Work done by the force in moving through the displacement of D⃗ is scalar product W ={\rm{\: }}\overrightarrow {{\rm{F\: }}}$${\rm{\: \vec D}}$   = FD cos θ

The scalar product of two vectors is a dot product.

Two vectors can be added by using the triangle law of vectors

If two vectors such as $\overrightarrow {{\rm{AB}}}$  and  $\overrightarrow {{\rm{BC}}}$ are representing the two sides of a triangle ABC, then the third side AC closing the other side of the triangle in opposite direction represents the sum of two vectors both in magnitude and vectors.

$\overrightarrow {{\rm{AC}}}$=  $\overrightarrow {{\rm{AB}}}$  + $\overrightarrow {{\rm{BC}}}$

By using parallelogram law of vectors

Statement: "If two vectors are acting simultaneously at a point and these are represented in magnitude and direction by any two adjacent side of a triangle taken in order then the remaining side of the triangle gives the magnitude and direction at their resultant taken in opposite sides.

Explanation

Let ${\rm{\vec P\: and\: \vec Q}}$ be the two vectors acting simultaneously at a point O and these are represented in magnitude and direction by two adjacent sides AB and BC of a triangle ABC then the side AC of a triangle ABC represent their resultant in magnitude and direction as shown in figure

i.e.$\frac{{\overrightarrow {{\rm{AC}}} }}{{{\rm{\vec R}}}} = \frac{{\overrightarrow {{\rm{AB}}} }}{{{\rm{\vec P}}}} + \frac{{\overrightarrow {{\rm{BC}}} }}{{{\rm{\vec Q}}}}$

$\mathop\to\limits_{\rm{R}} = {\rm{\: }}\mathop\to\limits_{\rm{A}} + {\rm{\: }}\mathop\to\limits_{\rm{B}} {\rm{\: }} + {\rm{\: }}\mathop\to\limits_{\rm{C}} + {\rm{\: }}\mathop\to\limits_{\rm{D}} {\rm{\: }}$