# Scalars and Vectors

**Scalar quantities**

Those physical quantities that have magnitude and but no any direction are known as scalar quantities. Example : Mass, Time etc. They can be added or subtracted by algebric method.

**Vector Quantities**

Those physical quantities that have both magnitude and direction are known as vector quantities. Example : Velocity, Weight etc. They can be added or subtracted by vector method.

**Representation of a vector**

Vector quantity can be represented by a straight line with an arrow head. The length of straight line defines the magnitude and arrow head defines the direction. Example: Velocity is a vector quantity. So, it can be written as \(\overrightarrow{v}\) .

**Types of Vector**

**Parallel Vectors**

Two vectors are called as parallel vectors if both of them acts along same direction.**Equal Vectors**

Two parallel vectors having same magnitude are known as equal vectors.

\(\overrightarrow{A} = \overrightarrow{B}\)**Opposite Vectors**

If two vectors have same magnitude but opposite direction, then those vectors are known as opposite vectors.**Co linear Vectors**

Two vectors are known as co linear vectors if both of them acts along same line.**Co planer Vectors**

Number of vectors lying on the same plane are known as co planer vectors.**Null Vectors**

If the magnitude of vector is zero, then it is known as null vector.**Unit Vector**

Vectors having magnitude one is known as unit vectors.

**Addition of Vectors**

When two or more vectors are added , we get a single value called resultant vector. The process of finding the resultant vector is also called the composition of vectors. They are of two types:

- Triangle Law of Vector
- Parallelogram Law of Vector

**Triangle Law of Vector**

It states that "If two vectors acting simultaneously on a body be represented in magnitude and direction by two sides of a triangle taken in same order, then the resultant vector can be represented by third side taken in opposite order."

**Proof**

Let us consider two vectors \(\overrightarrow{P}\) and \(\overrightarrow{Q}\) acting simultaneously on a body at the angle \(\theta\) between them.Then these vectors \(\overrightarrow{P}\) and \(\overrightarrow{Q}\) can be represented in magnitude and direction of side AB and BC of \(\triangle ABC\) taken in same order as shown in the figure. Then the resultant vector \(\overrightarrow{R}\) can be represented by third side AC taken in opposite site order. Then ,

\(\overrightarrow{R} = \overrightarrow{P} + \overrightarrow{Q}\)

We draw a perpendicular CD on the produced part of AB. Then, from \(\triangle CBD\) we get,

\(Cos \theta = \frac{BD}{BC} = \frac{BD}{Q}\)

\(or, BD = Q Cos \theta ............(i)\)

And,

\(Sin \theta = \frac{CD}{BC} = \frac{CD}{Q}\)

\(or, CD = Q Sin \theta ............(ii)\)

Applying pythagoras theorem to \(\triangle CAD\), we get

\(AC^{2} = AD^{2} + CD^{2}\)

\(or, AC^{2} = (AB+BD)^{2} + CD^{2}\)

\(or, R^{2} = (P + Q Cos \theta)^{2} + (Q Sin \theta)^{2}\)

\(or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} Sin^{2} \theta + Q^{2} Cos ^{2} \theta\)

\(or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} (Sin^{2} \theta + Cos^{2} \theta )\)

\(or,R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2}\)

\( R^{2} = \sqrt{P^{2} + 2 PQ Cos \theta + Q^{2}}\)

**Direction of \(\overrightarrow{R}\)**

Let, be the angle made by resultant vector \(\overrightarrow{R}\) with the direction of vector \(\overrightarrow{P}\). Then from \(\triangle CAD\), we get

\(Tan \alpha = \frac{CD}{AD} = \frac{CD}{AB+BD}\)

\(or, Tan \alpha = \frac{Q Sin \theta}{P+Q Cos \theta}\)

\(\alpha = tan^{-1}(\frac{Q Sin \theta}{P+Q Cos \theta})\)

**Parallelogram Law of Vectors**

It states that "If two vectors acting simultaneously on a body be represented in magnitude and direction by two adjacent sides of a parallelogram, then the resultant vector can be represented by the diagonal of this parallelogram passing throught that point."

**Proof:**

Let us consider two vectors \(\overrightarrow{P}\) and \(\overrightarrow{Q}\) acting simultaneously on a body are represented in both magnitude and direction by two adjacent sides of a parallelogram OACB as shown in the figure. Then from the parallelogram law of vectors, the resultant vector \(\overrightarrow{R}\) can be represented in magnitude and direction by diagonal OC.

Let us draw a perpendicular CD on the produced part of OA. Let, \(\measuredangle{BOA}= \theta\) then, \(\measuredangle{CAD}= \theta\).Now from \(\triangle CAD\), we get

\(Cos \theta = \frac{AD}{AC} = \frac{AD}{Q}\)

\(AD = Q Cos \theta ............(i)\)

And,

\(Sin \theta = \frac{CD}{AC} = \frac{CD}{Q}\)

\(or, CD = Q Sin \theta ............(ii)\)

By applying pythagorous theorum on \(\triangle OCD\), we get

\(OC^{2} = OD^{2} + CD^{2}\)

\(or, OC^{2} = (OA+AD)^{2} + CD^{2}\)

\(or, R^{2} = (P + Q Cos \theta)^{2} + (Q Sin \theta)^{2}\)

\(or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} Sin^{2} \theta + Q^{2} Cos ^{2} \theta\)

\(or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} (Sin^{2} \theta + Cos^{2} \theta )\)

\(or,R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2}\)

\( R^{2} = \sqrt{P^{2} + 2 PQ Cos \theta + Q^{2}}\)

This equation gives the magnitude of resultant vector \(\overrightarrow{R}\).

**Direction of \(\overrightarrow{R}\)**

If the resultant vector \(\overrightarrow{R}\) makes an angle of alpha with the direction of vector \(\overrightarrow{P}\), Then from \(\triangle OCD\) we get,

\(Tan \alpha = \frac{CD}{OD} = \frac{CD}{OA+AD}\)

\(or, Tan \alpha = \frac{Q Sin \theta}{P+Q Cos \theta}\)

\(\alpha = tan^{-1}(\frac{Q Sin \theta}{P+Q Cos \theta})\)

This equatipm gives the direction of resultant vector \(\overrightarrow{R}\).

**Special cases**

**Case I**

If the vectors \(\overrightarrow{P}\) and \(\overrightarrow{Q}\) are parellel then \(\theta = 0 \degree\).

\(R = \sqrt{P^{2} + 2 PQ Cos 0 \degree + Q^{2}\)

\( = \sqrt{P^{2} + 2 PQ+ Q^{2}\)

\(= \sqrt{(P+Q)^{2}}\)

\(= P + Q\)

And,

\(\alpha = tan^{-1}(\frac{Q Sin 0 \degree}{P + Q Cos 0 \degree})\)

\(= tan^{-1}(\frac{0}{P + Q})\)

\(= tan^{-1}(0)\)

\(= 0 \degree\)

**Case II**

If the vectors \(\overrightarrow{P}\) and \(\overrightarrow{Q}\) are perpendicular then \(\theta = 90 \degree\).

\(R = \sqrt{P^{2} + 2 PQ Cos 90 \degree + Q^{2}\)

\( = \sqrt{P^{2} + Q^{2}\)

\(R^{2} = P^{2} + Q^{2}\)

And,

\(\alpha = tan^{-1}(\frac{Q Sin 90 \degree}{P + Q Cos 90 \degree})\)

\(= tan^{-1}(\frac{Q}{P + 0})\)

\(= tan^{-1} \frac{Q}{P}\)

**Case III**

If the vectors \(\overrightarrow{P}\) and \(\overrightarrow{Q}\) are parellel but opposite in direction then \(\theta = 180 \degree\).

\(R = \sqrt{P^{2} + 2 PQ Cos 180 \degree + Q^{2}\)

\( = \sqrt{P^{2} - 2 PQ+ Q^{2}\)

\(= \sqrt{(P-Q)^{2}}\)

\(= P - Q\)

And,

\(\alpha = tan^{-1}(\frac{Q Sin 180 \degree}{P + Q Cos 180 \degree})\)

\(= tan^{-1}(\frac{0}{P - Q})\)

\(= tan^{-1}(0)\)

\(= 0 \degree\)

**Polygon Law of Vectors**

It states that, "If the number of vectors acting simultaneously on a body be represented in magnitude and direction by the sides of a polygon taken in same order, then resultant vector can be represented by the closing side of this polygon taken in opposite order.

**Proof**

Let us consider the vectors \(\overrightarrow{A}\), \(\overrightarrow{B}\), \(\overrightarrow{C}\) and \(\overrightarrow{D}\) are acting simultaneously on a body as shown in the figure. Then, these vectors can be represented in magnitude and direction by the sides MN, NO, OP and PQ of polygon MNOPQ taken in same order as shown in figure.

Hence, from polygon law of vectors, the resultant vector \(\overrightarrow{R}\) can be represented by the closing side MQ taken in opposite order. That is,

\(\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} + \overrightarrow{D}\)

**Resolution of a Vector**

The phenomenon of breaking a single vector into different components is called resolution of vector. Generally, there are three components of vector along x-axis, y-axis and z-axis direction.

Let us consider a vector \(\overrightarrow{F}\) acting along OC as shown in the figure. Point O is the origin of co ordinate axis. Let us draw two perpendiculars AC and BC on x-axis and y-axis respectively. Let, \(\measuredangle{COA}=\theta\)

Now, From \(\triangle COA\) we get,

\(Cos \theta = \frac{OA}{OC} = \frac{OA}{F}\)

\(or, OA = F Cos \theta\)

\(or, F_{x} = F Cos \theta\)

Similarly,

\(Sin \theta = \frac{AC}{OC} =\frac{OB}{P}\)

\(or, OB = P Sin \theta\)

\(or, F_{y} = F Sin \theta\)

**Multiplication of Vector and Scalar**

When the vector quantity is multiplied by scalar quantity, then it gives the new vector.

For example, when mass is multiplied with velocity, then momentum is formed.

**Scalar Product or Dot Product**

If the product of two vectors gives the scalar quantity then that type of product is called scalar product. If the vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) acting at an angle \(\theta\), then their scalar product is,

\(\overrightarrow{A}\) . \(\overrightarrow{B}\) = \(AB Cos \theta\)

That means the scalar product is the product of the magnitude of first vector and component of second vector along the first vector.

For example,

Work = F.s Cos \theta

\(= \overrightarrow{F} . \overrightarrow{s} \)

**Properties of Scalar Product**

- The scalar product of two vectors depends on the angle between them,

If \(theta = 0 \degree\) , then

\(\overrightarrow{A} . \overrightarrow{B} = AB Cos \theta = AB Cos 0 \degree = AB\)

If \(theta = 90 \degree\) , then

\(\overrightarrow{A} . \overrightarrow{B} = AB Cos \theta = AB Cos 90 \degree = 0\) - It obeys the commulative law.

\(\overrightarrow{A}.\overrightarrow{B}=\overrightarrow{B}. \overrightarrow{A}\) - It obeys distributive law.

\(\overrightarrow{A}. (\overrightarrow{B}+\overrightarrow{C}) = \overrightarrow{A}.\overrightarrow{B} + \overrightarrow{A}.\overrightarrow{C}\) - The scalar product of a vector by itself becomes the square of it's magnitude.

\(\overrightarrow{A} . \overrightarrow{A} = A A Cos 0 \degree = A^{2}\)

**Vector Product or Cross Product**

If the product of two vectors gives vector quantity then that type of product is called the vector product.

If vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are acting at an angle \(\theta\), then their vector product is given by,

\(\overrightarrow{A} \times \overrightarrow{B} = AB Sin \theta \hat{n}\)

where, \(\hat{n}\) is the unit vector. The direction of vector product is given by the right hand thumb rule.

\(Torque = Fr Sin \theta\)

\(= \overrightarrow{F} \times \overrightarrow{r}\)

**Properties of Vector Product**

- The scalar product of two vectors depends on the angle between them,

If \(theta = 0 \degree\) , then

\(\overrightarrow{A} . \overrightarrow{B} = AB Sin \theta = AB Sin 0 \degree = 0\)

If \(theta = 90 \degree\) , then

\(\overrightarrow{A} . \overrightarrow{B} = AB Sin \theta = AB Sin 90 \degree = AB\) - It does not obey the commulative law.

\(\overrightarrow{A}.\overrightarrow{B}= -(\overrightarrow{B}. \overrightarrow{A})\) - It obeys distributive law.

\(\overrightarrow{A} \times (\overrightarrow{B}+\overrightarrow{C}) = \overrightarrow{A} \times \overrightarrow{B} + \overrightarrow{A} \times \overrightarrow{C}\) - The scalar product of a vector by itself becomes zero.

\(\overrightarrow{A} . \overrightarrow{A} = A A Sin 0 \degree = 0\)