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# Physics - 11 Complete Syllabus Notes

### Chapter 2 - Vector

Scalar Quantities::
The physical quantities having only one magnitude are called scalar quantities.
eg. distance,speed etc.

Vector Quantities:

The physical quantities having magnitude and direction and obeying the law of vector addition are called vector quntities.
eg: displacement, velocity, etc.

A vector is represented graphically by straight line with arrow head. The length of the straight line represent magnitude and arrow head represent direction.

Terms regarding in vector:

i) Equal Vector:
Those vector are said to be equal vector if they have same magnitude and direction.

ii) parallel vector:
Vector acting in same direction with equal or unequal magnitude are called parallel vector.

iii)Anti-Parallel Vector:
Vector acting in opposite direction are called Anti- Parallel vector.

iv)Colinear Vector:
Vector acting in same line are called collinear vector.

v) Coplaner Vector:
Vector acting in same plane is called coplaner vector.

vi) Unit vector:
A vector having unit magnitude is called unit vector.
In certesian coordinate system x-axis, y-axis and z-axis are represented by i^, j^ and k^ respectively where i ^, j^ and k^ are unit vectors.

Statement: If two vectors acting at a point are represents at both in magnitude and direction by two adjacent side of parallelogram. Then the diagonal from that point represent their resultant both in magnitude and direction.

Explanation:
Let two vector $\overrightarrow{P}$ and $\overrightarrow{Q}$ are represented at both in magnitude and direction by two adjecent size $\overrightarrow{OA}$ and $\overrightarrow{OB}$ of a parallelogram OACB as shown in figure ii) then from parallelogram law of vector addition the diagonal $\overrightarrow{OC}$ represent their resultant $\overrightarrow{R}$ .

Let $\theta$ be the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ and $\beta$ be the angle between $\overrightarrow{R}$ and $\overrightarrow{P}$.

Magnitude of Resultant:
From C draw perpendicular, CD and OA produced.
In $\Delta$ ODC,
OC 2 = OD 2 + CD 2
OC 2 = (OA + AD) 2 + CD 2
R 2 = (P + Q COS $\theta$) 2 + (Q Sin $\theta$) 2
On solving with R,
we get,
$R=\sqrt{p^{2}&space;+&space;2PQcos\Theta&space;+Q^{2}}&space;......i)$

Direction Of Resultant:
In $\Delta$ ODC,
$tan\beta$ = $\frac{CD}{OD}$
$tan\beta$ = $\frac{CD}{OA + AD}$
$tan\beta&space;=&space;\frac{Qsin\Theta&space;}{p&space;+&space;Q&space;cos\Theta}$
$\therefore&space;\beta&space;=&space;Tan^{-1}[\frac{Qsin\Theta&space;}{P&space;+&space;Q&space;cos\Theta&space;}]....ii)$

eqn i) represent the magnitude of resultant and eqn ii) represent the direction of resultant with $\overrightarrow{P}$.

Special Cases:

i) WHEN $\theta$ = 00

Magnitude of Resultant:
$R=\sqrt{p^{2}+2pqcos0^{0}+q^{2}}$
$R=\sqrt{p^{2}+2pq+q^{2}}$
$R=\sqrt{(p+q)^{2}}$
$R=(p+q)$
$R=(p+q)=Maximum&space;Value$

Direction of Resultant:
$\beta&space;=&space;Tan^{-1}\left&space;(&space;\frac{Qsin\theta}{P&space;+&space;Qcos0^{0}}&space;\right&space;)$
$\therefore&space;\beta&space;=0$
i.e. the direction of resultant is in the direction of given vector (either P or Q vector)

ii) WHEN $\theta$ = 900

Magnitude of Resultant:
$R=\sqrt{p^{2}+2pqcos90^{0}+q^{2}}$
$R=\sqrt{(p^{2}+q^{2})}$

Direction of Resultant:
$\beta&space;=&space;Tan^{-1}\left&space;(&space;\frac{Qsin90^{0}}{P&space;+&space;Qcos90^{0}}&space;\right&space;)$
$\therefore&space;\beta&space;=Tan^{-1}&space;\frac{Q}{P}&space;&space;&space;with&space;\overrightarrow{P}$

iii) WHEN $\theta$ = 1800

Magnitude of Resultant:
$R=\sqrt{p^{2}+2pqcos180^{0}+q^{2}}$
$R=\sqrt{(p-q)^{2}}$
$R=P-Q$ or, $R=P-Q$

Direction of Resultant:
The direction of the resultant is in the direction of greater vector.

Statement: If two vector are represented both in magnitude and direction by two sides of a triangle taken in same order than the closing side ( third side ) of triangle taken in opposite order represents theirresultant both in magnitude and direction..

Explanation:
Let two vector $\overrightarrow{P}$ and $\overrightarrow{Q}$ are represented at both in magnitude and direction by two side $\overrightarrow{OA}$ and $\overrightarrow{AC}$ of a triangle OAC as shown in figure ii) then from trianglem law of vector addition the closing side $\overrightarrow{OC}$ represent their resultant $\overrightarrow{R}$ .

Let $\theta$ be the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ and $\beta$ be the angle between $\overrightarrow{R}$ and $\overrightarrow{P}$.

Magnitude of Resultant:
From C draw perpendicular, CD and OA produced.
In $\Delta$ ODC,
OC 2 = OD 2 + CD 2
OC 2 = (OA + AD) 2 + CD 2
R 2 = (P + Q COS $\theta$) 2 + (Q Sin $\theta$) 2
On solving with R,
we get,
$R=\sqrt{p^{2}&space;+&space;2PQcos\Theta&space;+Q^{2}}&space;......i)$

Direction Of Resultant:
In $\Delta$ ODC,
$tan\beta$ = $\frac{CD}{OD}$
$tan\beta$ = $\frac{CD}{OA + AD}$
$tan\beta&space;=&space;\frac{Qsin\Theta&space;}{p&space;+&space;Q&space;cos\Theta}$
$\therefore&space;\beta&space;=&space;Tan^{-1}[\frac{Qsin\Theta&space;}{P&space;+&space;Q&space;cos\Theta&space;}]....ii)$
eqn i) represent the magnitude of resultant and eqn ii) represent the direction of resultant with $\overrightarrow{P}$.

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