Class 11 Vectors Note

Unit 1- Mechanics

Chapter 2- Vectors

Scalar quantity

The physical quantities which have magnitude but not direction are called scalar quantities or scalar. Scalar quantities can be added or subtracted according to rule of algebra. Example: Mass, time, distance, speed, density, energy, temperature, pressure, charge, gravitational potential, electric potential energy etc.

Vector quantity

The physical quantities which have both magnitude and direction are called vector quantity or vectors. Vector quantities can be added or subtracted according to the rule of vector additions. Example: displacement, velocity, acceleration, force, area, weight, electric field weight, magnetic field, gravitational field , momentum, torque etc

   A vector is graphically represented by a straight line with arrow at one end. The direction of arrow represents the direction of the vector and Length of the line represents the magnitude of the vector.

A vector is represented in figure below:

Q. Is it necessary that a quantity having both magnitude and direction be always a vector?

Ans:  No, for a quantity to be vector it should not be added or subtracted according to the rule of algebra. But even current has both magnitude and direction, it can be added according to rule of algebra. So it is scalar quantity.

Types of vectors:

Unit Vector:

A vector having unit magnitude is called unit vector. It is denoted by $\widehat{A}$given by:
$\widehat{A}=\frac{\overrightarrow{A}}{\overrightarrow{\left|A\right|}}=\frac{\overrightarrow{A}}{A} where A=\overrightarrow{\left|A\right|}=magnitude of vector\overrightarrow{A}.$
It is generally used to indicate the direction of vector.
Note:
If Vector $\overrightarrow{A}=\overrightarrow{i}x+\overrightarrow{j}y+\overrightarrow{k}z$ then magnitude of vector $\overrightarrow{A} is denoted by \overrightarrow{\left|A\right|} or A and given by:$
$\overrightarrow{\left|A\right|} =A=\sqrt{x^2+y^2+z^2}$
Q. Find the unit vector of $\overrightarrow{A}=3\widehat{i}+4\widehat{j}+5\widehat{k}$
solution,
given
$\overrightarrow{A}=3\widehat{i}+4\widehat{j}+5\widehat{k}$
then
$\overrightarrow{\left|A\right|}=A=\sqrt{3^2+4^2+5^2}=\sqrt{50}=5\sqrt{2}$
so unit vector of |$\overrightarrow{A}$|=$\widehat{A}$=$\frac{\overrightarrow{A}}{\overrightarrow{\left|A\right|}}$
=$\frac{3\widehat{i}+4\widehat{j}+5\widehat{k}}{5\sqrt{2}}$
$\mathit{\therefore }\widehat{A}=\frac{3}{5\sqrt{2}}\widehat{i}+\frac{4}{5\sqrt{2}}\widehat{j}+\frac{1}{\sqrt{2}}\widehat{k}$

Null Vector

A vector which has direction but no magnitude (i.e magnitude is zero) is called null vector. The initial point and terminal point of a null vector consider ( i.e same)

Parallel Vector

Two or more vector having same direction on are called parallel vectors.

In parallel vector, the angle between two vectors is zero.

Anti-parallel Vectors:

Two or more vectors having opposite direction are called anti-parallel vectors.

In parallel vector, the angle between 2 vectors is 180°

Equal Vectors:

The vectors having same magnitude and direction are called equal vectors.

|$\overrightarrow{\left|A\right|}|=|\overrightarrow{\left|B\right|}$|

Negative Vectors:

The vectors having same magnitude but acting in opposite are called negative vector.

Coplanar vector:

Two or more vector laying same plane are called coplanar vector.

Co-initial vector:

Those vector whose initial points are same such vector are called co-initial vector.

Co-terminal vector:

 Those vector whose end point are same, such vector are called co-terminal vector.

Addition of vector (important)

Triangle Law of Vector Addition:

It takes that, “if two vector in magnitude and direction are represented by two sides of triangle taken in order than third side of the triangle represents vector in opposite order.”

Let two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are represented by two sides OX and XY of triangle OXY taken in order and their resultant vector $\overrightarrow{R}$ is represented by third side OY taken in opposite order.
Magnitude
To find magnitude, Let us produce OX and draw Perpendicular on it at M from point X.
Now, In right angle, $△$ MOY
$O{Y}^2= O{M}^2+Y{M}^2$
$O{Y}^2= (OM+XM{)}^2+Y{M}^2$
$R^2 = (A +XM{)}^2 + Y{M}^2 \dots \dots \dots \dots \dots (i)$
Again, In Right angle $△$XYM
$Sin\theta =\frac{YM}{XM}$
$or, KM=XYSin\theta =BSin\theta ⇢\left(ii\right)$
and
$Cos\theta =\frac{ XM }{XY}$
$or, XM=XYCos\theta$
$\therefore XM=BCos\theta ⇢\left(iii\right)$
Using equation (i), (ii), and (iii)
we get:
$R^2=(A+BCos\theta {)}^2+(Sin\theta {)}^2$
$R^2= A^2+2ABCos\theta +B^{2}Co{s}^2\theta +B^{2}Si{n}^2\theta$
$R^2= A^2+2ABCos\theta +B^2(Co{s}^2\theta +Si{n}^2\theta )$
$R^2= A^2+2ABCos\theta +B^2$
$R^2=\sqrt{ A^2+B^2+2ABCos\theta }⇢\left(iv\right)$
This equation( iv) gives the magnitude of the resultants of vector $\overrightarrow{A } and \overrightarrow{B}$
Direction
Let the resultants $\overrightarrow{R }$ makes angle α with vector $\overrightarrow{A }$
Now,
In Right angle $△$OMX,