Dated: 30-10-2024

Introduction of Vectors

There are some quantities which can be described completely by magnitude.
These are called scalar quantities.
Some quantities cannot be completely described using only magnitude and we also need direction.
There are called vector quantities.

Representation

These are represented as arrows directed in the direction of the action, while the magnitude is represented by the length of the arrow.

If \(A\) and \(B\) are the endpoints of this vector then it can be written as:

\[\vec{v} = \vec{AB}\]

and the magnitude can be determined by

\[\lvert \vec{v} \rvert = \lvert \vec{AB} \rvert\]

Unit Vector

It is a vector with magnitude \(1\).

\[\hat{v} = \frac{\vec{v}}{\lvert \vec{v} \rvert}\]

Addition of Vectors

Imagine we have 2 vectors \(\vec{OP}\) and \(\vec{OQ}\) then the vector which comes as the result of \(\vec{OP} + \vec{OQ}\) is constructed by joining the tail of either vector with the head of other one and then constructing a vector whose tail connects with tail of first vector and head with head of second vector.
This new vector is called resultant vector.

Equal Vectors

Two vectors are equal if they have same magnitude and same direction.

Opposite Vectors

Two vectors are called opposite vectors if they have same magnitude but opposite direction.

Parallel Vectors

Two vectors are called parallel vectors if they got same direction but magnitudes can be different or same.

Here \(c\) is a constant.

If \(\hat{i}, \hat{j} \text{ and } \hat{z}\) are the unit vectors then using vector addition, we can represent a point using a vector relative to the origin.

Subtraction of Vectors

Similar to vector addition, subtraction also works in similar way except that the direction of the vector being subtracted, is flipped.

Scalar or Dot Product

\[\vec{A} \cdot \vec{B} = \lvert \vec{A} \rvert \lvert \vec{B} \rvert \cos (\theta)\]

Where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

\[\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\]

If \(\vec{A} \perp \vec{B}\) then \(\vec{A} \cdot \vec{B} = 0\)
If \(\vec{A} \parallel \vec{B}\) then \(\vec{A} \cdot \vec{B} = 1\)

\[\vec{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{z}\]

\[\vec{B} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{z}\]

Then \(\(\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3\)\)

Angle between 2 Vectors

From the definition of dot product, the \(\theta\) can be found out as follows

\[\theta = \cos^{-1} \left( \frac{\vec{A} \cdot \vec{B}}{\lvert \vec{A} \rvert \lvert \vec{B} \rvert} \right)\]

Vector Projection

\[\cos(\theta) = \frac{\lvert \vec{OC} \rvert}{\lvert \vec{OB} \rvert}\]

\[\lvert \vec{OB} \rvert \cdot \cos(\theta) = \lvert \vec{OC} \rvert\]

\[\frac{\lvert \vec{OB} \rvert \cdot \rvert \vec{OA} \lvert \cdot \cos(\theta)}{\rvert \vec{OA} \lvert} = \lvert \vec{OC} \rvert\]

\[\frac{\vec{OB} \cdot \vec{OA}}{\lvert \vec{OA} \rvert} = \lvert \vec{OC} \rvert\]

\[\vec{OB} \cdot \hat{OA} = \lvert \vec{OC} \rvert\]

This number \(\lvert \vec{OB} \rvert \cos(\theta)\) is called the scalar component of \(\vec{OB}\) in the direction of \(\vec{OA}\).

Cross Product

\[\vec{A} \times \vec{B} = \lvert \vec{A} \rvert \lvert \vec{B} \rvert \sin(\theta) \hat{n}\]

According to right hand rule, curl your fingers from 1st vector towards the 2nd vector.
The thumb will point in the direction of \(\hat{n}\) which is perpendicular to the plane¹ in which both \(\vec{A}\) and \(\vec{B}\) are present.

\[\vec{B} \times \vec{A} = - \lvert \vec{A} \rvert \lvert \vec{B} \rvert \sin(\theta) \hat{n}\]

\[\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}\]

Area of a Parallelogram

\[\text{Area} = \text{base} \times \text{height}\]

\[\because \text{height} = \lvert \vec{B} \rvert \cdot \sin(\theta)\]

\[\because \text{base} = \lvert \vec{A} \rvert\]

\[\therefore \text{Area} = \lvert \vec{A} \rvert \cdot \lvert \vec{B} \rvert \sin(\theta)\]

\[= \lvert \vec{A} \times \vec{B} \rvert\]

\(\vec{A} \times \vec{B}\) From Components

If

\[\vec{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\]

\[\vec{B} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\]

then after some multiplication, we get

\[\vec{A} \times \vec{B} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1)\]

which can be written in more compact form as

\[= \hat{i} \left| \begin{matrix} a_2 & a_3 \\ b_2 & b_3 \end{matrix} \right| - \hat{j} \left| \begin{matrix} a_1 & a_3 \\ b_1 & b_3 \end{matrix} \right| + \hat{k} \left| \begin{matrix} a_1 & a_2 \\ b_1 & b_2 \end{matrix} \right| \]

which can further be written as

\[= \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix} \right| \]