Distance = Scalar

Velocity = Vectors

Mathematical Description.

Magnitude & direction.

directed line segment is called vector.

$\overrightarrow{A B} =\vec{a}$

$|\overrightarrow{A B}| =\text { mag. of } \overrightarrow{A B}$

Equal vectors: $|\overrightarrow{A B}|=|\overrightarrow{C D}|$ and $\overrightarrow{A B}$ and $\overrightarrow{C D}$ have same direction.

Magnitude and direction NOT by its location

If $|\vec{a}|=1$, unit vector.

If $|\vec{a}|=0$, zero vector.

Vectors are added in a particular way known as triangle law.

$\therefore$ $\overrightarrow{A B} + \overrightarrow{B C} = \overrightarrow{A C}$

1. $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ ( Commutativity )

1. $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ ( Commutativity )

If $\vec{a} + ( \vec{b} + \vec{c} ) = ( \vec{a} + \vec{b} ) + \vec{c}$ ( Associative )

$k \vec{a}$, $k$ scalar.

$k \vec{a}$ is a vector in the same direction as $\vec{a}$ but $k$ times longer tan $\vec{a}$, if $k>0$ If $k<0, k \vec{a}$ is in the opp. dir. but $k$ longer than $\vec{a}$.

Note :-

$\text { i) } k(\vec{a}+\vec{b}) =k \vec{a}+k \vec{b} \quad \text { distributivity. }$

$\text { ii) }(k+l) \vec{a} =k \vec{a}+l \vec{a}$

$\text { iii) } k(l \vec{a}) =(k l) \vec{a}=l(k \vec{a})$

Recall, If $|\vec{a}|=1, \vec{a}$ is a unit vector

If $\overrightarrow{A B}$ is any vector. $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}$ is a unit vector in the direction of $\overrightarrow{A B}$

1) $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=$ ?

$\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}$

$\Rightarrow \overrightarrow{A C}+\overrightarrow{C A}=0$

ii) Consider a regular hexagon $A B C D E F$.

Let $\overrightarrow{A B}=\vec{a}$

$\overrightarrow{B C}=\vec{b}$

Let $\overrightarrow{A B}=\vec{a}$

$\overrightarrow{B C}=\vec{b}$

Find the other sides in terms

$\vec{a}$ and $\vec{b}$

$\overrightarrow{A C} =\vec{a}+\vec{b}$

$A D | B C$&$A D =2 B C \quad \text { (Property of hexagon) }$

$\overrightarrow{C D} =\overrightarrow{C A}+\overrightarrow{A D}$

$=-(\vec{a}+\vec{b})+2 \vec{b}$

$=\vec{b}-\vec{a}$

$\overrightarrow{D E} =-\vec{a}$

$\overrightarrow{E F} =-\vec{b}$

$\overrightarrow{F A} =\vec{a}-\vec{b}$

$\overrightarrow{A C} =\vec{a}+\vec{b}$

$\therefore \overrightarrow{A D} =\overrightarrow{A C}+\overrightarrow{C D}$

$=\vec{a}+\vec{b}+\vec{c}$

A unit vector along ox is denoted by $\vec{i}$ "i"

A unit vector along oy is denoted by $\vec{j}$ "j"

Almost, $k \lambda$ Along oy, lj Almost, $k \lambda$

Along oy, lj

$\vec{r}=\overrightarrow{O P}=a i+b j$

$|\vec{r}|=|\overrightarrow{O P}|=\sqrt{a^2+b^2}$

$\overrightarrow{O P}$ is called the position vector of the point $P$

P(a, b) . $\quad \overrightarrow{O P}=a i+b j$

$\overrightarrow{O P} =\overrightarrow{O P^{\prime}}+\overrightarrow{P^{\prime} P} =a i+b_j+c k$

let $A(-1,1,4), B(8,0,0), C(5,-2,5)$

$\overrightarrow{A B} =\overrightarrow{O B}-\overrightarrow{O A}$

$=8 i-(-i+j+4 k) =9 i-j-4 k$

$|\overrightarrow{A B}| =\sqrt{81+1+16}=\sqrt{98}$

$\overrightarrow{B C} =\overrightarrow{O C}-\overrightarrow{O B}$

$=5 i-2 j+5 k-8 i =-3 i-2 j+5 k$

Let $A, B$ $\&$ $C$ be three points in space whose position vectors are $2 i-j+k, i-3 j-5 k$ $3 i-4 j-4 k$. Prove: $A, B, C$ are the vertices of a right angled triangle

$\overrightarrow{A B} =\overrightarrow{O B}-\overrightarrow{O A}$

$=(i-3 j-5 k)-(2 i-j+k)$

$=-i-2 j-6 k$

$\overrightarrow{B C} =2 i-j+k$

$\overrightarrow{C A} =-i+3 j+5 k$

$|\overrightarrow{A B}|^2=1+4+36=41$

$|\overrightarrow{B C}|^2=4+1+1=6$

$|\overrightarrow{C A}|^2=1+9+25=35$

$|\overrightarrow{A B}|^2=|\overrightarrow{B C}|^2+|\overrightarrow{C A}|^2$

$\therefore \triangle A B C$ is right angled