$\vec{A}=A_x \hat{\imath}+A_y \hat{\jmath}+A_3 \hat{k}$

$\vec{B}=B_x \hat{\imath}+B_y \hat{\jmath}+B_3 \hat{k}$

$\vec{A}+\vec{B}=\left(A_x+B_x\right) \hat{\imath}+\left(A_y+B_y\right) \hat{\jmath}$

$\text { eg } 2 \vec{A}-3 \vec{B} \quad+\left(A_3+B_3\right) \hat{k}$

$=2\left(A_x \hat{\imath}+A_y \hat{\jmath}+A_z \hat{k}\right)$

$\quad-3\left(B_x \hat{\imath}+B_y \hat{\jmath}+B_z \hat{k}\right)$

$=\left(2 A_x-3 B_x\right) \hat{\imath}+\left(2 A_y-3 B_y\right) \hat{\jmath}$

$+\left(2 A_z-3 B_3\right) \hat{k}$

Sum of 2 vectors is a vector. But we cannot say the same for a product.

2 ways of defining product of vectors.

Define 2 different types of products of vectors.

Both product satisfy distributive law of multiplication.

i.e. product of $\vec{A}$ and $(\vec{B}+\vec{C})$

= (product of $\vec{A}$ and $\vec{B})$ + (product of $\vec{A}$ and $\vec{C}$)

Scalar product of $\vec{A}$ $\text{and}$ $\vec{B}$ given by multiplying 3 quantities

a) $|\vec{A}|$

b) $|\vec{B}|$

c) cosine $\theta$

$\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos \theta$

a) scalar product is commutative

$\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}$

$|\vec{A}||\vec{B}| \cos \theta$

$|\vec{B}||\vec{A}| \cos \theta$

b) $\vec{A} \cdot \vec{B}$ is a scalar, it can be +ve or -ve.

sign will defind on $\angle \theta$.

if $\theta$ between 0 and $\frac{\pi}{2}$

$\vec{A} \cdot \vec{B}$ +ve

$\theta$ lies $\frac{\pi}{2}$ and $\pi$

$\vec{A} \cdot \vec{B} \rightarrow$ -ve

b) If $\vec{A} \cdot \vec{B}=0$, then

$|\vec{A}|=0$ or $|\vec{B}|=0$ or $\cos \theta=0$

$\vec{A} \perp \vec{B}$

$\vec{A}$ is orthogonal to $\vec{B}$

Scaler product of 2 unit vectors will be cosine of angle between them.

$\vec{A} \cdot(\vec{B}+\vec{C})=(\vec{A} \cdot \vec{B})+(\vec{A} \cdot \vec{C})$

unit vectors $\hat{\imath}$, $\hat{\jmath}$, $\hat{k}$

along Cartision axes

$i \cdot \hat{\imath}=1$

$\hat{\jmath} \cdot \hat{\jmath}=1,$

$\hat{k} \cdot \hat{k}=1$

$\hat{\imath} \cdot \hat{\jmath}=0$

$\hat{\imath} \cdot \hat{k}=0,$

$\hat{\jmath} \cdot \hat{k}=0$

$\vec{A} =A_x \hat{\imath}+A_y \hat{\jmath}+A_3 \hat{k}$

$\vec{B} =B_x \hat{\imath}+B_y \hat{\jmath}+B_3 \hat{k}$

$\vec{A} \cdot \vec{B} =\left(A_x \imath+A_y \hat{\jmath}+A_3 \hat{k}\right) \cdot\left(B_x \hat{\imath}+B_y \hat{\jmath}+B_3 \hat{k}\right)$

$\vec{A} \cdot \vec{B}=A_x B_x+A_y B_y+A_z B_z$

$\vec{C}=\vec{A}-\vec{B}$

$|\vec{C}|=|\vec{A}-\vec{B}|$

$\vec{C} \cdot \vec{C} =(\vec{A}-\vec{B}) \cdot(\vec{A}-\vec{B})$

$\vec{C} \cdot \vec{C}=\vec{A} \cdot \vec{A}+ \vec{B} \cdot \vec{B}-2 \vec{A} \cdot \vec{B}$

$|\vec{C}|^2=|\vec{A}|^2+|\vec{B}|^2-2|\vec{A}||\vec{B}| \cos (\theta)$

$C^2=A^2+B^2-2 A B \cos \theta$

$\cos (\vec{A}, \vec{B})=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}$

Work Done by a force Power due to force dot product of 2 vectors.

Work Done =$\vec{F} \cdot \vec{r}$

Power =$\vec{F} \cdot \vec{v}$

$\vec{A}=A_x \hat{\imath}+A_y \hat{\jmath}+A_z \hat{k}$

unit vector $\hat{e_A}$ along $\vec{A}$

$\text { unit vector: } \hat{e}_A=\frac{\vec{A}}{|\vec{A}|}$

$|\vec{A}|=\sqrt{A_x^2+A_y^2+A_z^2}$

$\hat{e}_A$ = $A_x \hat{\imath}$ + $A_y \hat{j}$ + $A_z \hat{k}$

Component of A along a director $\hat{e}_B \leftarrow$ unit vector

$\vec{A} \cdot \hat{e}_B \rightarrow$ component of $\vec{A}$ along $\vec{B}$

component of $\vec{A} \perp \hat{e}_B$

$\vec{A} \cdot \hat{e_B} \rightarrow$ component of $\vec{A}$ along $\hat{e}_B$

$\left(\vec{A} \cdot \hat{e_B}\right) \hat{e_B} \rightarrow$ vector component of $\vec{A}$ along $\hat{e_B}$

$\vec{A}-(\vec{A} \cdot \hat{e_B}) \hat{e_B} \rightarrow$ Component of $\vec{A} \perp \hat{e_B}$

Composition of vector along direction

$\rightarrow$ by taking dot product of vector with unit vector along that direction

Vector product of 2 vectors cross product

Vector product of 2 vectors is a vector.

$\vec{A} \times \vec{B}$

cross defined as a vector normal to the plane containing $\vec{A}$ and $\vec{B}$

Magnitude of $\vec{A} \times \vec{B}$

$\vec{C}=|\vec{A}||\vec{B}| \sin \theta$

$\vec{C}=\hat{e}_c|\vec{A}||\vec{B}| \sin \theta$

Direction of $\vec{C}$ determined by right hand thread rule.

Direction of thumb points towards $\vec{A} \times \vec{B}$

$\vec{A} \times \vec{B}$ point in opposite direction to $\vec{B} \times \vec{A}$

$\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$

$\vec{A} \times(\vec{B}+\vec{C})=(\vec{A} \times \vec{B})+(\vec{A} \times \vec{C})$

$\vec{A} \times \vec{A}=0$

2 parallel vectors, $\vec{A}$ and $\vec{B}$

contesion axes (3D) $\rightarrow$ Right Handed

$\hat{i}, \hat{j}, \hat{k}$

$\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=k \times \hat{k}=0$

$\hat{\imath} \times \hat{\jmath}=\hat{k}$

$\hat{j} \times \hat{k}=\hat{i}$

$\hat{k} \times \hat{i}=\hat{j}$

$\hat{\imath} \times \hat{\jmath}=\hat{k}$

$\hat{j} \times \hat{k}=\hat{i}$

$\hat{k} \times \hat{i}=\hat{j}$

$\hat{j} \times \hat{i} =-\hat{k}$

$\hat{i} \times \hat{k} =-\hat{j}$

$\hat{k} \times \hat{j} =-\hat{i}$

cyclic order $\rightarrow$ +ve

anticyclic order $\rightarrow$ -ve

$\vec{A} =A_x \hat{i}+A_y \hat{j}+A_z \hat{k}$

$\vec{B} =B_x \hat{i}+B_y \hat{j}+B_z \hat{k}$

$\vec{A} \times \vec{B} =(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}) \times (B_x \hat{i}+B_y \hat{j}+B_z \hat{k})$

= $(A_y B_z-A_z B_y) \hat{i}+(A_z B_x-A_x B_z) \hat{j} + (A_x B_y-A_y B_x) \hat{k}$

= $\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array}\right|$

$\text { a) } |\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin (\angle A, B)$

area of parallelogram with sides $\vec{A}$ and $\vec{B}$

Parallelopiped , A,B,C

Volumne of parallelopiped = V

$V = (\vec{A} \times \vec{B}) \cdot \vec{C} = (\vec{B} \times \vec{C}) \cdot \vec{A}$

= $(\vec{C} \times \vec{A} ) \cdot \vec{B}$

Torque or moment of a force about a point 0

draw a vector from 0 to the line of action of $\vec{F} = \vec{r}$

Moment of force $\vec{F}$ about 0 = $\vec{r} \times \vec{F}$

point charge moving with a velocity $\vec{V}$ in 0 magnetic field $\vec{B}$

Force on charge is in the direction of $\vec{V} \times {B}$

To show 2 vectors are parallel to each other,

cross product = 0

Eg. Given $\vec{B}=x \hat{i} + 3 \hat{j}$

$\vec{C}=2 \hat{i}+y \hat{j}$

a) Find x and y such that $\vec{B}$ and $\vec{C}$ are $\perp \vec{D}=5 \hat{i}+6 \hat{j}$

b) Show that for there value of xady, $\vec{B} || \vec{C}$

(a) $\vec{B} \perp \vec{D} \rightarrow 5 \hat{i}+6 \hat{j}$

$\vec{B} \perp \vec{D} \Rightarrow \vec{B} \cdot \vec{D}=0$

$\Rightarrow 5 x+18=0 \Rightarrow x=-\frac{18}{5}$

10 + 6 y = 0

$y=\frac{-10}{6}=-\frac{5}{3}$

$\vec{B} = -\frac{18}{5} \hat{c}+3 \hat{j}$

$\vec{C}=2 \hat{c}-\frac{5}{3} \hat{j}$

b) To show $\vec{B} || \vec{C}$, takes cross product.

$\vec{A}=5 \hat{i}+10 \hat{j}$

$\Rightarrow$ Find component of $\vec{A}$ along $\vec{B}$ where $\vec{B}=3 \hat{i}+4 \hat{j}$

b) Find the component of $\vec{A} \perp \vec{B}$.

a) First find $\hat{e}_B$.

$\vec{A} \cdot \hat{e}_B \rightarrow$ component of $\vec{A}$ along $\vec{B}$

b) $\vec{A}-(\vec{A} \cdot \vec{B}) \vec{e}_B \leftarrow \cdot$ component of $\vec{A} \perp \vec{B}$