$\vec{a} \cdot \vec{b}:=|\vec{a}||\vec{b}| \cdot \cos (\theta)$

i) $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$

ii) If one of $\vec{a}$ & $\vec{b}$ is zero, $\vec{a} \cdot \vec{b}=0$.

iii) if $\theta=\pi / 2, \vec{a} \cdot \vec{b}=0$

We say $\vec{a}$ & $\vec{b}$ are perpendicular

$\vec{a} \perp \vec{b}$.

iv) $\vec{a} \cdot \vec{a}=|a|^2$

$\therefore \sqrt{\vec{a} \cdot \vec{a}} \rightarrow$ length of the vector $\vec{a}$.

$If |\vec{a}|=6,|\vec{b}|=7, \quad \theta=60^{\circ}, then$

$\vec{a} \cdot \vec{b}=6 \times 7 \times \frac{1}{2}=21 \text {. }$

$i \cdot i=1$

$j \cdot j=1$

$k \cdot k=1$

$\vec{a} \cdot \vec{b}=|\vec{b}| \cdot|\vec{a}| \cos \theta$

$=|\vec{b}| \cdot$ [length of the projection of $\vec{a}$ on $\vec{b}$]

$\cos \theta=\frac{O M}{|\vec{a}|}$

$\Rightarrow O M=|\vec{a}| \cos \theta$

$\vec{a}.\vec{b}$= length $\vec{b} \times$ length of the projection of $\vec{a}$ on $\vec{b}$

$|\vec{a}| \cos \theta \times \frac{\vec{b}}{|\vec{b}|}$ is called

the projection vector of $\vec{a}$ along $\vec{b}$

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

$\vec{a} \cdot(\vec{b}+\vec{c}) =\text { length } of \vec{a} \times \text { length } of \text { the proj. }$

$\vec{b}(\vec{b}+\vec{c})$

$=|\vec{a}| \times O B_1$

$=|\vec{a}| \times\left(O A_1+A_1 B_1\right)$

$=|\vec{a}|OA_1 + |\vec{a}|A_1B_1$

$=|\vec{a}|\times$ length of the projection $\vec{a}$ on $\vec{a} + |\vec{a}|\times$ length of projection $\vec{c}$ on $\vec{a}$

ii) $(\vec{a}+\vec{b}) \cdot(\vec{c}+\vec{d})$

$=\vec{a} \cdot(\vec{c}+\vec{d})+\vec{b}(\vec{c}+\vec{d})$

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

iii)

$|\vec{a}+\vec{b}|^2 =(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})$

$=|a|^2+|b|^2+2 \vec{a} \cdot \vec{b}$

$\quad|\vec{a}|^2=\vec{a} \cdot \vec{a}$

iv) $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2$

v) $\vec{a} \cdot(k \vec{b})=k(\vec{a} \cdot \vec{b})=k \vec{a} \cdot \vec{b}$

vi) $\vec{a} \cdot(\vec{b}-\vec{c})=\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}$

vii) $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{\vec{a} \cdot \vec{b}}{\sqrt{\vec{a} \cdot \vec{a}} \cdot \sqrt{\vec{b} \cdot \vec{b}}}$

Consider

$\vec{a} =a_1 i+a_2 j$

$\vec{b}$ =b_ 1 i+b_2 j $

$\vec{a} \cdot \vec{b} =\left(a_1 i+a_2 j\right) \cdot\left(b_1 i+b_2 j\right)$

$=a_1 i \cdot\left(b_1 i+b_2 j\right)+a_2 j \cdot\left(b_1 i+b_2 j\right)$

$=a_1 b_1+a_2 b_2$

In general,

$\left(a_1 i+a_2 j+a_3 k\right) \cdot\left(b_1 i+b_2 j+b_3 k\right)$

$=a_1 b_1+a_2 b_2+a_3 b_3$

Ex $\quad$ $\vec{a} =5 i+3 j-2 k$, $\vec{b} =8 i-9 j+6 k$

$\vec{a} \cdot \vec{b} =40-27-12=1$

$\text { Ex. } \vec{a} =3 i+j+5 \vec{k}$

Verify: $|\vec{a}|^2=\vec{a} \cdot \vec{a}$

• Ex $\quad$ $\vec{a}=\frac{i}{2}+j \text {and} \vec{b}=2 i-j+k$

Pure : $\vec{a} \perp \vec{b}$

$\vec{a} \cdot \vec{b}=1-1+0=0$

Ex $\vec{a}=i-j-k$, $\vec{b}=2 i+j+2 k$

Find the angle between them

$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{2-1-2}{\sqrt{3} \sqrt{9}}=\frac{-1}{\sqrt{27}}$

Ex. $\vec{a}=2 i-j+k$, $\vec{b}=3 i+4 j-k$

Find the unit vector that is perpendicular to both $\vec{a}$ and $\vec{b}$.

Let $\vec{n}=x_i+y_j+z k$

since $\vec{a} \cdot \vec{n}=0$ and $\vec{b} \cdot \vec{n}=0$,

$2 x-y+z=0$, $3 x+4 y-z=0$ .

$\frac{x}{-3}=\frac{y}{5}=\frac{z}{11} \quad(=\lambda) \quad \begin{array}{r}x=-3 \lambda \\ y=5 \lambda \\ z=11 \lambda\end{array}$

$\therefore \text { Required Unit vector }=\frac{\vec{n}}{|\vec{n}|} =\frac{-3 i+5 j+11 k}{\sqrt{9+25+12 \mid}}=\frac{-3 i+5 j+11 k}{\sqrt{155}}$

$\vec{F}=\text { Force vector }$

$\overrightarrow{A B}=\text { displacement. }$

Work done by the

$\text { force }=\vec{F} \cdot \overrightarrow{A B}$

Ex. A particle acted on by two force vectors

$\vec{F}_1=2 i+j-3 k$

and $\vec{F}_2=3 i-2 j+2 k$, is displaced from $(1,3,-1)$ to $(4,-1,2)$. Find the work done by the forces

$\vec{F} =5 i-j-k$

$\vec{d} =3 i-4 j+3 k$

$W \cdot D =\vec{F} \cdot \vec{d}$

$=15+4-3=16 \text { unit }$

Prove: $|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2=2\left(|\vec{a}|^2+|\vec{b}|^2\right)$

Soln. $|\vec{a}+\vec{b}|^2 =(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})$

$=|\vec{a}|^2+|\vec{b}|^2+2(\vec{a} \cdot \vec{b})$

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

$\text { LHS } =2|\vec{a}|^2+2|\vec{b}|^2$

$=\text { RHS. }$

• Ex: Let $|\vec{a}|=3, \quad|\vec{b}|=4$

$\text { If }|\vec{a}+\vec{b}|=1, \quad|\vec{a}-\vec{b}|=\text { ? }$

Soln. By the above identity,

$1+|\vec{a}-\vec{b}|^2=2\left(3^2+4^2\right)=50$

$|\vec{a}-\vec{b}|=\sqrt{49}=7$

Some inequality.

Schwartz inequality.

Triangle inequality.

$|\vec{a} \cdot \vec{b}| =|| \vec{a}|| \vec{b}|\cdot \cos \theta|$

$\leq|\vec{a}| \mid \vec{b}$

$|\vec{a}+\vec{b}|^2 =(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})$

$=\vec{a} \cdot(\vec{a}+\vec{b})+\vec{b} \cdot(\vec{a}+\vec{b})$

$=|\vec{a}|^2+|\vec{b}|^2+2(\vec{a} \cdot \vec{b})$

$\leq|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}|$

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

$|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ Triangle inequality.

If equality holds, (ie) if

$\mid \vec{a}+\vec{b}|=| \vec{a}|+| \vec{b} \mid \text {, then }$

$|A C|=|A B|+|B C|$

(ie) $A, B, C$ are collinear point.