mathematics-class-12-unit-10-chapter-03-vectors-huyqsd9dmey

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Scalar product of two vectors </primary>
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- $\vec{a} \cdot \vec{b}:=|\vec{a}||\vec{b}| \cdot \cos (\theta)$

- 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 $\vec{a}$.

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<footer style = "font-size: 18px"> $\rightarrow$ Vectors lecture 3 $\rightarrow$ <span style = "color:blue"> Scalar product of two vectors </span> $\rightarrow$ Example 1 $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- $ 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$

- $i \cdot j=0=j \cdot k=k \cdot i$

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

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

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<footer style = "font-size: 18px">Vectors lecture 3 $\rightarrow$ Scalar product of two vectors $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 1 $\rightarrow$ Properties (1/3) </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Properties (1/3) </primary>
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- 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 $\vec{a}$

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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Properties (1/3) </span> $\rightarrow$ Properties (2/3) $\rightarrow$ Properties (3/3) </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Properties (2/3) </primary>
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- 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) $\vec{a} \cdot(k \vec{b})=k(\vec{a} \cdot \vec{b})=k \vec{a} \cdot \vec{b}$

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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Properties (1/3) $\rightarrow$ <span style = "color:blue"> Properties (2/3) </span> $\rightarrow$ Properties (3/3) $\rightarrow$ A formula to find dot product </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> A formula to find dot product </primary>
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- 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 $

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<footer style = "font-size: 18px">Properties (2/3) $\rightarrow$ Properties (3/3) $\rightarrow$ <span style = "color:blue"> A formula to find dot product </span> $\rightarrow$ Examples $\rightarrow$ Examples </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Examples </primary>
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- <ins>Ex</ins> $\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: $|\vec{a}|^2=\vec{a} \cdot \vec{a}$

- - <ins>Ex</ins> $\quad $ $\vec{a}=\frac{i}{2}+j \text {and}   \vec{b}=2 i-j+k$
 
- Pure : $\vec{a} \perp \vec{b}$
- $\vec{a} \cdot \vec{b}=1-1+0=0$

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

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

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<footer style = "font-size: 18px">Properties (3/3) $\rightarrow$ A formula to find dot product $\rightarrow$ <span style = "color:blue"> Examples </span> $\rightarrow$ Examples $\rightarrow$ An application work done by a force </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Examples </primary>
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- <ins>Ex.</ins> $ \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}} $

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<footer style = "font-size: 18px">A formula to find dot product $\rightarrow$ Examples $\rightarrow$ <span style = "color:blue"> Examples </span> $\rightarrow$ An application work done by a force $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> An application work done by a force </primary>
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- $\vec{F}=\text { Force vector }$
- $\overrightarrow{A B}=\text { displacement. }$

- 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 }$

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<footer style = "font-size: 18px">Examples $\rightarrow$ Examples $\rightarrow$ <span style = "color:blue"> An application work done by a force </span> $\rightarrow$ Problem-1 $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- 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 $

- $=\text { RHS. }$

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<footer style = "font-size: 18px">Examples $\rightarrow$ An application work done by a force $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Triangle inequality </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- - <ins>Ex:</ins> 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 $

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<footer style = "font-size: 18px">An application work done by a force $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Triangle inequality $\rightarrow$ Triangle inequality special case </footer>

### <Success style="font-size:25px"> Vectors L - 3 </Success>
### <primary style="font-size:24px"> Triangle inequality </primary>
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- 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.

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Triangle inequality </span> $\rightarrow$ Triangle inequality special case $\rightarrow$ Thank you </footer>