directed line segment : magnitude & direction

$|\overrightarrow{A B}| =\text { length of vector }$

$=\text { magnitude of } \overrightarrow{AB}$

equal, zero, Unit

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

$\quad \overrightarrow{D B}=\vec{a}-\vec{b}$

$\quad \quad i \quad j \quad k$

$\overrightarrow{O P}=a i+b j+c k$

$\text { Position vector of the point } P$.

$\overrightarrow{P Q} =\overrightarrow{O Q} - \overrightarrow{O P}$

$P$ - mid point of $A C$

$Q$ - mid point of $D B$.

Show that $\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}$

$=4 \overrightarrow{P Q}$

$\text { L.H.S. } =\overrightarrow{O B}-\overrightarrow{O A}+\overrightarrow{O D}-\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}+\overrightarrow{O D}-\overrightarrow{O C}$

$=2[\overrightarrow{O B}+\overrightarrow{O D}]-2[\overrightarrow{O A}+\overrightarrow{O C}]$

$=2[2 \overrightarrow{O Q}]-2[2 \overrightarrow{O P}], [\frac{\overrightarrow{O B}+\overrightarrow{O D}}{2}=\overrightarrow{O Q}] (?)$

$=4[\overrightarrow{O Q}-\overrightarrow{O P}]$

$=4 \overrightarrow{P Q}$

Let P & Q be given two points

Let $R$ divide $P Q$

in the ratio $l: m$

$R$ is given by$\left(\frac{m x_1+l x_2}{m+l}, \frac{m y_1+l y_2}{m+l}, \frac{m z_1+l z_2}{m+l}\right)$

$\overrightarrow{O R}=\frac{1}{m+l}\left[m\left(x_1 i+y_1 j+z_1 k\right)+l\left(x_2 i+y_2 j+z_2 k\right)\right]$

$=\frac{1}{m+l}\left[m \vec{r}_1+l \vec{r}_2\right]$

In particular, take m=l:

$R$ becomes mid point of $\overrightarrow {P Q}$

Then

$\overrightarrow{O R}=\frac{1}{2}\left[\vec{r}_1+\vec{r}_2\right]$

$l=\cos \alpha$ ,$m=\cos \beta$ ,$n=\cos \gamma$

$\text { [l, m, n direction cosines of the point.] }$

$l^2+m^2+n^2=1$

$a=l \cdot\left|\overrightarrow{O P}\right|$ , $b=m|\overrightarrow{O P}|$ ,$C=n|\overrightarrow{O P}|$

$[\text { a } b c \text { are called direction vats of the points }]$

$\text P(2,-1,2)$

$|\overrightarrow{O P}|=\sqrt{4+1+4}=3$

$\cos \alpha=2 / 3$

$\cos \beta=-1 / 3$

$\cos \gamma=2 / 3 .$

$\therefore i \cos \alpha+j \cos \beta+k \cos \gamma$ is the unit vector along $\overrightarrow{O P}$ :

i) To find the equation of the line passing through 0

Let $\vec{u}$ is the unit vector along the line $L$.

$\overrightarrow{O P}=\vec{r}$

$\vec{r}=k \vec{u} , k$ is a scalar

ii) To find equation n of $L$ passing though $P(a, b, c)$

$\vec{r}=\overrightarrow{O P}+\overrightarrow{P Q}$

$\vec{r}=\vec{d}+\alpha \vec{u}$

where $\vec{u}$ is the unit vector along $L$.

iii) To find the equation of L passing through two point P & Q

$\vec{r}-\vec{d}=\alpha \vec{u}$

$\overrightarrow{\vec{r}}-\vec{a}=\alpha\left[\frac{\vec{b}-\vec{a}}{|\vec{b}-\vec{a}|}\right]$

$\vec{r}-\vec{b}=k\left[\frac{\vec{a}-\vec{b}}{\mid \vec{a}-\vec{b}|}\right]$

$\vec{r}=\frac{k \vec{a}}{|\vec{a}-\vec{b}|}+\left[1-\frac{k}{|\vec{a}-\vec{b}|}\right] \vec{b}$

$\vec{r}=t \vec{a}+(1-t) \vec{b} \leftarrow t= \frac{k}{|\vec{a}-\vec{b}|}$

$\vec{r}=\alpha \vec{a}+\beta \vec{b}$

where $\alpha+\beta=1$

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

$P(9,3,-2)$ & $Q(4,5,-1)$

Find the equation line passing the $P \& Q$

$\vec{r}=(4 i+5 j-k)+k(5 i-2 j-k)$

Vector equation of a plane

Case (i)

Passing the origin & parallel to $\vec{a}$ & $\vec{b}$

$\vec{r}=\overrightarrow{O P} =\overrightarrow{O A}+\overrightarrow{O B}$

$=\alpha \vec{a}+\beta \vec{b}$

$\vec{r} =\alpha \vec{a}+\beta \overrightarrow{ b}$

Case(ii).

Passing through C & parallel to vectors $\vec{a}$ & $\vec{b}$

$\vec{r} =\overrightarrow{O C}+\overrightarrow{C P}$

$\text { (io) } \vec{r} =\vec{c}+\alpha \vec{a}+\beta \vec{b}$

Case (iii)

Check in following

The equation of required plane is $\vec{r}=\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}$ where $\gamma=1-\alpha-\beta$. $\vec{r}=\alpha \vec{a}+\beta \vec{b}+\gamma\vec{c}$

$\vec{r}= \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}$

$\alpha+\beta+\gamma=1$