If $P(x, y, z)$ divides $\overline{AB}$ joining $A (x_1, y_1, z_1)$ and $B (x_2,y_2,z_2)$ externally in ratio $m:n$ i.e. $\dfrac{PA}{PB}=\dfrac{m}{n}, then$

$x=\dfrac{mx_2-nx_1}{m-n}, \quad y= \frac{my_2-ny_1}{m-n} \quad z= \dfrac{mz_2-nz_1}{m-n}$

$\to$ It has two cases:

(a) For $m > n,$ $\dfrac{PA}{PB} > 1$ i.e. $A-B-P$

(b) For $m< n, \dfrac{PA}{PB} < 1$ i.e. $P-A-B$

Example:

Let $A( 7, 5, -2)$ and $B(3, 1, 4)$ be two points. $P(x, y, z)$ divides $\overline{AB}$ externally in ratio $5:3$ then

$\begin{aligned} x&= \dfrac{mx_2-nx_1}{m-n}= \dfrac{5 \times3-3\times7}{5-3}=\dfrac{15-21}{2} \\&=\dfrac{-6}{2}=-3 \end{aligned}$

$\begin{aligned} y&= \frac{my_2-ny_1}{m-n}= \frac{5 \times1-3\times5}{5-3}=\frac{5-15}{2}\\&=\frac{-10}{2}=-5 \end{aligned}$

$\begin{aligned} z&= \dfrac{mz_2-nz_1}{m-n}= \dfrac{5 \times4-3(-2)}{5-3}=\dfrac{20+6}{2} \\&=\frac{26}{2}=13\end{aligned}$

$\therefore P(x, y, z) = (-3,-5, 13)$

Midpoint Formula

If $M(x, y, z)$ be midpoint of $\overline{AB}$ joining $A (x_1, y_1, z_1)$ and $B (x_2,y_2,z_2),$ then

$x=\dfrac{x_1+x_2}{2}, \quad y=\dfrac{y_1+y_2}{2}, \quad z=\dfrac{z_1+z_2}{2}, \quad$

Example:

Let $A(3, 7, 2)$ and $B(7, -5, 6)$ be two points in space and $M(x, y, z)$ be midpoint of $\overline{AB}$ , then

$x=\dfrac{x_1+x_1}{2}=\dfrac{3+7}{2}=5$

$y=\dfrac{7+(-5)}{2}=2$

$z=\dfrac{z_1+z_2}{2}=\frac{2+6}{2}=4$

$\therefore M(x, y, z)=(5, 2, 4)$

Division Formula by $\lambda :1$ ratio

If $P$ divides $\overline{AB}$ in ratio $\lambda :1$ then coordinates of $P$ are given by:

(i) $x=\dfrac{\lambda x_2+x_1}{\lambda+1}, \quad y=\dfrac{\lambda y_2+ y_1}{\lambda+1}, \quad z=\dfrac{\lambda z_2+z_1}{\lambda+1}$ (Internal Division)

(ii) $x=\dfrac{\lambda x_2-x_1}{\lambda-1}, \quad y=\dfrac{\lambda y_2- y_1}{\lambda-1}, \quad z=\dfrac{\lambda z_2-z_1}{\lambda-1}$ (External Division)

Note:

If $A (x_1,y_1,z_1)$ and $B (x_2,y_2,z_2)$ are distinct points then the coordinates of any point on $\overleftrightarrow{AB}$ except $A \text { and }B$ are given by

$\left( \dfrac{x_1+\mu x_2}{1+\mu}, \dfrac{y_1+\mu y_2}{1+ \mu},\dfrac{z_1+\mu z_2}{1+\mu} \right)$

$\mu \in R ,\quad \mu \neq -1$

Formulas to find Coordinates

If $A(x_1, y_1, z_1)$ , $B(x_2,y_2,z_2)$ and $C(x_3,y_3,z_3)$ are vertices of triangle $ABC$ , whose sides $BC, CA, AB$ are of lengths $a,b,c$ respectively, then

(i) Centroid $(G)=\left( \dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3} \right)$

(ii) Incentre $(I)=\left( \dfrac{ax_1+bx_2+cx_3}{a+b+c}, \dfrac{ay_1+by_2+cy_3}{a+b+c},\dfrac{az_1+bz_2+cz_3}{a+b+c}\right)$

(iii) Excentre

(a) (to A) $(I_1)=\left( \dfrac{-ax_1+bx_2+cx_3}{-a+b+c}, \dfrac{-ay_1+by_2+cy_3}{-a+b+c},\dfrac{-az_1+bz_2+cz_3}{-a+b+c}\right)$

(b) (to B) $(I_2)=\left( \dfrac{ax_1-bx_2+cx_3}{a-b+c}, \dfrac{ay_1-by_2+cy_3}{a-b+c},\dfrac{az_1-bz_2+cz_3}{a-b+c}\right)$

(c) (to C) $(I_3)=\left( \dfrac{ax_1+bx_2-cx_3}{a+b-c}, \dfrac{ay_1+by_2-cy_3}{a+b-c},\dfrac{az_1+bz_2-cz_3}{a+b-c}\right)$

$\to$ If $A,B,C$ are vertices of triangle $ABC$ and $P(x_x, y_1, z_1),$ $Q(x_2,y_2,z_2),$ and $R(x_3,y_3,z_3),$ are coordinates of midpoint of $\overline{BC},\overline{AC} \text { and } \overline{AB}$ respectively,then coordinate of Centroid of $(G)=\left( \dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3}\right)$

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