Division of a segment in a given ratio (vector and coordinate methods).
Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.
Knowing the coordinates of points $M_1(x_1, y_1, z_1)$ and $M_2(x_2, y_2, z_2)$ and the ratio $\lambda$ at which the point $M$ divides the directed segment $\overline{M_1M_2}$, we will find the coordinates of point $M$.
Let $O$ be the origin. Let's denote $\overline{OM_1}=r_1$, $\overline{OM_2}=r_2$, $\overline{OM}=r$. Since $$\overline{M_1M}=r-r_1, \overline{MM_2}=r_2-r,$$ then $r-r_1=\lambda(r_2-r)$, from which (since $\lambda\neq -1$) $$r=\frac{r_1+\lambda r_2}{1+\lambda}.$$
This obtained form provides the solution to the problem in vector form. Transitioning to coordinates in this formula, we get$$x=\frac{x_1+\lambda x_2}{1+\lambda}, y=\frac{y_1+\lambda y_2}{1+\lambda}, z=\frac{z_1+\lambda z_2}{1+\lambda}.$$
Examples.
2.57. The segment with endpoints at $A(3, -2)$ and $B(6, 4)$ is divided into three equal parts. Find the coordinates of the division points.
Solution.
Let $C(x_C, y_C)$ and $D(x_D, y_D)$ be the points dividing the segment $AB$ into three equal parts. Then $$\lambda_1=\frac{AC}{CB}=\frac{1}{2};$$ $$x_C=\frac{x_A+\lambda_1x_B}{1+\lambda_1}=\frac{3+\frac{1}{2}\cdot 6}{1+\frac{1}{2}}=4;$$
$$y_C=\frac{y_A+\lambda_1y_B}{1+\lambda_1}=\frac{-2+\frac{1}{2}\cdot 4}{1+\frac{1}{2}}=0.$$
Next, we find the coordinates of point $D$:
$$\lambda_2=\frac{AD}{DB}=\frac{2}{1}=2;$$ $$x_D=\frac{x_A+\lambda_2x_B}{1+\lambda_2}=\frac{3+2\cdot 6}{1+2}=5;$$
$$y_D=\frac{y_A+\lambda_2y_B}{1+\lambda_2}=\frac{-2+2\cdot 4}{1+2}=2.$$
Answer: $(4, 0)$ и $(5, 2).$
2.58. Determine the coordinates of the endpoints of a segment that is divided into three equal parts, given that the points are $C(2, 0, 2)$ and $D(5, -2, 0)$.
Solution:
Let $A(x_A, y_A, z_A)$ and $B(x_B, y_B, z_B)$ be the endpoints of the given segment.
We'll write down the formulas for finding the coordinates of point $C$ and substitute the known coordinates:
$$\lambda_1=\frac{AC}{CB}=\frac{1}{2};$$ $$x_C=\frac{x_A+\lambda_1x_B}{1+\lambda_1}\Rightarrow 2=\frac{x_A+\frac{1}{2}\cdot x_B}{1+\frac{1}{2}}=2\frac{x_A+\frac{1}{2}\cdot x_B}{3}\Rightarrow $$ $$\Rightarrow 3=x_A+\frac{1}{2}\cdot x_B;$$
$$y_C=\frac{y_A+\lambda_1y_B}{1+\lambda_1}\Rightarrow 0=\frac{y_A+\frac{1}{2}\cdot y_B}{1+\frac{1}{2}}\Rightarrow 0=y_A+\frac{1}{2}\cdot y_B;$$
$$z_C=\frac{z_A+\lambda_1z_B}{1+\lambda_1}\Rightarrow 2=\frac{z_A+\frac{1}{2}\cdot z_B}{1+\frac{1}{2}}=2\frac{z_A+\frac{1}{2}\cdot z_B}{3}\Rightarrow$$ $$\Rightarrow 3=z_A+\frac{1}{2}\cdot z_B.$$
Similar equations will be written for point $D$:
$$\lambda_2=\frac{AD}{DB}=\frac{2}{1}=2;$$ $$x_D=\frac{x_A+\lambda_2x_B}{1+\lambda_2}\Rightarrow 5=\frac{x_A+2\cdot x_B}{1+2}=\frac{x_A+2\cdot x_B}{3}\Rightarrow $$ $$\Rightarrow 15=x_A+2\cdot x_B;$$
$$y_D=\frac{y_A+\lambda_2y_B}{1+\lambda_2}\Rightarrow -2=\frac{y_A+2\cdot y_B}{1+2}\Rightarrow -6=y_A+2\cdot y_B;$$
$$z_D=\frac{z_A+\lambda_2z_B}{1+\lambda_2}\Rightarrow 0=\frac{z_A+2\cdot z_B}{1+2}\Rightarrow 0=z_A+2\cdot z_B.$$
Next, we will write the obtained equations for $x_A, x_B;$ $y_A, y_B;$ and $z_A, z_B$ pairwise as systems and solve them:
$$\left\{\begin{array}{lcl}x_A+\frac{1}{2}x_B=3\\x_A+2x_B=15\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}x_A=3-0,5x_B\\3-0,5x_B+2x_B=15\end{array}\right.\Rightarrow$$ $$\Rightarrow\left\{\begin{array}{lcl}x_A=3-0,5\cdot8=-1\\x_B=\frac{12}{1,5}=8\end{array}\right.$$
$$\left\{\begin{array}{lcl}y_A+\frac{1}{2}y_B=0\\y_A+2y_B=-6\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}y_B=-2y_A\\y_A-4y_A=-6\end{array}\right.\Rightarrow$$ $$\Rightarrow\left\{\begin{array}{lcl}y_B=-4\\y_A=2\end{array}\right.$$
$$\left\{\begin{array}{lcl}z_A+\frac{1}{2}z_B=3\\z_A+2z_B=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}-2z_B+0,5z_B=3\\z_A=-2z_B\end{array}\right.\Rightarrow$$ $$\Rightarrow\left\{\begin{array}{lcl}z_B=-2\\z_A=4\end{array}\right.$$
Thus, we obtained the coordinates of the endpoints of the segment $A(-1, 2, 4)$ and $B(8, -4, -2).$
Answer: $A(-1, 2, 4),$ $B(8, -4, -2).$
