Given two points $A(\vec{a})$ and $B(\vec{b})$, we want to find the position vectors of the points that divide the line segment $AB$ internally and externally in the given ratios. (1) Ratio is $3:1$. (2) Ratio is $2:5$.

1. Problem Description

Given two points $A(\vec{a})$ and $B(\vec{b})$, we want to find the position vectors of the points that divide the line segment $AB$ internally and externally in the given ratios.

(1) Ratio is $3:1$.

(2) Ratio is $2:5$.

2. Solution Steps

(1) For a ratio of $3:1$:

Internal division: The position vector $\vec{p}$ of the point that divides $AB$ internally in the ratio $m:n$ is given by:

$\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}$.

In our case, $m = 3$ and $n = 1$, so the position vector is:

$\vec{p} = \frac{1\vec{a} + 3\vec{b}}{3+1} = \frac{\vec{a} + 3\vec{b}}{4}$.

External division: The position vector $\vec{q}$ of the point that divides $AB$ externally in the ratio $m:n$ is given by:

$\vec{q} = \frac{-n\vec{a} + m\vec{b}}{m-n}$.

In our case, $m = 3$ and $n = 1$, so the position vector is:

$\vec{q} = \frac{-1\vec{a} + 3\vec{b}}{3-1} = \frac{-\vec{a} + 3\vec{b}}{2}$.

(2) For a ratio of $2:5$:

Internal division: The position vector $\vec{p}$ of the point that divides $AB$ internally in the ratio $m:n$ is given by:

$\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}$.

In our case, $m = 2$ and $n = 5$, so the position vector is:

$\vec{p} = \frac{5\vec{a} + 2\vec{b}}{2+5} = \frac{5\vec{a} + 2\vec{b}}{7}$.

External division: The position vector $\vec{q}$ of the point that divides $AB$ externally in the ratio $m:n$ is given by:

$\vec{q} = \frac{-n\vec{a} + m\vec{b}}{m-n}$.

In our case, $m = 2$ and $n = 5$, so the position vector is:

$\vec{q} = \frac{-5\vec{a} + 2\vec{b}}{2-5} = \frac{-5\vec{a} + 2\vec{b}}{-3} = \frac{5\vec{a} - 2\vec{b}}{3}$.

3. Final Answer

(1) Internal division: $\frac{\vec{a} + 3\vec{b}}{4}$

External division: $\frac{-\vec{a} + 3\vec{b}}{2}$

(2) Internal division: $\frac{5\vec{a} + 2\vec{b}}{7}$

External division: $\frac{5\vec{a} - 2\vec{b}}{3}$