Points X and Y have position vectors x = 2i − 3j and y = -i + 2j. Find the position vector of point M which lies on the line segment XY such that |XM| : | MY| is 8:2.

The position vector of a point M that divides the line segment joining points with position vectors $\vec{x}$ and $\vec{y}$ in the ratio $n:m$ (where $n$ corresponds to $\vec{x}$ and $m$ corresponds to $\vec{y}$) is given by:

$$\vec{m} = \frac{n\vec{x} + m\vec{y}}{n+m}$$

In our case, the ratio is $|\vec{XM}| : |\vec{MY}| = 4:1$. This means M is closer to Y. So, the ratio is $n=1$ for $\vec{x}$ and $m=4$ for $\vec{y}$.

$$\vec{m} = \frac{2\vec{i} - 3\vec{j} - 4\vec{i} + 8\vec{j}}{5}$$

Combine the $\vec{i}$ components and the $\vec{j}$ components:

$$\vec{m} = \frac{(2-4)\vec{i} + (-3+8)\vec{j}}{5}$$ $$\vec{m} = \frac{-2\vec{i} + 5\vec{j}}{5}$$ $$\vec{m} = -\frac{2}{5}\vec{i} + \frac{5}{5}\vec{j}$$ $$\vec{m} = -\frac{2}{5}\vec{i} + \vec{j}$$

The position vector of M is $-\frac{2}{5}\vec{i} + \vec{j}$.