If C divides AB externally in the ratio 3:2, find the position vector of C in terms of A and B.

The position vectors of points A, B, and C are $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ respectively.
Point C divides the line segment AB externally in the ratio $m:n = 3:2$.
This means $m=3$ and $n=2$.
The formula for the position vector $\mathbf{c}$ of a point that divides a line segment AB externally in the ratio $m:n$ is:

$$\mathbf{c} = \frac{n\mathbf{a} - m\mathbf{b}}{n - m}$$

Substitute $m=3$ and $n=2$ into the formula:

$$\mathbf{c} = \frac{2\mathbf{a} - 3\mathbf{b}}{2 - 3}$$

$$\mathbf{c} = \frac{2\mathbf{a} - 3\mathbf{b}}{-1}$$ $$\mathbf{c} = -(2\mathbf{a} - 3\mathbf{b})$$ $$\mathbf{c} = -2\mathbf{a} + 3\mathbf{b}$$

Rearranging the terms:

$$\mathbf{c} = 3\mathbf{b} - 2\mathbf{a}$$

The expression for $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ is: