44. Let \( C \) be the point on the line segment \( A B \) that is twice as far from \( B \) as it is from \( A \). If \( \mathbf{a}=\overrightarrow{O A}, \mathbf{b}=\overrightarrow{O B} \), and \( \mathbf{c}=\overrightarrow{O C} \), show that \( \mathbf{c}=\frac{2}{3} \mathbf{a}+\frac{1}{3} \mathbf{b} \).

Let \( C \) be the point on the segment \( AB \) such that the distance from \( C \) to \( B \) is twice the distance from \( C \) to \( A \). To express this mathematically, we can use the concept of section formulas. Denote the length \( AC \) as \( x \) and the length \( CB \) as \( 2x \). Thus, the total length of segment \( AB \) is \[ AB = AC + CB = x + 2x = 3x. \] The position vector \( \mathbf{c} \) of point \( C \) can be expressed using the parameters from points \( A \) and \( B \): \[ \mathbf{c} = \frac{CB}{AB} \cdot \mathbf{a} + \frac{AC}{AB} \cdot \mathbf{b}. \] Since \( CB = 2x \) and \( AC = x \), we find the proportional distances: \[ CB = 2x \quad \text{and} \quad AC = x. \] Thus, \[ \mathbf{c} = \frac{2x}{3x} \mathbf{a} + \frac{x}{3x} \mathbf{b} = \frac{2}{3} \mathbf{a} + \frac{1}{3} \mathbf{b}. \] This shows that \( \mathbf{c} = \frac{2}{3} \mathbf{a} + \frac{1}{3} \mathbf{b} \), confirming the required relation. The point \( C \) lies closer to \( A \) since it is twice as far from \( B \) compared to \( A \), providing a proportional division of the segment \( AB \).