1. ABCDEF is a regular hexagon in which \( \overrightarrow{B C} \) rep- resents \( \underline{b} \) and \( \overrightarrow{F C} \) represents \( 2 \underline{a} \). Express the vectors \( \overrightarrow{A B}, C D \) and \( \overrightarrow{B E} \) in terms of \( \underline{a} \) and \( \underline{b} \).

To express the vectors \( \overrightarrow{A B}, \overrightarrow{C D} \), and \( \overrightarrow{B E} \) in terms of \( \underline{a} \) and \( \underline{b} \), we first need to understand the geometry of the regular hexagon ABCDEF. 1. **Understanding the Hexagon:** - A regular hexagon can be inscribed in a circle, and its vertices can be represented in a coordinate system. - Let's place the hexagon in the Cartesian plane with the following coordinates: - \( A(1, 0) \) - \( B\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \) - \( C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \) - \( D(-1, 0) \) - \( E\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \) - \( F\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \) 2. **Given Vectors:** - \( \overrightarrow{BC} = \underline{b} \) - \( \overrightarrow{FC} = 2\underline{a} \) 3. **Finding the Vectors:** - We can express the vectors in terms of the coordinates of the points. ### Step 1: Calculate \( \overrightarrow{A B} \) \[ \overrightarrow{A B} = \overrightarrow{B} - \overrightarrow{A} = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) - (1, 0) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \] ### Step 2: Calculate \( \overrightarrow{C D} \) \[ \overrightarrow{C D} = \overrightarrow{D} - \overrightarrow{C} = (-1, 0) - \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \] ### Step 3: Calculate \( \overrightarrow{B E} \) \[ \overrightarrow{B E} = \overrightarrow{E} - \overrightarrow{B} = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = (-1, -\sqrt{3}) \] ### Expressing in terms of \( \underline{a} \) and \( \underline{b} \) Now, we need to express these vectors in terms of \( \underline{a} \) and \( \underline{b} \). 1. From the given \( \overrightarrow{BC} = \underline{b} \), we can find \( \overrightarrow{AB} \) and \( \overrightarrow{CD} \) in terms of \( \underline{b} \): - Since \( \overrightarrow{BC} = C - B \), we can express \( \overrightarrow{AB} \) as: \[ \overrightarrow{AB} = -\underline{b} \] 2. For \( \overrightarrow{CD} \): - Since \( \overrightarrow{CD} = D - C \), we can express it as: \[ \overrightarrow{CD} = -\underline{b} \] 3. For \( \overrightarrow{BE} \): - Since \( \overrightarrow{FC} = 2\underline{a} \) and \( \overrightarrow{BE} = E - B \), we can express it as: \[ \overrightarrow{BE} = -\underline{b} - 2\underline{a} \] ### Final Expressions Thus, the vectors in terms of \( \underline{a} \) and \( \underline{b} \) are: - \( \overrightarrow{A B} = -\underline{b} \) - \( \overrightarrow{C D} = -\underline{b} \) - \( \overrightarrow{B E} = -\underline{b} - 2\underline{a} \)