1. **Problem Statement:** ABCD is a parallelogram and EF is the midpoint of side BC. Given vectors \( \mathbf{B} = (b_1, b_2) \) and \( \mathbf{F} = (c_1, c_2) \), express \( \mathbf{AF} \) as a column vector. Also, write down relations between \( \mathbf{AE} \neq \mathbf{0} \) and \( \mathbf{AE} \to \), and between \( \mathbf{CO} \) and \( \mathbf{FB} \). 2. **Formula and Rules:** - Vector from point A to point F is \( \mathbf{AF} = \mathbf{F} - \mathbf{A} \). - Midpoint E of BC means \( \mathbf{E} = \frac{\mathbf{B} + \mathbf{C}}{2} \). - In parallelogram ABCD, opposite sides are equal and parallel. 3. **Step-by-step Solution:** **Step 1:** Express \( \mathbf{AF} \) as a column vector. - Since \( \mathbf{F} = (c_1, c_2) \) and \( \mathbf{A} = (a_1, a_2) \), $$\mathbf{AF} = \mathbf{F} - \mathbf{A} = \begin{pmatrix} c_1 - a_1 \\ c_2 - a_2 \end{pmatrix}$$ **Step 2:** Relations between \( \mathbf{AE} \neq \mathbf{0} \) and \( \mathbf{AE} \to \). - \( \mathbf{AE} \neq \mathbf{0} \) means vector \( \mathbf{AE} \) is not the zero vector. - \( \mathbf{AE} \to \) indicates the direction of vector \( \mathbf{AE} \). - Since E is midpoint of BC, \( \mathbf{E} = \frac{\mathbf{B} + \mathbf{C}}{2} \), so $$\mathbf{AE} = \mathbf{E} - \mathbf{A} = \frac{\mathbf{B} + \mathbf{C}}{2} - \mathbf{A}$$ - This vector points from A towards midpoint E on BC. **Step 3:** Relation between \( \mathbf{CO} \) and \( \mathbf{FB} \). - \( \mathbf{CO} \) is vector from C to O (origin), so \( \mathbf{CO} = \mathbf{O} - \mathbf{C} = -\mathbf{C} \). - \( \mathbf{FB} \) is vector from F to B, so \( \mathbf{FB} = \mathbf{B} - \mathbf{F} \). - Since F is midpoint of BC, \( \mathbf{F} = \frac{\mathbf{B} + \mathbf{C}}{2} \), then $$\mathbf{FB} = \mathbf{B} - \frac{\mathbf{B} + \mathbf{C}}{2} = \frac{\mathbf{B} - \mathbf{C}}{2}$$ - Note that \( \mathbf{CO} = -\mathbf{C} \), so \( \mathbf{FB} = \frac{\mathbf{B} - \mathbf{C}}{2} \). **Final answers:** - \( \mathbf{AF} = \begin{pmatrix} c_1 - a_1 \\ c_2 - a_2 \end{pmatrix} \) - \( \mathbf{AE} = \frac{\mathbf{B} + \mathbf{C}}{2} - \mathbf{A} \), nonzero vector pointing from A to midpoint E. - \( \mathbf{FB} = \frac{\mathbf{B} - \mathbf{C}}{2} \), and \( \mathbf{CO} = -\mathbf{C} \). This completes the solution for the first problem.