1. **Problem Statement:** We have a parallelogram OACB with vectors \(\vec{OA} = \vec{a}\) and \(\vec{OB} = \vec{b}\). We need to express the vectors \(\vec{AO}, \vec{BC}, \vec{AB}, \vec{CO}\) in terms of \(\vec{a}\) and \(\vec{b}\). 2. **Recall Vector Rules:** - The vector from point P to Q is \(\vec{PQ} = \vec{OQ} - \vec{OP}\). - In a parallelogram, opposite sides are equal and parallel. 3. **Calculate each vector:** **(a) Vector \(\vec{AO}\):** \[ \vec{AO} = \vec{OO} - \vec{OA} = \vec{0} - \vec{a} = -\vec{a} \] **(b) Vector \(\vec{BC}\):** Points B and C are vertices of the parallelogram. Since \(\vec{OB} = \vec{b}\) and \(\vec{OC} = \vec{a} + \vec{b}\) (because C is diagonally opposite to O via A and B), \[ \vec{BC} = \vec{OC} - \vec{OB} = (\vec{a} + \vec{b}) - \vec{b} = \vec{a} \] **(c) Vector \(\vec{AB}\):** \[ \vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a} \] **(d) Vector \(\vec{CO}\):** \[ \vec{CO} = \vec{OO} - \vec{OC} = \vec{0} - (\vec{a} + \vec{b}) = -\vec{a} - \vec{b} \] 4. **Summary of answers:** - \(\vec{AO} = -\vec{a}\) - \(\vec{BC} = \vec{a}\) - \(\vec{AB} = \vec{b} - \vec{a}\) - \(\vec{CO} = -\vec{a} - \vec{b}\) These results use the vector subtraction rule and properties of parallelograms to express each vector in terms of \(\vec{a}\) and \(\vec{b}\).