Let's tackle each part step by step. --- Given: - \(\vec{x} = \overrightarrow{OA}\) - \(\vec{y} = \overrightarrow{OB}\) - \(\vec{z} = \overrightarrow{OC}\) ### a) \(\overrightarrow{DB}\) Let's find coordinates for \(D\) and \(B\) in terms of \(\vec{x}, \vec{y}, \vec{z}\): - \(O\) is the origin. - \(B\) is at \(\vec{y}\). - \(D\) is along \(\vec{y}\) and \(\vec{z}\): so \(D = O + \vec{y} + \vec{z}\). So, \[ \overrightarrow{DB} = \overrightarrow{OB} - \overrightarrow{OD} = \vec{y} - (\vec{y} + \vec{z}) = -\vec{z} \] --- ### b) \(\overrightarrow{DO}\) \[ \overrightarrow{DO} = \overrightarrow{OO} - \overrightarrow{OD} = \vec{0} - (\vec{y} + \vec{z}) = -\vec{y} - \vec{z} \] --- ### c) \(\overrightarrow{EA}\) \(E\) is at the center of the top face (\(EFGH\)), so its coordinates are: - \(E = O + \vec{x} + \vec{z}\) - \(A = O + \vec{x}\) So, \[ \overrightarrow{EA} = \overrightarrow{OA} - \overrightarrow{OE} = \vec{x} - (\vec{x} + \vec{z}) = -\vec{z} \] But actually, to get from \(E\) to \(A\): \[ \overrightarrow{EA} = \overrightarrow{OA} - \overrightarrow{OE} = \vec{x} - (\vec{x} + \vec{z}) = -\vec{z} \] --- ### d) Area of parallelogram \(OCEB\) The area of a parallelogram defined by vectors \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a} \times \vec{b}|\). \(OCEB\) is defined by vectors: - \(\overrightarrow{OC} = \vec{z}\) - \(\overrightarrow{OB} = \vec{y}\) So, area is: \[ \text{Area} = |\vec{z} \times \vec{y}| \] --- ### **Final Answers:** a) \(\overrightarrow{DB} = -\vec{z}\) b) \(\overrightarrow{DO} = -\vec{y} - \vec{z}\) c) \(\overrightarrow{EA} = -\vec{z}\) d) Area of parallelogram \(OCEB = |\vec{z} \times \vec{y}|\)