1. Problem: Given parallelogram OABC with origin at O, vectors \(\vec{OA} = \mathbf{p}\) and \(\vec{OC} = \mathbf{q}\). M is midpoint of OB, N divides AB in ratio 3:2. Find vectors \(\vec{OB}\), \(\vec{CM}\), \(\vec{MN}\), and extension point D where CB and ON meet. 2. Find \(\vec{OB}\): Since OABC is a parallelogram, \(\vec{OB} = \vec{OA} + \vec{OC} = \mathbf{p} + \mathbf{q}\). 3. Find \(\vec{CM}\): - M is midpoint of OB, so \(\vec{OM} = \frac{\vec{OB}}{2} = \frac{\mathbf{p} + \mathbf{q}}{2}\). - Vector \(\vec{CM} = \vec{OM} - \vec{OC} = \frac{\mathbf{p} + \mathbf{q}}{2} - \mathbf{q} = \frac{\mathbf{p} - \mathbf{q}}{2}\). 4. Find \(\vec{MN}\): - \(\vec{MN} = \vec{MO} + \vec{ON}\). - \(\vec{MO} = -\vec{OM} = -\frac{\mathbf{p} + \mathbf{q}}{2}\). - N divides AB in ratio 3:2, so \(\vec{ON} = \vec{OA} + \frac{3}{5} \vec{AB} = \mathbf{p} + \frac{3}{5}(\vec{OB} - \vec{OA}) = \mathbf{p} + \frac{3}{5}(\mathbf{p} + \mathbf{q} - \mathbf{p}) = \mathbf{p} + \frac{3}{5} \mathbf{q}\). - Thus, \(\vec{MN} = -\frac{\mathbf{p} + \mathbf{q}}{2} + \mathbf{p} + \frac{3}{5} \mathbf{q} = \frac{5\mathbf{p} + \mathbf{q}}{10}\). 5. Find \(\vec{OD}\) where D is intersection of extended CB and ON: - Parametrize \(\vec{OD} = \vec{OC} + \lambda \vec{CD} = \mathbf{q} + \lambda(\vec{D} - \vec{C})\). - Vector \(\vec{CB} = \vec{OB} - \vec{OC} = (\mathbf{p} + \mathbf{q}) - \mathbf{q} = \mathbf{p}\). - Vector \(\vec{ON} = \mathbf{p} + \frac{3}{5} \mathbf{q}\) from above. - Vector \(\vec{ON} = \mu \vec{OD} = \mu(\mathbf{q} + \lambda (\vec{CD}))\) but more straightforward is to solve for \(\lambda\) such that lines meet. - Using parametric forms: - Point on CB: \(\vec{OC} + t \vec{CB} = \mathbf{q} + t \mathbf{p}\) - Point on ON extended: \(s \vec{ON} = s \left(\mathbf{p} + \frac{3}{5}\mathbf{q}\right)\) - Set equal for intersection: \[ \mathbf{q} + t \mathbf{p} = s \mathbf{p} + \frac{3s}{5} \mathbf{q} \] - Equate coefficients: \[ t \mathbf{p} - s \mathbf{p} = \frac{3s}{5} \mathbf{q} - \mathbf{q} \] \[ (t - s) \mathbf{p} = \left( \frac{3s}{5} - 1 \right) \mathbf{q} \] - For \(\mathbf{p}\) and \(\mathbf{q}\) independent, coefficients zero: \[ t - s = 0 \Rightarrow t = s \] \[ \frac{3s}{5} - 1 = 0 \Rightarrow s = \frac{5}{3} \] - So \(t = \frac{5}{3}\). - Position vector \(\vec{OD} = \mathbf{q} + t \mathbf{p} = \mathbf{q} + \frac{5}{3} \mathbf{p} = \frac{5}{3} \mathbf{p} + \mathbf{q}\). Final answers: - \(\vec{OB} = \mathbf{p} + \mathbf{q}\) - \(\vec{CM} = \frac{\mathbf{p} - \mathbf{q}}{2}\) - \(\vec{MN} = \frac{5\mathbf{p} + \mathbf{q}}{10}\) - \(\vec{OD} = \frac{5}{3} \mathbf{p} + \mathbf{q}\)