1. **State the problem:** We have parallelogram OPQT with position vectors \( \overrightarrow{OP} = \mathbf{a} \) and \( \overrightarrow{OT} = \mathbf{b} \). Point K lies on PQ such that \( PK:KQ = 3:1 \). Lines OK and TQ are extended to meet at X. We must find \( \overrightarrow{OX} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \). 2. **Find vector \( \overrightarrow{PQ} \):** Since OPQT is a parallelogram, \( \overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{OT} = \mathbf{a} + \mathbf{b} \) so \[ \overrightarrow{PQ} = \overrightarrow{Q} - \overrightarrow{P} = (\mathbf{a} + \mathbf{b}) - \mathbf{a} = \mathbf{b} \] 3. **Position vector of K:** K divides PQ into ratio 3:1 so \[ \overrightarrow{K} = \overrightarrow{P} + \frac{3}{3+1} \overrightarrow{PQ} = \mathbf{a} + \frac{3}{4} \mathbf{b} \] 4. **Parametric form of line OK:** Vector \( \overrightarrow{O} = \mathbf{0} \). We define parameter \( \lambda \) such that \[ \overrightarrow{OX} = \lambda \overrightarrow{OK} = \lambda \left( \mathbf{a} + \frac{3}{4} \mathbf{b} \right) \] 5. **Parametric form of line TQ:** Point T is at \( \mathbf{b} \) and Q at \( \mathbf{a} + \mathbf{b} \) so line TQ can be parameterized as \[ \overrightarrow{TX} = \mu \overrightarrow{TQ} = \mu (\mathbf{a} + \mathbf{b} - \mathbf{b}) = \mu \mathbf{a} \] Thus \[ \overrightarrow{OX} = \overrightarrow{OT} + \overrightarrow{TX} = \mathbf{b} + \mu \mathbf{a} \] 6. **Since X lies on both lines: equate \( \overrightarrow{OX} \) expressions:** \[ \lambda \left( \mathbf{a} + \frac{3}{4} \mathbf{b} \right) = \mathbf{b} + \mu \mathbf{a} \] Equating components along \( \mathbf{a} \) and \( \mathbf{b} \): \[ \lambda = \mu \quad \text{(coefficient of } \mathbf{a} ) \] \[ \frac{3}{4} \lambda = 1 \quad \Rightarrow \quad \lambda = \frac{4}{3} \] So \[ \mu = \lambda = \frac{4}{3} \] 7. **Find \( \overrightarrow{OX} \):** Substitute \( \lambda = \frac{4}{3} \) into expression for \( \overrightarrow{OX} \) on line TQ: \[ \overrightarrow{OX} = \mathbf{b} + \frac{4}{3} \mathbf{a} = \frac{4}{3} \mathbf{a} + \mathbf{b} \] **Final answer:** \[ \boxed{ \overrightarrow{OX} = \frac{4}{3} \mathbf{a} + \mathbf{b} } \] This gives the position vector of point X in terms of \( \mathbf{a} \) and \( \mathbf{b} \) in simplest form.