1. **Problem statement:** Given trapezium OABC with AB \parallel OC, OC = 4 AB, D on OA with OD : DA = 3 : 1, E on OC with OE : EC = 1 : 3, express vectors ED and CB in terms of a and c, then find expressions for EX and CX, and determine scalars h and k where EX = hED and CX = kCB. 2. **Express vectors:** (i) Find \( \overrightarrow{ED} \): - Position vector of D on OA: \( \overrightarrow{OD} = \frac{3}{4} \overrightarrow{OA} = \frac{3}{4} a \) - Position vector of E on OC: \( \overrightarrow{OE} = \frac{1}{4} \overrightarrow{OC} = \frac{1}{4} c \) - Vector \( \overrightarrow{ED} = \overrightarrow{OD} - \overrightarrow{OE} = \frac{3}{4} a - \frac{1}{4} c \) (ii) Find \( \overrightarrow{CB} \): - Since AB \parallel OC and OC = 4 AB, let \( \overrightarrow{AB} = x \), then \( \overrightarrow{OC} = 4x \) - Vector \( \overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} \) - Note \( \overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = a + x \), and \( \overrightarrow{OC} = c \) - Since \( c = 4x \), then \( x = \frac{c}{4} \) - So \( \overrightarrow{CB} = (a + \frac{c}{4}) - c = a - \frac{3}{4} c \) 3. **Express EX and CX:** (i) \( \overrightarrow{EX} = h \overrightarrow{ED} = h \left( \frac{3}{4} a - \frac{1}{4} c \right) = \frac{3h}{4} a - \frac{h}{4} c \) (ii) \( \overrightarrow{CX} = k \overrightarrow{CB} = k \left( a - \frac{3}{4} c \right) = k a - \frac{3k}{4} c \) 4. **Find h and k:** - Since X lies on both lines, \( \overrightarrow{OX} = \overrightarrow{OE} + \overrightarrow{EX} = \overrightarrow{OC} + \overrightarrow{CX} \) - Substitute: \( \overrightarrow{OE} + \overrightarrow{EX} = \frac{1}{4} c + \frac{3h}{4} a - \frac{h}{4} c = \frac{3h}{4} a + \left( \frac{1}{4} - \frac{h}{4} \right) c \) \( \overrightarrow{OC} + \overrightarrow{CX} = c + k a - \frac{3k}{4} c = k a + \left( 1 - \frac{3k}{4} \right) c \) - Equate coefficients of a and c: \( \frac{3h}{4} = k \) and \( \frac{1}{4} - \frac{h}{4} = 1 - \frac{3k}{4} \) - From first: \( k = \frac{3h}{4} \) - Substitute into second: \( \frac{1}{4} - \frac{h}{4} = 1 - \frac{3}{4} \times \frac{3h}{4} = 1 - \frac{9h}{16} \) - Multiply both sides by 16: \( 4 - 4h = 16 - 9h \) - Rearrange: \( -4h + 9h = 16 - 4 \Rightarrow 5h = 12 \Rightarrow h = \frac{12}{5} = 2.4 \) - Then \( k = \frac{3}{4} \times 2.4 = 1.8 \) **Final answers:** (i) \( \overrightarrow{ED} = \frac{3}{4} a - \frac{1}{4} c \) (ii) \( \overrightarrow{CB} = a - \frac{3}{4} c \) (iii) \( \overrightarrow{EX} = \frac{3h}{4} a - \frac{h}{4} c = \frac{18}{20} a - \frac{12}{20} c \) (iv) \( \overrightarrow{CX} = k a - \frac{3k}{4} c = 1.8 a - 1.35 c \) (v) Scalars: \( h = \frac{12}{5} \), \( k = \frac{9}{5} \)