Additional Mathematics 0606 / Vectors / 3.463.46: Vector ratio in triangle OABOABOAB0606/22/M/J/23 — Question 10 · 8 marksMark as done · Save for later · Show all solutionsIn triangle OABOABOAB, CCC is the mid-point of OAOAOA, and DDD on CBCBCB satisfies CD→:DB→=2:3\overrightarrow{CD}:\overrightarrow{DB}=2:3CD:DB=2:3. Given OC→=c\overrightarrow{OC}=\mathbf{c}OC=c and CB→=b\overrightarrow{CB}=\mathbf{b}CB=b, the point EEE on ABABAB satisfies OE→=mOD→\overrightarrow{OE}=m\overrightarrow{OD}OE=mOD and AE→=nAB→\overrightarrow{AE}=n\overrightarrow{AB}AE=nAB.Find two expressions for OE→\overrightarrow{OE}OE and hence find AEEB\dfrac{AE}{EB}EBAE.[8]▸ Answer▸ Official mark scheme← 3.19: Collinear points and resultant of bearing vectors3.4: Particle path and section formula in a triangle →