Additional Mathematics 0606 / Vectors / 3.443.44: Vector section formula; collinear points0606/13/M/J/25 — Question 10 · 8 marksMark as done · Save for later · Show all solutions(a)Points AAA, BBB, CCC are collinear with AC→=13AB→\overrightarrow{AC}=\tfrac{1}{3}\overrightarrow{AB}AC=31AB. Given OA→=a\overrightarrow{OA}=\mathbf{a}OA=a and OB→=b\overrightarrow{OB}=\mathbf{b}OB=b, find OC→\overrightarrow{OC}OC in terms of a\mathbf{a}a and b\mathbf{b}b.[3]▸ Answer(b)Line OAOAOA is extended to DDD with OA:AD=2:7OA:AD=2:7OA:AD=2:7. Point EEE on CDCDCD has OE→=mb\overrightarrow{OE}=m\mathbf{b}OE=mb. Find mmm.[5]▸ Answer▸ Official mark scheme← 3.45: Vector geometry: lines PQPQPQ and OBOBOB3.37: Parallelogram vector ratio →
