Additional Mathematics 0606 / Vectors / 3.453.45: Vector geometry: lines PQPQPQ and OBOBOB0606/12/M/J/25 — Question 11 · 9 marksMark as done · Save for later · Show all solutions(a)In triangle OABOABOAB, OA→=a\overrightarrow{OA}=\mathbf{a}OA=a and OB→=b\overrightarrow{OB}=\mathbf{b}OB=b. Point PPP on OAOAOA has OP→=34OA→\overrightarrow{OP}=\tfrac{3}{4}\overrightarrow{OA}OP=43OA. Point QQQ on ABABAB has AQ→=13AB→\overrightarrow{AQ}=\tfrac{1}{3}\overrightarrow{AB}AQ=31AB. Line PQPQPQ meets OBOBOB at RRR with OR→=mb\overrightarrow{OR}=m\mathbf{b}OR=mb and PR→=nPQ→\overrightarrow{PR}=n\overrightarrow{PQ}PR=nPQ.Find OR→\overrightarrow{OR}OR in terms of a\mathbf{a}a, b\mathbf{b}b and nnn.[6]▸ Answer(b)Hence find the values of mmm and nnn.[3]▸ Answer▸ Official mark scheme← 3.21: Displacement RQ and linear dependence of vectors3.44: Vector section formula; collinear points →