Additional Mathematics 0606 / Vectors / 3.423.42: Vector line XYZXYZXYZ in triangle OABOABOAB0606/13/O/N/23 — Question 11 · 9 marksMark as done · Save for later · Show all solutions(a)In triangle OABOABOAB, OA→=a\overrightarrow{OA}=\mathbf{a}OA=a, OB→=b\overrightarrow{OB}=\mathbf{b}OB=b. The line XYZXYZXYZ satisfies OX→=45b\overrightarrow{OX}=\tfrac{4}{5}\mathbf{b}OX=54b, AY→=13AB→\overrightarrow{AY}=\tfrac{1}{3}\overrightarrow{AB}AY=31AB, AZ→=na\overrightarrow{AZ}=n\mathbf{a}AZ=na, and YZ→=mXY→\overrightarrow{YZ}=m\overrightarrow{XY}YZ=mXY.Show that XY→=23a−715b\overrightarrow{XY}=\dfrac{2}{3}\mathbf{a}-\dfrac{7}{15}\mathbf{b}XY=32a−157b.[3]▸ Answer(b)Find YZ→\overrightarrow{YZ}YZ in terms of mmm, a\mathbf{a}a and b\mathbf{b}b.[1]▸ Answer(c)Find YZ→\overrightarrow{YZ}YZ in terms of nnn, a\mathbf{a}a and b\mathbf{b}b.[2]▸ Answer(d)Hence find the values of mmm and nnn.[3]▸ Answer▸ Official mark scheme← 3.43: Vector geometry: intersection ZZZ of OYOYOY and AXAXAX3.47: Vector geometry in triangle OABOABOAB →
