ABCD is a parallelogram. Points $N$ and $M$ lie on $CO$ and $BC$, respectively, such that $$CN:NO=CM:MB=3:1.$$ If $\vec{AB}=\vec{a}$ and $\vec{AD}=\vec{b}$, then what $\vec{MN}$ is equal to?
We can write $\vec{MN}$ as $$\vec{MN}=\vec{CN}-\vec{CM}.$$ Now $$CN=\dfrac{3}{4}\vec{CO}=\dfrac{3}{4}.\dfrac{1}{2}\vec{CA}=\dfrac{3}{8}\vec{CA}=-\dfrac{3}{8}\vec{AC}=-\dfrac{3}{8}\left(\vec{a}+\vec{b}\right)$$ $$\vec{CM}=\dfrac{3}{4}\vec{CB}=-\dfrac{3}{4}\vec{BC}=-\dfrac{3}{4}\vec{b}$$ We got that $$\vec{MN}=-\dfrac{3}{8}\left(\vec{a}+\vec{b}\right)-\dfrac{3}{4}\vec{b}=-\dfrac{3}{8}\vec{a}-\dfrac{9}{8}\vec{b}$$ The given answer in my book is $$\vec{MN}=\dfrac{3}{8}\left(\vec{b}-\vec{a}\right)$$