$ABC$ is a random triangle. $M$ is from $BC$ and $CM:MB=3:5$. Point $P$ divides $AM$, $AP:PM=2:3$. $BP$ intersects $AC$ in $Q$. Find $AQ:QC$ by using vectors.
I know how to solve the problem with Minelay or with the centre of mass but I need a solution using the vectors. Personally I strart with basic vectors $\overrightarrow {CB}$ and $\overrightarrow {CA}$ which form my space.
Fix one point $O$ in the plane. (The forthcoming relations do not depend on the choice.)
By definition, the point $M$ satisfies $$ \overrightarrow {OM} = \frac 38\overrightarrow {OB} + \frac 58\overrightarrow {OC} \ . $$ Then $$ \overrightarrow {OP} = \frac 35\overrightarrow {OA} + \frac 25\overrightarrow {OM} = \frac 35\overrightarrow {OA} + \frac 25\cdot \frac 38\overrightarrow {OB} + \frac 25\cdot \frac 58\overrightarrow {OC} \ . $$ Now for some $\lambda\in \Bbb R$ we have that $$ \overrightarrow {OQ} = \lambda \overrightarrow {OP}+ (1-\lambda) \overrightarrow {OB} \ , $$ and this expression does not depend on $\overrightarrow {OB}$ when written in terms of the vectors $\overrightarrow {OA}$, $\overrightarrow {OB}$, $\overrightarrow {OC}$. This gives the equation in $\lambda$ $$ \frac 25\cdot \frac 38\lambda +(1-\lambda)=0\ , $$ the solution is $\lambda = 20/17$, we plug it in, obtain $$ \overrightarrow {OP} = \frac{20}{17} \left( \frac 35\overrightarrow {OA} + \underbrace{\frac 25\cdot \frac 58}_{=1/4}\overrightarrow {OC} \right) \ . $$ As seen, the special value of $\lambda$ and the whole previous step is not needed, it is enough to use the weights $\frac 35$ and $\frac 14$, bring them to a common denominator, $\frac 35=\frac{12}{20}$ and $\frac 14=\frac{5}{20}$ (having the sum...), and the proportion is the one of the numerators.
Check using Menelaos for $\Delta ACM$ and the line $QMB$: $$ \frac{QA}{QC}\cdot \frac{BC}{BM}\cdot \frac{PM}{PA} =1\ , $$ explicitly $$ \frac{QA}{QC}\cdot \frac{8}{5}\cdot \frac{3}{2} =1\ . $$