The point A has position vector <3, 2> and point B has one of <1, 3>. Find the position vector of the point that divides AB in the ratio 4:3.
I have attempted this question by manipulating the position vector values of A and B, but to no avail. Could someone show me how these sorts of questions are approached, as I am unacquainted with these sorts of vector questions?
I would also like to note that I have attempted to work backward from the answer, which is $<13/7, 18/7>$ but I simply do not understand how they obtained those numbers; I was able to come up with the numbers with the ones at hand by do not know what they represent. If A is multiplied by $<4, 3>$, one obtains $18/7$ and if B is multiplied by $<3, 4>$, $13/7$ is obtained, I believe this may be coincidence though.
The goal is to find a point $C$ on the line segment $AB$ such that $|AC|:|CB|=4:3$.
Equivalently, we want $C$ on $AB$ to be such that $|AC|=4s$ and $|CB|=3s$, for some $s > 0$.
Then $$|AC|:|AB|=\frac{|AC|}{|AB|}=\frac{|AC|}{|AC|+|CB|}=\frac{4s}{4s+3s}=\frac{4s}{7s}=\frac{4}{7}=4:7$$ hence we must have \begin{align*} \vec{C} &=\vec{A}+{\small{\frac{4}{7}}}\vec{AB}\\[4pt] &=\langle{3,2}\rangle+{\small{\frac{4}{7}}}\langle{-2,1}\rangle\\[4pt] &=\langle{{\small{\frac{13}{7}}},{\small{\frac{18}{7}}}}\rangle\\[4pt] \end{align*} so $C=\left({\large{\frac{13}{7}}},{\large{\frac{18}{7}}}\right)$.