Finding a position vector I am given position vectors: $\vec{OA} = i - 3j$ and $\vec{OC}=3i-j$.
And asked to find a position vector of the point that divides the line $\vec{AC}$ in the ratio $-2:3$. 
So I found the vector $\vec{AC}$, and it is $2i+2j$. Then, if the point of interest is $L$, position vector $\vec{OL} = \vec{OA} + \lambda\vec{AC}$. Where $\lambda$ is the appropriate scalar we have to apply. But I am a bit confused with the negative ratio. I tried to apply the logic of positive ratios like $3:2$. In this case, we would split the line into 5 "portions" and we would be applying a ratio of $\frac{3}{5}$. I interpret negative ratio, in the following manner. I first split the $\vec{AC}$ into 5 "portions", then I move two portions outside of the line, to the left; then I move 2 portions back and 1 in. If that makes any sense at all. Therefore the ratio to be applied should be $\frac{1}{5}$. However, I am not getting a required result.
 A: I think that one can imagine it like this: L -------- A ---- C, i.e., the length LA is 2 "units", while the total length LC is 3 "units".
A: HINT....the point dividing the line AB in the ratio $\lambda:\mu$ is $$\frac{\mu a+\lambda b}{\lambda+\mu}$$
A: Let the position vector of the point say $D$ be $\vec{OD}=ai+bj$ then we have
$$\vec{AD}=\vec{OD}-\vec{OA}=ai+bj-(i-3j)=(a-1)i+(b+3)j$$
$$\implies |\vec{AD}|=\sqrt{(a-1)^2+(b+3)^2}$$
$$\vec{CD}=\vec{OD}-\vec{OC}=ai+bj-(3i-j)=(a-3)i+(b+1)j$$
$$\implies |\vec{CD}|=\sqrt{(a-3)^2+(b+1)^2}$$
Now, the point $D$ lies on the (extended) line $AC$ then we have $\vec{AD}\parallel \vec{CD}$ then the ratio of corresponding coefficients is constant hence  $$\frac{a-1}{a-3}=\frac{b+3}{b+1}$$ 
$$\implies b=a-4\tag 1$$
Since, the point $D$ divides the line $AC$ in a ratio $-2:3$ or $2:3$ externally. Hence, we have $$\frac{|\vec{AD}|}{|\vec{CD}|}=\frac{2}{3}\implies 9|\vec{AD}|^2-4|\vec{CD}|^2=0$$ setting all the corresponding values, we get 
$$9((a-1)^2+(b+3)^2)-4((a-3)^2+(b+1)^2)=0$$
setting $b=a-4$ from (1),
$$9((a-1)^2+(a-4+3)^2)-4((a-3)^2+(a-4+1)^2)=0$$$$9(a-1)^2-4(a-3)^2=0\implies 5a^2+6a-27=0$$
$$(a+3)(5a-9)=0\implies a=-3, \ \frac{9}{5} $$
Substituting the values of $a$ in (1), we get corresponding values of $b$ as $b=-7$ & $b=-\frac{11}{5}$ respectively.
Hence, the position vector of the point $D$ is $$\color{red}{\vec{OD}=-3i-7j}$$
or $$\color{red}{\vec{OD}=\frac{9}{5}i-\frac{11}{5}j}$$
