Finding the position vector of a point which is related to two other known vectors? 
I have been stuck with this problem for so long. I have absolutely no idea how to find the position vector of N. I tried finding the lengths of NC, and AN, but only in vain. I don't think the moduli are going to help me in any way. 
I just need help with the method to use to find the position vector of N.
Please help.
 A: With $AN = 2NC$, this means $N$ lies two-thirds of the distance between $A$ and $C$, so it's co-ordinates would be those of $\vec{OA}$ plus two-thirds of $\vec{AC}$ (which is the difference between $\vec{OC}$ and $\vec{OA}$ since $\vec{AC} = \vec{AO} + \vec{OC} = \vec{OC} - \vec{OA}$). This gives
$$\begin{equation}\begin{aligned}
\vec{ON} & = \vec{OA} + \frac{2}{3}\left(\vec{OC} - \vec{OA}\right) \\
& = \frac{1}{3}\vec{OA} + \frac{2}{3}\vec{OC} \\
& = \left(\frac{1}{3} + \frac{8}{3}\right)\mathbf{i} - \frac{6}{3}\mathbf{j} + \left(\frac{-1}{3} + \frac{4}{3}\right)\mathbf{k} \\
& = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
An alternative way to determine the result is to use a parametric equation for $\vec{OP}$ for any point $P$ on $\vec{AC}$ being
$$\begin{equation}\begin{aligned}
\vec{OP} & = \vec{OA} + t\left(\vec{AC}\right) \\
& = (1 + 3t)\mathbf{i} + (-3t)\mathbf{j} + (-1 + 3t)\mathbf{k}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
with $\vec{OA}$ corresponding to $t = 0$ and $\vec{OC}$ corresponding to $t = 1$. Thus, $\vec{ON}$ corresponds to $t = \frac{2}{3}$ giving
$$\begin{equation}\begin{aligned}
\vec{ON} & = \left(1 + 3\left(\frac{2}{3}\right)\right)\mathbf{i} + \left(-3\left(\frac{2}{3}\right)\right)\mathbf{j} + \left(-1 + 3\left(\frac{2}{3}\right)\right)\mathbf{k} \\
& = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This, of course, gives the same result as \eqref{eq1A}.
Since you only ask for this result, so I trust you can finish the rest of the problem on your own.
A: Note that since $N$ lies on $AC$ between $A$ and $C$, $AN=2NC$ implies that $\vec{AN}=\frac{2}{3}\vec{AC}.\;$ Thus we have
$\begin{align} \vec{ON} &= \vec{OA} + \vec{AN}\\
&= \vec{OA} + \frac{2}{3}\vec{AC}\\
&= (\mathbf{i}-\mathbf{k}) + \frac{2}{3}(3\mathbf{i}-3\mathbf{j}+3\mathbf{k})\\
&= (\mathbf{i}-\mathbf{k}) + (2\mathbf{i}-2\mathbf{j}+2\mathbf{k})&= \boxed{3\mathbf{i}-2\mathbf{j}+\mathbf{k}}\\
\end{align}$
Equivalently, $AN=2NC$ also implies that $\vec{CN}=\frac{1}{3}\vec{CA}$, and then
$\begin{align} \vec{ON} &= \vec{OC} + \vec{CN}\\
&= \vec{OC} + \frac{1}{3}\vec{CA}\\
\end{align}$
which yields the same result.
