# 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.

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.

• Thanks for the prompt help! – Ramana Sep 27 at 9:17

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{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}\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{aligned} \vec{OP} & = \vec{OA} + t\left(\vec{AC}\right) \\ & = (1 + 3t)\mathbf{i} + (-3t)\mathbf{j} + (-1 + 3t)\mathbf{k} \end{aligned}\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{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}\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.

• Thanks a lot, John. I'll have to spend more time understanding the second method. Interesting. – Ramana Sep 27 at 9:16
• @Ramana You're welcome. The second method is basically just a generalization of the first one, using a parameter $t$ to represent any vector from the origin to a point along $\vec{AC}$. – John Omielan Sep 27 at 13:16