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Vectors_problem_1

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.

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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.

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  • $\begingroup$ Thanks for the prompt help! $\endgroup$ – Ramana Sep 27 at 9:17
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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.

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  • $\begingroup$ Thanks a lot, John. I'll have to spend more time understanding the second method. Interesting. $\endgroup$ – Ramana Sep 27 at 9:16
  • $\begingroup$ @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 $\vec{AC}$. $\endgroup$ – John Omielan Sep 27 at 13:16

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