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I was wondering how I would be able to find point p, shown in the following graphic:

enter image description here

We have a triangle in 3d space defined by the points $a, b$ and $c$. We want to find a point $p$ on the line segment between $a$ and $b$. However, this point is on a line that is perpendicular to $ab$ and the point $c$ also happens to be on this line. The image above describes the point $p$ perfectly well.

I will share the method I plan on trying to solve this problem now and check me if my terminology or ideas are incorrect if you wish:

1)Get the normal vector $n_1$ by crossing $(c-a)$ with $(c-b)$. Normalize n1?

2)Get the "second" normal vector $n_2$ by crossing $n_1$ with $(b-a)$. This second normal vector should be parallel to the line $pc$. Normalize $n_2$?

3)Create the parametric equation for the line $pc$ using $c$ as the reference point. Make sure this parametric equation is using normalized vectors?

4)Create the parametric equation for the line $ab$ using $a$ or $b$ as the reference point. Make sure this parametric equation is using normalized vectors?

5)Make these two parametric equations equal to each-other and solve for a $t$ value which can be used to plug back into either equation to get the $x, y$ and $z$ coordinates of $p$. For this step, I am looking at this problem to guide me through this process: Find intersection of two 3D lines

6)I'm not sure how, but I'm pretty sure that that the dot product of $ab$ and $pc$ is equal to $0$ because the two lines are perpendicular -- $(ab)\cdot (pc) = 0$. I think this dot product might also be used in solving this problem, or, perhaps it is unnecessary, perhaps only making the two lines intersect is enough.

Thank you for reading and thank you for any of your responses!

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Your solution is overkill. Notice that $P$ is the projection of $C$ onto $\overline{AB}$. Do you think you can figure it out from there?

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  • $\begingroup$ Ah, wow, thanks. I should have it from here -- if I don't, I'll ask for you, but I should have it now, so... thank you CulDeVu! I'm new here so if there is a way I can give you points let me know. $\endgroup$ – doobie Jan 4 '17 at 23:20
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True credit for the answer to this question goes to CulDeVu, and I am simply following up with an image that makes his correct answer more explicit.

Answer to Question

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  • $\begingroup$ You can use MathJax instead of an image; for example, $$\vec{p} = \vec{a} + \left ( \vec{b} - \vec{a} \right ) \frac{ \left ( \vec{c} - \vec{a} \right ) \cdot \left ( \vec{b} - \vec{a} \right ) } { \left ( \vec{b} - \vec{a} \right ) \cdot \left ( \vec{b} - \vec{a} \right ) }$$. $\endgroup$ – Nominal Animal Jan 5 '17 at 4:24

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