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Given three points $a, b, c\in \mathbb R^3$, how do we find the equation of the plane passing through these three points?

Original Question : Let $p : \mathbb{R} \to \mathbb{R}^3$ be a twice differentiable parametric curve, which is nowhere straight. For arbitrary h find the equation of the plane that goes through the three points $p(t)$, $p(t+h)$, $p(t−h)$. Then find the equation of the limit plane that one obtains from letting $h\to 0$. Verify that the tangent and normal unit vectors at $p(t)$ lie on that plane. I just want to know how to form the equation of the plane using the three coordinates provided. (Please no complete answer to the above question.)

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  • $\begingroup$ Nice question, +1. I added the "differential geometry" tag. $\endgroup$ – Robert Lewis Sep 28 '17 at 5:11
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    $\begingroup$ @RobertLewis is the answer $$r=a+(b-a)t+(c-a)s$$? $\endgroup$ – mathnoob123 Sep 28 '17 at 5:13
  • $\begingroup$ @RobertLewis Sorry that I made another changes to the question and took away the tag. The first version is just too misleading, the OP does not want ppl to solve that DG question.... $\endgroup$ – user99914 Sep 28 '17 at 5:13
  • $\begingroup$ @FaiqRaees Have you tried searching for "equation of plane passing through 3 points"? Are you looking for an implicit equation? A parameterization? $\endgroup$ – Kyle Miller Sep 28 '17 at 5:32
  • $\begingroup$ @KyleMiller please look at the previous edited version of the question $\endgroup$ – mathnoob123 Sep 28 '17 at 5:33
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The parametric equation of the plane passing through the three points $p(t)$, $p(t+h)$, $p(t−h)$ is $$Q_{t,h}(r,s):=p(t)+\frac{p(t+h)-p(t)}{h}\cdot r+\frac{p(t-h)-p(t)}{-h}\cdot s$$

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  • $\begingroup$ Can you please look at the previously edited version of the question? And respond in regards to that? $\endgroup$ – mathnoob123 Sep 28 '17 at 5:23
  • $\begingroup$ @Faiq Raees I have re-edited the question, otherwise it is hard to understand what' s going on. $\endgroup$ – Robert Z Sep 28 '17 at 5:39
  • $\begingroup$ Can you please tell why divide by h? Because I have learnt that if a,b,c are the vector points then the plane passing through them have the equation $$r=a+(b-a)s+(c-a)t$$ $\endgroup$ – mathnoob123 Sep 28 '17 at 5:39
  • $\begingroup$ Because $r$ and $s$ are free parameters. Note that $Q_{t,h}(r,s):=p(t)+({p(t+h)-p(t)})\cdot r+({p(t-h)-p(t)})\cdot s$ is another parametric equation of the same plane. $\endgroup$ – Robert Z Sep 28 '17 at 5:42
  • $\begingroup$ Can you answer one last question? Is dividing by h essential to solve this question? Is it done to calculate the limit plane? On that note, can you clearing what a limit plane is? $\endgroup$ – mathnoob123 Sep 28 '17 at 5:44
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First, assume that the 3 points are not co-linear. The normal to the plane is then nontrivial: $\mathbf n = (\mathbf b - \mathbf a) \times (\mathbf c - \mathbf a)$.

Let $\mathbf r$ be a point on the plane. The equation of the plane is:

$$ (\mathbf r - \mathbf a)\cdot \mathbf n = 0, $$

or explicitly:

$$ (\mathbf r - \mathbf a)\cdot \big( \mathbf (\mathbf b - \mathbf a) \times (\mathbf c - \mathbf a) \big)= 0. $$

Intuition: $(\mathbf r - \mathbf a)$ is a vector parallel to the plane, thus its dot product with the plane's normal has to vanish.

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  • $\begingroup$ I did not see the original question, so my answer is to the general question $\endgroup$ – MMM Sep 28 '17 at 5:47

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