Talk:Distance from a point to a plane

Point not at the origin

What if the origin is not (x,y,z) (0,0,0)?

I don't think it's in the scope of the article. I could use some help understanding things as they are, but if you can find it for the origin you can shift things to other locations. Kinda like this:

Vector x=//the point you're looking for

Plane shifted=(plane_in-x);//shift the plain so the point you're looking for is on the origin relative to the plain.

Vector result=shifted.closest_to_origin()+x;//compute for our translated plain then shift it back.

Anyone else seeing this feel free to delete my post especially if some time has gone by. I was just trying to help the original poster.

Space fountain (talk) 18:20, 19 April 2013 (UTC)[reply]

I think the general case of a point not at the origin should be added to the end, what's already in the article should be oriented to both the nearest point and its distance rather than just nearest point, and the article should be moved to Distance from a point to a plane. That topic is more likely to be of interest to readers. Loraof (talk) 15:54, 22 September 2014 (UTC)[reply]

Done. Loraof (talk) 16:35, 23 September 2014 (UTC)[reply]

Why did user Wcherowi remove my appendage to this page ?

He/ She claims the algebra is incorrect, it is correct. He/ She claims this my appendage is original research and not in good faith. My post is certainly original as far as including something as of yet not mentioned on the page, which is a very useful derivation, and it's certainly in good faith. — Preceding unsigned comment added by Zmilne (talk • contribs) 01:23, 8 December 2014 (UTC)[reply]

If you look more closely at my edit summary you will notice that it says your edit was done in good faith. Wikipedia's definition of original research (WP:OR) is very wide and if you can not provide a citation to a reliable secondary source, a derivation such as this is considered original research and will be removed. My claim about the incorrectness of the algebra was an inference. After you eliminated the variable x from D, when taking the partial derivatives you do not obtain three equations in three unknowns as you only have two unconstrained variables and I do not know how you obtained the results you claimed. A correct method via Lagrange multipliers was in an earlier version of this article, and I'll copy it here:

In Euclidean 3-space we will find the point on an arbitrary plane that is closest to the origin using the method of Lagrange multipliers.

First, let us start with an arbitrary plane, ax + by + cz = d. The distance, L, from the origin to a point (x,y,z) on the plane is given by:

$L={\sqrt {x^{2}+y^{2}+z^{2}}}.$

Therefore the function that we want to minimize is: $f(x,y,z)={\sqrt {x^{2}+y^{2}+z^{2}}}.$

Our one constraint on x, y, and z is that the point (x,y,z) must lie on the given plane. Thus, we define g(x,y,z) = ax + by + cz - d.

Next we define a new function with a Lagrange multiplier, $\lambda$

$f^{*}=f(x,y,z)-\lambda g(x,y,z)={\sqrt {x^{2}+y^{2}+z^{2}}}-\lambda (ax+by+cz-d).$

Take the partial of $f^{*}$ with respect to x, y, and z and set each to zero.

${\frac {\partial f^{*}}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}+z^{2}}}}-\lambda a=0$

${\frac {\partial f^{*}}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}+z^{2}}}}-\lambda b=0$

${\frac {\partial f^{*}}{\partial z}}={\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}-\lambda c=0$

Now each partial includes a $\lambda$ and a ${\sqrt {x^{2}+y^{2}+z^{2}}}$ term.

If we solve each equation for $\lambda {\sqrt {x^{2}+y^{2}+z^{2}}}$ and set them equal to one another

we can find the relation:

${\frac {a}{x}}={\frac {b}{y}}={\frac {c}{z}}.$

From this we can obtain y and z as functions of x:

$y={\frac {bx}{a}}$ and $z={\frac {cx}{a}}.$

Substitute these for y and z in the equation of the plane and solve for x to obtain:

$x={\frac {ad}{a^{2}+b^{2}+c^{2}}}.$

With this x you can solve for y and z:

$y={\frac {bd}{a^{2}+b^{2}+c^{2}}},$

and

$z={\frac {cd}{a^{2}+b^{2}+c^{2}}}.$

Hence the point on the plane closest to the origin is:

$(x,y,z)={\Big (}{\frac {ad}{a^{2}+b^{2}+c^{2}}},{\frac {bd}{a^{2}+b^{2}+c^{2}}},{\frac {cd}{a^{2}+b^{2}+c^{2}}}{\Big )}$

and the distance is given by:

$L={\sqrt {x^{2}+y^{2}+z^{2}}}={\frac {|d|}{\sqrt {a^{2}+b^{2}+c^{2}}}}.$

Notice also that the author of the above was careful about explaining what was being done, something that was missing in your addition. This was taken out of the article, I think, because the computation wasn't considered to be germane to the topic of the article (only the final result is). I would consider replacing it, but only if a citation could be found (this should not be difficult since this is a standard technique and I am sure that several textbooks would use this problem as an example). If you find a reliable secondary source for your calculation (it is possible to do the problem your way, just not the way you claimed to finish it) then I would be happy to reinstate and support a corrected version, but until then I will just revert again for the same reasons I gave before. Bill Cherowitzo (talk) 06:27, 8 December 2014 (UTC)[reply]

OK. Your explanation of why you removed my appendage makes sense to me. I was rushed and just wanted to throw something up here before I ran out of time to do so. I will say that my solution is far more simple than the Lagrange multiplier method, and is therefore more elegant. The reason I can't find a citation is because the quick search methods (Google scholar, scopus, etc..) don't readily yield one for this derivation.

Additionally, to counter your argument that there are only 2 equations and 3 variables in the end I say that this is incorrect: once you've eliminated one of the variables from "d" (using the equation for the plane) you can create two new independent equations by taking the partials with respect to the remaining two variables respectively; set these equal to zero. This gives you two equations, I agree. The third equation is the original constraint: the equation of the plane. It all works out in the end ;) — Preceding unsigned comment added by Zmilne (talk • contribs) 00:55, 10 December 2014 (UTC)[reply]

Original work vs. substitutions into cited formulae

Irked by the lack of formulae for arbitrary points, I've added a section (source cited) for the closest-point and distance formula for arbitrary points and n-dimensional hyperplanes. I've taken the liberty of substituting n=3 into the cited equations immediately afterwards to get the equations for $R^{3}$ , but daren't copy or cite my substitution elsewhere in the article in case this counts as original work. I understand the policy against original work clearly, e.g., in the case or original derivations of results or chains of argument, but at some level of mathematical literacy sanity must kick in to allow simple substitutions into (or trivial rearrangement of) well-known or cited formulae. Does anyone have any thoughts on this slippery slope? - Frentos (talk) 10:20, 12 January 2016 (UTC)[reply]

I feel your pain, although I would say that such substitutions fall under WP:SYNTH rather than WP:OR since the result reaches a conclusion that is not directly stated in the source. I tend to be a bit hard-nosed about OR in math articles, but I see more grey areas with synthesis. Using some source's words or formulae to draw your own conclusions can not be allowed on WP, but to object to plugging in to a formula seems to violate the raison d'etre of having the formula in the first place. Certainly there are editors who can not perform simple computations correctly, but these mistakes are easily taken care of. On the other hand, should I, as an editor, feel obligated to verify everyone else's manipulations? Doesn't doing this require some skill (expertise) on the part of the editor? This would appear to run counter to the general philosophy behind WP policies. I don't have a ready answer to this issue and will continue to deal with it on a case by case basis. As to the current article, since the result was obtained by other means (and, theoretically, sourced) I do not find this synthesis objectionable (and it is also true that if you looked hard enough you could find a source that actually plugged the value in). Bill Cherowitzo (talk) 18:44, 12 January 2016 (UTC)[reply]