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In geometry 3D, let $A(1,2,1)$, $B(-2,1,3)$, $C(2,-1,1)$, $D(0,3,1)$ be four points. Write the equation of the planes $(P)$ which passes through the points $A$, $B$ and equidistant from the two points $C$ and $D$.

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closed as not a real question by Thomas, Norbert, Did, tomasz, J. M. Oct 7 '12 at 13:25

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Wouldn't the midpoint of CD just be your third point after A and B? –  cheepychappy Oct 2 '12 at 14:45
This question needs to be edited or closed. As it stands it doesn't make sense if the equation one plane is being asked for or the equation of two planes. –  Thomas Oct 2 '12 at 16:48

4 Answers 4

up vote 2 down vote accepted

The equation of any plane passing through A$(1,2,1)$ is $a(x-1)+b(y-2)+c(z-1)=0-->(1)$

As it passes through B$(-2,1,3), a(-2-1)+b(1-2)+c(3-1)=0\implies b=2c-3a$

Now the perpendicular distance of C$(2,-1,1)$ from $a(x-1)+b(y-2)+c(z-1)=0$ is $$\pm\frac{a-3b}{\sqrt{a^2+b^2+c^2}}=\pm\frac{10a-6c}{\sqrt{a^2+b^2+c^2}}$$

and that of D$(0,3,1)$ from $a(x-1)+b(y-2)+c(z-1)=0$ is $$\pm\frac{-a+b}{\sqrt{a^2+b^2+c^2}}=\pm\frac{2c-4a}{\sqrt{a^2+b^2+c^2}}$$

$\implies 10a-6c=\pm(2c-4a)$

Consider both cases to express $b,c$ in terms of $a$ and their values in $(1)$

Taking '+', $ 10a-6c=2c-4a\implies 14a=8c\implies\frac a 4=\frac c 7=d$(say),


$4d(x-1)+2d(y-2)+7d(z-1)=0\implies 4(x-1)+2(y-2)+7(z-1)=0$ as $d \ne 0$

Taking '-', $10a-6c=-(2c-4a)\implies 6a=4c, \frac c 3= \frac a 2= e $(say),

$a=2e,c=3e, b=2c-3a=2(3e)-3(2e)=0$

$2e(x-1)+0(y-2)+3e(z-1)=0\implies 2(x-1)+3(z-1)=0$ as $e \ne 0$

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Please check your solution. –  minthao_2011 Oct 2 '12 at 15:38
@minthao_2011, could you please elaborate? –  lab bhattacharjee Oct 2 '12 at 15:47
Dear Lab bhattacharjee, We have $(P): a(x-1)+b(y-2)+c(z-1)=0$. Because, the distance from the point $C$ to the plane $(P)$ equal to the distance from the point $D$ to the plane $(P)$, therefore \begin{equation*} \dfrac{|a - 3b|}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{|-a + b|}{\sqrt{a^2 + b^2 + c^2}} \end{equation*} and then \begin{equation*} |a - 3b| = |-a + b|. \end{equation*} From here, we get $b = 0$, or $a = 2b$. –  minthao_2011 Oct 2 '12 at 16:20

The plane passes through the points $A(1,2,1)$, $B(-2,1,3)$ and the average of $C(2,-1,1)$ and $D(0,3,1)$, that is $E(1,1,1)$. The normal the plane would be $$ (A-E)\times(B-E)=(0,1,0)\times(-3,0,2)=(2,0,3) $$ The equation of the plane would then be $$ (x,y,z)\cdot(2,0,3)=(1,1,1)\cdot(2,0,3)=5 $$ that is $$ 2x+3z=5 $$

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If you have two points $C(2,-1,1)$ and $D(0,3,1)$ and you consider the set consisting of all the points that have the same distance to $C$ and $D$, then you get a plane. There are several ways to do this. One way would be to consider:

$$ (x - 2)^2 + (y + 1)^2 + (z - 1)^2 = (x - 0)^2 + (y - 3)^2 + (z - 1)^2. $$

You can "solve" (reduce) this equation and get an equation of a plane.

If this plane contains the points $A$ and $B$, then you are done. If it does not contain $A$ or $B$, then no plane satisfies the conditions listed.

If you meant that you want the plane that contains $A$, $B$, and the point equidistant from $C$ and $D$, then you would just find the midpoint (as suggested in the comments) of the line segment $CD$. Then you have reduced the problem to finding the equation of a plane that passes through three points.

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Please check your solution. –  minthao_2011 Oct 2 '12 at 14:49
@minthao_2011: I am not sure that I understand... –  Thomas Oct 2 '12 at 14:50
There is still a plane. –  minthao_2011 Oct 2 '12 at 14:52
@minthao_2011: I still don't get what you are trying to say. You want an equation of the plane... so just one plane. Try to work on the equation that I have in my answer... –  Thomas Oct 2 '12 at 14:53
The second plane passes through the points $A$, $B$ and parallel to the line $CD$. –  minthao_2011 Oct 2 '12 at 14:58

Let $M(1,1,1)$ be midpoint of the segment $CD$. There are two needing planes:

First plane passes the three points $A$, $B$, $M$.

We have $\overrightarrow{AB} = (-3, -1, 2) $, $\overrightarrow{AM} = (0, -1, 0) $. The crossproduct of the two vectors $\overrightarrow{AB} $ and $\overrightarrow{AM}$ is $(2, 0, 3)$. Therefore, equation of this plane is $$2x + 3z - 5 = 0.$$

Second plane passes the two points $A$, $B$ and parallel to the line $CD$. We have $\overrightarrow{AB} = (-3, -1, 2) $, $\overrightarrow{CD} = (-2, 4, 0) $. The crossproduct of the two vectors $\overrightarrow{AB} $ and $\overrightarrow{CD}$ is $(-8, -4, -14)$ The equation of this plane is \begin{equation*} -8(x-1) - 4(y - 2) - 14(z - 1) = 0 \Leftrightarrow 4x + 2y + 7z - 15 = 0. \end{equation*}

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Again, you might want to edit your question so that it is clear what you are asking about. The way you have worded things now, it sounds like you are looking for the equation of one plane. But with this answer it seems like you want two planes. –  Thomas Oct 2 '12 at 15:44
I have just edited. –  minthao_2011 Oct 2 '12 at 16:04

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