The equation of the plane [closed]

Question

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 $D$.

Answer 1

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),

$a=4d,c=7d,b=2c-3a=2(7d)-3(4d)=2d$

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

Answer 2

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 $$ 2x+3z=5 $$

Answer 3

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.

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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.

Answer 4

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*}