How to build 2 planes from known 3 points and distance between them?

Question

There're 2 parallel planes $P_1, P_2$ and the distance between them is 2. $P_1$ goes through points $A=(2,0,3)$ and $B=(0,0,6)$ and $P_2$ goes through point $C=(-2,0,2)$. Find the equations of the planes.

I thought of the following although the solution seems not correct and too long. First this is the visualization that I made and I hope is correct:

First we can find $\underline u$ the normal vector of vectors $BA$ and $BC$ through cross product which is:

$$\underline u=<2,0,-3> \times <-2,0,-4>=<0,14,0>$$

Now we can find $\underline n_1$ the normal vector of plane $P_1$ from: $$\underline n_1=\underline u \times BA=<0,14,0> \times <2,0,-3>=<42,0,28>$$ Now we have the plane equation of $P_1: 42x+28z+d=0$ after dividing by $14$ becomes: $3x+2z+{d \over 14}=0$.

I understand the normal vectors for $P_1$ and $P_2$ are the same but $d$'s are different. I could calculate the $d$ from the distance formula: $$ D=\frac{|3x+2z+d|}{\sqrt{3^2+2^2}}=2 $$

But then how would I find the $d$ for $P_2$?

Answer 1

Your solution is wrong.

Look carefully at what you did. First you calculated vector $u$ orthogonal to the plane $ABC$. Then you calculated vector $n_1$ orthogonal to $u$ and $AB$. But, that vector lies in the plane $ABC$, so you have that the distance of $C$ from $P_1$ is the distance of $C$ from $AB$, and you can easily see that it is not equal to $2$.

What you can do is denote by $C'$ orthogonal projection of $C$ on the plane $P_1$. Let $n = C-C'$, $v= A - C'$, $w= B - C'$.

We know that $n\perp v, n\perp w$ and $\|n\| = 2$, i.e.

\begin{align}\langle n,v\rangle &= 0\\ \langle n,w\rangle &= 0\\ \langle n,n\rangle &= 4\end{align}

and if we write $C'=(x,y,z)$, the above becomes

\begin{align} (x-2)(x+2)+y^2+(z-3)(z-2)&=0\\ x(x+2)+y^2+(z-6)(z-2) &= 0 \\ (x+2)^2+y^2+(z-2)^2 &= 4.\end{align}

Finally, substitute \begin{align} x' &= x+2,\\ y' &= y,\\ z' &= z-2,\end{align} to get

\begin{align} x'^2+y'^2+z'^2& =4x'+z'\\ x'^2+y'^2+z'^2& =2x'+4z'\\ x'^2+y'^2+z'^2& =4\\ \end{align}

which can now easily be solved to get $C'=(-\frac 8 7, \pm\frac{12} 7, \frac{18}7)$. $P_1$ is the plane through $A$, $B$, $C'$, while $P_2$ is $P_1$ translated by $n$. Hopefully, you can write the equations yourself from here.

Notice that there are actually two solutions to the problem. How come? Well, geometrically, we have a sphere with center $C$ and radius $2$, and we are looking for tangent plane passing through line $AB$. There are two such planes (analogous to finding tangent line of a circle through a point outside of it).

Answer 2

First of all, the normal vector $n$ should be penpardicular to $AB=(2,0,-3)$, solving this we know that $n$ has the direction $l=(3a,b,2a)$ for $(a,b)\neq(0,0)$. Denote the segment for distance between two planes from the starting point $C$ and ending point $D$ by $CD=2*n=2*(3a/\sqrt{13a^2+b^2},b/\sqrt{13a^2+b^2},2a/\sqrt{13a^2+b^2})$, then $D=C+2l/\|l\|=(-2+6a/\sqrt{13a^2+b^2},2b/\sqrt{13a^2+b^2},2+4a/\sqrt{13a^2+b^2})$. On the other hand, the segment $CD$ should also be penpardicular to $DA$. If u now consider the inner product of $CD$ and $DA$, then it is an equation of $a$ and $b$ which equals zero (at most, u will solve a polynomial equation of $a$ and $b$ of order 4). Solving this you will find at the end the relation between $a$ and $b$, and I think the solution must not be unique, at least you can imagine that if you have one solution, you can always with help of symmetry to construct another solution.

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Update:write the inner product of $CD$ and $DA$ explicitely, we come to \begin{equation} 1872a^4+92a^2b^2-4b^4=0. \end{equation} Solving this we get $b=\pm 6a$. Therefore the normal vector of the planes is $(\frac{3}{7},\pm\frac{6}{7},\frac{2}{7})$. With this in hand, the next steps are clear.