Find a plane $\Pi'$ given two points, a plane, and a prescribed distance.

I have to work this out:

Let $\Pi:2x-y+2z=4$, $P=(2,2,2)$ and $Q=(1,0,1)$. Find a plane $\Pi'$ containing $P$ and $Q$, and a point $R$ of $\Pi$ such that $d(P,R)=d(P,\Pi)$.

Now, observe that $Q$ is in $\Pi$, so we can find $d(P,\Pi)$ as follows:

\eqalign{ d\left( {P,\Pi } \right) &=& \frac{{\left| {\left( {Q - P} \right) \cdot N} \right|}}{{\left\| N \right\|}} \cr &=& \frac{{\left| {\left( {\left( {1,0,1} \right) - \left( {2,2,2} \right)} \right) \cdot \left( {2, - 1,2} \right)} \right|}}{{\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {2^2}} }} \cr &=&\frac{{\left| {\left( { - 1, - 2, - 1} \right) \cdot \left( {2, - 1,2} \right)} \right|}}{{\sqrt 9 }} \cr &=&\frac{{\left| { - 2 + 2 - 2} \right|}}{3} = \frac{2}{3} \cr}

Now I need to find $R$. But since $R$ is in $\Pi$, it must be that $R$ is the orthogonal projection of $P$ onto $\Pi$ so to speak, or, $R$ is the intersection of $\Pi$ with the line $$\mathscr L:\bar{x} =\lambda\cdot(2,-1,2)+(2,2,2),\lambda\in \Bbb R$$

Thus, I get \eqalign{ & x \in {\mathscr L} \Rightarrow x = (2\lambda + 2,2 - \lambda ,2\lambda + 2) \cr & x \in \Pi \Rightarrow 2x - y + 2z = 4 \cr} so

$$2\left( {2\lambda + 2} \right) - \left( {2 - \lambda } \right) + 2\left( {2\lambda + 2} \right) = 4$$

or $$\lambda = - \frac{2}{9}$$

Then $$R = \left( {\frac{{14}}{9},\frac{{20}}{9},\frac{{14}}{9}} \right)$$

So now I use as the normal vector of $\Pi'$ the cross product of $\bar{PR}$ and $\bar{PQ}$:

\eqalign{ N &=& PQ \times PR = \cr & =& \left( {Q - P} \right) \times \left( {R - P} \right) \cr & =& \left( { - 1, - 2, - 1} \right) \times \left( { - \frac{4}{9},\frac{2}{9}, - \frac{4}{9}} \right) \cr & =& - \frac{2}{9}\left[ {\left( { - 1, - 2, - 1} \right) \times \left( {2, - 1,2} \right)} \right] \cr & =& - \frac{2}{9}\left( { - 4 - 1,2 - 2,1 + 4} \right) = \left( {\frac{{10}}{9},0,-\frac{{10}}{9}} \right) \cr}

So $\Pi'$ ends up being $$\Pi':{\frac{{10}}{9}x - \frac{{10}}{9}z = 0}$$ or

$$\boxed{ \Pi ':x = z}$$

Is this OK? Can it be checked graphically? Is there another way to solve this?

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From $d(P,R)=d(P,\Pi)$,we see that $R$ is the orthogonal projection of $P$ to $\Pi$. Hence $\vec{PR}$ is a multiple of $\Pi$'s normal vector $(2, -1, 2)$. (But we don't need $R$ exactly, that's my simplification). The cross product with $\vec{QP} = (1, 2, 1)$ is $(-5, 0, 5)$, which (after checking against $P$) also leads to $\Pi'\colon x=y$. – Hagen von Eitzen Sep 3 '12 at 22:27

As you said $PR \perp \Pi$, and $R \in \Pi \cap \Pi'$.
Since $PR \subset \Pi'$ and $PR \perp \Pi$ you get $\Pi \perp \Pi'$ and hence the normal vector $(2,-1,2)$ of $\Pi$ is parallel to $\Pi'$.
But then the point $S$ so that $\vec{PS}=(2,-1,2)$ is in $\Pi'$.
You can easily find $S$, and then you can find the plane through $P,Q, S$ by standard methods.