Affinity which maps one plane to another

Question

I'm learning affine geometry, specifically affine transformations, and need help with the following exercise :

Let $P_1 : -x + z = 3$ and $P_2 : x - 4y + 3z = 9$ be two planes of $\mathbb{R^3}$.

$(1)$ Give affine frames of the form $\mathcal{R} = \{a_0, (\overrightarrow{a_0a_1}, \overrightarrow{a_0a_2})\}$ and $\mathcal{R'} = \{a_0', (\overrightarrow{a_0'a_1'}, \overrightarrow{a_0'a_2'})\}$ for the planes $P_1$ and $P_2$, resp.

$(2)$ Find an (invertible) affine transformation of the form $\mathbf{x'} = A \mathbf{x} + \mathbf{b}$ (where $A$ is the linear part and $\mathbf{b}$ the displacement vector) which send $P_1$ to $P_2$.

Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.

$(1)$ An affine frame of an affine space consists of a point $a_0$, that we call origin, and a linear basis of the associated vector space. From the parametric equation of the plane $P_1$

$$\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\3\end{pmatrix} + u\begin{pmatrix}0\\0\\1\end{pmatrix} + t\begin{pmatrix}0\\1\\0\end{pmatrix}.$$

we let $a_0 = (0, 0, 3)$ be the origin, $\overrightarrow{a_0a_1} = (0, 0, 1)^T$ and $\overrightarrow{a_0a_2} = (0, 1, 0)^T$. Hence, an affine frame for the plane $P_1$ is given by $\mathcal{R} = \{(0, 0, 3),((0, 0, 1)^T, (0, 1, 0)^T)\}$.

We proceed similarly for $P_2$ and find the affine frame $\mathcal{R'} = \{(9, 0, 0),((4, 1, 0)^T, (-3, 0, 1)^T)\}$.

---

$(2)$ Unfortunately, I don't know how to construct an affine transformation $f : P_1 \to P_2$. I think I should make use of part $(1)$ since, by definition, the affine transformation $f : P_1 \to P_2$ is uniquely determined by the image of the affine basis. Also, since the three points $a_0, a_1$ and $a_2$ form an affine coordinate system, I thought I could construct an affine transformation from $P_1$ to $P_2$ by sending the points $a_0$ to $a_0'$, $a_1$ to $a_1'$ and $a_2$ to $a_2'$. I don't know if this is the correct idea.

Any help would be appreciated.

Answer 1

Your idea is correct, and it leads to a system consisting of nine linear equations for twelve coordinates of $A$ and $\mathbf{b}$, so you’ll have to fix some three of these coordinates. For instance, maybe it is possible to put $\mathbf{b}=0$.

But I have an other idea, which is more simple for me. Put $r_1=(-1/3,0,1/3)$ and $r_2=(1/9,-4/9,3/9)$.

Then for $i=1,2$ holds

$$P_i=\{r\in\Bbb R^3: (r,r_i)=1\}.$$

Then it suffices to put $\mathbf{b}=0$ and pick as $A$ the adjoint to an (invertible) affine transformation which sends $r_2$ to $r_1$, because then $(Ar, r_2)=(r, A^*r_2)=(r,r_1)$. Can you find such a transformation?

$A^*(0,0,1)^T=(0,0,1)^T$, $A^*(0,1,0)^T=(0,1,0)^T$, $A^*(1,0,0)^T=(-3,4,0)^T.$

Answer 2

1) For the first question, you rightly found the frames of the two planes through their parametric equations, which is one of the possible methods.

2) Concerning the linear transformation, we shall note that each plane is characterized by $3$ indipendent parameters: the four coefficients, deducted a common multiplier.
The linear transformation has $9+3=12$ parameters to be determined: so there are many that can be chosen.

For instance, the common intersection line is clearly an invariant under the transformation.
If as the position vector of both of the frames we choose a point on the line, e.g. $(0,0,3)^T$, then we can put $\bf b=0$.
That means that we consider the origin placed there, and are going to find the transformation between the frames, or in other words that we are considering the transformation from $(x,y,z-3)$ to $(x',y',z'-3)$.

Then , if as vector of both of the frames we choose one parallel to the intersection line, e.g. $\bf u=(1,1,1)^T$, this shall not vary and we can look for a matrix that keeps $\bf u$ unchanged, while bringing one of the vectors of $P_1$ that you already found, e.g. $\bf v_1=(1,0,1)^T$, (*) into one of those of $P_2$, e.g. $\bf v_2=(4,1,0)^T$.
----
note (*): there is a typo in your parametric eq. for $P_1$
----

That is $$ \left( {\matrix{ 1 & 4 \cr 1 & 1 \cr 1 & 0 \cr } } \right) = {\bf A}\;\left( {\matrix{ 1 & 1 \cr 1 & 0 \cr 1 & 1 \cr } } \right) $$

Now, we normally want $\bf A$ to be a full rank matrix, so that it is invertible.
To achieve this we can include a third vector $\bf w_1$ and $\bf w_2$, each indipendent from the other two, which means that each one does not lie on the corresponding plane.
That also means that we can put ${\bf w_1}= {\bf w_2}={\bf w}$, with $\bf w$ being a vector not lying on either plane.
And more, it means that we can also put ${\bf w_1}= {\bf v_2}$ and ${\bf w_2}= {\bf v_1}$.

Thereafter we can compute ${\bf A}$ as ${\bf V_2}{\bf V_1^{-1}}$.
.

For example $$ \left( {\matrix{ 1 & 4 & 1 \cr 1 & 1 & 0 \cr 1 & 0 & 1 \cr } } \right) = \left( {\matrix{ 1 & { - 3} & 3 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right)\;\left( {\matrix{ 1 & 1 & 4 \cr 1 & 0 & 1 \cr 1 & 1 & 0 \cr } } \right) $$ which then returns $$ \eqalign{ & 0 = \left( {\matrix{ 1 & { - 4} & 3 \cr } } \right)\left( {\matrix{ {x'} \cr {y'} \cr {z' - 3} \cr } } \right) = \left( {\matrix{ 1 & { - 4} & 3 \cr } } \right)\left( {\matrix{ 1 & { - 3} & 3 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right)\;\left( {\matrix{ x \cr y \cr {z - 3} \cr } } \right) = \cr & = \left( {\matrix{ 1 & 0 & { - 1} \cr } } \right)\left( {\matrix{ x \cr y \cr {z - 3} \cr } } \right) \cr} $$

Or you can write $$ \eqalign{ & \left( {\matrix{ {x'} \cr {y'} \cr {z'} \cr } } \right) - \left( {\matrix{ 0 \cr 0 \cr 3 \cr } } \right) = \left( {\matrix{ 1 & { - 3} & 3 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right)\left( {\left( {\matrix{ x \cr y \cr z \cr } } \right) - \left( {\matrix{ 0 \cr 0 \cr 3 \cr } } \right)} \right)\quad \Rightarrow \cr & \Rightarrow \quad \left( {\matrix{ {x'} \cr {y'} \cr {z'} \cr } } \right) = \left( {\matrix{ 1 & { - 3} & 3 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right)\left( {\matrix{ x \cr y \cr z \cr } } \right) - \left( {\matrix{ 9 \cr 3 \cr { - 3} \cr } } \right) \cr} $$

Finally, concerning your idea to send 3 points of $P_1$ to 3 points of $P_2$, that is also a possible approach, working in homogeneous coordinates. So for example, putting $$ {\bf V}_{\,1} = \left( {\matrix{ 0 & { - 3} & 0 \cr 0 & 0 & 1 \cr 3 & 0 & 3 \cr } } \right)\quad {\bf V}_{\,{\bf 2}} = \left( {\matrix{ 0 & 9 & 1 \cr 0 & 0 & 1 \cr 3 & 0 & 4 \cr } } \right)\quad \Rightarrow \quad {\bf A} = {\bf V}_{\,{\bf 2}} \,{\bf V}_{\,1} ^{\, - \,{\bf 1}} = \left( {\matrix{ { - 3} & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 1 & 1 \cr } } \right) $$ then $$ \eqalign{ & 0 = \left( {\matrix{ 1 & { - 4} & 3 & { - 9} \cr } } \right)\left( {\matrix{ {x'} \cr {y'} \cr {z'} \cr 1 \cr } } \right) = \left( {\matrix{ 1 & { - 4} & 3 & { - 9} \cr } } \right)\left( {\matrix{ { - 3} & 1 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr } } \right)\;\left( {\matrix{ x \cr y \cr z \cr 1 \cr } } \right) = \cr & = \left( {\matrix{ { - 3} & 0 & 3 & { - 9} \cr } } \right)\;\left( {\matrix{ x \cr y \cr z \cr 1 \cr } } \right) = \left( {\matrix{ { - 1} & 0 & 1 & { - 3} \cr } } \right)\;\left( {\matrix{ x \cr y \cr z \cr 1 \cr } } \right) \cr} $$