Affine subspaces and parallelism

Question

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

We are given an affine space $\mathbb{A}^3$ with basis $(a_0, a_1, a_2, a_3)$ and two affine subspaces $P_1, P_2$ defined by the affine spans $P_1 = \text{aff}(c_0, c_1, c_2), \, P_2 = \text{aff}(c'_0, c'_1, c'_2)$ where

\begin{array}{c} c_0 = -3a_0 + a_1 + 0a_2 + 3a_3 \\ c_1 = -3a_0 + 3a_1 - a_2 + 2a_3 \\ c_2 = -6a_0 + 2a_1 + a_2 + 4a_3 \end{array}

and

\begin{array}{c} c'_0 = 2a_0 - a_1 - a_2 + a_3 \\ c'_1 = -a_0 + 2a_1 - a_2 + a_3 \\ c'_2 = -7a_0 + 0a_1 + 3a_2 + 5a_3 \end{array}

Show that $P_1, P_2$ are parallel planes.

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I'm sorry for the lack of effort on my part but I was not able to do much with this exercise. Since the notation being used can be troublesome I would like to make two remarks :

$(1)$ The basis $(a_0, a_1, a_2, a_3)$ of $\mathbb{A}^3$ can be written more formally as the affine frame $$\cal R = \{a_0, (\vec{a_0a_1}, \vec{a_0a_2}, \vec{a_0a_3})\}$$

where we arbitrary choose the point $a_0$ as the origin of the frame.

$(2)$ The affine span (or affine hull) $P_1 = \text{aff}(c_0, c_1, c_2)$ is the set of all affine combinations of the points $\{c_i\}_{i=0}^{2}$, that is

$$P_1 = \text{aff}(c_0, c_1, c_2) = \left\{ \sum_{i = 0}^{2} \alpha_i c_i : \alpha_i \in \mathbb{R} \, \text{and} \, \sum_{i = 0}^{2} \alpha_i = 1 \right\}.$$

EDIT :

The point $c_0$ is given as an affine combination of the points $a_i, i = 0, \ldots, 3$, where the coefficients $\lambda_i, i = 0, \ldots, 3$ are the barycentric coordinates of the point satisfying $$\sum_{i = 0}^{3} \lambda_i = 1.$$

We now wish to determine the coordinates of the point $c_0$ w.r.t. the affine frame $\cal R = \{a_0, (\overrightarrow{a_0a_1}, \overrightarrow{a_0a_2}, \overrightarrow{a_0a_3})\}$. We note that

$$c_0 = -3a_0 + a_1 + 0a_2 + 3a_3 \tag{1}$$ and

$$-3 + 1 + 0 + 3 = 1 \implies -3 = 1 - 1 - 0 - 3\tag{2}$$

Substituting $(2)$ in $(1)$ we get

\begin{align} c_0 &= (1 - 1 - 0 - 3)a_0 + a_1+0a_2 + 3a_3 \\ &= a_0 + 1(a_1 - a_0) + 0(a_2 - a_0) + 3(a_3 - a_0) \\ &= a_0 + 1(\overrightarrow{a_0a_1}) + 0(\overrightarrow{a_0a_2}) + 3(\overrightarrow{a_0a_3}). \end{align}

Hence, the coordinates of $c_0$ w.r.t. the affine frame $\mathcal{R}$ are $c_0 = (1, 0, 3)$. We proceed similarly for the points $c_1, c_2$ and get

$$c_1 = (3, -1, 2), \quad c_2 = (2, 1, 4).$$

Finally,

$$\overrightarrow{c_0c_1}=\begin{bmatrix}\phantom{-}3-1\\-1-0\\\phantom{-}2-3\end{bmatrix}=\begin{bmatrix}\phantom{-}2\\-1\\-1\end{bmatrix},\quad \overrightarrow{c_0c_2}= \begin{bmatrix}2 - 1\\ 1-0 \\ 4 - 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$

Answer 1

Hint:

In the affine frame $\mathcal R= \{a_0, (\vec{a_0a_1}, \vec{a_0a_2}, \vec{a_0a_3})\}$, show the vector planes $\bigl\langle\,\overrightarrow{c_0c_1},\,\overrightarrow{c_0c_2}\,\bigr\rangle$ and $\bigl\langle\,\overrightarrow{c'_0c'_1},\,\overrightarrow{c'_0c'_2}\,\bigr\rangle$ are the same.

For this, calculate their coordinates: $$\overrightarrow{c_0c_1}=\begin{bmatrix}3-1\\-1-0\\\phantom{-}2-3\end{bmatrix}=\begin{bmatrix}\phantom{-}2\\-1\\-1\end{bmatrix},\quad \overrightarrow{c_0c_2}=\begin{bmatrix}1\\1\\1\end{bmatrix},\quad \text{&c.}$$ and show each augmented matrix $\begin{bmatrix}\overrightarrow{c\phantom{'}_{\!0}c_1},\,\overrightarrow{c_0c\phantom{'}_{\!2}},\,\overrightarrow{c'_0c'_i}\end{bmatrix}$ has rank $2$.