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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.


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,

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

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  • $\begingroup$ There's a problem with the coefficients: they do not all sum up to $1$. $\endgroup$ – Bernard Aug 14 '17 at 19:00
  • $\begingroup$ @Bernard Typo fixed. Thank you. $\endgroup$ – user347616 Aug 14 '17 at 19:11
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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 $2$.

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  • $\begingroup$ Thank you very much for your answer. I'm getting different coordinates for the vectors $\overrightarrow{c_0c_1}$ and $\overrightarrow{c_0c_2}$. I have edit my post to show you how I worked out their coordinates. Would you mind verifying if my work is correct? You possibly misread the coefficients when writing your answer. I still don't understand why showing that the augmented matrices have rank $2$ means that the planes are parallel. Maybe you could help me understand this part? I can certainly do the computations. I am mostly interested to understand why this is true. $\endgroup$ – user347616 Aug 15 '17 at 17:58
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    $\begingroup$ @Elix: you're right, either I miscalculated of I misread. I've updated the answer. For the parallelism, it's the definition: the vector space attached to an affine subspace, once you've chosen an origin is called its direction (one shows it does not depend on the chosen origin) and two affine spaces which have the same direction are said to be parallel (more generally, one also says they're parallel if the direction of one of them is contained in the direction of the other — somewhat like a line and a plane can be parallel un 3-space). Linear algebra implies they have no intersection. $\endgroup$ – Bernard Aug 15 '17 at 18:17

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