My geometry textbook states that the vectors $(a, b, c)^T$ and $k(a, b, c)^T$ represent the same line for any non-zero $k$; in other words, two such vectors related by an overall scaling are considered equivalent. It then goes on to state that an equivalence class of vectors under this equivalence relationship is known as a homogeneous vector.

The Wikipedia page for homogeneous coordinates states that for non-zero elements of $\mathbb{R}^3$, $(x_1, y_1, z_1) \sim (x_2, y_2, z_3)$ is defined to mean there is a non-zero $\lambda$ so that $(x_1, y_1, z_1) = (\lambda x_2, \lambda y_2, \lambda z_2)$. Then $\sim$ is an equivalence relation.

  1. I'm confused as to how this could be an equivalence relationship, since equivalence relations are binary relations, whereas this is a triple relation?

  2. If this is an equivalence relation, then what is the relation in set theoretic notation? I have the following: If $X = \{(a, b, c)^T \mid a, b, c \in \mathbb{R} \}$, then the relation on $X$ is $R = \{ (x, y, z)^T \mid (x, y, z)^T = k(a, b, c)^T \ \text{for some $k \in (\mathbb{R} \setminus \{ 0 \}) $} \}$. Is this correct? If not, then what's wrong with it?

I would greatly appreciate it if people could please take the time to clarify this.

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    $\begingroup$ There are TWO triples being related, so it is binary. $\endgroup$ – Randall May 9 at 11:05
  • $\begingroup$ @Randall Wikipedia gives the following definition of binary relation: In mathematics, a binary relation over two sets $A$ and $B$ is a set of ordered pairs $(a, b)$ consisting of elements $a$ of $A$ and elements $b$ of $B$; in short, it is a subset of the Cartesian product $A \times B$. Can you please elaborate on how what you're saying is in agreement with said definition? $\endgroup$ – The Pointer May 9 at 11:07
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    $\begingroup$ We say $R$ is a relation over set $X$, when $R\subseteq X\times X$. $\endgroup$ – Graham Kemp May 9 at 11:20
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    $\begingroup$ $R=\{\langle\langle a,b,c\rangle^\top,k\langle a,b,c\rangle^\top\rangle\mid \langle a,b,c\rangle\in \Bbb R^3, k\in(\Bbb R\smallsetminus\{0\})\}$ $\endgroup$ – Graham Kemp May 9 at 11:32
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    $\begingroup$ Sorry. Allow me to rephrase: Yes, I agree your relation is also a correct as I find them to be equivalent. So if you can work with that, you have a representation of the equivalence relation. $\endgroup$ – Graham Kemp May 10 at 6:56

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