Working on the book: Robert Messer. "Linear algebra - The gateway to mathematics" (p. 60)

Notice that the difference between any two distinct vectors $rv + x$ and $sv + x$ on this line is $(rv + x)-(sv + x) = (r-s)v$, a non-zero multiple of v. So if an arbitrary line is given in $\mathbb{R^2}$ or $\mathbb{R^3}$, any point on the line can be used as the translation vector $x$, and the difference between any two distinct vectors on the line can be used as the direction vector $v$. Once these vectors are determined, the line can be written as the set $\{rv + x\mid r \in \mathbb{R}\}$.

Quick Example Write the plane in $\mathbb{R^3}$ through the points $(1, 2, 3)$, $(0,2,1)$, and $(3, —1, 1)$ as a set in the form indicated above.

Choosing $\mathbf{x} = (1, 2, 3)$, and noting $(0,2,1)$, and $(3, —1, 1)$ do not lie on the same line, I would write the set as: $$ \{r(0,2,1) + s(3,-1,1) + (1, 2, 3) \mid r,s \in \mathbb{R}\} $$

Specifically, I would like to know:

  • Is my solution correct ?
  • What's the concept behind subtracting two scalars $(r-s)v$ in the author's explanation ?
  • What's the connection behing the subtraction of two vectors and finding a direction vector?

EDIT: corrected typo where a plane was defined (instead of a line). Last sentence of quote from book.


I am doing some major guesswork here due to the text being quite confusing. (Where does $w$ come from? Why is $\{rv+sw+x|r,s\in\mathbb R\}$ being classified as a line, when it is clearly a plane?) So here is my attempt to clear your confusion.

Q1. Your solution is incorrect. You need to subtract $\bf x$ from the other vectors.

Q2. Here is also a point I'm confused about. The author is clearly talking about a line in that section of the text. Defining a line as the set $\{kv+x|k\in\mathbb R\}$, we see that we need a direction vector $v$ and a translation vector $x$. The translation vector $x$ defines a starting point of the line, and the direction vector $v$ extends the line in that direction (forwards and backwards). The subtraction $(rv+x)-(sv+x)=(r-s)v$ shows that the direction vector $v$ and the translation vector $x$ are independent, so $x$ can be chosen as any point on the line.

Q3. Subtracting two position vectors gives you the direction vector from one of the points to the other.

Now, to understand the construction of the plane $\{rv+sw+x|r,s\in\mathbb R\}$, we see that there are two direction vectors $v,w$ and a translation vector $x$. The direction vectors $v,w$ span our plane.

To see that $v,w$ are found by subtracting vectors, consider three given (non-collinear) points. They form a triangle. Choose one vertex as the starting point. We calculate the direction vectors from our starting point to the other two vertices. These two vectors are precisely $v,w$.

Finally, the plane $\mathbb R^3$ passing through the points $(1,2,3),(0,2,1),(3,-1,1)$ is given by $$\{r(-1,0,-2)+s(2,-3,-2)+(1,2,3)|r,s\in\mathbb R\}$$

Substituting $(r,s) = (0,0), (1,0),(0,1)$ shows that the plane above does pass through our points.

  • $\begingroup$ Thank you, @player3236. You're right. I mistyped $\{rv + sw + x \mid r,s \in \mathbb{R}\}$ in place of $\{rv + x \mid r,s \in \mathbb{R}\}$. I will correct it. Could you tell me why I need to subtract x from both vectors in Q1 ? $\endgroup$
    – F. Zer
    Sep 28 '20 at 18:27
  • $\begingroup$ $(0,2,1)$ and $(3,-1,1)$ are position vectors, that is, the vector from the origin to that point. If $\bf x$ was not subtracted, the spanning vectors would originate from the origin, but not the point represented by $\bf x$. $\endgroup$
    – player3236
    Sep 28 '20 at 18:30
  • $\begingroup$ Excellent. Thank you. When you say "the direction vector 𝑣 and the translation vector 𝑥 are independent, x can be chosen as any point on the line", could you elaborate a little bit on the concept of "independence" ? I cannot quite get what you are trying to convey. $\endgroup$
    – F. Zer
    Sep 28 '20 at 18:34
  • 1
    $\begingroup$ Yes. It is a shorthand for $(r=0, s=0)$ or $(r= 1, s=0)$ or $(r = 0, s=1)$. $\endgroup$
    – player3236
    Sep 28 '20 at 18:37
  • 1
    $\begingroup$ Independent, as in not depending on/related to one another. No matter which translation vector $x$ you chose, the direction vector $v$ stays the same. $\endgroup$
    – player3236
    Sep 29 '20 at 3:29

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