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My professor wrote:

Let $\begin{pmatrix} a\\ b\\ c \end{pmatrix}\in \mathbb{R}^3$ then $L = \operatorname{Span}\{\begin{pmatrix} a\\ b\\ c \end{pmatrix}\} =$ line.

My question is: Why $\operatorname{Span}\{\begin{pmatrix} a\\ b\\ c \end{pmatrix}\}$ is a line? could someone explain this to me please? My background is that I know the slope-intercept form, two-points form and the point-slope form equations of the line but still I can not see why the above column vector spans a line.

Thanks!

Mr.Gandalf Sauron
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Brain
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    you seem to be talking about lines in the 2D plane: "the slope-intercept form, two-points form and the point-slope form equations" – 311411 Apr 06 '22 at 17:29
  • @311411 sorry edited thank you! – Brain Apr 06 '22 at 17:31
  • but the line we have is in $\mathbb{R}^3$ and the given is just a vector. So, what are the equations of a line in $\mathbb{R}^3$ ? – Brain Apr 06 '22 at 17:36
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    Think of the vector as an arrow. If you take all real multiples of it, what happens to it? – John Douma Apr 06 '22 at 17:38
  • My feeling is that you are due to study the key idea of "span" today. When you do, you will see that the origin $(0,0,0)$ is relevant. Perhaps these will help: https://math.stackexchange.com/questions/1720226 - - - https://math.stackexchange.com/questions/1556068 – 311411 Apr 06 '22 at 17:39
  • Perhaps to understand "span" you should also review "linear combination" https://math.stackexchange.com/questions/2952863 – 311411 Apr 06 '22 at 17:42

1 Answers1

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You can try and google the definition of a line in the $3D$ space .

First of all what is $\text{span}(\{(a,b,c)^{T}\}$ ? .

As a set it is precisely $\{t(a,b,c)\,:t\in\mathbb{R}\}$.

Now compare this to the case of $\mathbb{R}^{2}$. Say you had $(a,b)$ instead. Then the span would be $\{t(a,b)=(ta,tb):t\in\mathbb{R}\}$.

Compare this with the technique you actually use to draw a line with a ruler and pencil. Once you are given a point and a direction , or two points, you join them using the ruler and extend it on both sides. Isn't it analogous to considering a point of starting point $O$ with it's position vector given by $\vec{p}$ and a direction given by $\vec{v}$ and just saying that all the points in your line are the same as those with coordinates $\vec{p}+t\vec{v}\,\,$ for all $-\infty<t<\infty$? .

The idea is precisely the same for higher dimensions.

A line in a higher dimension passing through two points $\vec{p}$ and $\vec{q}$ is as a set precisely equal to $\{\vec{p}+t(\vec{q}-\vec{p}):t\in\mathbb{R}\}$.

This is why they are saying that it is a line.

You can also see my answer here

Mr.Gandalf Sauron
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