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Is it possible to found the cartesian equation of the line generated by the vector $\mathbf{d} = [-2,1]$ ? AFAIK I need at least two points so I'm using the unitary vector of $\mathbf{d}$ as the other point, calculating the slope and using $y-y_1 = m(x-x_1)$

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The approach you are using seems valid. –  ShawnD Aug 24 '11 at 16:46
    
In fact, you can use the coordinates of $\mathbf{d}$, and use the origin as your second point. –  Arturo Magidin Aug 24 '11 at 17:27
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2 Answers 2

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The line generated by $\vec{d}$ is all points of the form $(-2t,t)$ where $t \in \mathbb{R}$. When $t=1$ we get the point $(-2,1)$ and when $t=0$ we get $(0,0).$ As a result, the line has slope $-1/2$ and $y$-intercept $0$ and its equation is $y=-(1/2) x$.

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Thanks for the reply Shawn, so the cartesian equation can be seen as a parametric equation? –  Randolf R-F Aug 24 '11 at 16:47
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Yes. In linear algebra, the line generated by a vector is by definition everything in the set spanned by the vector, which is by definition all things of the form $t$ times the vector. –  ShawnD Aug 24 '11 at 16:50
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The question turns on the use of the words "line" and "generate". I suspect what confused you was that these are taken from two different contexts.

The term "generate" is from the language of vector spaces, where the subspace generated by a set of vectors is the set of all linear combinations of those vectors. Thus, the subspace generated by the vector $(-2,1)$ is as Shawn writes, the set of all vectors of the form $t(-2,1)$.

However, one doesn't usually talk about the line generated by a vector. The term "line" is from the language of geometry, where one thinks of a line uniquely determined by two points, or perhaps by a point and a direction, but not by one point.

This is related to the fact that in linear algebra, the zero vector plays a special role, and the subspace generated by a set of vectors always includes the zero vector, whereas in geometry the origin plays no special role, is just an arbitrary reference point for coordinates and obviously isn't included in all lines.

There is an analogue to the origin-independent world of geometry in linear algebra: affine combinations and affine subspaces. Whereas $n$ linearly independent vectors determine an $n$-dimensional linear subspace of a vector space, $n+1$ affinely independent vectors determine an $n$-dimensional affine subspace. The term "generate" is sometimes used in this context, though not as widely as with respect to linear subspaces; so in a certain sense you were right to think that it takes two vectors to "generate" a line.

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Thanks joriki. Do you know any book maybe more soft than fraleigh's linear algebra (with I'm currently using 'cause was the used where I get my bachelors degree). I'm a lawyer that (for weird reasons) build software and likes maths and want to self-study them as much as I can. I don't know if the book election was fine but maybe you could provide me a more useful reference. –  Randolf R-F Aug 24 '11 at 17:14
    
@Randolf: Sorry, I'd like to, but I studied theoretical physics in German, so whatever linear algebra I know I didn't learn from books that would likely be accessible to you -- unless you happen to speak German... (by the way, I see you're from Bogotá -- I happen to be sharing a flat with a student from Bogotá who's in Berlin for the summer to learn German :-) –  joriki Aug 24 '11 at 20:12
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