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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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    In fact, you can use the coordinates of $\mathbf{d}$, *and* use the origin as your second point.2011-08-24

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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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    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.2011-08-24
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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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    @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 :-)2011-08-24