My Math Forum Equation of a line in higher dimensions

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 June 14th, 2013, 02:23 PM #1 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Equation of a line in higher dimensions I'm not really sure where this thread belongs, but hopefully the mods will move it to the right place. Here it is: What is the general formula of a line in an n-dimensional coordinate system?
June 14th, 2013, 07:06 PM   #2
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Re: Equation of a line in higher dimensions

Quote:
 Originally Posted by eddybob123 I'm not really sure where this thread belongs, but hopefully the mods will move it to the right place. Here it is: What is the general formula of a line in an n-dimensional coordinate system?
Assuming you have two distinct points in an n-dimensional euclidean space, let $\overarrow{a}$ be a position vector from the origin to one of the points and let $\overarrow{b}$ be a position vector from the origin to the other point. Then an equation for the line through these two points is

$\overarrow{r}=\overarrow{a}+t(\overarrow{b}-\overarrow{a})$

 June 15th, 2013, 06:11 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Equation of a line in higher dimensions You cannot have an equation like "y= ax+ b" for a line in more than two dimensions. If your space has n dimensions, then to give one dimensional line you need to reduce by n-1 dimensions. For example, 3x+ 2y- z= 4 and 5x+ y+ z= 1 are, separately, equations of planes. One equation reduces by one dimension and 3- 1= 2 dimensions. The (x, y, z) that satisfy both equations lie on the intersection of the two planes, a 3- 2= 1 dimensional line. You will also see lines written as "linked equations": 3x+ y+ 2z- 1= 2x+ y+ z+ 2= x- 2y- z+ 1. That is the same as the two equations 3x+ y+ 2z- 1= 2x+ y- z+ 1 and 2x+ y+ z+ 2= x- 2y- z+ 1. (3x+ y+ 2z- 1= x- 2y- z+ 1 is not a third equation- it is equivalent to the other 2). Finally, we can have "parametric equations" in which we write each of x, y, z, or whatever other variables we have, depending on some parameter, t, say. We can argue that we then have n equations in the n+ 1 (n coordinates of the n dimensional space and t) so that we have n+1- n= 1 dimensions, or simply note that the n coordinates depend on the one parameter, t, so the set is one dimensional. And, of course, we can use those parametric formulas as components in a "position" vector as mrtwhs suggests.
 June 17th, 2013, 11:04 AM #4 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: Equation of a line in higher dimensions OK, how about this: given a point (a1, a2, a3,..., a n) in an n-dimensional coordinate system that forms a line with the origin, then what is the equation of an (n-1) dimensional "hyperplane" that is perpendicular to the line at the origin? For example, if n=3, and a line crosses the origin, then what is the equation of the 2d plane that is perpendicular to that line at the origin?
 June 17th, 2013, 11:15 AM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Equation of a line in higher dimensions Why "n" points? Two points determine a line in any dimension. If $(a_1, a_2, ..., a_n)$ is a single point then it determines a line through the origin. If $(x_1, x_2, ..., x_n)$ is any point on the plane perpendicular to that line, containing the origin, then the dot product of the vectors $$, the vector from the origin to $(a_1, a_2, ..., a_n)$, and $$, the vector from the origin to $(x_1, x_2, ..., x_n>$, is 0. That is $lt;a_1, a_2, ..., a_n=>\cdot== a_1x_1+ a_2x_2+ ...+ a_nx_n= 0$. That is the equation of the plane, perpendicular to the line, containing the origin.

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# What is the equation of a line through origin in nth dimensions

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