# How to define the new vector in my case?

1. Aug 26, 2011

### mellon

If I have a vector:

origin = new Vector(x, y)

then, I would like to get N sub vectors, each one of the N sub vector is got by adding origin vector with a new vector value like:

The result I would like to achieve is showing in the image below:

http://i.stack.imgur.com/ZvH1j.png"

I would like to get N sub vectors which are in the same horizontal level and spaced by "L" length,

My questions are:

1. what is the new vector (Vector(m, n)) m, n values I should use to get all the sub vectors ? (N is dynamic value)

I end up with something like:

for(var i=0; i<N; i++){

//But, how to select the m, n value ??
var subVector = origin.add(new Vector(m, n));

}

2. How the choose the new vector (Vector(m, n)) m, n value to make sure the distance between two neighboring sub vectors is L as showed in the image ?

Last edited by a moderator: Apr 26, 2017
2. Aug 26, 2011

### HallsofIvy

Are you talking about Linear Algebra or programming? It looks to me like you do not have a "vector" and "subvectors" but that you start with a line line and want to find position vectors of points on that line, with distance between them equal to L.

If, in some coordinate system, you line is given by parametric equation x= at+ b, y= ct+ d, and you have a given starting point on the line, with postion vector $\ve{v}_0$, then a vector pointing along the line is $a\vec{i}+ b\vec{j}$. That has length $\sqrt{a^2+ b^2}$ so that $(1/\sqrt{a^2+ b^2})(a\vec{i}+ b\vec{j})$ is a unit vector.

For simplicity, let $a'= a/\sqrt{a^2+ b^2}$ and $b'= b/\sqrt{a^2+ b^2}$ so the unit vector in the direction of the line is $a'\vec{i}+ b'\vec{j}$. n points on that line, starting with $\vec{v}_0$, with distance L between them, are given by $\vec{v}_0+ Li(a'\vec{i}+ b'\vec{j})$ with i ranging from 0 to n-1.