# Tracking Points

Can you give the coordinates of the vertices of the fifth point in
the patterm on this 3D grid?

## Problem

Copies of the line segment AB that joins the points (0,0,0) to (1,1,1) are linked together (end-to-end) in a chain starting at (0,0,0). What would the coordinates of the ends of the tenth segment be?

If a second line segment (CD) starts at (0,0,0) and ends at (1,3,2) and copies of this are made and joined end-to-end to it, what would the coordinates of the ends of the 2nd segment be? The tenth segment? The nth segment?

What if the chain of the line segment AB described in the first paragraph started at (1,0,-1)?

Where would the ends of the nth segment be this time?

Where would the ends of the nth segment be if the first segment started at (a,b,c)?

What if copies of the second line segment, CD, started at (-3,0,2) and were placed end to end. Where would the ends of the nth segment be?

If the chain started at (a,b,c) where would the ends of the nth segment be?

If you now make lots of copies of the two line segments and decide to place them end-to-end alternately, investigate the end points of the nth segment for different starting points.

If a second line segment (CD) starts at (0,0,0) and ends at (1,3,2) and copies of this are made and joined end-to-end to it, what would the coordinates of the ends of the 2nd segment be? The tenth segment? The nth segment?

What if the chain of the line segment AB described in the first paragraph started at (1,0,-1)?

Where would the ends of the nth segment be this time?

Where would the ends of the nth segment be if the first segment started at (a,b,c)?

What if copies of the second line segment, CD, started at (-3,0,2) and were placed end to end. Where would the ends of the nth segment be?

If the chain started at (a,b,c) where would the ends of the nth segment be?

If you now make lots of copies of the two line segments and decide to place them end-to-end alternately, investigate the end points of the nth segment for different starting points.

## Getting Started

The second segment will stretch from (1,1,1) to (2,2,2), the third from(2,2,2) to (3,3,3) and so on. So the tenth segment will have its ends at (9,9,9) and (10,10,10).

Establishing a rule for what is happening to each of the coordinates in turn is best achieved by thinking of the effect at seeing just a 2-D image.

In the last section there are a number of options to consider.

## Student Solutions

Andre (age 15),
from Romania, tackled this problem. Here's his solution, which uses
vectors:

In the Cartesian system of axes, the
vector $\overrightarrow{AB}$, with the origin at (0,0,0) and end
point at (1,1,1) could be written as:
$\overrightarrow{AB}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ where
$\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are the unit vectors of
the axes Ox, Oy and Oz respectively.

Now, linking end to end 10 vectors
$\overrightarrow{AB}$ comes to multiplying the vector
$\overrightarrow{AB}$ by 10, so that one obtains a vector with the
origin at (0,0,0) and end point at (10,10,10).

For the vector $\overrightarrow{CD}$:
$\overrightarrow{CD}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}$. Adding up
2 such vectors, one easily sees that the resulting vector has the
origin at (0,0,0) and the end point at (2,6,4). Adding up 10 such
vectors, the origin remains the same and the end point is (10, 30,
20). For $n$ vectors, the end point is $(n, 3n, 2n)$.

If the starting point is (1,0,-1) and the
vector to be drawn is
$\mathbf{v}=n\mathbf{i}+n\mathbf{j}+n\mathbf{k}$, then the end
point will be at $(1+n, 0+n, -1+n)$, i.e.\ at $(n+1, n,
n-1)$.

If the origin is at $(a,b,c)$, the end
point is at $(n+a, n+b, n+c)$. If the vector is
$\mathbf{u}=n\mathbf{i}+3n\mathbf{j}+2n\mathbf{k}$, and its origin
is at (-3,0,2), its end point will be at $(n-3, 3n, 2n+2)$. If for
the same vector the origin is at $(a,b,c)$, its end point will be
at $(n+a, 3n+b, 2n+c)$.

If the vector of interest is now
$\mathbf{u}+\mathbf{v}=2n\mathbf{i}+ 4n\mathbf{j}+3n\mathbf{k}$, in
the general case of the origin at $(a,b,c)$, the end point will be
at $(2n+a,4n+b, 3n+c)$.

Stephen (from Framwellgate School,
Durham) solved the problem and went on to think about what happens
if you alternate segments. Here is what he sent us :

With AB followed by CD (if we count AB as segment 1 and CD as segment 2) we can see that 2 segments makes (1,1,1)+(1,3,2)=(2,4,3) therefore the nth segment when n is even is (n,2n,3n/2) because we need to divide it by 2, as (2,4,3) is 2 segments. For n segments when n is odd, n-1 is even so take the n-1th segment from last coordinates to give (n-1,2(n-1),3(n-1)/2) and add (1,1,1), because this is the first segment and therefore will be the last segment on odd number of segments (n) - to get: (n,2n-1,(3n-1)/2) if starting from (a,b,c) then we just need to add this coordinate to the two solutions.

## Teachers' Resources

By viewing the 3-D coordinates as if you could not see the third dimension you can extend what is happening in the 2D case and this is explored in the problem Coordinate Patterns first published in September 2004.