6572
/www/nrich/html/content/id/6572/
1
2011-02-01T00:00:01
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Suppose that I am given a large supply of basic vectors $b_1=(2,1)$
and $b_2=(0,1)$.<br></br>
<br></br>
Starting at the origin, I take a 2-dimensional 'vector walk' where
each step is either a $b_1$ vector or a $b_2$ vector, either
forwards or backwards.<br></br>
<br></br>
<span style="font-weight: bold;">Investigate the possible
coordinates for the end destinations of my walk.</span><br></br>
<br></br>
Can you find any other pairs of basic vectors which yield exactly
the same set of destinations?<br></br>
<br></br>
Can you find any pairs of basic vectors which yield none of these
destinations? <br></br>
<br></br>
<br></br>
<br></br>
<div class="framework">NOTES AND BACKGROUND<br></br>
<br></br>
In more formal mathematics, the points visited by such a vector
walk would be called a <span style="font-style: italic;">lattice</span> and the two basic
vectors would be called the <span style="font-style: italic;">generators</span> . Lattices which
repeat themselves are structurally interesting; the symmetry
properties of such lattices are important in both pure mathematics
and its applications to, for example, the properties of
crystals.</div>
<br></br>
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<br></br>
$\mathbf{P} = n\begin{pmatrix}2\\1\end{pmatrix} +
m\begin{pmatrix}0\\1\end{pmatrix} =
\begin{pmatrix}2n\\n+m\end{pmatrix}$<br></br>
<br></br>
The end destination tells you how many of each vector were used,
but does not include cancelled steps: from the $x$ coordinate we
can deduce how many net $b_1$ steps there were; this then allows us
to determine how many net $b_2$ steps there must have been.<br></br>
<br></br>
The vectors can together reach any point with integer $y$
coordinate and even $x$ coordinate: Thus, the set of possible
destinations are the points, and only the points, with coordiantes
$\begin{pmatrix}2p\\q\end{pmatrix}$ for any integers $p, q$.<br></br>
<br></br>
Two other base vectors which give rise to exactly the same set of
points are<br></br>
$$<br></br>
\begin{pmatrix}2\\0\end{pmatrix}
\,,\begin{pmatrix}0\\1\end{pmatrix}<br></br>
$$<br></br>
<br></br>
Imagine now that we have two other more general base vectors<br></br>
$$<br></br>
\mathbf{a} = \begin{pmatrix} x_a \\ y_a\end{pmatrix} \quad
\mathbf{b} = \begin{pmatrix} x_b \\y_b\end{pmatrix}<br></br>
$$<br></br>
If these are to reach exactly the same set of points we must have,
for any integers $N$ and $M$ that<br></br>
$$<br></br>
N\begin{pmatrix} x_a \\ y_a\end{pmatrix} +M\begin{pmatrix} x_b\\
y_b\end{pmatrix}\mathbf{b} = \begin{pmatrix} 2p \\
q\end{pmatrix}<br></br>
$$<br></br>
Clearly to reach these, and only these points, I would need both
$a$ and $b$ to have even $x$ component. Furthermore, if I can reach
the points $(2, 0)$ and $(0, 1)$ with combinations of my base
vectors then I would be able to reach all of the points $(2n, m)$
by adding together $n$ and $m$ lots of these combinations.<br></br>
<br></br>
Algebra quickly becomes very complicated, as there are too many
variable to give me anything to 'solve'.<br></br>
<br></br>
So, look at some more particular cases. Look at a base vector with
an $x$-component of $4$.<br></br>
$$<br></br>
\begin{pmatrix} 4 \\ 0\end{pmatrix}<br></br>
$$<br></br>
Clearly, to be able to reach the point $(2, 0)$ the other base
vector would need to be $\begin{pmatrix} \pm 2 \\ 0\end{pmatrix}$,
but this would not allow us to reach the point $(0, 1)$. Thus, any
other possible pairs of base vectors would have to have non-zero
$x$ and $y$ components.<br></br>
A suitable pair would be<br></br>
$$<br></br>
\mathbf{a}= \begin{pmatrix} 4 \\ 1\end{pmatrix}\,, \mathbf{b}
=\begin{pmatrix}2 \\ 1\end{pmatrix}<br></br>
$$<br></br>
To see why, note that the combinations $\mathbf{a}-\mathbf{b}$ and
$2\mathbf{b}-\mathbf{a}$ take us to the points $(2, 0)$ and $(0,
1)$.<br></br>
Another suitable pair would be<br></br>
$$ \mathbf{a}= \begin{pmatrix}6 \\ 2\end{pmatrix}\,, \mathbf{b}
=\begin{pmatrix}4 \\ 1\end{pmatrix} $$ To see why, note that the
combinations $2\mathbf{b}-\mathbf{a}$ and $2\mathbf{a}-3\mathbf{a}$
take us to the points $(2, 0)$ and $(0, 1)$.<br></br>
<br></br>
<br></br>
For base vectors which yield none of the points in the
original lattice, we would need irrational components, because if I
had a base vector with rational components $\frac{n}{m}$ and
$\frac{p}{q}$ then taking $2mq$ lots of this vector would take us
to the point $(2qn, 2pm)$, which is in our original lattice of
points. <br></br>
<br></br>
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<h3>Why do this problem?</h3>
<p><a href="http://nrich.maths.org/public/viewer.php?obj_id=6572&part=">This problem</a> encourages students to think about vectors as representing a movement from one point to another. The need for coordinate representation of points will emerge automatically and the problem naturally requires an interplay between geometry and algebra.</p>
<h3>Possible approach</h3>
<div> </div>
<div>The problem is available as a handout <a class="doclink" href="/content/id/6572/Vector%20Walk.doc">here</a>.</div>
<div> </div>
<div>Set students the challenge to investigate possible end points when combining steps of vectors $b_1$ and $b_2$ in a vector walk. Some students will prefer to work algebraically while others will wish to represent the problem geometrically; by encouraging students to work in groups with others who have different preferred methods, rich mathematical thinking can emerge.</div>
<div>Students should aim to describe geometrically the set of points which can be made by combining the two vectors.</div>
<br></br>
<div>Once students have successfully described the set of points made from combinations of $b_1$ and $b_2$, set them the two challenges - to find other pairs of basic vectors which yield the same possibilities, and to find a pair of basic vectors which will <span style="font-weight: bold;">never</span> lead to the points found in the first part of the question.</div>
<br></br>
<h3>Key questions</h3>
<div>What do the points you can reach with $b_1$ and $b_2$ have in common?</div>
<div>Can you describe the resulting set of points geometrically (i.e. describe them clearly without algebra)?</div>
<div> </div>
<br></br>
<h3>Possible extension</h3>
<p><a href="http://nrich.maths.org/6632&part=">Polygon Walk</a> explores vector walks which form polygons around the origin.</p>
<h3>Possible support</h3>
<div>Work systematically combining $b_1$ steps with $b_2$ steps, recording the points visited.</div>
<div>Investigate the effect of changing the order in which the steps are taken.</div>
<p><br></br>
<br></br>
</p></mdoxml>
<mdoxml version="1.0">Try creating some walks from combinations of $b_1$ and $b_2$. Do
you notice anything about the destinations you reach? Can you
explain what you find?<br></br>
<br></br>
What is the simplest horizontal step you can make by combining
$b_1$ and $b_2$ steps? What about vertically?<br></br>
<br></br>
<span style="font-style: italic;">This problem builds on GCSE level
vector work, and provides a foundation for concepts met in the
later Core A Level modules.</span> <br></br></mdoxml>
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