Transformations Part 3:
We now take up the question of reflections across three lines. There are several cases, and we
will start with the easy ones. Not much that is done in these lecture notes could be considered a
formal proof, but they will give you an idea of how a formal proof could be established with a
little more time and care. It will give you an idea of the major technique, which uses the Second
Rotation Theorem and the Second Translation Theorem to move the lines to make it easier to see
what is going on. The following chart outlines the cases to be considered.
Case
Description
Results
A
Three lines all the same line
Reflection: Two reflections result in the identity;
the third results in a simple reflection.
B
Three lines all concurrent
Reflection: If l, m, and n are concurrent, examine
. Since
is a rotation, by the second
rotation theorem we know that there is a line k such
that
so that
C
Three lines all parallel
Reflection: If l, m, and n are all parallel, examine
. The three lines all share a common
perpendicular t. Since
is a translation, by the
second translation theorem we know that there is a
line k z t such that
so that as above
.
D
At least two of the three
lines intersect.
This requires a new kind of isometry, our last. It is
called a glide reflection.
Definition: Let A and B be two points on a line l. The composition
is called a glide
reflection. It is the result of a translation followed by a reflection in the line
.
Note that a glide reflection is determined by the two points A and B, so it can be denoted simply
by GAB without ambiguity.
It can be shown that
.
Below we will walk through a demonstration that the composition of three reflections across
lines that intersect pairwise is a glide reflection. First, note that in terms of reflections, a glide
reflection is defined by reflections across a pair of parallel lines (to accomplish the translation)
followed by a reflection across a line perpendicular to the parallel lines.
Now suppose three lines intersect as below. I have labeled them I, II, and III to reflect the order
of the reflections: the isometry here is
. We look at the rotation accomplished by
and use the second rotation theorem to find two lines, IV and V, such that they accomplish
the same rotation.
Note that
accomplishes the same rotation as
line V to be perpendicular to line III. Now
, but the trick is that we have chosen
.
Next, we look at the rotation accomplished by reflections through line V and then line III. We
can use the second rotation theorem to replace these lines by others, specifically, lines VI and
VII. This time, we cleverly choose line VII to be perpendicular to IV.
So now,
The original
reflections have been replaced by equivalent reflections with lines we have cleverly chosen. So,
the final result is that we have “moved the lines” to a set of parallel lines with one mutual
perpendicular.
The result is a glide reflection.
The same technique of “moving the lines” can be used to show that, for example, the product of
two rotations is either a rotation or a translation. Suppose we have two rotations, represented
below by
. We can replace lines j, k, l, and m respectively with lines jN, kN, lN and
mN such that the rotations are the same but lines jN and mN correspond.
Then,
. Since reflections over jNand mN
will result in the identity (since they are the same line) the reflections over kN and lN result in a
rotation (as in the sketch above) or a translation (if kN and lN ) happen to be parallel.
Similar procedures can be used to decide the results of a composition of two translations or a
translation and a reflection.
We can sum up by saying that every isometry, being the product of at most three reflections, is
either:
1.
2.
3.
4.
A reflection;
A translation;
A rotation; or
A glide reflection.