Chapter 14.7, Problem 31WE

To determine

To prove: The composition $H_B\circ H_A$ is a translation.

Explanation of Solution

Given information: It is given that $l_1\perp l_2$ , $l_3\perp l_2$ , $R_1$ , $R_2$ , and $R_3$ denote reflections in $l_1$ , $l_2$ , and $l_3$ . The figure is given below:

  

Proof:

The complete proof is shown in the table below:

	Statements		Reasons
1.	$H_A=R_2\circ R{1}_{}$	1.	As $R_1$ and $R_2$ are perpendicular to each other. So, the angle between the object and the image will be $180°$ .
2.	$H_B=R_3\circ R{2}_{}$	2.	Similarly, as $R_3$ and $R_2$ are perpendicular. So, the angle between the object and the image will be $180°$ .
3.	$H_B\circ H_A=\left(R_3\circ R_2\right)\circ \left(R_2\circ R_1\right)$	3.	From the definition that forming composites of transformations is an associative operation.
4.	$H_B\circ H_A=\left(R_3\circ \left(R_2\circ R_2\right)\right)\circ R_1$	4.	Composition is associative
5.	$H_B\circ H_A=\left(R_3\circ I\right)\circ R_1$	5.	Now as the composite of half-turn to itself is an identity.
6.	$H_B\circ H_A=R_3\circ R_1$	6.	Now as the composite of half-turn to itself is an identity.
7.	$H_B\circ H_A$ is a translation.	7.	As the reflection of an object in two parallel lines is translation, therefore, it is also a translation.

Hence, it is proved that $H_B\circ H_A$ is a translation.