Chapter 14.7, Problem 29WE

a

To determine

To describe in words the distance and direction of each translation.

Answer to Problem 29WE

  ${\text{R}}_k\circ {\text{R}}_j$ translates from 2 unit right and ${\text{R}}_j\circ {\text{R}}_k$ translates from 2 unit left

Explanation of Solution

Given information :

j and k are vertical lines 1 unit apart.

  

Theorem 14-7: A Composite of reflections in two parallel lines is a translation. The translation glides all point through twice the distance from the first line of reflection to the second.

When lines j and k are parallel, ${\text{R}}_k\circ {\text{R}}_j$ translates point through twice the distance between the lines.

Now, $\begin{array}{l}{\text{R}}_k\circ {\text{R}}_j\\ \left({\text{R}}_k\circ {\text{R}}_j\right)\circ \left({\text{R}}_j\circ {\text{R}}_k\right)\\ {\text{R}}_k\left({\text{R}}_k\circ {\text{R}}_j\right)\circ \left({\text{R}}_j\circ {\text{R}}_k\right)\end{array}$ translates = 2 (distance from j to k)

= 2 x 1

= 2 unit right

Similarly,

  ${\text{R}}_j\circ {\text{R}}_k$ translates = 2 (distance from j to k)

= 2 x 1

= 2 unit left

Hence,

  ${\text{R}}_k\circ {\text{R}}_j$ translates from 2 unit rightand ${\text{R}}_j\circ {\text{R}}_k$ translates from 2 unit left

b

To determine

To show that $\left({\text{R}}_k\circ {\text{R}}_j\right)$ and $\left({\text{R}}_j\circ {\text{R}}_k\right)$ are inverse.

Answer to Problem 29WE

  $\left({\text{R}}_k\circ {\text{R}}_j\right)$ and $\left({\text{R}}_j\circ {\text{R}}_k\right)$ are inverse.

Explanation of Solution

Given information :

  $\left({\text{R}}_k\circ {\text{R}}_j\right)$ and $\left({\text{R}}_j\circ {\text{R}}_k\right)$ both are translations.

Formation of composites of transformations is an associative operation.

  $\begin{array}{l}\left(R_k\circ R_j\right)\circ \left(R_j\circ R_k\right)=R_k\circ \left(R_j\circ R_j\right)\circ R_k\\ \left(R_k\circ R_j\right)\circ \left(R_j\circ R_k\right)=R_k\circ I\circ R_k\\ \left(R_k\circ R_j\right)\circ \left(R_j\circ R_k\right)=R_k\circ R_k\\ \left(R_k\circ R_j\right)\circ \left(R_j\circ R_k\right)=I\\ \left(R_k\circ R_j\right)={\left(R_j\circ R_k\right)}^{-1}\end{array}$

Hence,

  $\left({\text{R}}_k\circ {\text{R}}_j\right)$ and $\left({\text{R}}_j\circ {\text{R}}_k\right)$ both are inverse.