Chapter 8.8, Problem 8E

To determine

To write: matrix for the given prism.

Answer to Problem 8E

  $\left[\begin{array}{rrrrrrrr}\hfill \begin{array}{l}-3\\ -2\\ 2\end{array}& \hfill \begin{array}{l}-3\\ 1\\ 2\end{array}& \hfill \begin{array}{l}-3\\ 1\\ -2\end{array}& \hfill \begin{array}{l}-3\\ -2\\ -2\end{array}& \hfill \begin{array}{l}4\\ -2\\ -2\end{array}& \hfill \begin{array}{l}4\\ -2\\ 2\end{array}& \hfill \begin{array}{l}4\\ 1\\ 2\end{array}& \hfill \begin{array}{l}4\\ 1\\ -2\end{array}\end{array}\right]$

Explanation of Solution

Given information:

  

Calculation:

The figure shows that the points of $A$ and $B$ are same except for $y$ -coordinates. Similarly, $A$ and $D$ are same except for $z$ -coordinates. The points of $C\text{ and }A$ differs in the $y$ - as well as $z$ - coordinates. The $y$ -coordinate of $A$ is similar to that of $E\text{ and }F$ whereas $z$ -coordinate of $A$ is similar to that of $F\text{ and }G$ .

Now evaluated $\overrightarrow{AH}$ as follows:

  $\begin{array}{c}\overrightarrow{AH}=\langle 4-\left(-3\right),1-\left(-2\right),-1-2\rangle \left[\text{Using }\overrightarrow{ab}=\left(x_a,y_a,z_a\right)-\left(x_b,y_b,z_b\right)\right]\\ =\langle 7,3,-4\rangle \end{array}$

The coordinates of $B,C,D,E,F\text{ and }G$ are evaluated as follows:

  $\begin{array}{c}B\langle -3,-2,2\rangle +\langle 0,3,0\rangle =\langle -3,1,2\rangle \\ C\langle -3,-2,2\rangle +\langle 0,3,-4\rangle =C\langle -3,1,-2\rangle \\ D\langle -3,-2,2\rangle +\langle 0,0,-4\rangle =D\langle -3,-2,-2\rangle \\ E\langle -3,-2,2\rangle +\langle 7,0,-4\rangle =E\langle 4,-2,-2\rangle \\ F\langle -3,-2,2\rangle +\langle 7,0,0\rangle =F\langle 4,-2,2\rangle \\ G\langle -3,-2,2\rangle +\langle 7,3,0\rangle =G\langle 4,1,2\rangle \end{array}$

Thus the matrix for the given prism will be as follows:

  $\left[\begin{array}{rrrrrrrr}\hfill \begin{array}{l}-3\\ -2\\ 2\end{array}& \hfill \begin{array}{l}-3\\ 1\\ 2\end{array}& \hfill \begin{array}{l}-3\\ 1\\ -2\end{array}& \hfill \begin{array}{l}-3\\ -2\\ -2\end{array}& \hfill \begin{array}{l}4\\ -2\\ -2\end{array}& \hfill \begin{array}{l}4\\ -2\\ 2\end{array}& \hfill \begin{array}{l}4\\ 1\\ 2\end{array}& \hfill \begin{array}{l}4\\ 1\\ -2\end{array}\end{array}\right]$