Chapter 8.3, Problem 42E

To determine

To state: whether the goal of having every atom in the molecule to be equidistant is achieved by giving reason.

Answer to Problem 42E

No, because the evaluated distance are not equal.

Explanation of Solution

Given information:

The position of the molecules are as follows:

  $\begin{array}{l}A=\left(2,0,0\right)\\ B=\left(1,\sqrt{3},0\right)\\ C=\left(1,\frac{1}{3},\frac{2\sqrt{2}}{3}\right)\end{array}$

Calculation:

Use distance formula to evaluate $\left|\overrightarrow{AB}\right|,\left|\overrightarrow{BC}\right|\text{ and }\left|\overrightarrow{AC}\right|$ as shown below.

  $\begin{array}{c}\left|\overrightarrow{AB}\right|=\sqrt{{\left(x_B-x_A\right)}^2+{\left(y_B-y_A\right)}^2+{\left(z_B-z_A\right)}^2}\\ =\sqrt{{\left(1-2\right)}^2+{\left(\sqrt{3}-0\right)}^2+{\left(0-0\right)}^2}\left[\text{Put the point coordinates}\right]\\ =\sqrt{4}\\ =2\end{array}$

  $\begin{array}{c}\left|\overrightarrow{BC}\right|=\sqrt{{\left(x_C-x_B\right)}^2+{\left(y_C-y_B\right)}^2+{\left(z_C-z_B\right)}^2}\\ =\sqrt{{\left(1-1\right)}^2+{\left(13-\sqrt{3}\right)}^2+{\left(\frac{2\sqrt{2}}{3}-0\right)}^2}\left[\text{Put the point coordinates}\right]\\ =\frac{\sqrt{366\sqrt{3}}}{3}\\ =1.69\end{array}$

  $\begin{array}{c}\left|\overrightarrow{AB}\right|=\sqrt{{\left(x_C-x_A\right)}^2+{\left(y_C-y_A\right)}^2+{\left(z_C-z_A\right)}^2}\\ =\sqrt{{\left(1-2\right)}^2+{\left(13-0\right)}^2+{\left(\frac{2\sqrt{2}}{3}-0\right)}^2}\left[\text{Put the point coordinates}\right]\\ =\sqrt{2}\\ =1.41\end{array}$

Since, the distances are not equal. Thus, the goal of having every atom in the molecule to be equidistant is not achieved.