Problem:
Find the cross product of the following vectors:
$\left(
\begin{array}{c}
 \frac{53}{7} \\
 -\frac{58}{7} \\
 \frac{67}{7} \\
\end{array}
\right)$ and
$\left(
\begin{array}{c}
 -\frac{24}{7} \\
 -\frac{3}{7} \\
 \frac{68}{7} \\
\end{array}
\right)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Compute the following cross product}: \\
 \, \left(\frac{53}{7},-\frac{58}{7},\frac{67}{7}\right)\, \times \, \left(-\frac{24}{7},-\frac{3}{7},\frac{68}{7}\right)\,  \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{53}{7},-\frac{58}{7},\frac{67}{7}\right)\,  \text{and }\, \left(-\frac{24}{7},-\frac{3}{7},\frac{68}{7}\right)\, : \\
 \left(
\begin{array}{ccc}
 \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
 \frac{53}{7} & -\frac{58}{7} & \frac{67}{7} \\
 -\frac{24}{7} & -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
 \left| 
\begin{array}{ccc}
 \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
 \frac{53}{7} & -\frac{58}{7} & \frac{67}{7} \\
 -\frac{24}{7} & -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
  \text{= }\left| 
\begin{array}{ccc}
 \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
 \frac{53}{7} & -\frac{58}{7} & \frac{67}{7} \\
 -\frac{24}{7} & -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
\begin{array}{ccc}
 \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
 \frac{53}{7} & -\frac{58}{7} & \frac{67}{7} \\
 -\frac{24}{7} & -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
\begin{array}{cc}
 -\frac{58}{7} & \frac{67}{7} \\
 -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right| +\left(-\hat{\text{j}}\right) \left| 
\begin{array}{cc}
 \frac{53}{7} & \frac{67}{7} \\
 -\frac{24}{7} & \frac{68}{7} \\
\end{array}
\right| +\hat{\text{k}} \left| 
\begin{array}{cc}
 \frac{53}{7} & -\frac{58}{7} \\
 -\frac{24}{7} & -\frac{3}{7} \\
\end{array}
\right| : \\
  \text{= }\hat{\text{i}} \left| 
\begin{array}{cc}
 -\frac{58}{7} & \frac{67}{7} \\
 -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right| +\left(-\hat{\text{j}}\right) \left| 
\begin{array}{cc}
 \frac{53}{7} & \frac{67}{7} \\
 -\frac{24}{7} & \frac{68}{7} \\
\end{array}
\right| +\hat{\text{k}} \left| 
\begin{array}{cc}
 \frac{53}{7} & -\frac{58}{7} \\
 -\frac{24}{7} & -\frac{3}{7} \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \hat{\text{i}} \left| 
\begin{array}{cc}
 -\frac{58}{7} & \frac{67}{7} \\
 -\frac{3}{7} & \frac{68}{7} \\
\end{array}
\right| =\hat{\text{i}} \left(\left(-\frac{58}{7}\right)\, \times \, \frac{68}{7}-\frac{67 (-3)}{7\ 7}\right)=\frac{\hat{\text{i}} (-3743)}{49}=\fbox{$-\frac{3743 \hat{\text{i}}}{49}$}: \\
  \text{= }\fbox{$-\frac{3743 \hat{\text{i}}}{49}$}+\left(-\hat{\text{j}}\right) \left| 
\begin{array}{cc}
 \frac{53}{7} & \frac{67}{7} \\
 -\frac{24}{7} & \frac{68}{7} \\
\end{array}
\right| +\hat{\text{k}} \left| 
\begin{array}{cc}
 \frac{53}{7} & -\frac{58}{7} \\
 -\frac{24}{7} & -\frac{3}{7} \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 -\hat{\text{j}} \left| 
\begin{array}{cc}
 \frac{53}{7} & \frac{67}{7} \\
 -\frac{24}{7} & \frac{68}{7} \\
\end{array}
\right| =-\hat{\text{j}} \left(\frac{53\ 68}{7\ 7}-\frac{67 (-24)}{7\ 7}\right)=\frac{-\hat{\text{j}} 5212}{49}=\fbox{$-\frac{5212 \hat{\text{j}}}{49}$}: \\
  \text{= }\frac{-3743 \hat{\text{i}}}{49}+\fbox{$-\frac{5212 \hat{\text{j}}}{49}$}+\hat{\text{k}} \left| 
\begin{array}{cc}
 \frac{53}{7} & -\frac{58}{7} \\
 -\frac{24}{7} & -\frac{3}{7} \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \hat{\text{k}} \left| 
\begin{array}{cc}
 \frac{53}{7} & -\frac{58}{7} \\
 -\frac{24}{7} & -\frac{3}{7} \\
\end{array}
\right| =\hat{\text{k}} \left(\frac{53 (-3)}{7\ 7}-\left(-\frac{58}{7}\right)\, \left(-\frac{24}{7}\right)\right)=\frac{\hat{\text{k}} (-1551)}{49}=\fbox{$-\frac{1551 \hat{\text{k}}}{49}$}: \\
  \text{= }\frac{-3743 \hat{\text{i}}}{49}-\frac{5212 \hat{\text{j}}}{49}+\fbox{$-\frac{1551 \hat{\text{k}}}{49}$} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
  \text{= }-\frac{3743 \hat{\text{i}}}{49}-\frac{5212 \hat{\text{j}}}{49}-\frac{1551 \hat{\text{k}}}{49} \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{3743 \hat{\text{i}}}{49}-\frac{5212 \hat{\text{j}}}{49}-\frac{1551 \hat{\text{k}}}{49}=\, \left(-\frac{3743}{49},-\frac{5212}{49},-\frac{1551}{49}\right)\, : \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \, \left(-\frac{3743}{49},-\frac{5212}{49},-\frac{1551}{49}\right)\,  \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/cross_product_w_steps/4034.txt'}