Problem:
Find the cross product of the following vectors:
$\left(
\begin{array}{c}
 -4 \\
 -4 \\
 3 \\
\end{array}
\right)$ and
$\left(
\begin{array}{c}
 -3 \\
 10 \\
 6 \\
\end{array}
\right)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Compute the following cross product}: \\
 \, (-4,-4,3)\, \times \, (-3,10,6)\,  \\
\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 }\, (-4,-4,3)\,  \text{and }\, (-3,10,6)\, : \\
 \left(
\begin{array}{ccc}
 \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
 -4 & -4 & 3 \\
 -3 & 10 & 6 \\
\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}} \\
 -4 & -4 & 3 \\
 -3 & 10 & 6 \\
\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}} \\
 -4 & -4 & 3 \\
 -3 & 10 & 6 \\
\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}} \\
 -4 & -4 & 3 \\
 -3 & 10 & 6 \\
\end{array}
\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
\begin{array}{cc}
 -4 & 3 \\
 10 & 6 \\
\end{array}
\right| +\left(-\hat{\text{j}}\right) \left| 
\begin{array}{cc}
 -4 & 3 \\
 -3 & 6 \\
\end{array}
\right| +\hat{\text{k}} \left| 
\begin{array}{cc}
 -4 & -4 \\
 -3 & 10 \\
\end{array}
\right| : \\
  \text{= }\hat{\text{i}} \left| 
\begin{array}{cc}
 -4 & 3 \\
 10 & 6 \\
\end{array}
\right| +\left(-\hat{\text{j}}\right) \left| 
\begin{array}{cc}
 -4 & 3 \\
 -3 & 6 \\
\end{array}
\right| +\hat{\text{k}} \left| 
\begin{array}{cc}
 -4 & -4 \\
 -3 & 10 \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \hat{\text{i}} \left| 
\begin{array}{cc}
 -4 & 3 \\
 10 & 6 \\
\end{array}
\right| =\hat{\text{i}} ((-4)\, \times \, 6-3\ 10)=\hat{\text{i}} (-54)=\fbox{$-54 \hat{\text{i}}$}: \\
  \text{= }\fbox{$-54 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
\begin{array}{cc}
 -4 & 3 \\
 -3 & 6 \\
\end{array}
\right| +\hat{\text{k}} \left| 
\begin{array}{cc}
 -4 & -4 \\
 -3 & 10 \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 -\hat{\text{j}} \left| 
\begin{array}{cc}
 -4 & 3 \\
 -3 & 6 \\
\end{array}
\right| =-\hat{\text{j}} ((-4)\, \times \, 6-3 (-3))=-\hat{\text{j}} (-15)=\fbox{$15 \hat{\text{j}}$}: \\
  \text{= }-54 \hat{\text{i}}+\fbox{$15 \hat{\text{j}}$}+\hat{\text{k}} \left| 
\begin{array}{cc}
 -4 & -4 \\
 -3 & 10 \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \hat{\text{k}} \left| 
\begin{array}{cc}
 -4 & -4 \\
 -3 & 10 \\
\end{array}
\right| =\hat{\text{k}} ((-4)\, \times \, 10-(-4)\, (-3))=\hat{\text{k}} (-52)=\fbox{$-52 \hat{\text{k}}$}: \\
  \text{= }-54 \hat{\text{i}}+15 \hat{\text{j}}+\fbox{$-52 \hat{\text{k}}$} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
  \text{= }-54 \hat{\text{i}}+15 \hat{\text{j}}-52 \hat{\text{k}} \\
\end{array}
 \\
 
\begin{array}{l}
 -54 \hat{\text{i}}+15 \hat{\text{j}}-52 \hat{\text{k}}=\, (-54,15,-52)\, : \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \, (-54,15,-52)\,  \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/cross_product_w_steps/564.txt'}