Answer
\[{\mathbf{u}} \times {\mathbf{v}} = \left\langle {6,11,5} \right\rangle {\text{ and }}{\mathbf{u}} \times {\mathbf{v}} = \left\langle { - 6, - 11, - 5} \right\rangle \]
Work Step by Step
\[\begin{gathered}
{\mathbf{u}} = \left\langle {2,3, - 9} \right\rangle ,\,\,\,{\mathbf{v}} = \left\langle { - 1,1, - 1} \right\rangle \hfill \\
\hfill \\
{\text{Calculate }}{\mathbf{u}} \times {\mathbf{v}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
2&3&{ - 9} \\
{ - 1}&1&{ - 1}
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
3&{ - 9} \\
1&{ - 1}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
2&{ - 9} \\
{ - 1}&{ - 1}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
2&3 \\
{ - 1}&1
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( { - 3 + 9} \right){\mathbf{i}} - \left( { - 2 - 9} \right){\mathbf{j}} + \left( {2 + 3} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = 6{\mathbf{i}} + 11{\mathbf{j}} + 5{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left\langle {6,11,5} \right\rangle \hfill \\
\hfill \\
{\text{Calculate }}{\mathbf{v}} \times {\mathbf{u}} \hfill \\
{\mathbf{v}} \times {\mathbf{u}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 1}&1&{ - 1} \\
2&3&{ - 9}
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
1&{ - 1} \\
3&{ - 9}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 1}&{ - 1} \\
2&{ - 9}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 1}&1 \\
2&3
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( { - 9 + 3} \right){\mathbf{i}} - \left( {9 + 2} \right){\mathbf{j}} + \left( { - 3 - 2} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = - 6{\mathbf{i}} - 11{\mathbf{j}} - 5{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left\langle { - 6, - 11, - 5} \right\rangle \hfill \\
\end{gathered} \]
