Answer
\[{\mathbf{u}} \times {\mathbf{v}} = \left\langle { - 2, - 4, - 4} \right\rangle {\text{ and }}{\mathbf{u}} \times {\mathbf{v}} = \left\langle {2,4,4} \right\rangle \]
Work Step by Step
\[\begin{gathered}
{\mathbf{u}} = \left\langle { - 4,1,1} \right\rangle ,\,\,\,{\mathbf{v}} = \left\langle {0,1, - 1} \right\rangle \hfill \\
\hfill \\
{\text{Calculate }}{\mathbf{u}} \times {\mathbf{v}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 4}&1&1 \\
0&1&{ - 1}
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{ - 1}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 4}&1 \\
0&{ - 1}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 4}&1 \\
0&1
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( { - 1 - 1} \right){\mathbf{i}} - \left( {4 - 0} \right){\mathbf{j}} + \left( { - 4 - 0} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = - 2{\mathbf{i}} - 4{\mathbf{j}} - 4{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left\langle { - 2, - 4, - 4} \right\rangle \hfill \\
\hfill \\
{\text{Calculate }}{\mathbf{v}} \times {\mathbf{u}} \hfill \\
{\mathbf{v}} \times {\mathbf{u}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
0&1&{ - 1} \\
{ - 4}&1&1
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
1&{ - 1} \\
1&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
0&{ - 1} \\
{ - 4}&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
0&1 \\
{ - 4}&1
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( {1 + 1} \right){\mathbf{i}} - \left( {0 - 4} \right){\mathbf{j}} + \left( {0 + 4} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = 2{\mathbf{i}} + 4{\mathbf{j}} + 4{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left\langle {2,4,4} \right\rangle \hfill \\
\end{gathered} \]
