Question: In Exercises 17-22, determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set.

20. \(\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)\)

A set \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}, \ldots ,{{\bf{u}}_p}} \right\}\) is called anorthonormal set when it is an orthogonal set of unit vectors.

Consider that \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),{\rm{ }}{\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)\).

Check whether the set of the vector is orthogonal, as shown below:

\(\begin{array}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)\\ = \left( { - \frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \frac{1}{3}\left( {\frac{2}{3}} \right) + 0\\ = - \frac{2}{9} + \frac{2}{9}\\ = 0\end{array}\)

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is an orthogonal set because \({\bf{u}} \cdot {\bf{v}} = 0\).

Check whether the set of vectors is orthonormal as shown below:

\(\begin{array}{c}{\left\| {\bf{u}} \right\|^2} = {\bf{u}} \cdot {\bf{u}}\\ = {\left( { - \frac{2}{3}} \right)^2} + {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{2}{3}} \right)^2}\\ = \left( {\frac{4}{9}} \right) + \left( {\frac{1}{9}} \right) + \left( {\frac{4}{9}} \right)\\ = \frac{9}{9}\\ = 1\\{\left\| {\bf{v}} \right\|^2} = {\bf{v}} \cdot {\bf{v}}\\ = {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{2}{3}} \right)^2} + 0\\ = \frac{1}{9} + \frac{4}{9}\\ = \frac{5}{9}\end{array}\)

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not an orthonormal set.

Normalize the vectors to produce the orthonormal set as shown below:

\(\begin{array}{c}\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\frac{1}{{\sqrt 1 }}\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\frac{1}{{\sqrt {\frac{5}{9}} }}\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)} \right\}\\ = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\frac{3}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)} \right\}\\ = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\\0\end{array}} \right)} \right\}\end{array}\)

Therefore, the orthonormal set is \(\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\\0\end{array}} \right)} \right\}\).