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.

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

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

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

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

\(\begin{array}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\0\\{\frac{1}{2}}\end{array}} \right)\\ = \frac{1}{3}\left( { - \frac{1}{2}} \right) + \frac{1}{3}\left( 0 \right) + \frac{1}{3}\left( {\frac{1}{2}} \right)\\ = - \frac{1}{6} + 0 + \frac{1}{6}\\ = 0\end{array}\)

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:

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

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:

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

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