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.

22. \(\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {18} }}}\\{\frac{4}{{\sqrt {18} }}}\\{\frac{1}{{\sqrt {18} }}}\end{array}} \right),{\rm{ }}\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}\\0\\{ - \frac{1}{{\sqrt 2 }}}\end{array}} \right),{\rm{ }}\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{ - \frac{2}{3}}\end{array}} \right)\)

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

Consider that, \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {18} }}}\\{\frac{4}{{\sqrt {18} }}}\\{\frac{1}{{\sqrt {18} }}}\end{array}} \right),{\rm{ }}{\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}\\0\\{ - \frac{1}{{\sqrt 2 }}}\end{array}} \right),\) and \({\bf{w}} = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{ - \frac{2}{3}}\end{array}} \right)\).

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

\(\begin{array}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {18} }}}\\{\frac{4}{{\sqrt {18} }}}\\{\frac{1}{{\sqrt {18} }}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}\\0\\{ - \frac{1}{{\sqrt 2 }}}\end{array}} \right)\\ = \left( {\frac{1}{{\sqrt {18} }}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right) + \frac{4}{{\sqrt {18} }}\left( 0 \right) + \frac{1}{{\sqrt {18} }}\left( { - \frac{1}{{\sqrt 2 }}} \right)\\ = \frac{1}{6} - 0 - \frac{1}{6}\\ = 0\end{array}\)

\(\begin{array}{c}{\bf{u}} \cdot {\bf{w}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {18} }}}\\{\frac{4}{{\sqrt {18} }}}\\{\frac{1}{{\sqrt {18} }}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{ - \frac{2}{3}}\end{array}} \right)\\ = \left( {\frac{1}{{\sqrt {18} }}} \right)\left( { - \frac{2}{3}} \right) + \frac{4}{{\sqrt {18} }}\left( {\frac{1}{3}} \right) + \frac{1}{{\sqrt {18} }}\left( { - \frac{2}{3}} \right)\\ = - \frac{{\sqrt 2 }}{9} + \frac{{2\sqrt 2 }}{9} - \frac{{\sqrt 2 }}{9}\\ = 0\end{array}\)

And,

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

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

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

\(\begin{array}{c}{\left\| {\bf{u}} \right\|^2} = {\bf{u}} \cdot {\bf{u}}\\ = {\left( {\frac{1}{{\sqrt {18} }}} \right)^2} + {\left( {\frac{4}{{\sqrt {18} }}} \right)^2} + {\left( {\frac{1}{{\sqrt {18} }}} \right)^2}\\ = \left( {\frac{1}{{18}}} \right) + \left( {\frac{{16}}{{18}}} \right) + \left( {\frac{1}{{18}}} \right)\\ = 1\end{array}\)

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

And,

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

It is observed that \(\left\{ {{\bf{u}},{\bf{v}},{\bf{w}}} \right\}\) is an orthonormal set.