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.

21. \(\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {20} }}}\\{\frac{3}{{\sqrt {20} }}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{3}{{\sqrt {10} }}}\\{\frac{{ - 1}}{{\sqrt {20} }}}\\{\frac{{ - 1}}{{\sqrt {20} }}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}0\\{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{array}} \right)\)

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

Consider that \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {20} }}}\\{\frac{3}{{\sqrt {20} }}}\end{array}} \right),{\rm{ }}{\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{3}{{\sqrt {10} }}}\\{ - \frac{1}{{\sqrt {20} }}}\\{ - \frac{1}{{\sqrt {20} }}}\end{array}} \right)\), and \({\bf{w}} = \left( {\begin{array}{*{20}{c}}0\\{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{array}} \right)\).

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

\(\begin{array}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {20} }}}\\{\frac{3}{{\sqrt {20} }}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{\frac{3}{{\sqrt {10} }}}\\{ - \frac{1}{{\sqrt {20} }}}\\{ - \frac{1}{{\sqrt {20} }}}\end{array}} \right)\\ = \left( {\frac{1}{{\sqrt {10} }}} \right)\left( {\frac{3}{{\sqrt {10} }}} \right) + \frac{3}{{\sqrt {20} }}\left( { - \frac{1}{{\sqrt {20} }}} \right) + \frac{3}{{\sqrt {20} }}\left( { - \frac{1}{{\sqrt {20} }}} \right)\\ = \frac{3}{{10}} - \frac{3}{{20}} - \frac{3}{{20}}\\ = 0\end{array}\)

\(\begin{array}{c}{\bf{u}} \cdot {\bf{w}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {20} }}}\\{\frac{3}{{\sqrt {20} }}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}0\\{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{array}} \right)\\ = \left( {\frac{1}{{\sqrt {10} }}} \right)\left( 0 \right) + \frac{3}{{\sqrt {20} }}\left( { - \frac{1}{{\sqrt 2 }}} \right) + \frac{3}{{\sqrt {20} }}\left( {\frac{1}{{\sqrt 2 }}} \right)\\ = - \frac{{3\sqrt {10} }}{{\sqrt {20} }} + \frac{{3\sqrt {10} }}{{\sqrt {20} }}\\ = 0\end{array}\)

And,

\(\begin{array}{c}{\bf{v}} \cdot {\bf{w}} = \left( {\begin{array}{*{20}{c}}{\frac{3}{{\sqrt {10} }}}\\{ - \frac{1}{{\sqrt {20} }}}\\{ - \frac{1}{{\sqrt {20} }}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}0\\{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{array}} \right)\\ = \left( {\frac{3}{{\sqrt {10} }}} \right)\left( 0 \right) - \frac{1}{{\sqrt {20} }}\left( { - \frac{1}{{\sqrt 2 }}} \right) - \frac{1}{{\sqrt {20} }}\left( {\frac{1}{{\sqrt 2 }}} \right)\\ = \frac{{\sqrt {10} }}{{\sqrt {20} }} - \frac{{\sqrt {10} }}{{\sqrt {20} }}\\ = 0\end{array}\)

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:

\(\begin{array}{c}{\left\| {\bf{u}} \right\|^2} = {\bf{u}} \cdot {\bf{u}}\\ = {\left( {\frac{1}{{\sqrt {10} }}} \right)^2} + {\left( {\frac{3}{{\sqrt {20} }}} \right)^2} + {\left( {\frac{3}{{\sqrt {20} }}} \right)^2}\\ = \left( {\frac{1}{{10}}} \right) + \left( {\frac{9}{{20}}} \right) + \left( {\frac{9}{{20}}} \right)\\ = 1\end{array}\)

\(\begin{array}{c}{\left\| {\bf{v}} \right\|^2} = {\bf{v}} \cdot {\bf{v}}\\ = {\left( {\frac{3}{{\sqrt {10} }}} \right)^2} + {\left( { - \frac{1}{{\sqrt {20} }}} \right)^2} + {\left( { - \frac{1}{{\sqrt {20} }}} \right)^2}\\ = \frac{9}{{10}} + \frac{1}{{20}} + \frac{1}{{20}}\\ = 1\end{array}\)

And,

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

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