Question

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.

18. \(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right),{\rm{ }}\left( {\begin{array}{*{20}{c}}0\\{ - 1}\\0\end{array}} \right)\)

Short answer

The set of vectors \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not orthogonal because \({\bf{u}} \cdot {\bf{v}} \ne 0\).

Step-by-step solution

Definition of orthonormal sets

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

Check whether the set of vectors are orthogonal

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

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

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

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not an orthogonal set because \({\bf{u}} \cdot {\bf{v}} \ne 0\).