91Ó°ÊÓ

Let the vectors \({{\bf{v}}_1} = \left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),{{\bf{v}}_2} = \left( {\begin{aligned}{{}{}}3\\{\frac{3}{2}}\\{\frac{3}{2}}\end{aligned}} \right)\)from Exercise 3.

Compute \(\left\| {{{\bf{v}}_1}} \right\|\) and \(\left\| {{{\bf{v}}_2}} \right\|\) as shown below:

\(\begin{aligned}{}\left\| {{{\bf{v}}_1}} \right\| &= \sqrt {{2^2} + {{\left( { - 5} \right)}^2} + {1^2}} \\ &= \sqrt {4 + 25 + 1} \\ &= \sqrt {30} \\\left\| {{{\bf{v}}_2}} \right\| &= \sqrt {{3^2} + {{\left( {\frac{3}{2}} \right)}^2} + {{\left( {\frac{3}{2}} \right)}^2}} \\ &= \sqrt {9 + \left( {\frac{9}{4}} \right) + \left( {\frac{9}{4}} \right)} \\ &= \sqrt {\frac{{54}}{4}} \\ &= \sqrt {\frac{{27}}{2}} \\ &= \frac{{3\sqrt 6 }}{2}\end{aligned}\)

Obtain an orthonormal basis as shown below:

\(\begin{aligned}{}\left\{ {\frac{{{{\bf{v}}_1}}}{{\left\| {{{\bf{v}}_1}} \right\|}},\frac{{{{\bf{v}}_2}}}{{\left\| {{{\bf{v}}_2}} \right\|}}} \right\} &= \left\{ {\frac{1}{{\sqrt {30} }}\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\frac{1}{{\frac{{3\sqrt 6 }}{2}}}\left( {\begin{aligned}{{}{}}3\\{\frac{3}{2}}\\{\frac{3}{2}}\end{aligned}} \right)} \right\}\\ &= \left\{ {\left( {\begin{aligned}{{}{}}{\frac{2}{{\sqrt {30} }}}\\{\frac{{ - 5}}{{\sqrt {30} }}}\\{\frac{1}{{\sqrt {30} }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{\frac{2}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\end{aligned}} \right)} \right\}\end{aligned}\)

Hence, an orthonormal basis is\(\left\{ {\left( {\begin{aligned}{{}{}}{\frac{2}{{\sqrt {30} }}}\\{\frac{{ - 5}}{{\sqrt {30} }}}\\{\frac{1}{{\sqrt {30} }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{\frac{2}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\end{aligned}} \right)} \right\}\).