• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q29E

Expert-verified Found in: Page 217 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider the orthonormal vectors ${\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{1}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{2}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{3}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{4}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{5}}}$ in ${{\mathbf{ℝ}}}^{{\mathbf{10}}}$ . Find the length of the vector $\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{=}}{\mathbf{7}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{1}}}{\mathbf{-}}{\mathbf{3}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{2}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{3}}}{\mathbf{+}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{4}}}{\mathbf{-}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{5}}}$ .

The length of the vector $\stackrel{\to }{x}=7{\stackrel{\to }{u}}_{1}-3{\stackrel{\to }{u}}_{2}+2{\stackrel{\to }{u}}_{3}+{\stackrel{\to }{u}}_{4}-{\stackrel{\to }{u}}_{5}$ is $8$.

See the step by step solution

## Step by Step Solution Step 1: Properties

Consider an orthonormal vector ${\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{2},{\stackrel{\to }{u}}_{3},{\stackrel{\to }{u}}_{4},{\stackrel{\to }{u}}_{5}$ in ${\mathrm{ℝ}}^{10}$ such that $\stackrel{\to }{x}=7{\stackrel{\to }{u}}_{1}-3{\stackrel{\to }{u}}_{2}+2{\stackrel{\to }{u}}_{3}+{\stackrel{\to }{u}}_{4}-{\stackrel{\to }{u}}_{5}$.

The property of dot product is defined as follows:

1. $\left(\stackrel{\to }{x}+\stackrel{\to }{y}\right){·}\stackrel{\to }{w}{=}\stackrel{\to }{x}{·}\stackrel{\to }{w}{+}\stackrel{\to }{y}{·}\stackrel{\to }{w}$
2. $\left(c\stackrel{\to }{x}\right){·}\stackrel{\to }{w}{=}{c}\left(\stackrel{\to }{x}·\stackrel{\to }{w}\right)$
3. ${\stackrel{\to }{u}}_{{i}}{·}{\stackrel{\to }{u}}_{{j}}{=}\left\{\begin{array}{cc}0& ifi\ne j\\ 1& ifi=j\end{array}\right\}$

## Step 2: Determine the length of the vector x→

Simplify the equation ${||\stackrel{\to }{x}||}^{2}=\stackrel{\to }{x}·\stackrel{\to }{x}$ as follows:

role="math" localid="1659539563887" ${||\stackrel{\to }{x}||}^{2}=\stackrel{\to }{x}·\stackrel{\to }{x}\phantom{\rule{0ex}{0ex}}{||\stackrel{\to }{x}||}^{2}=\left\{7{\stackrel{\to }{u}}_{1}-3{\stackrel{\to }{u}}_{2}+2{\stackrel{\to }{u}}_{3}+{\stackrel{\to }{u}}_{4}-{\stackrel{\to }{u}}_{5}\right\}·\left\{7{\stackrel{\to }{u}}_{1}-3{\stackrel{\to }{u}}_{2}+2{\stackrel{\to }{u}}_{3}+{\stackrel{\to }{u}}_{4}-{\stackrel{\to }{u}}_{5}\right\}\phantom{\rule{0ex}{0ex}}{||\stackrel{\to }{x}||}^{2}=\left\{49\left({\stackrel{\to }{u}}_{1}·{\stackrel{\to }{u}}_{1}\right)-21\left({\stackrel{\to }{u}}_{1}·{\stackrel{\to }{u}}_{2}\right)+14\left({\stackrel{\to }{u}}_{1}·{\stackrel{\to }{u}}_{3}\right)+7\left({\stackrel{\to }{u}}_{1}·{\stackrel{\to }{u}}_{4}\right)-7\left({\stackrel{\to }{u}}_{1}·{\stackrel{\to }{u}}_{5}\right)\phantom{\rule{0ex}{0ex}}-21\left({\stackrel{\to }{u}}_{2}·{\stackrel{\to }{u}}_{1}\right)+9\left({\stackrel{\to }{u}}_{2}·{\stackrel{\to }{u}}_{2}\right)-6\left({\stackrel{\to }{u}}_{2}·{\stackrel{\to }{u}}_{3}\right)-3\left({\stackrel{\to }{u}}_{2}·{\stackrel{\to }{u}}_{4}\right)+3\left({\stackrel{\to }{u}}_{2}·{\stackrel{\to }{u}}_{5}\right)\phantom{\rule{0ex}{0ex}}+14\left({\stackrel{\to }{u}}_{3}·{\stackrel{\to }{u}}_{1}\right)-6\left({\stackrel{\to }{u}}_{3}·{\stackrel{\to }{u}}_{2}\right)+4\left({\stackrel{\to }{u}}_{3}·{\stackrel{\to }{u}}_{3}\right)+2\left({\stackrel{\to }{u}}_{3}·{\stackrel{\to }{u}}_{4}\right)-2\left({\stackrel{\to }{u}}_{3}·{\stackrel{\to }{u}}_{5}\right)\phantom{\rule{0ex}{0ex}}+7\left({\stackrel{\to }{u}}_{4}·{\stackrel{\to }{u}}_{1}\right)-3\left({\stackrel{\to }{u}}_{4}·{\stackrel{\to }{u}}_{2}\right)+2\left({\stackrel{\to }{u}}_{4}·{\stackrel{\to }{u}}_{3}\right)+\left({\stackrel{\to }{u}}_{4}·{\stackrel{\to }{u}}_{4}\right)-\left({\stackrel{\to }{u}}_{4}·{\stackrel{\to }{u}}_{5}\right)\phantom{\rule{0ex}{0ex}}-7\left({\stackrel{\to }{u}}_{5}·{\stackrel{\to }{u}}_{1}\right)+3\left({\stackrel{\to }{u}}_{5}·{\stackrel{\to }{u}}_{2}\right)-2\left({\stackrel{\to }{u}}_{5}·{\stackrel{\to }{u}}_{3}\right)-\left({\stackrel{\to }{u}}_{5}·{\stackrel{\to }{u}}_{4}\right)+\left({\stackrel{\to }{u}}_{5}·{\stackrel{\to }{u}}_{5}\right)\phantom{\rule{0ex}{0ex}}\right\}$

By the property of dot product, simplify the equation as follows:

${||\stackrel{\to }{x}||}^{2}=49+9+4+1+1\phantom{\rule{0ex}{0ex}}{||\stackrel{\to }{x}||}^{2}=64\phantom{\rule{0ex}{0ex}}||\stackrel{\to }{x}||=8$

Hence, the length of the vector $\stackrel{\to }{x}=7{\stackrel{\to }{u}}_{1}-3{\stackrel{\to }{u}}_{2}+2{\stackrel{\to }{u}}_{3}+{\stackrel{\to }{u}}_{4}-{\stackrel{\to }{u}}_{5}$ is $8$. ### Want to see more solutions like these? 