91影视

Consider an orthonormal vector ${\stackrel{鈫�}{u}}_1,{\stackrel{鈫�}{u}}_2,{\stackrel{鈫�}{u}}_3,{\stackrel{鈫�}{u}}_4,{\stackrel{鈫�}{u}}_5$in ${\mathrm{鈩�}}^{10}$such that $\stackrel{鈫�}{x}=7{\stackrel{鈫�}{u}}_1-3{\stackrel{鈫�}{u}}_2+2{\stackrel{鈫�}{u}}_3+{\stackrel{鈫�}{u}}_4-{\stackrel{鈫�}{u}}_5$.

The property of dot product is defined as follows:

1. $\left(\stackrel{鈫�}{x}+\stackrel{鈫�}{y}\right)路\stackrel{鈫�}{w}=\stackrel{鈫�}{x}路\stackrel{鈫�}{w}+\stackrel{鈫�}{y}路\stackrel{鈫�}{w}$
2. $\left(c\stackrel{鈫�}{x}\right)路\stackrel{鈫�}{w}=c\left(\stackrel{鈫�}{x}路\stackrel{鈫�}{w}\right)$
3. ${\stackrel{鈫�}{u}}_i路{\stackrel{鈫�}{u}}_j=\left\{\begin{array}{cc}0& ifi鈮�j\\ 1& ifi=j\end{array}\right\}$
   

Simplify the equation ${||\stackrel{鈫�}{x}||}^2=\stackrel{鈫�}{x}路\stackrel{鈫�}{x}$as follows:

role="math" localid="1659539563887" $$\begin{gathered} {||\stackrel{鈫�}{x}||}^2=\stackrel{鈫�}{x}路\stackrel{鈫�}{x} \\ {||\stackrel{鈫�}{x}||}^2=\left\{7{\stackrel{鈫�}{u}}_1-3{\stackrel{鈫�}{u}}_2+2{\stackrel{鈫�}{u}}_3+{\stackrel{鈫�}{u}}_4-{\stackrel{鈫�}{u}}_5\right\}路\left\{7{\stackrel{鈫�}{u}}_1-3{\stackrel{鈫�}{u}}_2+2{\stackrel{鈫�}{u}}_3+{\stackrel{鈫�}{u}}_4-{\stackrel{鈫�}{u}}_5\right\} \\ {||\stackrel{鈫�}{x}||}^2=\left\{49\left({\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{u}}_1\right)-21\left({\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{u}}_2\right)+14\left({\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{u}}_3\right)+7\left({\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{u}}_4\right)-7\left({\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{u}}_5\right) \\ -21\left({\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{u}}_1\right)+9\left({\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{u}}_2\right)-6\left({\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{u}}_3\right)-3\left({\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{u}}_4\right)+3\left({\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{u}}_5\right) \\ +14\left({\stackrel{鈫�}{u}}_3路{\stackrel{鈫�}{u}}_1\right)-6\left({\stackrel{鈫�}{u}}_3路{\stackrel{鈫�}{u}}_2\right)+4\left({\stackrel{鈫�}{u}}_3路{\stackrel{鈫�}{u}}_3\right)+2\left({\stackrel{鈫�}{u}}_3路{\stackrel{鈫�}{u}}_4\right)-2\left({\stackrel{鈫�}{u}}_3路{\stackrel{鈫�}{u}}_5\right) \\ +7\left({\stackrel{鈫�}{u}}_4路{\stackrel{鈫�}{u}}_1\right)-3\left({\stackrel{鈫�}{u}}_4路{\stackrel{鈫�}{u}}_2\right)+2\left({\stackrel{鈫�}{u}}_4路{\stackrel{鈫�}{u}}_3\right)+\left({\stackrel{鈫�}{u}}_4路{\stackrel{鈫�}{u}}_4\right)-\left({\stackrel{鈫�}{u}}_4路{\stackrel{鈫�}{u}}_5\right) \\ -7\left({\stackrel{鈫�}{u}}_5路{\stackrel{鈫�}{u}}_1\right)+3\left({\stackrel{鈫�}{u}}_5路{\stackrel{鈫�}{u}}_2\right)-2\left({\stackrel{鈫�}{u}}_5路{\stackrel{鈫�}{u}}_3\right)-\left({\stackrel{鈫�}{u}}_5路{\stackrel{鈫�}{u}}_4\right)+\left({\stackrel{鈫�}{u}}_5路{\stackrel{鈫�}{u}}_5\right) \\ \right\} \end{gathered}$$

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

$$\begin{gathered} {||\stackrel{鈫�}{x}||}^2=49+9+4+1+1 \\ {||\stackrel{鈫�}{x}||}^2=64 \\ ||\stackrel{鈫�}{x}||=8 \end{gathered}$$

Hence, the length of the vector$\stackrel{鈫�}{x}=7{\stackrel{鈫�}{u}}_1-3{\stackrel{鈫�}{u}}_2+2{\stackrel{鈫�}{u}}_3+{\stackrel{鈫�}{u}}_4-{\stackrel{鈫�}{u}}_5$ is $8$.