91影视

Consider a vector $\stackrel{鈫�}{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$and $\stackrel{鈫�}{x}=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]$.

If the vector $\stackrel{\mathbf{鈫�}}{\mathbf{x}}$ is perpendicular $\stackrel{\mathbf{鈫�}}{\mathbf{y}}$then $\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{鈰�}\stackrel{\mathbf{鈫�}}{\mathbf{y}}\mathbf{=}\mathbf{0}$.

As$\stackrel{鈫�}{v}$ is perpendicular to $\stackrel{鈫�}{x}$, by the definition of orthogonal $\stackrel{鈫�}{x}鈰�\stackrel{鈫�}{v}=0$.

Simplify the equation$\stackrel{鈫�}{x}鈰�\stackrel{鈫�}{v}=0$ as follows.

$\begin{array}{c}\stackrel{鈫�}{x}鈰�\stackrel{鈫�}{v}=0\\ \left[\begin{array}{cccc}{\stackrel{鈫�}{x}}_1& {\stackrel{鈫�}{x}}_2& {\stackrel{鈫�}{x}}_3& {\stackrel{鈫�}{x}}_4\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=0\\ {\stackrel{鈫�}{x}}_1+2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4=0\\ {\stackrel{鈫�}{x}}_1=鈭�\left\{2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4\right\}\end{array}$

Substitute the value $鈭�\left\{2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4\right\}$for $\stackrel{鈫�}{x_1}$in the equation $\stackrel{鈫�}{x}=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]$as follows.

$\begin{array}{l}\stackrel{鈫�}{x}=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]\\ \stackrel{鈫�}{x}=\left[\begin{array}{c}鈭�\left\{2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4\right\}\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]\end{array}$

Therefore, any vector in the form$\begin{array}{l}\\ \stackrel{鈫�}{x}=\left[\begin{array}{c}鈭�\left\{2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4\right\}\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]\end{array}$ is perpendicular to $\stackrel{鈫�}{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$.

Simplify the equation$\stackrel{鈫�}{x}=\left[\begin{array}{c}鈭�\left\{2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4\right\}\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]$ as follows.

$\begin{array}{c}\stackrel{鈫�}{x}=\left[\begin{array}{c}鈭�\left\{2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3+4{\stackrel{鈫�}{x}}_4\right\}\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_4\end{array}\right]\\ =\left[\begin{array}{c}鈭�2{\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_2\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}鈭�3{\stackrel{鈫�}{x}}_3\\ 0\\ {\stackrel{鈫�}{x}}_3\\ 0\end{array}\right]+\left[\begin{array}{c}鈭�4{\stackrel{鈫�}{x}}_4\\ 0\\ 0\\ {\stackrel{鈫�}{x}}_4\end{array}\right]\\ \stackrel{鈫�}{x}={\stackrel{鈫�}{x}}_2\left[\begin{array}{c}鈭�2\\ 1\\ 0\\ 0\end{array}\right]+{\stackrel{鈫�}{x}}_3\left[\begin{array}{c}鈭�3\\ 0\\ 1\\ 0\end{array}\right]+{\stackrel{鈫�}{x}}_4\left[\begin{array}{c}鈭�4\\ 0\\ 0\\ 1\end{array}\right]\end{array}$

Therefore, the vector $\stackrel{鈫�}{x}鈭�span\left\{\left[\begin{array}{c}鈭�2\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}鈭�3\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}鈭�4\\ 0\\ 0\\ 1\end{array}\right]\right\}$.

Hence, the vector$\stackrel{鈫�}{x}$ is spanned by $\left\{\left(\begin{array}{cccc}鈭�2& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}鈭�3& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}鈭�4& 0& 0& 1\end{array}\right)\right\}$which is perpendicular to$\stackrel{鈫�}{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$ .