91影视

Consider a set $span\left\{{\stackrel{鈫�}{v}}_1=\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right],{\stackrel{鈫�}{v}}_2=\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right]\right\}$spanned ${\mathrm{鈩�}}^3{\mathrm{鈩�}}^3$and a vector$\stackrel{鈫�}{x}=\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]$.

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

By the theorem of orthogonal basis, the orthogonal${\stackrel{鈫�}{u}}_1$is defined as follows.

${\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||}$

Substitute the value$\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]$for${\stackrel{鈫�}{v}}_1$in the equation${\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||}$as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||} \\ {\stackrel{鈫�}{u}}_1=\frac{\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]}{||\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]||} \\ {\stackrel{鈫�}{u}}_1=\frac{1}{\sqrt{2^2+3^2+6^2}}\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right] \\ {\stackrel{鈫�}{u}}_1=\frac{1}{7}\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_1=\frac{1}{7}\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right] \\ {\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}2/7\\ 3/7\\ 6/7\end{array}\right] \end{gathered}$$

Similarly, the orthogonal is defined as follows.

${\stackrel{鈫�}{u}}_2=\frac{{\stackrel{鈫�}{v}}_2}{||{\stackrel{鈫�}{v}}_2||}$

Substitute the value$\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]$for${\stackrel{鈫�}{v}}_2$in the equation${\stackrel{鈫�}{u}}_2=\frac{{\stackrel{鈫�}{v}}_2}{||{\stackrel{鈫�}{v}}_2||}$as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{{\stackrel{鈫�}{v}}_2}{||{\stackrel{鈫�}{v}}_2||} \\ {\stackrel{鈫�}{u}}_2=\frac{\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right]}{||\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right]||} \\ {\stackrel{鈫�}{u}}_2=\frac{1}{\sqrt{{\left(3\right)}^2+{\left(-6\right)}^2+{\left(2\right)}^2}}\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\frac{1}{7}\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{1}{7}\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}3/7\\ -6/7\\ 2/7\end{array}\right] \end{gathered}$$

Therefore, the orthogonal basis is$span\left\{{\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}2/7\\ 3/7\\ 6/7\end{array}\right],{\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}3/7\\ -6/7\\ 2/7\end{array}\right]\right\}$.

${\stackrel{鈫�}{u}}_2$By the theorem of orthogonal projection, the orthogonal projection of vector$\stackrel{鈫�}{x}$ is defined as follows.

$pro{j}_v\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{u}}_1.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_1+\left({\stackrel{鈫�}{u}}_2.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_2$

Substitute the value $\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]$for ${\stackrel{鈫�}{u}}_1,\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right]$for${\stackrel{鈫�}{u}}_2$ and$\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]$ for$\stackrel{鈫�}{x}$ in the equation$pro{j}_v\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{u}}_1.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_1+\left({\stackrel{鈫�}{u}}_2.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_2$ as follows.

$$\begin{gathered} pro{j}_v\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{u}}_1.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_1+\left({\stackrel{鈫�}{u}}_2.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_2 \\ pro{j}_v\stackrel{鈫�}{x}=\left(\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right].\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]\right)\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]+\left(\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right].\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]\right)\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \\ pro{j}_v\stackrel{鈫�}{x}=\left(\frac{2}{7}.49+\frac{3}{7}.49+\frac{6}{7}.49\right)\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]+\left(\frac{3}{7}.49-\frac{6}{7}.49+\frac{2}{7}.49\right)\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \\ pro{j}_v\stackrel{鈫�}{x}=77\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]-7\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} pro{j}_v\stackrel{鈫�}{x}=77\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]-7\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \\ pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{ccc}22& 33& 66\end{array}\right]-\left[\begin{array}{ccc}3& -6& 2\end{array}\right] \\ pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{ccc}19& 39& 64\end{array}\right] \end{gathered}$$

Hence, the orthogonal projection of the vector$\stackrel{鈫�}{x}$ is$pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{ccc}19& 39& 64\end{array}\right]$ .