91影视

For any given vector $\stackrel{鈫�}{y}$, the reflection of it on any vector $\stackrel{鈫�}{z}$ is given by:

${\text{ref}}_L\left(\stackrel{鈫�}{y}\right)=2{\text{proj}}_L\left(\stackrel{鈫�}{y}\right)-\stackrel{鈫�}{y}$

Where, the projection of the given vector will be:

${\text{proj}}_L\left(\stackrel{鈫�}{y}\right)=\left(\frac{\stackrel{鈫�}{y}路\stackrel{鈫�}{z}}{\stackrel{鈫�}{z}路\stackrel{鈫�}{z}}\right)\stackrel{鈫�}{z}$

Given, the vector of the line onto which the reflection will fall upon is $\stackrel{鈫�}{z}=\left[\begin{array}{c}4\\ 3\end{array}\right]$.

Let us find the projection first as follows:

$$\begin{gathered} {\text{proj}}_L\left(\stackrel{鈫�}{y}\right)=\left(\frac{\begin{array}{cc}{\displaystyle \stackrel{鈫�}{y}}& \left[\begin{array}{c}4\\ 3\end{array}\right]\end{array}}{\begin{array}{cc}{\displaystyle \left[\begin{array}{c}4\\ 3\end{array}\right]}& \left[\begin{array}{c}4\\ 3\end{array}\right]\end{array}}\right)\left[\begin{array}{c}4\\ 3\end{array}\right] \\ =\left(\frac{{\displaystyle 4y_1+3y_2}}{16+9}\right)\left[\begin{array}{c}4\\ 3\end{array}\right] \\ =\frac{1}{25}\left[\begin{array}{c}4\left(4y_1+3y_2\right)\\ 3\left(4y_1+3y_2\right)\end{array}\right] \end{gathered}$$

On further evaluation, we get:

$$\begin{gathered} {\text{proj}}_L\left(\stackrel{鈫�}{y}\right)=\frac{1}{25}\left[\begin{array}{c}4\left(4y_1+3y_2\right)\\ 3\left(4y_1+3y_2\right)\end{array}\right] \\ =\frac{1}{25}\left[\begin{array}{c}16y_1+12y_2\\ 12y_1+9y_2\end{array}\right] \\ \frac{1}{25}\left[\begin{array}{cc}16& 12\\ 12& 9\end{array}\right]\stackrel{鈫�}{y} \end{gathered}$$

Now, the reflection of vector$\stackrel{鈫�}{y}$ on the vector localid="1660900077985" $\stackrel{鈫�}{z}$ will be:

$$\begin{gathered} {\text{ref}}_L\left(\stackrel{鈫�}{y}\right)=2{\text{proj}}_L\left(\stackrel{鈫�}{y}\right)-\stackrel{鈫�}{y} \\ =2\left\{\frac{1}{25}\left[\begin{array}{cc}16& 12\\ 12& 9\end{array}\right]\stackrel{鈫�}{y}\right\} \\ =2\left\{\frac{{\displaystyle 1}}{{\displaystyle 25}}\left[\begin{array}{cc}16& 12\\ 12& 9\end{array}\right]-1\right\}\stackrel{鈫�}{y} \\ 2\left\{\frac{{\displaystyle 1}}{{\displaystyle 25}}\left[\begin{array}{cc}16& 12\\ 12& 9\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right\}\stackrel{鈫�}{y} \end{gathered}$$

Proceeding the solution as follows:

role="math" localid="1660901336914" $$\begin{gathered} {\text{ref}}_L\left(\stackrel{鈫�}{y}\right)=\left\{\frac{{\displaystyle 2}}{{\displaystyle 25}}\left[\begin{array}{cc}16& 12\\ 12& 9\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right\}\stackrel{鈫�}{y} \\ =\begin{array}{c}\left[\begin{array}{cc}\frac{32}{25}-1& \frac{24}{25}\\ \frac{24}{25}& \frac{18}{25}-1\end{array}\right]\end{array}\stackrel{鈫�}{y} \\ \begin{array}{c}=\left[\begin{array}{cc}\frac{7}{25}& \frac{24}{25}\\ \frac{24}{25}& -\frac{7}{25}\end{array}\right]\end{array}\stackrel{鈫�}{y} \\ =\frac{1}{25}\left[\begin{array}{cc}7& 24\\ 24& -7\end{array}\right]\stackrel{鈫�}{y} \end{gathered}$$

Thus, the matrix of the reflection obtained is $A=\frac{1}{25}\left[\begin{array}{cc}7& 24\\ 24& -7\end{array}\right]$.

Hence, this is the required answer.