Let us consider the vector (x, y, z).

The vector is first rotated by \(\displaystyle{90}^{\circ}\) along x - axis and then scaled up by 5 times.

Ratation: In homogeneous coordinates, the matrix for a rotetion about the origin through a given angle \(\displaystyle\theta\ {i}{s}\ {B}={\left[\begin{matrix} \cos{\theta}&- \sin{\theta}&{0}\\ \sin{\theta}& \cos{\theta}&{0}\\{0}&{0}&{1}\end{matrix}\right]}\)

The vector (x, y, z) is rotated by \(\displaystyle{90}^{\circ}\) along x-axis then the vector becomes

\(\displaystyle{\left[\begin{matrix}{ \cos{{90}}^{\circ}-}{ \sin{{90}}^{\circ}{0}}\\{ \sin{{90}}^{\circ} \cos{{90}}^{\circ}}\\{0}{0}{1}\end{matrix}\right]}{\left[\begin{matrix}{x}\\{y}\\{z}\end{matrix}\right]}\)

\(\displaystyle={\left[\begin{matrix}{0}-{1}{0}\\{1}{0}{0}\\{0}{0}{1}\end{matrix}\right]}{\left[\begin{matrix}{x}\\{y}\\{z}\end{matrix}\right]}\)

\(\displaystyle={\left[\begin{matrix}-{y}\\{x}\\{z}\end{matrix}\right]}\)

Now, the vector is scaled up by 5 times.

Scaling: In homogeneous coordinates, a scaling about the origin with scale factors c in the x-direction and c in the y-direction is computed by multiplying the vector by the matrix

\(\displaystyle{A}={\left[\begin{matrix}{c}{0}{0}\\{0}{c}{0}\\{0}{0}{1}\end{matrix}\right]}\)

Thus, the vector becomes

\(\displaystyle{\left[\begin{matrix}{5}{0}{0}\\{0}{5}{0}\\{0}{0}{1}\end{matrix}\right]}{\left[\begin{matrix}{x}\\{y}\\{z}\end{matrix}\right]}\)

\(\displaystyle={\left[\begin{matrix}-{5}{y}\\{5}{x}\\{z}\end{matrix}\right]}\)

Now, given that the reduced vector is \(\displaystyle{\left({15},-{10},{20}\right)}\). Therefore, we have

\(\displaystyle{\left[\begin{matrix}-{5}{y}\\{5}{x}\\{z}\end{matrix}\right]}={\left[\begin{matrix}{15}\\-{10}\\{20}\end{matrix}\right]}\)

By the equality of two matrices we have

\(\displaystyle{\left\lbrace\begin{matrix}-{5}{y}={15}\\{5}{x}=-{10}\\{z}={20}\end{matrix}\right.}={\left\lbrace\begin{matrix}{y}=-{3}\\{x}=-{2}\\{z}={20}\end{matrix}\right.}\)

Thus, the original vector is (-2, -3, 20)

The vector is first rotated by \(\displaystyle{90}^{\circ}\) along x - axis and then scaled up by 5 times.

Ratation: In homogeneous coordinates, the matrix for a rotetion about the origin through a given angle \(\displaystyle\theta\ {i}{s}\ {B}={\left[\begin{matrix} \cos{\theta}&- \sin{\theta}&{0}\\ \sin{\theta}& \cos{\theta}&{0}\\{0}&{0}&{1}\end{matrix}\right]}\)

The vector (x, y, z) is rotated by \(\displaystyle{90}^{\circ}\) along x-axis then the vector becomes

\(\displaystyle{\left[\begin{matrix}{ \cos{{90}}^{\circ}-}{ \sin{{90}}^{\circ}{0}}\\{ \sin{{90}}^{\circ} \cos{{90}}^{\circ}}\\{0}{0}{1}\end{matrix}\right]}{\left[\begin{matrix}{x}\\{y}\\{z}\end{matrix}\right]}\)

\(\displaystyle={\left[\begin{matrix}{0}-{1}{0}\\{1}{0}{0}\\{0}{0}{1}\end{matrix}\right]}{\left[\begin{matrix}{x}\\{y}\\{z}\end{matrix}\right]}\)

\(\displaystyle={\left[\begin{matrix}-{y}\\{x}\\{z}\end{matrix}\right]}\)

Now, the vector is scaled up by 5 times.

Scaling: In homogeneous coordinates, a scaling about the origin with scale factors c in the x-direction and c in the y-direction is computed by multiplying the vector by the matrix

\(\displaystyle{A}={\left[\begin{matrix}{c}{0}{0}\\{0}{c}{0}\\{0}{0}{1}\end{matrix}\right]}\)

Thus, the vector becomes

\(\displaystyle{\left[\begin{matrix}{5}{0}{0}\\{0}{5}{0}\\{0}{0}{1}\end{matrix}\right]}{\left[\begin{matrix}{x}\\{y}\\{z}\end{matrix}\right]}\)

\(\displaystyle={\left[\begin{matrix}-{5}{y}\\{5}{x}\\{z}\end{matrix}\right]}\)

Now, given that the reduced vector is \(\displaystyle{\left({15},-{10},{20}\right)}\). Therefore, we have

\(\displaystyle{\left[\begin{matrix}-{5}{y}\\{5}{x}\\{z}\end{matrix}\right]}={\left[\begin{matrix}{15}\\-{10}\\{20}\end{matrix}\right]}\)

By the equality of two matrices we have

\(\displaystyle{\left\lbrace\begin{matrix}-{5}{y}={15}\\{5}{x}=-{10}\\{z}={20}\end{matrix}\right.}={\left\lbrace\begin{matrix}{y}=-{3}\\{x}=-{2}\\{z}={20}\end{matrix}\right.}\)

Thus, the original vector is (-2, -3, 20)