A vector is first rotated by displaystyle{90}^{circ} along x-axis and then scaled up by 5 times is equal to displaystyle{left({15},-{10},{20}right)}. What was the original vector

Question
Vectors and spaces
asked 2021-02-06
A vector is first rotated by \(\displaystyle{90}^{\circ}\) along x-axis and then scaled up by 5 times is equal to \(\displaystyle{\left({15},-{10},{20}\right)}\). What was the original vector

Answers (1)

2021-02-07
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)
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