 Mylo O'Moore

2021-02-06

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

Let us consider the vector (x, y, z).
The vector is first rotated by ${90}^{\circ }$ 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
The vector (x, y, z) is rotated by ${90}^{\circ }$ along x-axis then the vector becomes
$\left[\begin{array}{c}\mathrm{cos}{90}^{\circ }-\mathrm{sin}{90}^{\circ }0\\ \mathrm{sin}{90}^{\circ }\mathrm{cos}{90}^{\circ }\\ 001\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
$=\left[\begin{array}{c}0-10\\ 100\\ 001\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
$=\left[\begin{array}{c}-y\\ x\\ z\end{array}\right]$
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
$A=\left[\begin{array}{c}c00\\ 0c0\\ 001\end{array}\right]$
Thus, the vector becomes
$\left[\begin{array}{c}500\\ 050\\ 001\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
$=\left[\begin{array}{c}-5y\\ 5x\\ z\end{array}\right]$
Now, given that the reduced vector is $\left(15,-10,20\right)$. Therefore, we have
$\left[\begin{array}{c}-5y\\ 5x\\ z\end{array}\right]=\left[\begin{array}{c}15\\ -10\\ 20\end{array}\right]$
By the equality of two matrices we have
$\left\{\begin{array}{c}-5y=15\\ 5x=-10\\ z=20\end{array}=\left\{\begin{array}{c}y=-3\\ x=-2\\ z=20\end{array}$
Thus, the original vector is (-2, -3, 20)

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