Question #293623

Find the transformation of the equation 12𝑥2 − 2𝑦2 + 𝑧2 = 2𝑥𝑦 if the origin is kept fixed and the axes are rotated in such a way that the direction ratios of the new axes are 1, −3, 0; 3, 1, 0; 0, 0, 1


1
Expert's answer
2022-02-14T11:25:38-0500

Solution:

The normalized axes are



e1=110(1,3,0),e2=110(3,1,0),e3=(0,0,1)\mathbf{e}'_1 = \frac{1}{\sqrt{10}} (1, -3, 0), \hspace{3mm} \mathbf{e}'_2 = \frac{1}{\sqrt{10}} (3, 1, 0), \hspace{3mm} \mathbf{e}'_3 = (0, 0, 1)

or



(e1e2e3)=(11031003101100001)(e1e2e3)\begin{pmatrix} \mathbf{e}'_1 \\ \mathbf{e}'_2 \\ \mathbf{e}'_3 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} & 0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0\\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} \mathbf{e}_1 \\ \mathbf{e}_2 \\ \mathbf{e}_3 \end{pmatrix}

- the basis in the rotated frame. The transformation of coordinates is given by

(xyz)=(11031003101100001)1(xyz)\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} & 0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0\\ 0 & 0 & 1 \\ \end{pmatrix}^{-1} \begin{pmatrix} x \\ y \\ z \end{pmatrix}


so



(xyz)=(11031003101100001)(xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} & 0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0\\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}{x=x103y10y=3x10+y10z=z\bigg\{\begin{matrix} x &=& \frac{x'}{\sqrt{10}} - \frac{3y'}{\sqrt{10}} \\ y &=& \frac{3x'}{\sqrt{10}} + \frac{y'}{\sqrt{10}} \\ z &=& z' \end{matrix}12x22y2+z2=2xy12x^2 -2y^2+z^2=2xy12(x3y)2102(3x+y)210+z2=2(x3y)(3x+y)1012\frac{(x' - 3y')^2}{10} - 2\frac{(3x' +y')^2}{10} + z'^2 = 2\frac{(x' - 3y')(3x' +y')}{10}





12(x26xy+9y2)2(9x2+6xy+y2)++10z2=2(3x28xy3y2)12(x'^2 - 6x'y'+9y'^2) - 2(9x'^2 +6x'y' + y'^2)+ \\ + 10z'^2 = 2(3x'^2 -8x'y'-3y'^2)




12x268xy+112y2+10z2=0-12x'^2 -68x'y' +112y'^2 +10z'^2 = 0


Answer:


5z2+56y26x2=34xy5z^2 +56y'^2-6x'^2=34x'y'

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