Question

Find the transformation of the eqn 12x^2-2y^2+z^2=2xy if the origin is kept fixed and the axes are rotated in such a way that the direction ratio of the new axes are (1,-3,0);3,1,0;0,0,1

Answer 1

The normalized axes are

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

or

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

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

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

so

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} & 0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0\\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}$$

$$\bigg\{\begin{matrix} x &=& \frac{x'}{\sqrt{10}} - \frac{3y'}{\sqrt{10}} \\ y &=& \frac{3x'}{\sqrt{10}} + \frac{y'}{\sqrt{10}} \\ z &=& z' \end{matrix}$$

$$12x^2 -2y^2+z^2=2xy$$

$$12\frac{(x' - 3y')^2}{10} - 2\frac{(3x' +y')^2}{10} + z'^2 = 2\frac{(x' - 3y')(3x' +y')}{10}$$

$$12(x'^2 - 6x'y'+9y'^2) - 2(9x'^2 +6x'y' + y'^2)+ \\ + 10z'^2 = 2(3x'^2 -8x'y'-3y'^2)$$

$$-12x'^2 -68x'y' +112y'^2 +10z'^2 = 0$$

Answer:

$$5z^2 +56y'^2-6x'^2=34x'y'$$