Question

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.

Answer 1

The new normalized basis vectors are

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

using

$$\sqrt{1^2 + 3^2 + 0^2} = \sqrt{10}$$

The coordinates transform with the transpose matrix

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

hence

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

$$\begin{cases} x &= &\frac{1}{\sqrt{10}} (x' - 3y')\\ y &= &\frac{1}{\sqrt{10}} (3x' + y') \\ z &= &z' \end{cases}$$

$$12x^2 - 2 y^2 + z^2 =2xy$$ $$\frac{12}{10}(x'-3y')^2 -\frac{2}{10} (3x'+y')^2+z'^2 =\frac{2}{10} (x'-3y')(3x'+y')$$ $$\frac{6}{5}(x'^2 -6x'y' + 9y'^2) - \frac{1}{5} (9x'^2 + 6x'y' + y'^2) + z'^2 = \frac{1}{5} (3x'^2 -8x'y' -3y'^2)$$ $$-\frac{6}{5}x'^2 +\frac{56}{5}y'^2 + z'^2 = \frac{34}{5}x'y'$$ $$-6x'^2 + 56y'^2 + 5z'^2 = 34x'y'$$