Answer to Question #89741 in Analytic Geometry for Anjali

Question #89741
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
1
Expert's answer
2019-05-15T08:40:40-0400

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}\n\\mathbf{e}'_3 = (0, 0, 1)"

or


"\\begin{pmatrix}\n \\mathbf{e}'_1 \\\\ \\mathbf{e}'_2 \\\\ \\mathbf{e}'_3\n\\end{pmatrix} = \n\\begin{pmatrix}\n\\frac{1}{\\sqrt{10}} & \\frac{-3}{\\sqrt{10}} & 0 \\\\\n\\frac{3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}} & 0\\\\\n0 & 0 & 1 \\\\\n\\end{pmatrix}\n\\begin{pmatrix}\n \\mathbf{e}_1 \\\\ \\mathbf{e}_2 \\\\ \\mathbf{e}_3\n\\end{pmatrix}"

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


"\\begin{pmatrix}\n x' \\\\ y' \\\\ z'\n\\end{pmatrix} =\\begin{pmatrix}\n\\frac{1}{\\sqrt{10}} & \\frac{-3}{\\sqrt{10}} & 0 \\\\\n\\frac{3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}} & 0\\\\\n0 & 0 & 1 \\\\\n\\end{pmatrix}^{-1}\n\\begin{pmatrix}\n x \\\\ y \\\\ z\n\\end{pmatrix}"


so


"\\begin{pmatrix}\n x \\\\ y \\\\ z\n\\end{pmatrix} =\\begin{pmatrix}\n\\frac{1}{\\sqrt{10}} & \\frac{-3}{\\sqrt{10}} & 0 \\\\\n\\frac{3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}} & 0\\\\\n0 & 0 & 1 \\\\\n\\end{pmatrix}\n\\begin{pmatrix}\n x' \\\\ y' \\\\ z'\n\\end{pmatrix}"

"\\bigg\\{\\begin{matrix}\nx &=& \\frac{x'}{\\sqrt{10}} - \\frac{3y'}{\\sqrt{10}} \\\\\ny &=& \\frac{3x'}{\\sqrt{10}} + \\frac{y'}{\\sqrt{10}} \\\\\nz &=& z'\n\\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'"


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