Hitting a square centered on {0, 0} with a hanger and then with a stretcher corn

Question

Hitting a square centered on {0, 0} with a hanger and then with a stretcher corners = {{-1, -1}, {1, -1}, {1, 1}, {-1, 1}, {-1, -1}}: ranger = 2.5: Clear[hitplotter, matrix]: hitplotter[matrix_]: s {Graphics[{Black, Thickness[0.02], Line[Table[matrix. corners[[k]], {k, 1, Length[corners]}]]}], Table[Graphics[{PointSize[0.035], Red, Point[matrix. corners[[k]]]}], {k, 1, Length[corners]}]}: Show[hitplotter[IdentityMatrix[2]], PlotRange - > {{-ranger, ranger}, {-ranger, ranger}}, Axes - > True, AspectRatio - > Automatic, AxesLabel - > {"x", "y"}] s = Random[Real, {0.1 Pi, 0.45 Pi}]: {perpframe[1], perpfrane[2]} = N[{{Cos[s], Sin[s]}, {Cos[s + Pi/2], Sin[s + Pi/2]}}]: hanger = Transpose[{perpframe[1], perpframe[2]}]: MatrixForm[hanger] s = Random[Real, {0.1 Pi, 0.45 Pi}]j {perpframe[1], perpframe[2]} = N[{{Cos[s], Sin[s]}, {Cos[s + pi/2], Sin[s + Pi/2]}}]: hanger = Transpose[{perpframe[1], perpframe[2]}]: MatrixForm[hanger] (0.868208 -0.4962 0.4962 0.868208) Set up a stretcher matrix as follows: xstretch = 1.5: ystretch = 0.8: stretcher = DiagonalMatrix[{xstretch, ystretch}]: MatrixForm[stretcher] (1.5 0 0 0. 0.8) Pul A - stretcher. hanger A = stretcher. hanger: MatrixForm[A] (1.30231 -0.7443 0.39696 0.694567) Hit the square with A: Show[hitplotter[A], PlotRange - > {{-ranger, ranger}, {-ranger, ranger}}, Axes - > True, AspectRatio - > Automatic, AxesLabel - > {"x", "y")] Show[hitplotter[A], PlotRange - > {{-ranger, ranger}, {-ranger, ranger}}, Axes - > True, AspectRatio - > Automatic, AxesLabel - > {"x", "y"}] Why did you get a parallelogram instead of an ordinary rectangle? Use the information in the code defining A to help to measure the area inside this, parallelogram. We get a parallelogram instead of a rectangle beam se the shape is hung on two perpframes. The shape is also a stretched square. So simply, the shape is a stretched and rotated square, the matrix A will come out to be ({1.5(0.868) -0.75}, {0.4 0.8(0.86)}] which will not make a shape with 90 degrees. This shape is rotated before it is stretched. If you are doing a lot of unpleasant calculation, then you are not doing this problem the easiest way.

Explanation / Answer

Hi,

Your answers looks a bit dubious. You have mentioned that the structure is a stretched and rotated square. You have repititively mentioned that it is a square, whereas your final matrix says it will not make a shape with 90 degrees. So, if the parallelogram is a square, it must give a matrix that has opposite angles not only equal (as in a parallelogram) but also equal to 90 degrees (as in a square).

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