ShearingMatrix[θ,v,n]
给出这样一个矩阵,它沿向量 v 的方向剪切 θ 弧度,法向量是 n.
ShearingMatrix
ShearingMatrix[θ,v,n]
给出这样一个矩阵,它沿向量 v 的方向剪切 θ 弧度,法向量是 n.
更多信息
- ShearingMatrix 根据剪切要求来生成矩阵而不改变坐标原点.
- ShearingMatrix 根据面积或体积转化要求生成相应的矩阵,且矩阵的行列式为1.
- 在二维图形中,ShearingMatrix 将三角形转变为矩形. ShearingMatrix[θ,{1,0},{0,1}] 向右倾斜 θ 角.
- 在三维图形中,ShearingMatrix 的剪切方式类似沿向量 v 方向旋转 θ 角剪切一副卡片,而卡片的法向量是 n.
范例
打开所有单元 关闭所有单元基本范例 (2)
范围 (5)
ShearingMatrix[θ, {1, 0}, {0, 1}]//MatrixFormShearingMatrix[θ, {0, 1}, {1, 0}]//MatrixFormShearingMatrix[θ, {1, 0, 0}, {0, 0, 1}]//MatrixFormShearingMatrix[θ, {1, 0, 0}, {0, 1, 0}]//MatrixFormShearingMatrix[θ, {1, 1}, {1, -1}]//MatrixFormgr = {Rectangle[], AbsolutePointSize[10], Opacity[1], {Magenta, Point[{0, 0}]}, {Green, Point[{1, 1}]}};Graphics[{{Opacity[.35], Blue, gr}, GeometricTransformation[{Opacity[.85], Red, gr}, ShearingMatrix[Pi / 6, {1, 0}, {0, 1}]]}]gr = {Cuboid[], AbsolutePointSize[10], Opacity[1], {Magenta, Point[{0, 0, 0}]}, {Green, Point[{1, 1, 1}]}};Graphics3D[{{Opacity[.35], Blue, gr}, GeometricTransformation[{Opacity[.85], Red, gr}, ShearingMatrix[Pi / 4, {1, 0, 0}, {-1, 1, 0}]]}, Boxed -> False]应用 (2)
Plot3D[-Sin[x y], {x, 0, Pi}, {y, 0, Pi}, BoxRatios -> Automatic, PlotRange -> All] /. gc_GraphicsComplex :> GeometricTransformation[gc, ShearingMatrix[-Pi / 4, {1, 1, 0}, {0, 0, 1}]]Shearings[θ_, n_] :=
With[{pairs = Subsets[Table[ UnitVector[n, j], {j, n}], {2}]}, Map[ShearingMatrix[θ, #[[1]], #[[2]]]&, Join[pairs, Reverse /@ pairs]]];Map[ MatrixForm, Shearings[θ, 2]]Map[Framed@Graphics[GeometricTransformation[Rectangle[], #], PlotRange -> {{0, 1.6}, {0, 1.6}}]&, Shearings[Pi / 6, 2]]Map[ MatrixForm, Shearings[θ, 3]]Map[Graphics3D[GeometricTransformation[Cuboid[], #]]&, Shearings[Pi / 6, 3]]Map[ MatrixForm, Shearings[θ, 4]]属性和关系 (4)
Det[ ShearingMatrix[θ, {1, 0, 0}, {0, 0, 1}]]ShearingMatrix[θ,v,n] 的求逆可以通过 ShearingMatrix[-θ,v,n] 来求:
ShearingMatrix[θ, {1, 0, 0}, {0, 0, 1}].ShearingMatrix[-θ, {1, 0, 0}, {0, 0, 1}]ShearingMatrix[θ,v,n] 的求逆也可以通过 ShearingMatrix[θ,-v,n] 来求:
ShearingMatrix[θ, {1, 0, 0}, {0, 0, 1}].ShearingMatrix[θ, -{1, 0, 0}, {0, 0, 1}]MatrixPower[ShearingMatrix[a, {1, 0}, {0, 1}], n] == ShearingMatrix[ArcTan[n Tan[a]], {1, 0}, {0, 1}]可能存在的问题 (3)
t1 = ShearingMatrix[θ, {1, 0, 0}, {0, 0, 1}];
t2 = ShearingMatrix[ϕ, {1, 0, 0}, {-1, 1, 0}];t1.t2 - t2.t1//SimplifyLimit[ShearingMatrix[a, {1, 0, 0}, {0, 0, 1}], a -> Pi / 2]ShearingMatrix[a, {1, 0}, {1, 1}] == ShearingMatrix[a, Projection[{1, 0}, Cross[{1, 1}]], {1, 1}]技术笔记
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- 构造矩阵
文本
Wolfram Research (2007),ShearingMatrix,Wolfram 语言函数,https://reference.wolfram.com/language/ref/ShearingMatrix.html.
CMS
Wolfram 语言. 2007. "ShearingMatrix." Wolfram 语言与系统参考资料中心. Wolfram Research. https://reference.wolfram.com/language/ref/ShearingMatrix.html.
APA
Wolfram 语言. (2007). ShearingMatrix. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/ShearingMatrix.html 年
BibTeX
@misc{reference.wolfram_2026_shearingmatrix, author="Wolfram Research", title="{ShearingMatrix}", year="2007", howpublished="\url{https://reference.wolfram.com/language/ref/ShearingMatrix.html}", note=[Accessed: 03-November-2026]}
BibLaTeX
@online{reference.wolfram_2026_shearingmatrix, organization={Wolfram Research}, title={ShearingMatrix}, year={2007}, url={https://reference.wolfram.com/language/ref/ShearingMatrix.html}, note=[Accessed: 03-November-2026]}