用正交变换解空间曲线曲面题目

例1. 已知#card=math&code=A%3D%28a_%7Bij%7D%29&id=zkbtS)是三阶正交矩阵。求证：曲面

是一个球面。求出球面的球心坐标与半径。

解：设

即

与坐标变换公式相比较，这就是说，在坐标系之下，球面%5E2%3D1#card=math&code=%5Csum_%7Bi%3D1%7D%5E3%20%28X_i%20%2B%20b_i%29%5E2%3D1&id=weH4w) 被变换到坐标系之下的，以方程(1)给出的曲面。由于是正交矩阵，所以方程(1)给出的曲面也是球面，半径显然是1, 而球心坐标从点%5E%5Ctop#card=math&code=%28-b_1%2C%20-b_2%2C%20-b_3%29%5E%5Ctop&id=k1QMY)被变换到%5E%5Ctop%20%3D%20-A%5E%5Ctop%20(b_1%2C%20b_2%2C%20b_3)%5E%5Ctop#card=math&code=-A%5E%7B-1%7D%28b_1%2C%20b_2%2C%20b_3%29%5E%5Ctop%20%3D%20-A%5E%5Ctop%20%28b_1%2C%20b_2%2C%20b_3%29%5E%5Ctop&id=HWvWv).

例2. 设有向量%2C%20%5C%2C%20i%3D1%2C2#card=math&code=v_i%20%3D%20%28a_i%2C%20b_i%2C%20c_i%29%2C%20%5C%2C%20i%3D1%2C2&id=mJJNF). 两者线性无关。求证：曲面

(a_2%20x%20%2B%20b_2%20y%20%2B%20c_2%20z)%3D1%0A#card=math&code=%28a_1%20x%20%2B%20b_1%20y%2B%20c_1%20z%29%28a_2%20x%20%2B%20b_2%20y%20%2B%20c_2%20z%29%3D1%0A&id=qhznK)

是一个柱面，母线的方向与平行。

证：如果向量正交，则令

(a_1%20x%20%2B%20b_1%20y%20%2B%20c_1%20z)%2C%20%5C%5C%0AY%20%26%3D(1%2F%7Cv_2%7C)(a_2%20x%20%2B%20b_2%20y%20%2B%20c_2%20z)%2C%20%5C%5C%0AZ%20%26%3D%20z%2C%0A%5Cend%7Bcases%7D%0A#card=math&code=%5Cbegin%7Bcases%7D%0AX%20%26%3D%281%2F%7Cv_1%7C%29%28a_1%20x%20%2B%20b_1%20y%20%2B%20c_1%20z%29%2C%20%5C%5C%0AY%20%26%3D%281%2F%7Cv_2%7C%29%28a_2%20x%20%2B%20b_2%20y%20%2B%20c_2%20z%29%2C%20%5C%5C%0AZ%20%26%3D%20z%2C%0A%5Cend%7Bcases%7D%0A&id=UOs3H)

由于矩阵

是正交矩阵，所以上述坐标变换是正交变换。原方程则变成, 这是坐标系中的柱面，母线方向是-轴的方向，即与平行。

但只是线性无关，不一定正交。为此先来Schmidt正交化：令

%7D%7B(w_1%2C%20w_1)%7Dv_1%20%5CLongrightarrow%20v_2%20%3D%20w_2%20%2B%20kw_1%2C%20%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0Aw_1%20%26%3D%20v_1%2C%20%5C%5C%0Aw_2%20%26%3D%20v_2%20-%20%5Cfrac%7B%28v_2%2C%20w_1%29%7D%7B%28w_1%2C%20w_1%29%7Dv_1%20%5CLongrightarrow%20v_2%20%3D%20w_2%20%2B%20kw_1%2C%20%0A%5Cend%7Baligned%7D%0A&id=Dz7a3)

其中%7D%7B(w_1%2C%20w_1)%7D#card=math&code=k%3D%5Cfrac%7B%28v_2%2C%20w_1%29%7D%7B%28w_1%2C%20w_1%29%7D&id=kAhum). 此时作坐标代换

(x%2C%20y%2C%20z)%5E%5Ctop%20%3D%20(1%2F%7Cw_1%7C)(a_1%20x%20%2B%20b_1%20y%20%2B%20c_1%20z)%2C%20%5C%5C%0AY%20%3D%20(1%2F%7Cw_2%7C)(x%2C%20y%2C%20z)%5E%5Ctop%2C%20%5C%5C%0AZ%20%3D%20z.%20%0A%5Cend%7Bcases%7D%0A#card=math&code=%5Cbegin%7Bcases%7D%0AX%20%3D%20%281%2F%7Cw_1%7C%29%28x%2C%20y%2C%20z%29%5E%5Ctop%20%3D%20%281%2F%7Cw_1%7C%29%28a_1%20x%20%2B%20b_1%20y%20%2B%20c_1%20z%29%2C%20%5C%5C%0AY%20%3D%20%281%2F%7Cw_2%7C%29%28x%2C%20y%2C%20z%29%5E%5Ctop%2C%20%5C%5C%0AZ%20%3D%20z.%20%0A%5Cend%7Bcases%7D%0A&id=XES1W)

那么%5Cbegin%7Bpmatrix%7D%20x%20%5C%5C%20y%20%5C%5C%20z%20%5Cend%7Bpmatrix%7D%20%3D%20%7Cw_2%7CY%20%2B%20k%7Cw_1%7CX#card=math&code=a_2%20x%20%2B%20b_2%20y%20%2B%20c_2%20z%20%3D%20v_2%20%5Cbegin%7Bpmatrix%7D%20x%20%5C%5C%20y%20%5C%5C%20z%20%5Cend%7Bpmatrix%7D%20%3D%20%28w_2%20%2B%20kw_1%29%5Cbegin%7Bpmatrix%7D%20x%20%5C%5C%20y%20%5C%5C%20z%20%5Cend%7Bpmatrix%7D%20%3D%20%7Cw_2%7CY%20%2B%20k%7Cw_1%7CX&id=td1aY). 此时原曲面方程变为

%3D1.%0A#card=math&code=%7Cw_1%7CX%20%28%7Cw_2%7CY%2Bk%7Cw_1%7CX%29%3D1.%0A&id=gEfAC)

这是柱面方程，母线平行于-轴方向，即平行于的方向。又

%3Dw_1%20%5Ctimes%20w_2%2C%0A#card=math&code=v_1%20%5Ctimes%20v_2%20%3D%20w_1%20%5Ctimes%20%28w_2%20%2B%20kw_1%29%3Dw_1%20%5Ctimes%20w_2%2C%0A&id=m6Qfy)

于是这个柱面的母线也平行于的方向。

由于上面的坐标变换是正交变换，所以原命题成立。

另证：曲面方程(4)等价于

其中实数. 这个方程组的解是上述两个平面的交线。这条交线是直线，方向向量是. 当遍历全部非零实数时，这些直线的并集，给出了一个柱面。

例3. 求实数, 使得椭圆抛物面%3D2z#card=math&code=x%5E2%20%2B%20%28y%5E2%2F2%29%3D2z&id=C1MRY)与平面的交线是一个圆。并求出圆心坐标与半径。

解：在平面内，以直线为新的-轴，以直线为新的-轴，再以原来的-轴为-轴。这样就得到了新的坐标系. 具体而言，我们做正交的坐标变换

此时平面变成, 椭圆抛物面%3D2z#card=math&code=x%5E2%20%2B%20%28y%5E2%2F2%29%3D2z&id=XP9bc)变成

%5E2%7D%7B1%2Bk%5E2%7D%2B%5Cfrac%7BY%5E2%7D%7B2%7D%3D2%5Cfrac%7B-kX%2BZ%7D%7B%5Csqrt%7B1%2Bk%5E2%7D%7D.%0A#card=math&code=%5Cfrac%7B%28X%2BkZ%29%5E2%7D%7B1%2Bk%5E2%7D%2B%5Cfrac%7BY%5E2%7D%7B2%7D%3D2%5Cfrac%7B-kX%2BZ%7D%7B%5Csqrt%7B1%2Bk%5E2%7D%7D.%0A&id=XKmns)

它们的交线的方程是

%5E2%7D%7B1%2Bk%5E2%7D%2B%5Cfrac%7BY%5E2%7D%7B2%7D%3D2%5Cfrac%7BZ%7D%7B%5Csqrt%7B1%2Bk%5E2%7D%7D.%0A%5Cend%7Bcases%7D%0A#card=math&code=%5Cbegin%7Bcases%7D%0AX%3D0%2C%20%5C%5C%0A%5Cfrac%7B%28kZ%29%5E2%7D%7B1%2Bk%5E2%7D%2B%5Cfrac%7BY%5E2%7D%7B2%7D%3D2%5Cfrac%7BZ%7D%7B%5Csqrt%7B1%2Bk%5E2%7D%7D.%0A%5Cend%7Bcases%7D%0A&id=NOieY)

这是一个圆当且仅当. 圆的半径是. 圆心坐标是%3D(0%2C%200%2C%20%5Csqrt%7B2%7D)%2C#card=math&code=%28X%2C%20Y%2C%20Z%29%3D%280%2C%200%2C%20%5Csqrt%7B2%7D%29%2C&id=k9zkP) 即%3D(%5Cpm%201%2C%200%2C%201)#card=math&code=%28x%2Cy%2Cz%29%3D%28%5Cpm%201%2C%200%2C%201%29&id=KR86c).

例4. 假设#card=math&code=A%3D%28a_%7Bij%7D%29&id=qTTwv)是三阶正交矩阵. 设

有坐标变换，这里是右手直角坐标系。

(1) 三维欧几里得空间中的单位球面

求证：在上述坐标变换下变成.

证：因为(x%2C%20y%2C%20z)%5E%5Ctop%20%3D%20(X%2C%20Y%2C%20Z)A%5E%5Ctop%20A%20(X%2C%20Y%2C%20Z)#card=math&code=x%5E2%20%2B%20y%5E2%20%2B%20z%5E2%20%3D%20%28x%2C%20y%2C%20z%29%28x%2C%20y%2C%20z%29%5E%5Ctop%20%3D%20%28X%2C%20Y%2C%20Z%29A%5E%5Ctop%20A%20%28X%2C%20Y%2C%20Z%29&id=z1rII).

(2)用

表示三维空间的Laplace算子 (Laplacian). 用

表示新坐标系之下的Laplace算子。求证.

证：记%5E%5Ctop%20f%20%3D%20(%5Cpartial_x%20f%2C%20%5Cpartial_y%20f%2C%20%5Cpartial%20_z%20f)%5E%5Ctop#card=math&code=%5Cpartial_x%20%3D%20%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%2C%20%5Cquad%20%28%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z%29%5E%5Ctop%20f%20%3D%20%28%5Cpartial_x%20f%2C%20%5Cpartial_y%20f%2C%20%5Cpartial%20_z%20f%29%5E%5Ctop&id=tANdT). 因此%20(%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z)%5E%5Ctop#card=math&code=%5CDelta%20%3D%20%28%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z%29%20%28%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z%29%5E%5Ctop&id=afJbU).

因为, 链式法则给出, 所以

%5E%5Ctop%20%3D%20A%20(%5Cpartial_X%2C%20%5Cpartial_Y%2C%20%5Cpartial_Z)%5E%5Ctop.%0A#card=math&code=%28%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z%29%5E%5Ctop%20%3D%20A%20%28%5Cpartial_X%2C%20%5Cpartial_Y%2C%20%5Cpartial_Z%29%5E%5Ctop.%0A&id=gL2XD)

所以

%20(%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z)%5E%5Ctop%20%3D%20%20(%5Cpartial_X%2C%20%5Cpartial_Y%2C%20%5Cpartial_Z)A%5E%5Ctop%20A(%5Cpartial_X%2C%20%5Cpartial_Y%2C%20%5Cpartial_Z)%5E%5Ctop%3D%5CDelta_1%2C%0A#card=math&code=%5CDelta%20%3D%20%28%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z%29%20%28%5Cpartial_x%2C%20%5Cpartial_y%2C%20%5Cpartial_z%29%5E%5Ctop%20%3D%20%20%28%5Cpartial_X%2C%20%5Cpartial_Y%2C%20%5Cpartial_Z%29A%5E%5Ctop%20A%28%5Cpartial_X%2C%20%5Cpartial_Y%2C%20%5Cpartial_Z%29%5E%5Ctop%3D%5CDelta_1%2C%0A&id=ExSHS)

因为是正交矩阵（所以. ）

应用：因为”求导数”是线性变换，所以是线性微分算子。我们称满足的函数是调和函数(harmonic function). 于是可以说“调和函数是Laplace算子的属于特征值0的特征向量“。

一种比较简单（但有趣）的调和函数是调和多项式。 Laplace算子把次多项式变为次多项式。全部调和多项式构成一个线性空间。更特别，考虑齐次调和多项式。全部次齐次(复系数)调和多项式组成一个线性空间. 例如就是全部线性函数. 它是3维的， 而是5维的，它有一个基

把中的多项式限制到，得到的函数称为球调和函数(spherical harmonic function). 由(1)可知，如果#card=math&code=f%28x%2C%20y%2C%20z%29&id=yvISk)是球调和函数，那么#card=math&code=f%28X%2C%20Y%2C%20Z%29&id=R8gtm)也是，这里#card=math&code=%28X%2C%20Y%2C%20Z%29&id=x9ySb) 由上面的坐标变换公式给出。