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谢之题解:一道综合性的解几题

Eufisky posted @ 2015年10月23日 04:30 in 谢惠民 with tags 解析几何 谢惠民 , 1150 阅读

谢惠民下册P238第21章的一个参考题:

证明与曲面$ax^2+by^2+cz^2=1(abc\neq0)$相切的三个互相垂直的平面的交点在球面$x^2+y^2+z^2=\frac1a+\frac1b+\frac1c$上.


证:(幸子)椭球面方程为$ax^2+by^2+cz^2=1(abc\neq0)$,则法向量${n_i} = \left( {a{x_i},b{y_i},c{z_i}} \right)$,切平面方程为$a{x_i}x + b{y_i}y + c{z_i}z = 1$.

 

设三个切点分别为${\alpha _i}\left( {{x_i},{y_i},{z_i}} \right)\left( {i = 1,2,3} \right)$,三平面交点为$(x,y,z)$.由三平面垂直可知

\[\overrightarrow {{n_i}} \cdot \overrightarrow {{n_j}} = {a^2}{x_i}{x_j} + {b^2}{y_i}{y_j} + {c^2}{z_i}{z_j} = 0\left( {i \ne j} \right).\]

原点到三切平面的距离分别为

\[\frac{1}{{\sqrt {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} }}\left( {i = 1,2,3} \right).\]

由几何关系(考虑长方体对角线)可知

\[{x^2} + {y^2} + {z^2} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} .\]

 

设$\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)$,则对任意的$i = 1,2,3$,有

\[1 = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right) = {t_i}\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right).\]

并且

 

\begin{align*}&{x^2} + {y^2} + {z^2} = \left( {\begin{array}{*{20}{c}}x&y&z\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{t_1}}&{{t_2}}&{{t_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{b{y_1}}&{c{z_1}}\\{a{x_2}}&{b{y_2}}&{c{z_2}}\\{a{x_3}}&{b{y_3}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)\\=& \sum\limits_{i = 1}^3 {{t_i}^2\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} = \sum\limits_{i = 1}^3 {\frac{{a{x_i}^2 + b{y_i}^2 + c{z_i}^2}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} \\=& a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right).\end{align*}

 

注意到对任意的$w$,记$w = \sum\limits_{i = 1}^3 {{s_i}{n_i}} $,则${s_i} = \frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}$,即\[w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} .\]

 

分别令$w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} $,得

\begin{align*}a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{a},\\b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{b},\\c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{c}.\end{align*}

因此有\[{x^2} + {y^2} + {z^2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.\]


解法二.记$d_i=\sqrt{a^2x_i^2+b^2y_i^2+c^2z_i^2},\ i=1,2,3$,则$\begin{pmatrix}\frac{ax_1}{d_1}&\frac{by_1}{d_1}&\frac{cz_1}{d_1}\\\frac{ax_2}{d_2}&\frac{by_2}{d_2}&\frac{cz_2}{d_2}\\\frac{ax_3}{d_3}&\frac{by_3}{d_3}&\frac{cz_3}{d_3}\end{pmatrix} $是正交矩阵,从而有

\[\frac{a^2x_1^2}{d_1^2}+\frac{a^2x_2^2}{d_2^2}+\frac{a^3x_3^2}{d_3^2}=1\Rightarrow \frac{ax_1^2}{d_1^2}+\frac{ax_2^2}{d_2^2}+\frac{ax_3^2}{d_3^2}=\frac{1}{a}.\]

类似得到另外两式, 相加便有

\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ax_1^2+by_1^2+cz_1^2}{d_1^2}+\frac{ax_2^2+by_2^2+cz_2^2}{d_2^2}+\frac{ax_3^2+by_3^2+cz_3^2}{d_3^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}.\]


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