2016年10月29日土曜日

数学 - 立体的な広がりの中の図形 - 空間図形 – 直線・平面・球の方程式(球あるいは球面の方程式、アポロニススの円の空間への拡張の一例)

1. $\begin{array}{l}{\left(x-2\right)}^{2}+{\left(y+3\right)}^{2}+{\left(z-1\right)}^{2}={5}^{2}\\ 点\left(\text{2,-3,1}\right)を中心とする半径\text{5}の球。\end{array}$

2. $\begin{array}{l}{\left(x+2\right)}^{2}+{\left(y+2\right)}^{2}+{\left(z-5\right)}^{2}={7}^{2}\\ 点\left(\text{-2,-2,5}\right)を中心とする半径\text{7}の球。\end{array}$

1. ${\left(x-3\right)}^{2}+{y}^{2}+{\left(z+2\right)}^{2}=36$

2. ${\left(x+4\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z-5\right)}^{2}=16$

3. $\begin{array}{l}\left(1,-2,-3\right),\\ {r}^{2}=1+4+9=14\\ {\left(x-1\right)}^{2}+{\left(y+2\right)}^{2}+{\left(z+3\right)}^{2}=14\end{array}$

4. $\begin{array}{l}{\left(x-r\right)}^{2}+{\left(y-r\right)}^{2}+{\left(z-r\right)}^{2}={r}^{2}\\ {\left(1-r\right)}^{2}+{\left(4-r\right)}^{2}+{\left(5-r\right)}^{2}={r}^{2}\\ 2{r}^{2}-20r+42=0\\ {r}^{2}-10r+21=0\\ \left(r-7\right)\left(r-3\right)=0\\ r=3,7\\ {\left(x-3\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z-3\right)}^{2}=9\\ {\left(x-7\right)}^{2}+{\left(y-7\right)}^{2}+{\left(z-7\right)}^{2}=49\end{array}$

5. $\begin{array}{l}{x}^{2}+{y}^{2}+{z}^{2}+Ax+By+Cz+D=0\\ 1+A+D=0\\ 4+2B+D=0\\ 9+3C+D=0\\ 1+4+9+A+2B+3C+D=0\\ D=-A-1\\ B=\frac{-D-4}{2}=\frac{A+1-4}{2}=\frac{A-3}{2}\\ C=\frac{-D-9}{3}=\frac{A+1-9}{3}=\frac{A-8}{3}\\ 14+A+A-3+A-8-A-1=0\\ A=-1\\ B=-2\\ C=-3\\ D=0\\ {x}^{2}+{y}^{2}+{z}^{2}-x-2y-3z=0\\ {\left(x-\frac{1}{2}\right)}^{2}+{\left(y-1\right)}^{2}+{\left(z-\frac{3}{2}\right)}^{2}=\frac{7}{2}\end{array}$

$\begin{array}{l}P\left(x,y,z\right)\\ {\left(x+2\right)}^{2}+{y}^{2}+{z}^{2}={2}^{2}\left({\left(x-1\right)}^{2}+{y}^{2}+{z}^{2}\right)\\ {x}^{2}+4x+4+{y}^{2}+{z}^{2}=4\left({x}^{2}-2x+1+{y}^{2}+{z}^{2}\right)\\ 3{x}^{2}+3{y}^{2}+3{z}^{2}-12x=0\\ {x}^{2}+{y}^{2}+{z}^{2}-4x=0\\ {\left(x-2\right)}^{2}+{y}^{2}+{z}^{2}={2}^{2}\\ \left(\text{2,0,0}\right)を中心とする半径\text{2}の球。\end{array}$

$\begin{array}{l}{|\stackrel{\to }{p}-\stackrel{\to }{a}|}^{2}=4{|\stackrel{\to }{p}-\stackrel{\to }{b}|}^{2}\\ \left(\stackrel{\to }{p}-\stackrel{\to }{a}\right)·\left(\stackrel{\to }{p}-\stackrel{\to }{a}\right)=4\left(\stackrel{\to }{p}-\stackrel{\to }{b}\right)·\left(\stackrel{\to }{p}-\stackrel{\to }{b}\right)\\ {|\stackrel{\to }{p}|}^{2}+{|\stackrel{\to }{a}|}^{2}-2\stackrel{\to }{a}·\stackrel{\to }{p}=4{|\stackrel{\to }{p}|}^{2}+4{|\stackrel{\to }{b}|}^{2}-8\stackrel{\to }{b}·\stackrel{\to }{p}\\ 3{|\stackrel{\to }{p}|}^{2}+2\left(\stackrel{\to }{a}-4\stackrel{\to }{b}\right)·\stackrel{\to }{p}-{|\stackrel{\to }{a}|}^{2}+4{|\stackrel{\to }{b}|}^{2}=0\\ 3{|\stackrel{\to }{p}|}^{2}-6\stackrel{\to }{c}·\stackrel{\to }{p}-{|\stackrel{\to }{a}|}^{2}+4{|\stackrel{\to }{b}|}^{2}=0\\ 3\left({|\stackrel{\to }{p}|}^{2}-2\stackrel{\to }{c}·\stackrel{\to }{p}\right)-{|\stackrel{\to }{a}|}^{2}+4{|\stackrel{\to }{b}|}^{2}=0\\ 3\left({|\stackrel{\to }{p}|}^{2}-2\stackrel{\to }{c}·\stackrel{\to }{p}+{|\stackrel{\to }{c}|}^{2}\right)-{|\stackrel{\to }{a}|}^{2}+4{|\stackrel{\to }{b}|}^{2}=3{|\stackrel{\to }{c}|}^{2}\\ 3{|\stackrel{\to }{p}-\stackrel{\to }{c}|}^{2}={|\stackrel{\to }{a}|}^{2}-4{|\stackrel{\to }{b}|}^{2}+3{|\stackrel{\to }{c}|}^{2}\\ {|\stackrel{\to }{a}|}^{2}-4{|\stackrel{\to }{b}|}^{2}+3{|\stackrel{\to }{c}|}^{2}\\ ={|\stackrel{\to }{a}|}^{2}-4{|\stackrel{\to }{b}|}^{2}+3{|\frac{4\stackrel{\to }{b}-\stackrel{\to }{a}}{3}|}^{2}\\ ={|\stackrel{\to }{a}|}^{2}-4{|\stackrel{\to }{b}|}^{2}+\frac{1}{3}\left(4\stackrel{\to }{b}-\stackrel{\to }{a}\right)·\left(4\stackrel{\to }{b}-\stackrel{\to }{a}\right)\\ ={|\stackrel{\to }{a}|}^{2}-4{|\stackrel{\to }{b}|}^{2}+\frac{1}{3}\left(16{|\stackrel{\to }{b}|}^{2}+{|\stackrel{\to }{a}|}^{2}-8\stackrel{\to }{a}·\stackrel{\to }{b}\right)\\ =\frac{1}{3}\left(4{|\stackrel{\to }{a}|}^{2}+4{|\stackrel{\to }{b}|}^{2}-8\stackrel{\to }{a}·\stackrel{\to }{b}\right)\\ =\frac{4}{3}{|\stackrel{\to }{a}-\stackrel{\to }{b}|}^{2}\\ 3{|\stackrel{\to }{p}-\stackrel{\to }{c}|}^{2}=\frac{4}{3}{|\stackrel{\to }{a}-\stackrel{\to }{b}|}^{2}\\ |\stackrel{\to }{p}-\stackrel{\to }{c}|=\frac{2}{3}|\stackrel{\to }{a}-\stackrel{\to }{b}|\\ |\stackrel{\to }{p}-\stackrel{\to }{c}|=r\end{array}$