【数学A】例題4.2.18：座標空間における点（One More）★★★

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% 例題A4.2.18：座標空間における点（One More）★★★
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座標空間内の3点$\mathrm{O}(0,0,0),\mathrm{A}(0,2,-2),\mathrm{B}(2,2,0)$からの距離がともに$2\sqrt{2}$である点Cの座標を求めよ．

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% 例題A4.2.18の解答（One More）★★★
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点Cの座標を$(x,y,z)$とする．

与えられた条件から，$\mathrm{OC}=\mathrm{AC}=\mathrm{BC}=2\sqrt{2}$

したがって，$\mathrm{OC}^2=\mathrm{AC}^2=\mathrm{BC}^2=8$

$\mathrm{OC}^2=\mathrm{AC}^2$より，$x^2+y^2+z^2=x^2+(y-2)^2+(z+2)^2$

整理すると，$y^2+z^2=(y-2)^2+(z+2)^2$

ゆえに，$y-z=2\cdots(\mathrm{i})$

$\mathrm{OC}^2=\mathrm{BC}^2$より，$x^2+y^2+z^2=(x-2)^2+(y-2)^2+z^2$

整理すると，$x^2+y^2=(x-2)^2+(y-2)^2$

ゆえに，$x+y=2\cdots(\mathrm{ii})$

さらに，$\mathrm{OC}^2=8$より，$x^2+y^2+z^2=8\cdots(\mathrm{iii})$

(i)，(ii)より，$x=2-y,z=y-2\cdots(\mathrm{iv})$

これを(iii)に代入すると，$(2-y)^2+y^2+(y-2)^2=8$

整理すると，$3y^2-8y=0$

これを解くと，$y=0,y=\frac{8}{3}$

よって，(iv)より，求める点Cの座標は，$(2,0,-2),(-\frac{2}{3},\frac{8}{3},\frac{2}{3})$

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% 問題A4.2.18：座標空間における点（One More）★★★
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座標空間において，$\mathrm{A}(3,2,4)$，$\mathrm{B}(4,3,0)$，$\mathrm{C}(5,4,5)$を頂点とする三角形は，直角三角形であることを示せ．

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% 問題A4.2.18の解答（One More）★★★
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\begin{align*}
&\mathrm{AB}^2=(4-3)^2+(3-2)^2+(0-4)^2=18,\\
&\mathrm{BC}^2=(5-4)^2+(4-3)^2+(5-0)^2=27,\\
&\mathrm{CA}^2=(3-5)^2+(2-4)^2+(4-5)^2=9
\end{align*}

$\mathrm{BC}^2=\mathrm{CA}^2+\mathrm{AB}^2$であるから，三平方の定理の逆より，$\triangle\mathrm{ABC}$は$\angle\mathrm{A}=90^{\circ}$の直角三角形である．$\blacksquare$

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% 例題A4.2.18の側注：座標空間における点（One More）★★★
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% 側注1
2乗した形を考えると扱いやすい．

% 側注2
$y^2,z^2$の項がなくなる．

% 側注3
$x^2,y^2$の項がなくなる．

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% 問題A4.2.18の側注：座標空間における点（One More）★★★
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% 側注1
直角である角を示す．

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% 例題A4.2.18の考え方・補足など：座標空間における点（One More）★★★
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% 考え方
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座標空間において，点$\mathrm{A}(x_1,y_1,z_1)$，点$\mathrm{B}(x_2,y_2,z_2)$があるとする．このとき，

\[
\mathrm{AB}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\]
が成り立つことを利用する．ここでは，$\mathrm{OC}=\mathrm{AC}$や$\mathrm{OC}=\mathrm{BC}$のままでは扱いにくいから，これと同値な条件である2乗した$\mathrm{OC}^2=\mathrm{AC}^2$や$\mathrm{OC}^2=\mathrm{BC}^2$の形を考える．

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座標空間において，点$\mathrm{A}(x_1,y_1,z_1)$，点$\mathrm{B}(x_2,y_2,z_2)$の距離ABは，

\[
\mathrm{AB}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\]