立方体の影の作図問題　解答(4)

■問題(4)

図のような一辺の長さが $3$\displaystyle{1}$ mの立方体がある．頂点 $3$\displaystyle{\text{A}}$ を原点に，対角線 $3$\displaystyle{\text{A}\text{G}}$ が $3$\displaystyle{z}$ 軸と重なり，平面 $3$\displaystyle{\text{A}\text{C}\text{G}\text{E}}$ が $3$\displaystyle{yz}$ 平面と一致し，頂点 $3$\displaystyle{\text{C}}$ の座標の $3$\displaystyle{y}$ 成分が正になるように，すなわち北側になるように，頂点 $3$\displaystyle{\text{E}}$ の座標の $3$\displaystyle{y}$ 成分が負になるように，すなわち南側になるように立方体を地表面に設置した(下図を参照せよ)．立方体の影をグラフ用紙に作図せよ．グラフ用紙には東西南北の方角を明記する．また， $3$\displaystyle{1}$ mの長さが $3$\displaystyle{5}$ cmになるように作図する．作図に必要な計算を記入しておくこと．

立方体

立方体を真上から見た状態
立方体を東から見た状態

■答

■解き方

地面に配置した立方体は下図のように $3$\displaystyle{ xy }$ 平面上に置いた立方体を $3$\displaystyle{z}$ 軸の周りに $3$\displaystyle{45^\circ }$ 回転させた後， $3$\displaystyle{x}$ 軸の周りに $3$\displaystyle{\angle \text{E}\text{A}\text{C}}$ 回転させたものである．

$3$\displaystyle{z}$ 軸に $3$\displaystyle{\angle \text{E}\text{A}\text{C}}$ 回転させる1次変換の表現行列を $3$\displaystyle{ \hspace{1}{R}_{z} }$ とすると

$3$\displaystyle{ \hspace{1}{R}_{z} = $ \begin{array} \cos 45^\circ & - \sin 45^\circ &0& \vspace{6}\\ \sin 45^\circ & \cos 45^\circ &0& \vspace{6}\\ 0&0&1& \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ 0&0&1& \vspace{6}\\ \end{array} $ }$

となる．

$3$\displaystyle{\angle \text{E}\text{A}\text{G}={\theta}}$ とする． $3$\displaystyle{x}$ 軸の周りに $3$\displaystyle{{\theta}}$ 回転させる1次変換の表現行列を $3$\displaystyle{ \hspace{1}{R}_{x} }$ とすると

$3$\displaystyle{ \hspace{1}{R}_{x} = $ \begin{array}1&0&0& \vspace{6}\\ 0& \cos {\theta} & - \sin {\theta} & \vspace{6}\\ 0& \sin {\theta} & \cos {\theta} & \vspace{6}\\ \end{array} $ }$

$3$\displaystyle{\sin {\theta}=\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} }}$ ， $3$\displaystyle{\cos {\theta}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}}$

よって

$3$\displaystyle{ \hspace{1}{R}_{x} = $ \begin{array}1&0&0& \vspace{6}\\ 0& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ 0& \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$

となる．

したがって $3$\displaystyle{z}$ 軸の周りに $3$\displaystyle{ 45^\circ }$ 回転させた後， $3$\displaystyle{x}$ 軸の周りに $3$\displaystyle{{\theta}}$ 回転させる1次変換の表現行列は $3$\displaystyle{ \hspace{1}{R}_{x} \hspace{1}{R}_{z} }$ となる．

$3$\displaystyle{ \hspace{1}{R}_{x} \hspace{1}{R}_{z} = $ \begin{array}1&0&0& \vspace{6}\\ 0& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ 0& \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ 0&0&1& \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$

1次変換する前の各頂点の座標は

$3$\displaystyle{\text{A} $ 0,0,0, $ }$ $3$\displaystyle{\text{E} $ 0,0,1 $ }$

$3$\displaystyle{\text{B} $ 1,0,0 $ }$ $3$\displaystyle{\text{F} $ 1,0,1 $ }$

$3$\displaystyle{\text{C} $ 1,1,0 $ }$ $3$\displaystyle{\text{G} $ 1,1,1 $ }$

$3$\displaystyle{\text{D} $ 0,1,0 $ }$ $3$\displaystyle{H $ 0,1,1 $ }$

である．よって1次変換後(問題文の図4，図5の状態)の各頂点の座標は

$3$\displaystyle{ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array}1& \vspace{6}\\ 0& \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{\text{B} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}},\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}},\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} $ }$

$3$\displaystyle{ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array}1& \vspace{6}\\ 1& \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ = $ \begin{array}0& \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ = $ \begin{array}0& \vspace{6}\\ \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{\text{C} $ 0,\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} },\frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} $ }$

$3$\displaystyle{ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array}0& \vspace{6}\\ 1& \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ = $ \begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{\text{D} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}},\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}},\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} $ }$

$3$\displaystyle{ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array}0& \vspace{6}\\ 0& \vspace{6}\\ 1& \vspace{6}\\ \end{array} $ = $ \begin{array}0& \vspace{6}\\ - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{\text{E} $ 0,- \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} },\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} $ }$

$3$\displaystyle{ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array}1& \vspace{6}\\ 0& \vspace{6}\\ 1& \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{\text{F} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}},- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}},\frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} $ }$

$3$\displaystyle{ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} &0& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} } & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ $ \begin{array}0& \vspace{6}\\ 1& \vspace{6}\\ 1& \vspace{6}\\ \end{array} $ = $ \begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{H $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}},- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}},\frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} $ }$

となる．

次に各頂点の影の位置を(3)で求めた表現行列を使って計算する．点 $3$\displaystyle{\text{A}}$ の影を点 $3$\displaystyle{{\text{A}}^{\prime }}$ と表すことにする．

$3$\displaystyle{ $ \begin{array}1&0\hspace{5}& - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&1& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \frac{\hspace{2} \sqrt{3}+2 \hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{B}}^{\prime } $ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}3\hspace{2}},\frac{\hspace{2} \sqrt{6}+\sqrt{2} \hspace{2}}{\hspace{2}6\hspace{2}},0 $ }$

$3$\displaystyle{ $ \begin{array}1&0\hspace{5}& - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&1& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ $ \begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ = $ \begin{array} - \frac{\hspace{2} 2\sqrt{2} \hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \frac{\hspace{2} \sqrt{3}+1 \hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{D}}^{\prime } $ - \frac{\hspace{2} 2\sqrt{2} \hspace{2}}{\hspace{2}3\hspace{2}},\frac{\hspace{2} \sqrt{6}+\sqrt{2} \hspace{2}}{\hspace{2}6\hspace{2}},0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&1& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2} - \sqrt{3}+2 \hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{F}}^{\prime } $ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}6\hspace{2}},\frac{\hspace{2} - \sqrt{6}+2\sqrt{2} \hspace{2}}{\hspace{2}6\hspace{2}},0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&1& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ $ \begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}} & \vspace{6}\\ \end{array} $ = $ \begin{array} - \frac{\hspace{2}5\hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2} - \sqrt{3}+2 \hspace{2}}{\hspace{2} 3\sqrt{2} \hspace{2}} & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{H}^{\prime } $ - \frac{\hspace{2} 5\sqrt{2} \hspace{2}}{\hspace{2}6\hspace{2}},\frac{\hspace{2} - \sqrt{6}+2\sqrt{2} \hspace{2}}{\hspace{2}6\hspace{2}},0 $ }$

以上求めたデータを用いて影を描くと下図のようになる．

ホーム>>カテゴリー分類>>行列>>線形代数>>線形代数に関する問題>>線形写像・1次変換>>立方体の影の作図問題>>(4)の解答

作成：学生スタッフ
最終更新日： 2025年4月25日

立方体の影の作図問題 解答(4)

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立方体の影の作図問題　解答(4)