立方体の影の作図問題　解答(5)

■問題(5)

太陽の位置を南南東の地表面から $3$\displaystyle{45^\circ }$ (仰角)の方向に変更する．この場合の(4)で説明した立方体の影を作図せよ．(4)と同様に答えよ．

■答

■解き方

(1)～(4)と同様にして

影の長さは $3$\displaystyle{z}$

$3$\displaystyle{u=- z\sin 22.5^\circ +x}$

$3$\displaystyle{v=z\cos 22.5^\circ +y}$

よって

$3$\displaystyle{X= $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ }$

各頂点の影の位置を求める．

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ $ \begin{array}0& \vspace{6}\\ 0& \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ = $ \begin{array}0& \vspace{6}\\ 0& \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{A}}^{\prime } $ 0,0,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \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} \sqrt{3} \hspace{2}}\sin 22.5^\circ & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{B}}^{\prime } $ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\sin 22.5^\circ ,\frac{\hspace{2} \sqrt{6} \hspace{2}}{\hspace{2}6\hspace{2}}+\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\cos 22.5^\circ ,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \vspace{6}\\ 0&0&0& \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} $ = $ \begin{array} - \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\sin 22.5^\circ & \vspace{6}\\ \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} }+\frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{C}}^{\prime } $ - \frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\sin 22.5^\circ ,\frac{\hspace{2} \sqrt{6} \hspace{2}}{\hspace{2}3\hspace{2}}+\frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\cos 22.5^\circ ,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \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} \sqrt{3} \hspace{2}}\sin 22.5^\circ & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{D}}^{\prime } $ - \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\sin 22.5^\circ ,\frac{\hspace{2} \sqrt{6} \hspace{2}}{\hspace{2}6\hspace{2}}+\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\cos 22.5^\circ ,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \vspace{6}\\ 0&0&0& \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} $ = $ \begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\sin 22.5^\circ & \vspace{6}\\ - \sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} }+\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{E}}^{\prime } $ - \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\sin 22.5^\circ ,- \frac{\hspace{2} \sqrt{6} \hspace{2}}{\hspace{2}3\hspace{2}}+\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\cos 22.5^\circ ,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \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} \sqrt{2} \hspace{2}}- \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\sin 22.5^\circ & \vspace{6}\\ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{F}}^{\prime } $ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\sin 22.5^\circ ,- \frac{\hspace{2} \sqrt{6} \hspace{2}}{\hspace{2}6\hspace{2}}+\frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\cos 22.5^\circ ,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \vspace{6}\\ 0&0&0& \vspace{6}\\ \end{array} $ $ \begin{array}0& \vspace{6}\\ 0& \vspace{6}\\ \sqrt{3} & \vspace{6}\\ \end{array} $ = $ \begin{array} - \sqrt{3}\sin 22.5^\circ & \vspace{6}\\ \sqrt{3}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{G}}^{\prime } $ - \sqrt{3}\sin 22.5^\circ ,\sqrt{3}\cos 22.5^\circ ,0 $ }$

$3$\displaystyle{ $ \begin{array}1&0& - \sin 22.5^\circ & \vspace{6}\\ 0&1& \cos 22.5^\circ & \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} \sqrt{2} \hspace{2}}- \frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\sin 22.5^\circ & \vspace{6}\\ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{6} \hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2} \sqrt{3} \hspace{2}}\cos 22.5^\circ & \vspace{6}\\ 0& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{{\text{H}}^{\prime } $ - \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\sin 22.5^\circ ,- \frac{\hspace{2} \sqrt{6} \hspace{2}}{\hspace{2}6\hspace{2}}+\frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}\cos 22.5^\circ ,0 $ }$

この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>>(5)の解答

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最終更新日： 2023年2月10日

立方体の影の作図問題 解答(5)

■問題(5)

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立方体の影の作図問題　解答(5)