基本的な1次変換の問題

■問題

次の回転行列 ${\displaystyle R\left(\theta \right)}$ を求め，点${\displaystyle \left(3,3\right)}$ が回転で移る座標を求めよ．

(1)$R\left(\frac{\pi }{3}\right)$   (2)$R\left(\frac{\pi }{2}\right)$   (3)$R\left(\frac{\pi }{6}\right)$   (4)$R\left(\frac{\pi }{4}\right)$

■答

(1) ${\displaystyle \left(\frac{3}{2}-\frac{3}{2}\sqrt{3},\frac{3}{2}\sqrt{3}+\frac{3}{2}\right)}$   (2) ${\displaystyle \left(-3,3\right)}$   (3) ${\displaystyle \left(\frac{3}{2}\sqrt{3}-\frac{3}{2},\frac{3}{2}+\frac{3}{2}\sqrt{3}\right)}$   (4) ${\displaystyle \left(0,3\sqrt{2}\right)}$

■解き方

$R\left(\frac{\pi }{3}\right){\displaystyle =\left(\begin{array}{cc}\cos \frac{\pi }{3}& -\sin \frac{\pi }{3}\\ \sin \frac{\pi }{3}& \cos \frac{\pi }{3}\end{array}\right)}$${\displaystyle =\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)}$${\displaystyle =\frac{1}{2}\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)}$

点${\displaystyle \left(3,3\right)}$ を原点を中心に${\displaystyle \frac{\pi }{3}}$ 回転すると，

$R\left(\frac{\pi }{3}\right){\displaystyle \left(\begin{array}{c}3\\ 3\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)}$${\displaystyle =\frac{1}{2}\left(\begin{array}{c}3-3\sqrt{3}\\ 3\sqrt{3}+3\end{array}\right)}$${\displaystyle =\left(\begin{array}{c}\frac{3}{2}-\frac{3}{2}\sqrt{3}\\ \frac{3}{2}\sqrt{3}+\frac{3}{2}\end{array}\right)}$

となり，点${\displaystyle \left(\frac{3}{2}-\frac{3}{2}\sqrt{3},\frac{3}{2}\sqrt{3}+\frac{3}{2}\right)}$ に移る．

(2)

$R\left(\frac{\pi }{2}\right){\displaystyle =\left(\begin{array}{cc}\cos \frac{\pi }{2}& -\sin \frac{\pi }{2}\\ \sin \frac{\pi }{2}& \cos \frac{\pi }{2}\end{array}\right)}$${\displaystyle =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)}$

点${\displaystyle \left(3,3\right)}$ を原点を中心に${\displaystyle \frac{\pi }{2}}$ 回転すると，

$R\left(\frac{\pi }{2}\right){\displaystyle \left(\begin{array}{c}3\\ 3\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)}$${\displaystyle =\left(\begin{array}{c}-3\\ 3\end{array}\right)}$

となり，点${\displaystyle \left(-3,3\right)}$に移る．

(3)

$R\left(\frac{\pi }{6}\right){\displaystyle =\left(\begin{array}{cc}\cos \frac{\pi }{6}& -\sin \frac{\pi }{6}\\ \sin \frac{\pi }{6}& \cos \frac{\pi }{6}\end{array}\right)}$${\displaystyle =\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)}$${\displaystyle =\frac{1}{2}\left(\begin{array}{cc}\sqrt{3}& -1\\ 1& \sqrt{3}\end{array}\right)}$

点${\displaystyle \left(3,3\right)}$ を原点を中心に${\displaystyle \frac{\pi }{6}}$ 回転すると，

$R\left(\frac{\pi }{6}\right){\displaystyle \left(\begin{array}{c}3\\ 3\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}\sqrt{3}& -1\\ 1& \sqrt{3}\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)}$${\displaystyle =\frac{1}{2}\left(\begin{array}{c}3\sqrt{3}-3\\ 3+3\sqrt{3}\end{array}\right)=\left(\begin{array}{c}\frac{3}{2}\sqrt{3}-\frac{3}{2}\\ \frac{3}{2}+\frac{3}{2}\sqrt{3}\end{array}\right)}$

となり，点${\displaystyle \left(\frac{3}{2}\sqrt{3}-\frac{3}{2},\frac{3}{2}+\frac{3}{2}\sqrt{3}\right)}$ に移る．

(4)

$R\left(\frac{\pi }{4}\right){\displaystyle =\left(\begin{array}{cc}\cos \frac{\pi }{4}& -\sin \frac{\pi }{4}\\ \sin \frac{\pi }{4}& \cos \frac{\pi }{4}\end{array}\right)}$${\displaystyle =\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)}$${\displaystyle =\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)}$

点${\displaystyle \left(3,3\right)}$ を原点を中心に${\displaystyle \frac{\pi }{4}}$ 回転すると，

$R\left(\frac{\pi }{6}\right){\displaystyle \left(\begin{array}{c}3\\ 3\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)}$ ${\displaystyle =\frac{1}{\sqrt{2}}\left(\begin{array}{c}3-3\\ 3+3\end{array}\right)}$ ${\displaystyle =\left(\begin{array}{c}0\\ \frac{6}{\sqrt{2}}\end{array}\right)}$ ${\displaystyle =\left(\begin{array}{c}0\\ 3\sqrt{2}\end{array}\right)}$

となり，点${\displaystyle \left(0,3\sqrt{2}\right)}$ に移る．

 

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作成：学生スタッフ

最終更新日：2023年2月10日