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Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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| (같은 사용자의 중간 판 7개는 보이지 않습니다) | |||
| 13번째 줄: | 13번째 줄: | ||
<math>v_{2}=(-\sqrt{3}/2, 1/2)</math> | <math>v_{2}=(-\sqrt{3}/2, 1/2)</math> | ||
| − | + | :<math> | |
\begin{array}{|rcl|} | \begin{array}{|rcl|} | ||
\hline | \hline | ||
| − | + | (1,0) & = & v_1 \\ | |
\hline | \hline | ||
| − | \left | + | \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right) & = & \sqrt{3} v_1+v_2 \\ |
\hline | \hline | ||
| − | \left | + | \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) & = & 2 v_1+\sqrt{3} v_2 \\ |
\hline | \hline | ||
| − | + | (0,1) & = & \sqrt{3} v_1+2 v_2 \\ | |
\hline | \hline | ||
| − | \left | + | \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right) & = & v_1+\sqrt{3} v_2 \\ |
\hline | \hline | ||
| − | \left | + | \left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right) & = & v_2 \\ |
\hline | \hline | ||
| − | + | (-1,0) & = & -v_1 \\ | |
\hline | \hline | ||
| − | \left | + | \left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right) & = & -\sqrt{3} v_1-v_2 \\ |
\hline | \hline | ||
| − | \left | + | \left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) & = & -2 v_1-\sqrt{3} v_2 \\ |
\hline | \hline | ||
| − | + | (0,-1) & = & -\sqrt{3} v_1-2 v_2 \\ | |
\hline | \hline | ||
| − | \left | + | \left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right) & = & -v_1-\sqrt{3} v_2 \\ |
\hline | \hline | ||
| − | \left | + | \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right) & = & -v_2 \\ |
\hline | \hline | ||
\end{array} | \end{array} | ||
| − | + | </math> | |
| − | |||
==역사== | ==역사== | ||
| 49번째 줄: | 48번째 줄: | ||
* http://www.google.com/search?hl=en&tbs=tl:1&q= | * http://www.google.com/search?hl=en&tbs=tl:1&q= | ||
| − | * [[ | + | * [[수학사 연표]] |
| 66번째 줄: | 65번째 줄: | ||
==관련된 항목들== | ==관련된 항목들== | ||
| + | * [[행렬과 선형사상]] | ||
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| 90번째 줄: | 76번째 줄: | ||
* https://docs.google.com/file/d/0B8XXo8Tve1cxZi1vYlpSZTVoRjQ/edit | * https://docs.google.com/file/d/0B8XXo8Tve1cxZi1vYlpSZTVoRjQ/edit | ||
| − | + | [[분류:선형대수학]] | |
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2020년 11월 12일 (목) 01:53 기준 최신판
개요
예
\(v_{1}=(1, 0)\)
\(v_{2}=(-\sqrt{3}/2, 1/2)\)
\[ \begin{array}{|rcl|} \hline (1,0) & = & v_1 \\ \hline \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right) & = & \sqrt{3} v_1+v_2 \\ \hline \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) & = & 2 v_1+\sqrt{3} v_2 \\ \hline (0,1) & = & \sqrt{3} v_1+2 v_2 \\ \hline \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right) & = & v_1+\sqrt{3} v_2 \\ \hline \left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right) & = & v_2 \\ \hline (-1,0) & = & -v_1 \\ \hline \left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right) & = & -\sqrt{3} v_1-v_2 \\ \hline \left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) & = & -2 v_1-\sqrt{3} v_2 \\ \hline (0,-1) & = & -\sqrt{3} v_1-2 v_2 \\ \hline \left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right) & = & -v_1-\sqrt{3} v_2 \\ \hline \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right) & = & -v_2 \\ \hline \end{array} \]
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들