数学 - Python - 解析学 - n次元空間 - ユークリッド空間 - ベクトルの外積、内積、スカラー倍、性質

学習環境

解析入門(上) (松坂和夫数学入門シリーズ 4) (松坂和夫(著)、岩波書店)の第10章(n次元空間)、10.1(ユークリッド空間)、問題10の解答を求めてみる。

1. $\begin{array}{l} - (b \times a) \\ = - (b_{2} a_{3} - b_{3} a_{2}, b_{3} a_{1} - b_{1} a_{3}, b_{1} a_{2} - b_{2} a_{1}) \\ = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) \\ = a \times b \end{array}$
2. $\begin{array}{l} a \times (b + b') \\ = (a_{2} (b_{3} + b_{3}') - a_{3} (b_{2} + b_{2}'), a_{3} (b_{1} + b_{1}') - a_{1} (b_{3} + b_{3}'), a_{1} (b_{2} + b_{2}') - a_{2} (b_{1} + b_{1}')) \\ = (a_{2} b_{3} + a_{2} b_{3}' - a_{3} b_{2} - a_{3} b_{2}', \\ a_{3} b_{1} + a_{3} b_{1}' - a_{1} b_{3} - a_{1} b_{3}', \\ a_{1} b_{2} + a_{1} b_{2}' - a_{2} b_{1} - a_{2} b_{1}') \\ = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) \\ + (a_{2} b_{3}' - a_{3} b_{2}, a_{3} b_{1}' - a_{1} b_{3}', a_{1} b_{2}' - a_{2} b_{1}') \\ = a \times b - a \times b' \end{array}$
3. $\begin{array}{l} (α a) \times b \\ = (α a_{2} b_{3} - α a_{3} b_{2}, α a_{3} b_{1} - α a_{1} b_{3}, α a_{1} b_{2} - α a_{2} b_{1}) \\ = α (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{2} b_{3}, a_{2} b_{2} - a_{2} b_{1}) \\ = α (a \times b) \\ (α a) \times b \\ = (a_{2} (α b_{3}) - a_{3} (α b_{2}), a_{3} (α b_{1}) - a_{1} (α b_{3}), a_{1} (α b_{2}) - a_{2} (α b_{1})) \\ = a \times (α b) \end{array}$
4. $\begin{array}{l} (a \times b) \cdot c \\ = (a_{2} b_{3} - a_{3} b_{2}) c_{1} + (a_{3} b_{1} - a_{1} b_{3}) c_{2} + (a_{1} b_{2} - a_{2} b_{1}) c_{3} \\ = a_{2} b_{3} c_{1} - a_{3} b_{2} c_{1} + a_{3} b_{1} c_{2} - a_{1} b_{3} c_{2} + a_{1} b_{2} c_{3} - a_{2} b_{1} c_{3} \\ a \cdot (b \times c) \\ = a_{1} (b_{2} c_{3} - b_{3} c_{2}) + a_{2} (b_{3} c_{1} - b_{1} c_{3}) + a_{3} (b_{1} c_{2} - b_{2} c_{1}) \\ = a_{2} b_{3} c_{1} - a_{3} b_{2} c_{1} + a_{3} b_{1} c_{2} - a_{1} b_{3} c_{2} - a_{2} b_{1} c_{3} \end{array}$
  
  よって、
  
  $(a \times b) \cdot c = a \cdot (b \times c)$
5. $\begin{array}{l} (a \times b) \times c \\ = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) \times c \\ = ((a_{3} b_{1} - a_{1} b_{3}) c_{3} - (a_{1} b_{2} - a_{2} b_{1}) c_{2}, \\ (a_{1} b_{2} - a_{2} b_{1}) c_{1} - (a_{2} b_{3} - a_{3} b_{2}) c_{3}, \\ (a_{2} b_{3} - a_{3} b_{2}) c_{2} - (a_{3} b_{1} - a_{1} b_{3}) c_{1}) \\ = (a_{3} b_{1} c_{3} - a_{1} b_{3} c_{3} - a_{1} b_{2} c_{2} + a_{2} b_{1} c_{2}, \\ a_{1} b_{2} c_{1} - a_{2} b_{1} c_{1} - a_{2} b_{3} c_{3} + a_{3} b_{2} c_{3}, \\ a_{2} b_{3} c_{2} - a_{3} b_{2} c_{2} - a_{3} b_{1} c_{1} + a_{1} b_{3} c_{1}) \\ (a \cdot c) b - (b \cdot c) a \\ = (a_{1} b_{1} c_{1} + a_{2} b_{1} c_{2} + a_{3} b_{1} c_{3}, \\ a_{1} b_{2} c_{1} + a_{2} b_{2} c_{2} + a_{3} b_{2} c_{3}, \\ a_{1} b_{3} c_{1} + a_{2} b_{3} c_{2} + a_{3} b_{3} c_{3}) - \\ (a_{1} b_{1} c_{1} + a_{1} b_{2} c_{2} + a_{1} b_{3} c_{3}, \\ a_{2} b_{1} c_{1} + a_{2} b_{2} c_{2} + a_{2} b_{3} c_{3}, \\ a_{3} b_{1} c_{1} + a_{3} b_{2} c_{2} + a_{3} b_{3} c_{3}) \\ = (a_{2} b_{1} c_{2} + a_{3} b_{1} c_{3} - a_{1} b_{2} c_{2} - a_{2} b_{3} c_{3}, \\ a_{1} b_{2} c_{1} + a_{3} b_{2} c_{3} - a_{2} b_{1} c_{1} - a_{2} b_{3} c_{3}, \\ a_{1} b_{3} c_{1} + a_{2} b_{3} c_{2} - a_{3} b_{1} c_{1} - a_{3} b_{2} c_{2}) \end{array}$
  
  よって、
  
  $(a \times b) \times c = (a \cdot c) b - (b \cdot c) a$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import symbols, Matrix print('10.') alpha = symbols('α') a = Matrix(symbols('a:3')).reshape(1, 3) b = Matrix(symbols('b:3')).reshape(1, 3) c = Matrix(symbols('c:3')).reshape(1, 3) class MyTestCase(TestCase): def test_a(self): self.assertEqual(a.cross(b).simplify(), (-b.cross(a)).simplify()) def test_b(self): self.assertEqual(a.cross(b + c).simplify(), (a.cross(b) + a.cross(c)).simplify()) self.assertEqual((a + b).cross(c).simplify(), (a.cross(c) + b.cross(c)).simplify()) def test_c(self): self.assertEqual((alpha * a).cross(b).simplify(), a.cross(alpha * b).simplify(), alpha * a.cross(b)) def test_d(self): self.assertEqual(a.cross(b).dot(c).expand(), a.dot(b.cross(c)).expand()) def test_e(self): self.assertEqual(a.cross(b).cross(c).simplify(), (a.dot(c) * b - b.dot(c) * a).simplify()) if __name__ == "__main__": main()

% ./sample10.py -v 10. test_a (__main__.MyTestCase) ... ok test_b (__main__.MyTestCase) ... ok test_c (__main__.MyTestCase) ... ok test_d (__main__.MyTestCase) ... ok test_e (__main__.MyTestCase) ... ok ---------------------------------------------------------------------- Ran 5 tests in 0.824s OK %

Kamimura's blog

2020年3月28日土曜日

数学 - Python - 解析学 - n次元空間 - ユークリッド空間 - ベクトルの外積、内積、スカラー倍、性質

0 コメント:

コメントを投稿