The Algebra Behind the Cross Product Magnitude
This expansion shows why the expression |A| 2 |B| 2 − (A · B) 2 is equal to the squared magnitude of the cross product: |A × B| 2 Let A = (a 1 , a 2 , a 3 ) and B = (b 1 , b 2 , b 3 ) Then: |A| 2 = a 1 2 + a 2 2 + a 3 2 |B| 2 = b 1 2 + b 2 2 + b 3 2 Therefore: |A| 2 |B| 2 = (a 1 2 + a 2 2 + a 3 2 )(b 1 2 + b 2 2 + b 3 2 ) Expanding: |A| 2 |B| 2 = a 1 2 b 1 2 + a 1 2 b 2 2 + a 1 2 b 3 2 + a 2 2 b 1 2 + a 2 2 b 2 2 + a 2 2 b 3 2 + a 3 2 b 1 2 + a 3 2 b 2 2 + a 3 2 b 3 2 Now expand the dot product. A · B = a 1 b 1 + a 2 b 2 + a 3 b 3 So: (A · B) 2 = (a 1 b 1 + a 2 b 2 + a 3 b 3 ) 2 Expanding: (A · B) 2 = a 1 2 b 1 2 + a 2 2 b 2 2 + a 3 2 b 3 2 + 2a 1 b 1 a 2 b 2 + 2a 1 b 1 a 3 b 3 + 2a 2 b 2 a 3 b 3 Now subtract: |A| 2 |B| 2 − (A · B) 2 The matching diagonal terms cancel: a 1 2 b 1 2 , a 2 2 b 2 2 , a 3 2 b 3 2 This leaves: |A| 2 |B| 2 − (A · B) 2 = a 1 2...