Geometric Bites

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...

The area of a parallelogram can be found using two vectors. Let the two vectors be: A and B If A is the base of the parallelogram, then the height is the perpendicular part of B. From the diagram: height = |B|sin(θ) Therefore: Area = base × height Area = |A||B|sin(θ) This is the standard formula for the area of a parallelogram formed by two vectors. Removing the Angle The formula Area = |A||B|sin(θ) uses the angle θ between the two vectors. But the angle is not always given. To remove the angle, start with the dot product identity: A · B = |A||B|cos(θ) Now square both sides: (A · B) 2 = |A| 2 |B| 2 cos 2 (θ) Using the trigonometric identity: cos 2 (θ) = 1 − sin 2 (θ) we get: (A · B) 2 = |A| 2 |B| 2 (1 − sin 2 (θ)) Expand the right-hand side: (A · B) 2 = |A| 2 |B| 2 − |A| 2 |B| 2 sin 2 (θ) Now rearrange: |A| 2 |B| 2 sin 2 (θ) = |A| 2 |B| 2 − (A · B) 2 The Area Identity Since Area = |A||B|sin(θ) s...