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...

Why the Line ax + by = 0 Passes Through the Point (−b, a) In ℝ² , the equation ax + by = 0 describes a line that is perpendicular to the vector (a, b) . This article explains exactly why—and why that line always passes through the point (−b, a) . 1. Start with the Vector (a, b) Consider the vector (a, b) . To find a line perpendicular to it, we need a vector whose dot product with (a, b) is zero. Try the vector (−b, a) : (a, b) · (−b, a) = a(−b) + b(a) = −ab + ab = 0 Therefore, (−b, a) is perpendicular to (a, b) . 2. Any Scalar Multiple Also Works If (−b, a) is perpendicular to (a, b) , then any multiple λ(−b, a) is also perpendicular: (a, b) · [λ(−b, a)] = λ[(a, b) · (−b, a)] = λ · 0 = 0 Let this perpendicular vector be (x, y) . Then (x, y) = λ(−b, a) Every point on the line comes from a particular choice of λ . 3. Converting to an Equation Since (x, y) is perpendicular to (a, b) , we have: (a, b) · (x, y) = 0 Expanding ...

The Dot Product Identity and the Cosine Rule in ℝ 3 In this article we derive the dot product identity A · B = |A| × |B| × cos(θ) and show how this identity leads directly to the cosine rule, using a combination of coordinate algebra and geometric interpretation. 1. Vectors in ℝ 3 Let the vectors be: A = (a 1 , a 2 , a 3 ) B = (b 1 , b 2 , b 3 ) Their difference is: A - B = (a 1 - b 1 , a 2 - b 2 , a 3 - b 3 ) The squared magnitude of this difference vector is: |A - B| 2 = (a 1 - b 1 ) 2 + (a 2 - b 2 ) 2 + (a 3 - b 3 ) 2 . 2. Expanding the Square of the Difference Expand each component: (a 1 - b 1 ) 2 = a 1 2 - 2a 1 b 1 + b 1 2 (a 2 - b 2 ) 2 = a 2 2 - 2a 2 b 2 + b 2 2 (a 3 - b 3 ) 2 = a 3 2 - 2a 3 b 3 + b 3 2 Adding these three expansions gives: |A - B| 2 = (a 1 2 + a 2 2 + a 3 2 ) + (b 1 2 + b 2 2 + b 3 2 ) - 2(a 1 b 1 + a 2 b 2 + a 3 b 3 ). Recognise the squared magnitudes: |A| 2 = a 1 2 + a 2 2 ...