Geometric Bites

Let A be a non-zero vector in ℝ³, and also regard its endpoint as the point through which the plane passes: A = (a 1 , a 2 , a 3 ) T ≠ (0, 0, 0) T . Let P be an arbitrary point on the plane: P = (x, y, z) T . The displacement from A to P is P − A = (x − a 1 , y − a 2 , z − a 3 ) T . Because the plane is perpendicular to A , the vector A is its normal vector. Every displacement P − A lying in the plane must therefore be perpendicular to A . Perpendicular vectors have a dot product of zero. Hence A · ( P − A ) = 0. Writing this condition in coordinates gives a 1 (x − a 1 ) + a 2 (y − a 2 ) + a 3 (z − a 3 ) = 0. Expanding the brackets: a 1 x − a 1 2 + a 2 y − a 2 2 + a 3 z − a 3 2 = 0. Move the squared terms to the right-hand side: a 1 x + a 2 y + a 3 z = a 1 2 + a 2 2 + a 3 2 . In vector notation, this becomes A · P = A · A = ‖ A ‖ 2 . ...

Let A, B and X be column vectors in ℝ³: A = a₁ a₂ a₃ , B = b₁ b₂ b₃ , X = x y z . BAᵀX = B(AᵀX). Matrix multiplication is associative, so AᵀX may be evaluated first. Since Aᵀ is a 1 × 3 row matrix and X is a 3 × 1 column matrix, their product is a scalar: AᵀX = a₁x + a₂y + a₃z. Therefore, BAᵀX = b₁ b₂ b₃ (a₁x + a₂y + a₃z). The quantity in parentheses is a scalar, so it multiplies every component of B: BAᵀX = b₁a₁x + b₁a₂y + b₁a₃z b₂a₁x + b₂a₂y + b₂a₃z b₃a₁x + b₃a₂y + b₃a₃z . Collect the coefficients of x, y and z into a matrix multiplying X: BAᵀX = b₁a₁ b₁a₂ b₁a₃ b₂a₁ b₂a₂ b...

Direction Cosines of a Unit Vector A vector in 3D can be written as v = (x, y, z). This vector points from the origin to the point (x, y, z). Its direction depends on how much it travels in the x-direction, the y-direction and the z-direction. Magnitude of the Vector The magnitude, or length, of v is |v| = √(x² + y² + z²). This comes from the 3D version of Pythagoras' theorem. The vector has three perpendicular components: x, y and z. Squaring them, adding them, and taking the square root gives the total length. Unit Vector A unit vector is a vector with length 1. To turn v into a unit vector, divide every component by the magnitude of v: v̂ = (1 / |v|)(x, y, z). So v̂ = (x / |v|, y / |v|, z / |v|). This new vector points in the same direction as v, but its length is exactly 1. The Dot Product The dot product has two important forms. Algebraic form: a · b = a₁b₁ + a₂b₂ + a₃b₃. Geometric form: a · b = |a||b|cos(θ). The algebraic form uses co...

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 ...