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...

Let v = p q r ≠ 0 0 0 The objective is to construct the standard 3 × 3 matrix [v]ₓ associated with v, without assuming its entries in advance. Begin with two requirements: Kᵀ = −K Kv = 0. The first condition requires skew-symmetry. The second requires v to lie in the null space of K. 1. Write the general skew-symmetric matrix Since Kᵀ = −K, the diagonal entries must be zero and entries reflected across the diagonal must have opposite signs. Therefore, K = 0 α β −α 0 γ −β −γ 0 At this stage, α, β and γ are unknown. 2. Require Kv = 0 0 α β −α 0 γ −β −γ 0 p q r ...

A skew-symmetric matrix is a square matrix whose transpose equals its negative. The construction below begins with an arbitrary 3 × 3 matrix, subtracts its transpose, and derives the complete general form without assuming the result in advance. K is skew-symmetric precisely when Kᵀ = −K. 1. Begin with an arbitrary matrix Let A = a b c d e f g h i and Aᵀ = a d g b e h c f i Transposition reflects the entries across the main diagonal: rows become columns and columns become rows. 2. Subtract the transpose Define K = A − Aᵀ. Subtract corresponding entries: K = a−a b−d c−g d−b e−e f−h g−c h−f i−i The diagonal entries cancel: ...

A matrix transforms input vectors into output vectors. Some independent directions survive, some are combined, and some may disappear completely into the zero vector. Span, dimension, column space, rank, null space and nullity describe this process precisely. input space → matrix transformation → column space 1. Span: all reachable combinations Given vectors v₁, v₂, …, vₖ, their span is the set of every vector that can be made by scaling and adding them: Span{v₁, v₂, …, vₖ} = {c₁v₁ + c₂v₂ + ⋯ + cₖvₖ : c₁, c₂, …, cₖ ∈ ℝ}. One non-zero direction spans a line through the origin. Two independent directions span a plane through the origin. Three independent directions in ℝ³ span all of ℝ³. 2. Independence and dimension Vectors are linearly independent when none of them can be constructed from the others. Each independent vector contributes a genuinely new direction. A dependent vector contributes no new...

Let v = (p, q, r) T be a vector in ℝ³. From its three components, form the 3 × 3 skew-symmetric matrix K = ( 0 −r q r 0 −p −q p 0 ) . The entries reflected across the main diagonal have opposite signs, while every diagonal entry is zero. Therefore, K T = −K. Multiplication by the vector Matrix-vector multiplication can be written as a linear combination of the columns of K. The first column is multiplied by p, the second by q, and the third by r: K v = p ( 0 r −q ) + q ( −r 0 p ) + r ( q −p 0 ) . Distributing p, q and r gives K v = ( 0 · p + (−r) · q + q · r r · p + 0 · q + (−p) · r −q · p + p · q + 0 · r ) . Combining the entries in each row produces ...

This proof demonstrates that subtracting the transpose of any real 3 × 3 matrix from the original matrix produces a 3 × 3 skew-symmetric matrix. Construction Let M = ( a b c d e f g h i ) Its transpose is M T = ( a d g b e h c f i ) Subtracting the transpose from the original matrix gives K = M − M T = ( 0 b − d c − g d − b 0 f − h g − c h − f 0 ) ...

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...

The associative, commutative and distributive laws are three of the most important structural rules in algebra. They explain how expressions may be grouped, reordered, expanded and simplified without changing their mathematical value. These laws are used throughout arithmetic, algebra, factorisation, equation solving and mathematical proof. Associative Law The associative law describes how terms may be grouped when the same operation is repeated. If an operation is associative, changing the placement of the brackets does not change the final value of the expression. For addition: a + (b + c) = (a + b) + c For example: 1 + (2 + 3) = (1 + 2) + 3 The associative law also applies to multiplication: a × (b × c) = (a × b) × c For example: 2 × (3 × 4) = (2 × 3) × 4 Subtraction is not associative because changing the grouping can change the result. a − (b − c) ≠ (a − b) − c For example: 1 − (2 − 3) ≠ (1 − 2) − 3 Commutative Law The commutative law describe...

These workings use compound angle identities to derive double angle, triple angle, and related trigonometric identities. Compound Angles, Extras 1. Deriving sin(2θ) sin(θ + θ) = sinθ cosθ + cosθ sinθ = 2sinθ cosθ = sin(2θ) 2. Deriving cos(2θ) cos(θ + θ) = cosθ cosθ - sinθ sinθ = cos 2 θ - sin 2 θ = cos(2θ) 3. Deriving cos(2θ) = 2cos²θ - 1 cos(2θ) = cos 2 θ - sin 2 θ = cos 2 θ - (1 - cos 2 θ) = cos 2 θ - 1 + cos 2 θ = 2cos 2 θ - 1 4. Deriving cos(2θ) = 1 - 2sin²θ cos(2θ) = cos 2 θ - sin 2 θ = 1 - sin 2 θ - sin 2 θ = 1 - 2sin 2 θ 5. Deriving sin(3θ) sin(2θ + θ) = sin(2θ)cosθ + cos(2θ)sinθ = 2sinθ cosθ cosθ + (1 - 2sin 2 θ)sinθ = 2sinθ cos 2 θ + sinθ - 2sin 3 θ = sinθ(2cos 2 θ + 1) - 2sin 3 θ = sinθ(2(1 - sin 2 θ) + 1) - 2sin 3 θ = sinθ(2 - 2sin 2 θ + 1) - 2sin 3 θ = sinθ(3 - 2sin 2 θ) - 2sin 3 θ = 3sinθ - 2sin 3 θ - 2sin 3 θ = 3sinθ - 4sin 3 θ 6. Deriving cos(3θ) cos(2θ + θ) = cos2θ cosθ - sin2...

These workings derive the laws of base-10 logarithms from exponent laws by converting between exponential form and logarithmic form. Assume a > 0 , b > 0 , and n ≠ 0 . Product Rule log(ab) = log a + log b Let 10 x = a Let 10 y = b Therefore: log 10 a = x log 10 b = y 10 x 10 y = ab 10 x+y = ab Therefore: log 10 (ab) = x + y Substituting: log 10 (ab) = log 10 a + log 10 b Therefore: log(ab) = log a + log b Quotient Rule log(a / b) = log a - log b Let 10 x = a Let 10 y = b Therefore: log 10 a = x log 10 b = y 10 x / 10 y = a / b 10 x-y = a / b Therefore: log 10 (a / b) = x - y Substituting: log 10 (a / b) = log 10 a - log 10 b Therefore: log(a / b) = log a - log b Power Rule log(a n ) = n log a Let log(a n ) = x Therefore: 10 x = a n Taking the n-th root of both sides: (10 x ) 1/n = (a n ) 1/n 10 x/n = a Therefore: log 10 a = x / n n log 10 a = x Theref...

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...

Two of the most important results in differential calculus are d/dx(ln x) = 1/x d/dx(aˣ) = aˣ ln a. These formulas are closely connected. One describes the derivative of the natural logarithm, while the other gives the derivative of an exponential function with positive base. Together, they reveal the deep relationship between logarithms, exponentials, inverse functions, and differentiation. 1. Derivative of the natural logarithm Begin with ln x = y. This is equivalent to eʸ = x. Differentiate both sides with respect to y: dx/dy = eʸ. Now invert this result: dy/dx = 1/eʸ. Since eʸ = x, substitute back: dy/dx = 1/x. Therefore, d/dx(ln x) = 1/x, for x > 0. So the derivative of the natural logarithm is the reciprocal of x. 2. Derivative of the exponential function aˣ Now let y = aˣ, where a > 0 and a ≠ 1. Take logarithms to base a: logₐ y = x. Now apply the change-of-...

Proving that the sum of three consecutive integers is divisible by 3. Algebraic proof questions in Edexcel International GCSE (Higher) often look difficult because they are written in words. The quickest way to handle them is to translate the words into standard algebraic forms that guarantee the number property you need (even, odd, multiple, consecutive, square, etc.). Once the translation is correct, the rest of the proof is usually straightforward simplification, factoring, and a clear final statement. This post gives a compact “toolkit” of the most common forms, presented in a table you can reuse, plus a small set of extras and techniques that frequently appear in Higher-tier proof questions. The core principle In an algebraic proof, represent the numbers so the required property is built in. For example: If a number is even, write it as 2n for some integer n. If a number is a multiple of 3, write it as 3n for some integer n. The phrase “for some integer n” matters...