Geometric Bites

2×2 Orthogonal Matrix Mastery — A Generalised Construction Orthogonal matrices in two dimensions reveal one of the cleanest structures in linear algebra. A 2×2 matrix is orthogonal when its columns (and rows) satisfy two conditions: They are perpendicular (their dot product is zero); They have unit length (their magnitude is one). This article presents a clear generalisation: any pair of perpendicular vectors with equal magnitude can be normalised to form an orthogonal matrix. 1. Begin with two perpendicular vectors Let the first vector be: (a, b) A perpendicular vector can be chosen as: (−b, a) Their dot products confirm orthogonality: (a, b) · (−b, a) = −ab + ab = 0 (a, −b) · (b, a) = ab − ab = 0 2. Compute their shared magnitude Both vectors have the same length: |(a, b)| = √(a² + b²) We can therefore normalise each one by dividing by √(a² + b²). 3. Form the matrix using the normalised vectors Place the two normalised vectors...

Orthogonal Matrices and Mutually Orthogonal Vectors Orthogonal matrices appear naturally throughout linear algebra, geometry, physics, and computer graphics. They preserve lengths, angles, and orientation, which makes them fundamental in describing rotations and rigid motions in three-dimensional space. This article provides a clear and carefully structured explanation of what orthogonal matrices are, why they matter, and how to verify that a given matrix is orthogonal. 1. Definition of an Orthogonal Matrix Let M be an n × n square matrix. M is called orthogonal if it satisfies: M M T = I Here: M T is the transpose of M. I is the identity matrix of the same size. Because of this property, every orthogonal matrix has a very useful consequence: M -1 = M T This means that the inverse of an orthogonal matrix is obtained simply by transposing it. This property is central to rigid-body transformations in 3D geometry and computer graphics. 2...

Linear Transformations in ℝ³ and 3×3 Matrices Matrices give us a compact way to describe linear transformations in three-dimensional space. A linear transformation is a mapping T : ℝ³ → ℝ³ that sends a point with position vector (x, y, z) to another point, according to a rule with two key properties. What Makes a Transformation Linear? A transformation T : ℝ³ → ℝ³ is called linear if, for all real numbers λ and all vectors (x, y, z) in ℝ³, T(λx, λy, λz) = λ T(x, y, z), and for all vectors (x₁, y₁, z₁) and (x₂, y₂, z₂) in ℝ³, T(x₁ + x₂, y₁ + y₂, z₁ + z₂) = T(x₁, y₁, z₁) + T(x₂, y₂, z₂). The point that (x, y, z) is sent to is called the image of (x, y, z) under T. The Standard Basis Vectors To find the matrix that represents a particular transformation, it is enough to know what happens to three special vectors, called the standard basis for ℝ³: î = (1, 0, 0) ĵ = (0, 1, 0) k̂ = (0, 0, 1) Once we know the images of î, ĵ and k̂, th...

This article presents the rules of logarithms using complete, line-by-line derivations. Every identity is built directly from its exponential origin, without shortcuts, matching the structure of formal handwritten algebra. 1. Definition We begin with fundamental exponent facts: a⁰ = 1 ⇒ logₐ(1) = 0 a¹ = a ⇒ logₐ(a) = 1 Say: aᵐ = p Then, by definition: logₐ(p) = m Raise both sides of aᵐ = p to the power 1/m (with m ≠ 0 ): p^(1/m) = a Therefore: logₚ(a) = 1/m Since m = logₐ(p) , we obtain: logₐ(p) = 1 / logₚ(a) 2. Product Rule — Full Derivation Say: aᵐ = p and aⁿ = q Multiply: aᵐ · aⁿ = p · q Using index addition: a^(m+n) = p · q Taking logarithms: logₐ(p · q) = m + n Substitute: logₐ(p · q) = logₐ(p) + logₐ(q) 3. Quotient Rule — Full Derivation Say: aᵐ = p and aⁿ = q Divide: aᵐ / aⁿ = p / q Index subtraction gives: a^(m−n) = p / q Taking logarithms: logₐ(p / q) = m − n So: log...

Finding the Inverse of a 2x2 Matrix from Scratch This post shows a complete, step-by-step derivation of the inverse of a 2x2 matrix. Everything is expressed using stable, browser-safe ASCII formatting so the layout displays correctly on all devices and all templates. FIRST PART. Start with the matrix equation: A = [[a, b], [c, d]] A^(-1) = [[w, x], [y, z]] Goal: A * A^(-1) = I This produces the column equations: [aw + by, cw + dy]^T = [1, 0]^T [ax + bz, cx + dz]^T = [0, 1]^T Which gives the four equations: aw + by = 1 cw + dy = 0 ax + bz = 0 cx + dz = 1 SECOND PART. Use the first two equations to find w. aw + by = 1 cw + dy = 0 Multiply: (ad)w + (bd)y = d (first eq multiplied by d) (bc)w + (bd)y = 0 (second eq multiplied by b) Subtract: (ad - bc)w = d w = d / (ad - bc) (ad - bc != 0) THIRD PART. Use the next pair to find x. ax + bz = 0 cx + dz = 1 Multiply: (ad)x + (bd)z = 0 (bc)x + (bd)z = b Subtract: (ad - bc)...

Converting the Vector Equation of a Line into Cartesian Form A straight line in three-dimensional space can be expressed using vectors. One important vector form is (𝐑 − 𝐀) × 𝐁 = 0 This equation states that the displacement vector from a fixed point 𝐀 to a general point 𝐑 is parallel to the direction vector 𝐁. Two non-zero vectors have a zero cross product precisely when they are parallel. From this fact, the Cartesian (symmetric) equation of the line can be derived. 1. Substituting Coordinate Vectors The general point on the line is written as 𝐑 = (x, y, z) The fixed point is 𝐀 = (x₁, y₁, z₁) The direction vector is 𝐁 = (l, m, n) Substituting these into the vector equation yields: ((x, y, z) − (x₁, y₁, z₁)) × (l, m, n) = 0 which simplifies to: (x − x₁, y − y₁, z − z₁) × (l, m, n) = 0 2. Using the Condition for a Zero Cross Product If two non-zero vectors have a zero cross product, then one is a scalar multiple of the other. T...

The Difference Between the Lines 𝐀 + t𝐁 and 𝐁 + t(𝐀 − 𝐁) A line in vector form is defined by two components: a base point that determines its position, and a direction vector that determines its orientation. Two expressions may involve the same vectors but still represent completely different lines when either the base point or the direction vector changes. The expressions L₁: 𝐀 + t𝐁 L₂: 𝐁 + t(𝐀 − 𝐁) provide a clear example of how distinct lines arise from different vector components. 1. Line L₁: 𝐀 + t𝐁 The expression 𝐀 + t𝐁 describes a line passing through the point represented by vector 𝐀 with direction vector 𝐁. As the real parameter t varies, the expression generates all points on the line. Base point: 𝐀 Direction vector: 𝐁 This is the line through 𝐀 directed along 𝐁. 2. Line L₂: 𝐁 + t(𝐀 − 𝐁) The expression 𝐁 + t(𝐀 − 𝐁) describes a different line. Its base point is 𝐁, and its direction vector is t...