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Deriving the Direction Cosines of a Unit Vector

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...
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The Associative, Commutative and Distributive Laws

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...
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Deriving Compound Angle Identities: Additional Trigonometric Proofs

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...
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Proofs of the Base-10 Logarithm Laws

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...
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The Algebra Behind the Cross Product Magnitude

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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...
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The Area of a Parallelogram: Angle Not Required

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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...
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The Derivative of aˣ and the Natural Logarithm

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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-...
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