Three-dimensional space - Vector Algebra in Three‑Dimensional Space

Linear Algebra in Three-Dimensional Space Introduction Linear algebra in three-dimensional space provides the mathematical framework for representing and manipulating objects in 3D environments. We work with vectors—quantities that have both direction and magnitude—and operations that combine them in meaningful ways. This framework is essential for physics, computer graphics, engineering, and many other fields. Vector Representation A vector in three-dimensional space is an ordered triple of real numbers: $\mathbf{v} = [v1, v2, v3]$. Each component represents the vector's extent along one of the three coordinate axes. Think of a vector as an arrow in space. The components $v1$, $v2$, and $v3$ tell you how far to move along the $x$-, $y$-, and $z$-axes respectively. For example, the vector $[3, 2, 1]$ means move 3 units along the $x$-axis, 2 units along the $y$-axis, and 1 unit along the $z$-axis. The magnitude (or length) of a vector $\mathbf{v}$ is denoted $\|\mathbf{v}\|$ and represents the distance from the origin to the point the vector reaches. The formula extends the Pythagorean theorem to three dimensions: $$\|\mathbf{v}\| = \sqrt{v1^2 + v2^2 + v3^2}$$ For example, if $\mathbf{v} = [3, 2, 1]$, then $\|\mathbf{v}\| = \sqrt{9 + 4 + 1} = \sqrt{14}$. Dot Product The dot product (also called scalar product) takes two vectors and produces a single number. For vectors $\mathbf{A} = [A1, A2, A3]$ and $\mathbf{B} = [B1, B2, B3]$, the dot product is: $$\mathbf{A} \cdot \mathbf{B} = A1B1 + A2B2 + A3B3$$ Example: If $\mathbf{A} = [1, 2, 3]$ and $\mathbf{B} = [4, 5, 6]$, then: $$\mathbf{A} \cdot \mathbf{B} = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32$$ Two Important Properties Property 1: The dot product of a vector with itself equals the square of its magnitude. $$\mathbf{A} \cdot \mathbf{A} = \|\mathbf{A}\|^2$$ This makes intuitive sense: $A1^2 + A2^2 + A3^2 = \|\mathbf{A}\|^2$. Property 2: Geometric interpretation — The dot product relates to the angle between two vectors. $$\mathbf{A} \cdot \mathbf{B} = \|\mathbf{A}\| \|\mathbf{B}\| \cos\theta$$ where $\theta$ is the angle between the vectors. This is a powerful relationship because it connects an algebraic computation to a geometric property. This has important implications: When $\theta = 0°$ (vectors point the same direction), $\cos(0°) = 1$, so $\mathbf{A} \cdot \mathbf{B}$ is maximized When $\theta = 90°$ (vectors are perpendicular), $\cos(90°) = 0$, so $\mathbf{A} \cdot \mathbf{B} = 0$ — perpendicular vectors always have dot product zero When $\theta = 180°$ (vectors point opposite), $\cos(180°) = -1$, so $\mathbf{A} \cdot \mathbf{B}$ is minimized (most negative) The dot product is useful for measuring how much two vectors align with each other. Cross Product The cross product is an operation that takes two vectors and produces a third vector perpendicular to both. For vectors $\mathbf{A} = [A1, A2, A3]$ and $\mathbf{B} = [B1, B2, B3]$: $$\mathbf{A} \times \mathbf{B} = [A2B3 - A3B2, \; A3B1 - A1B3, \; A1B2 - A2B1]$$ This formula looks complex, but it's systematic—each component follows the same pattern of products and differences. Example: If $\mathbf{A} = [1, 0, 0]$ and $\mathbf{B} = [0, 1, 0]$, then: $$\mathbf{A} \times \mathbf{B} = [(0)(0) - (0)(1), \; (0)(0) - (1)(0), \; (1)(1) - (0)(0)] = [0, 0, 1]$$ Notice that $[0, 0, 1]$ is perpendicular to both input vectors. Magnitude and Direction The magnitude of the cross product is: $$\|\mathbf{A} \times \mathbf{B}\| = \|\mathbf{A}\| \|\mathbf{B}\| \sin\theta$$ where $\theta$ is the angle between the vectors. This tells us that: The cross product has maximum magnitude when vectors are perpendicular ($\sin(90°) = 1$) The cross product is zero when vectors are parallel ($\sin(0°) = 0$) The direction of $\mathbf{A} \times \mathbf{B}$ follows the right-hand rule: point your right hand's fingers along $\mathbf{A}$, curl them toward $\mathbf{B}$, and your thumb points in the direction of $\mathbf{A} \times \mathbf{B}$. One crucial point: the cross product is not commutative. In fact, $\mathbf{B} \times \mathbf{A} = -(\mathbf{A} \times \mathbf{B})$ — swapping the order reverses the direction. <extrainfo> The cross product is a special operation that only works meaningfully in three (and seven) dimensions for a binary operation that produces a vector. This is a fascinating mathematical fact, but the "why" is beyond the scope of most courses at this level. </extrainfo> Vector Spaces and Bases A vector space over the real numbers is a set of objects (vectors) where you can add vectors together and multiply them by real numbers, with operations following standard rules (like commutativity and distributivity). Three-dimensional real space, denoted $\mathbb{R}^3$, is a vector space with dimension three. What is a Basis? A basis is a set of vectors with two critical properties: Linear independence: No vector in the basis can be written as a combination of the others Span the space: Every vector in $\mathbb{R}^3$ can be written as a linear combination of the basis vectors A basis for $\mathbb{R}^3$ must contain exactly three vectors (since the dimension is three). The Standard Basis The standard basis is the most familiar choice: $$\mathbf{e}1 = (1, 0, 0), \quad \mathbf{e}2 = (0, 1, 0), \quad \mathbf{e}3 = (0, 0, 1)$$ These three vectors align with the coordinate axes and form an orthonormal basis (perpendicular and unit length). Why Multiple Bases Exist Here's an important insight: the standard basis is not special in any mathematical sense. Any set of three linearly independent vectors forms a valid basis. For instance, you could rotate the coordinate system and get a different basis that works just as well. $$\mathbf{v}1 = (1, 1, 0), \quad \mathbf{v}2 = (1, -1, 0), \quad \mathbf{v}3 = (0, 0, 1)$$ (You can verify these are linearly independent). This rotated basis represents $\mathbb{R}^3$ just as completely as the standard basis. Why does this matter? Different bases are useful for different problems. In physics, choosing a basis aligned with the geometry of your problem can make calculations much simpler. However, the underlying geometry of $\mathbb{R}^3$ doesn't change—you're just using different "coordinate glasses" to look at it.