In this section, we will explore the fundamental concepts of vectors and vector spaces, which are essential building blocks for understanding more advanced topics in linear algebra and 3D mathematics. We will cover the following key concepts:

1. Definition of Vectors
2. Vector Operations
3. Vector Spaces
4. Basis and Dimension

A vector is a mathematical object that has both magnitude and direction. Vectors are often represented as arrows in a coordinate system, but they can also be represented as ordered lists of numbers.

Vectors are typically denoted by boldface letters (e.g., v) or with an arrow above the letter (e.g., \(\vec{v}\)). In component form, a vector in 3D space can be written as: \[ \vec{v} = \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} \]

Consider the vector \(\vec{v} = \begin{bmatrix} 3 \ 4 \ 5 \end{bmatrix}\). This vector has components 3, 4, and 5 along the x, y, and z axes, respectively.

Vectors can be added together by adding their corresponding components. \[ \vec{u} + \vec{v} = \begin{bmatrix} u_1 \ u_2 \ u_3 \end{bmatrix} + \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} u_1 + v_1 \ u_2 + v_2 \ u_3 + v_3 \end{bmatrix} \]

\[ \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} + \begin{bmatrix} 4 \ 5 \ 6 \end{bmatrix} = \begin{bmatrix} 1+4 \ 2+5 \ 3+6 \end{bmatrix} = \begin{bmatrix} 5 \ 7 \ 9 \end{bmatrix} \]

A vector can be multiplied by a scalar (a real number) by multiplying each component of the vector by the scalar. \[ c \cdot \vec{v} = c \cdot \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} c \cdot v_1 \ c \cdot v_2 \ c \cdot v_3 \end{bmatrix} \]

\[ 3 \cdot \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} = \begin{bmatrix} 3 \cdot 1 \ 3 \cdot 2 \ 3 \cdot 3 \end{bmatrix} = \begin{bmatrix} 3 \ 6 \ 9 \end{bmatrix} \]

The dot product of two vectors is a scalar obtained by multiplying corresponding components and summing the results. \[ \vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]

\[ \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \cdot \begin{bmatrix} 4 \ 5 \ 6 \end{bmatrix} = 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 = 4 + 10 + 18 = 32 \]

The cross product of two vectors in 3D space is a vector that is perpendicular to both original vectors. \[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \]

\[ \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \times \begin{bmatrix} 4 \ 5 \ 6 \end{bmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 4 & 5 & 6 \end{vmatrix} = \begin{bmatrix} -3 \ 6 \ -3 \end{bmatrix} \]

A vector space is a collection of vectors that can be added together and multiplied by scalars, satisfying certain axioms (e.g., associativity, commutativity, existence of an additive identity and inverses).

The set of all 3D vectors \(\mathbb{R}^3\) is a vector space.

A subspace is a subset of a vector space that is itself a vector space under the same operations.

The set of all vectors in \(\mathbb{R}^3\) that lie on a plane through the origin is a subspace of \(\mathbb{R}^3\).

A basis of a vector space is a set of vectors that are linearly independent and span the vector space.

The standard basis for \(\mathbb{R}^3\) is \(\left{ \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \right}\).

The dimension of a vector space is the number of vectors in a basis for the vector space.

The dimension of \(\mathbb{R}^3\) is 3.

Given vectors \(\vec{a} = \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix}\) and \(\vec{b} = \begin{bmatrix} 4 \ 0 \ -2 \end{bmatrix}\), compute \(\vec{a} + \vec{b}\).

Solution: \[ \vec{a} + \vec{b} = \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix} + \begin{bmatrix} 4 \ 0 \ -2 \end{bmatrix} = \begin{bmatrix} 2+4 \ -1+0 \ 3-2 \end{bmatrix} = \begin{bmatrix} 6 \ -1 \ 1 \end{bmatrix} \]

Find the dot product of \(\vec{u} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}\) and \(\vec{v} = \begin{bmatrix} 4 \ -5 \ 6 \end{bmatrix}\).

Solution: \[ \vec{u} \cdot \vec{v} = 1 \cdot 4 + 2 \cdot (-5) + 3 \cdot 6 = 4 - 10 + 18 = 12 \]

Determine if the set \(\left{ \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \right}\) is a basis for \(\mathbb{R}^3\).

Solution: To determine if the set is a basis, we need to check if the vectors are linearly independent and span \(\mathbb{R}^3\). We set up the matrix with these vectors as columns and row reduce it: \[ \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix} \] This matrix is in row echelon form and has three pivot columns, indicating that the vectors are linearly independent and span \(\mathbb{R}^3\). Therefore, the set is a basis for \(\mathbb{R}^3\).

In this section, we have covered the basics of vectors and vector spaces, including vector operations, the concept of vector spaces, and the importance of basis and dimension. These foundational concepts are crucial for understanding more advanced topics in linear algebra and 3D mathematics. In the next section, we will delve into matrices and determinants, which are essential tools for working with linear transformations and systems of linear equations.