3.11 Vectors: Algebraic Representation (HL)

1. 2-Dimensional Algebraic Vectors

A vector $\vec{u}$ is formally expressed algebraically as a column matrix containing its directional components: $\begin{pmatrix} a \\ b \end{pmatrix}$.

Magnitude: The absolute length of the vector evaluates utilizing Pythagorean distance:
$|\vec{u}| = \sqrt{a^2 + b^2}$.

Addition & Subtraction: Component rows are computed independently.
$\begin{pmatrix} a_1 \\ b_1 \end{pmatrix} \pm \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 \pm a_2 \\ b_1 \pm b_2 \end{pmatrix}$.

Scalar Multiplication: The scalar multiplier strictly distributes across all components.
$k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$. The resulting magnitude scales proportionally: $|k\vec{u}| = |k| |\vec{u}|$.

Equality: Two vectors are equal $\iff$ their corresponding components are strictly identical.
$\begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} \implies a_1 = a_2 \text{ and } b_1 = b_2$.

EXAMPLE 1 (Vector Magnitudes & Addition)

Given vectors $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$:

Magnitudes:
$|\vec{u}| = \sqrt{3^2 + 4^2} = \sqrt{25} = \mathbf{5}$
$|\vec{v}| = \sqrt{2^2 + 5^2} = \mathbf{\sqrt{29}}$

Addition:
$\vec{u} + \vec{v} = \begin{pmatrix} 3+2 \\ 4+5 \end{pmatrix} = \mathbf{\begin{pmatrix} 5 \\ 9 \end{pmatrix}}$

EXAMPLE 2 (Linear Combinations)

Using the same vectors $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$, evaluate $2\vec{u} + 3\vec{v}$:

$$\begin{aligned} 2\vec{u} + 3\vec{v} &= 2\begin{pmatrix} 3 \\ 4 \end{pmatrix} + 3\begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} 6 \\ 8 \end{pmatrix} + \begin{pmatrix} 6 \\ 15 \end{pmatrix} \\ &= \mathbf{\begin{pmatrix} 12 \\ 23 \end{pmatrix}} \end{aligned}$$

2. Unit Vectors and Base Notation

The Unit Vector: A unit vector $\hat{u}$ is a specialized vector possessing a magnitude of exactly $1$, pointing precisely in the same direction as $\vec{u}$. It is calculated by multiplying the vector by the reciprocal of its own magnitude:

$\hat{u} = \dfrac{1}{|\vec{u}|} \vec{u}$

EXAMPLE 3 (Scaling to a Target Magnitude)

Establish a vector $\vec{a}$ strictly parallel to $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ possessing a magnitude of exactly $20$.

Step 1: Formulate the unit vector $\hat{u}$ to establish the pure direction.
$|\vec{u}| = 5 \implies \hat{u} = \dfrac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 3/5 \\ 4/5 \end{pmatrix}$

Step 2: Multiply the unit vector by the targeted length of $20$.
$\vec{a} = 20\hat{u} = 20\begin{pmatrix} 3/5 \\ 4/5 \end{pmatrix} = \mathbf{\begin{pmatrix} 12 \\ 16 \end{pmatrix}}$

Alternative Logic: Since $|\vec{u}| = 5$, the target magnitude is exactly $4$ times larger ($20/5 = 4$). The solution instantly evaluates to $\vec{a} = 4\vec{u} = \begin{pmatrix} 12 \\ 16 \end{pmatrix}$.

Standard Base Vectors ($\mathbf{i}$ and $\mathbf{j}$)

Any 2D vector can be expressed strictly as a linear combination of the fundamental unit vectors tracing the x-axis and y-axis.

$\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (Horizontal unit vector)
$\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ (Vertical unit vector)

The column vector maps directly to base notation:
$\begin{pmatrix} a \\ b \end{pmatrix} = a\mathbf{i} + b\mathbf{j}$.

3. 3-Dimensional Vectors and Point Mapping

Expanding into 3D space introduces a third vertical $z$-axis component (following the right-hand rule). Vectors map algebraically as $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ or base notation $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$.

EXAMPLE 4 (3D Operations)

Evaluate operations strictly for $\vec{u} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \\ 5 \\ -4 \end{pmatrix}$.

Linear Combination:

$$\begin{aligned} 3\vec{u} + 2\vec{v} &= 3\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 2\begin{pmatrix} 2 \\ 5 \\ -4 \end{pmatrix} \\ &= \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} + \begin{pmatrix} 4 \\ 10 \\ -8 \end{pmatrix} = \mathbf{\begin{pmatrix} 7 \\ 16 \\ 1 \end{pmatrix}} \end{aligned}$$

3D Unit Vector $\hat{u}$:

$$\begin{aligned} |\vec{u}| &= \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \\ \hat{u} &= \dfrac{1}{\sqrt{14}}\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \mathbf{\begin{pmatrix} 1/\sqrt{14} \\ 2/\sqrt{14} \\ 3/\sqrt{14} \end{pmatrix}} \end{aligned}$$

Vector Mapping Between Points in 3D

The directional displacement vector spanning strictly from point $A(x_1, y_1, z_1)$ to point $B(x_2, y_2, z_2)$ is mathematically extracted by subtracting the position vector of the origin tail ($A$) from the position vector of the destination head ($B$):

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \\ z_2 - z_1 \end{pmatrix}$

The absolute geometric distance strictly between points $A$ and $B$ aligns with the vector's magnitude:
$|\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$