3.11 Vectors: Coordinate RepresentationHL ONLY

1. Two-Dimensional Vectors

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

Magnitude: The length of the vector is calculated using the Pythagorean theorem:
$|\vec{u}| = \sqrt{a^2 + b^2}$.

Addition & Subtraction: Components 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 distributes across all components:
$k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$. The magnitude scales proportionally: $|k\vec{u}| = |k| |\vec{u}|$.

Equality: Two vectors are equal if and only if their corresponding components are 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 $\vec{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$, find $2\vec{u} + 3\vec{v}$:

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

2. Unit Vectors and Base Notation

The Unit Vector: A unit vector $\hat{u}$ has a magnitude of $1$ and points in the same direction as $\vec{u}$. It is calculated by multiplying the vector by the reciprocal of its magnitude:

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

EXAMPLE 3 (Scaling to a Target Magnitude)

Find a vector $\vec{a}$ parallel to $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ with a magnitude of $20$.

Step 1: Find the unit vector $\hat{u}$ to determine the 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 target magnitude 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 $4$ times larger ($20/5 = 4$). The solution is $\vec{a} = 4\vec{u} = \begin{pmatrix} 12 \\ 16 \end{pmatrix}$.

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

Any 2D vector can be expressed as a linear combination of the unit vectors along 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 to base notation:
$\begin{pmatrix} a \\ b \end{pmatrix} = a\mathbf{i} + b\mathbf{j}$.

3. Three-Dimensional Vectors

Expanding into 3D space introduces a third component along the z-axis. Vectors are written as $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ or in base notation $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$.

EXAMPLE 4 (3D Operations)

Perform operations 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 Between Two Points in 3D

The vector from point $A(x_1, y_1, z_1)$ to point $B(x_2, y_2, z_2)$ is found by subtracting the position vector of $A$ from the position vector of $B$:

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

The distance between points $A$ and $B$ equals the magnitude of the vector: $$|\overrightarrow{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$