3.10 Vectors: Geometric RepresentationHL ONLY

1. Scalars versus Vectors

Quantities are classified into two categories:

Scalars: Have magnitude (size). Examples include age ($28$), length ($4\text{ m}$), and temperature ($25^\circ\text{C}$).

Vectors: Have both magnitude and direction. Examples include force ($7\text{ N}$ downwards) and velocity ($35\text{ m/s}$ towards southeast).

A vector is represented as an arrow. It is denoted by a bold letter ($\mathbf{u}$), a letter with an arrow ($\vec{u}$), or by its endpoints ($\overrightarrow{AB}$, where $A$ is the tail and $B$ is the head). The length of the vector defines its magnitude, denoted as $|\vec{u}|$ or $|\overrightarrow{AB}|.

Equality of Vectors & Unit Vectors

Two vectors are equal ($\vec{u} = \vec{v}$) $\iff$ they have the same magnitude and direction. A vector remains identical regardless of parallel translation.

In a parallelogram $ABCD$, the parallel lines establish vector equivalencies: $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$.

Unit Vector: A vector with a magnitude of $1$. To find a unit vector $\hat{u}$ parallel to and in the direction of $\vec{u}$, divide by its magnitude:

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

2. Geometric Vector Operations

Addition ($\vec{u} + \vec{v}$): Executed by placing vectors "head-to-tail". Moving from point $A$ to $B$, and then from $B$ to $C$, produces the vector from $A$ to $C$. Thus, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.

Subtraction ($\vec{u} - \vec{v}$): Executed by adding the opposite vector: $\vec{u} + (-\vec{v})$. When drawn "tail-to-tail", $\vec{u} - \vec{v}$ spans from the head of $\vec{v}$ to the head of $\vec{u}$.

The Opposite Vector ($-\vec{u}$): Has the same magnitude but reverses the direction. Using endpoint notation: $\overrightarrow{AB} = -\overrightarrow{BA}$.

The Zero Vector ($\vec{0}$): A vector with zero magnitude and no defined direction. It occurs when a vector cancels itself: $\overrightarrow{AB} - \overrightarrow{AB} = \overrightarrow{AA} = \vec{0}$.

Scalar Multiplication ($k\vec{u}$): For a positive scalar $k > 0$, the product $k\vec{u}$ scales the magnitude by $k$. If $k < 0$, the magnitude scales by $|k|$ and the direction reverses. Two vectors are parallel ($\vec{u} \parallel \vec{v}$) $\iff \vec{u} = k\vec{v}$ for a scalar $k \in \mathbb{R}$.

Collinearity: Three points $A$, $B$, and $C$ are collinear (lie on a straight line) $\iff \overrightarrow{AB} = k\overrightarrow{BC}$ for a scalar $k$.

EXAMPLE 1 (Vector Pathways in a Cube)

Consider a cube $ABCDEFGH$. Base vectors are assigned to edges from origin $A$:

$\vec{a} = \overrightarrow{AB}$
$\vec{b} = \overrightarrow{AD}$
$\vec{c} = \overrightarrow{AF}$

Parallel edges equate to these base vectors:
$\vec{a} = \overrightarrow{DC} = \overrightarrow{FG} = \overrightarrow{EH}$.

To express the diagonal vector $\overrightarrow{FC}$ in terms of $\vec{a}$, $\vec{b}$, and $\vec{c}$, map a path from $F$ to $C$.

Path:

$$\begin{aligned} \overrightarrow{FC} &= \overrightarrow{FG} + \overrightarrow{GB} + \overrightarrow{BC} \\ &= \overrightarrow{AB} + (-\overrightarrow{AF}) + \overrightarrow{AD} \\ &= \vec{a} - \vec{c} + \vec{b} \end{aligned}$$

3. Position Vectors & Coordinates

Any vector on a Cartesian plane can be translated to originate from the origin $O(0,0)$.

If a vector starts at $O$ and ends at $A(a,b)$, it is the position vector of $A$, denoted as $\overrightarrow{OA} = \begin{pmatrix} a \\ b \end{pmatrix}$.

For two coordinates $A$ and $B$, the displacement vector connecting them is calculated by subtracting their position vectors:

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$

EXAMPLE 2 (Cartesian Translation & Position)

Consider points $A(1, 2)$ and $B(4, 6)$. Determine the displacement vector $\overrightarrow{AB}$.

Step 1: Identify the position vectors from origin $O$:

$\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$
$\overrightarrow{OB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$

Step 2: Apply the subtraction formula:

$$\begin{aligned} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} 3 \\ 4 \end{pmatrix} \end{aligned}$$

This demonstrates the vector translates $3$ units horizontally and $4$ units vertically.