1E. Vector - Basics Explanation (2D)

Syntax

\vec{v} = [\begin{matrix} x \\ y \\ z \end{matrix}]

$\vec{P Q} = Q - P$ = Vector from Point P -> Point Q (Destination - Source)
$\vec{0}$ = $[\begin{matrix} 0 \\ 0 \end{matrix}]$

Alt notation

$\vec{v} = i + j + k$

$i$ = x component
$j$ = y-component
$k$ = z-component
Example:
$\vec{v} = 3 i + 4 j + 5 k = [\begin{matrix} 3 x \\ 4 y \\ 5 z \end{matrix}]$

Info

Addition

Addition chains the vectors.
Example:

$\vec{v_{1}} + \vec{v_{2}}$ = start point of $v_{1}$ to endpoint of $v_{2}$
$\vec{P R} = \vec{P Q} + \vec{Q R}$

Subtraction

Example: - $\vec{v_{1}}-\vec{v_{2}}$= endpoint of $v_{2}$ to endpoint of $v_1$ - $\vec{RQ}=\vec{PQ}-\vec{PR}$

Another way to see it:

$\vec{P Q} = \vec{P R} + \vec{R Q} ⟹ \vec{R Q} = \vec{P Q} - \vec{P R}$

Dot product

Link: Interactive

Vector projection derivation

Link: Interactive

$\vec{PS}$ is a scalar on $\vec{PR}$, so $\vec{PS} = c\times\hat{PR}$ (constant $\times$ $\vec{PR}$ unit vector) So to find $c$ (which is $\|\vec{PS}\|$),

\begin{aligned} \cos θ & = \frac{∥ \vec{P S} ∥}{∥ \vec{P Q} ∥} \\ ∥ \vec{P S} ∥ & = ∥ \vec{P Q} ∥ \cos θ \\ Substitute \cos θ & with dot product equation (\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} ∥ ∥ \vec{b} ∥}) \\ ∥ \vec{P S} ∥ & = ∥ \vec{P Q} ∥ (\frac{\vec{P S} \cdot \vec{P Q}}{∥ \vec{P S} ∥ ∥ \vec{P Q} ∥}) \\ ∥ \vec{P S} ∥ & = \frac{\vec{P S} \cdot \vec{P Q}}{∥ \vec{P S} ∥} \\ ∥ \vec{P S} ∥ & = \vec{P Q} \cdot \hat{P S} \\ So, \vec{P S} & = (\vec{P Q} \cdot \hat{P S}) \hat{P S} \end{aligned}