1. Vector - Basics (2D)

All notation is in 2D but it applies in 3D too

Points

Notation

$R^{2} =$ xy plane
$R^{3} =$ xyz plane

Let P = $(x_{1}, y_{1})$ , Q = $(x_{2}, y_{2})$ ,

Mid point

M_{P Q} = \frac{P + Q}{2} = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

Distance

d (P, Q) = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Vectors

$\vec{P Q} = Q - P = \vec{O Q} - \vec{O P}$

Let $\vec{u} = [\begin{matrix} u_{1} \\ u_{2} \end{matrix}]$ , $\vec{v} = [\begin{matrix} v_{1} \\ v_{2} \end{matrix}]$

Length

∥ \vec{a} ∥ = \sqrt{a_{1}^{2} + a_{2}^{2}}

Dot product (Info)

\vec{a} \cdot \vec{b} = a_{1} b_{1} + a_{2} b_{2}

If dot product > 0, angle between $\vec{a}$ and $\vec{b}$ is < $90^{\circ}$
If dot product = 0, angle between $\vec{a}$ and $\vec{b}$ is = $90^{\circ}$
If dot product < 0, angle between $\vec{a}$ and $\vec{b}$ is > $90^{\circ}$
Why? Because of the $\cos^{- 1}$

Angle between 2 vectors

θ = \cos^{- 1} (\frac{\vec{a} \cdot \vec{b}}{∥ \vec{a} ∥ ∥ \vec{b} ∥})

Special cases

Dot product result	Angle
-1	$180^{\circ}$
0	$90^{\circ}$
1	$0^{\circ}$

^ ONLY WORKS FOR UNIT VECTORS ^

For dot product 1 and -1 NEED to be unit vectors. For dot product 0, it doesn't matter

Checking if vectors are parallel/perpendicular

For parallel

$\vec{u} = c \vec{v}$
$\vec{u}$ must be a scalar of $\vec{v}$

\vec{u}

and

\vec{v}

are unit vectors, can use Dot product, Special cases

For perpendicular

$\vec{u} \cdot \vec{v} = 0$
See Dot product, Special cases

Making a normal vector

In 2 steps:

Swap $x$ and $y$
Negate either the $x$ or $y$
- Negate $x$ to rotate $90^{\circ}$ anti-clockwise
- Negate $y$ to rotate $90^{\circ}$ clockwise

Example

$\vec{d} = [\begin{matrix} 2 \\ 3 \end{matrix}]$

$\vec{n} = [\begin{matrix} 3 \\ - 2 \end{matrix}]$

Vector Projection (Info)

Project $\vec{b}$ onto $\vec{a}$ ,

General:

p r o j_{\vec{a}} \vec{b} = \frac{\vec{b} \cdot \vec{a}}{∥ \vec{a} ∥^{2}} \vec{a}

If $\vec{a}$ is a unit vector:

p r o j_{\vec{a}} \vec{b} = (\vec{b} \cdot \hat{a}) \hat{a}

Checking if moving towards a point

Only 1 point moving

Get a vector $\vec{Q P}$ ,
Find $\vec{Q P} \cdot \vec{v}$
At $Q$ , Result < 0, angle is < 90 $^{\circ}$ so moving towards
At $Q_{1}$ , Result = 0, angle is = 90 $^{\circ}$ . Neither moving towards/away. This is will only happen for a split second
At $Q_{2}$ , Result > 0, angle is > 90 $^{\circ}$ so moving away

2 points moving

Assume P is stationary **relative** to Q #todo-low

What it means by relative

Imagine P and Q are cars at a intersection. And you are sitting in car P.
From my perspective, Car Q moves towards me relative to my (Car P) current speed.

Checking if a moving point will hit line

Hit	Miss

Hit if the signs of these 2 are different

$\vec{Q P} \cdot \vec{n}$
$\vec{v} \cdot \vec{n}$
Or in code:

\vec{Q P} \cdot \vec{n} < 0! = \vec{v} \cdot \vec{n} < 0

Excalidraw - Full table