2. Vectors - Vector, Parametric equations (2D)

Linear equation

For reference, this is the normal way to represent an equation

Where

• $m$ = slope and
• $c$ = y-intercept

Find using $\overrightarrow{n}$ and 1 point (Proof)

Expanded out:

Vector equation

Notation

Let Line ($l$) be a line through $P=(x_0,y_0)$ with a direction $\overrightarrow{d}=\left[\begin{array}{c}a\\ b\end{array}\right]$

--- Question
Line $l$ goes through $P=(1,2)$ and has direction $\vec{d}=\begin{bmatrix}2\\5\end{bmatrix}$.
Find the *vector equation* and *3 points* on $l$
--- Answer
Find vector equation
$\begin{align}
\vec{x}&=P+t\vec{d} \\
\vec{x}&=(1,2)+t\begin{bmatrix}2\\5\end{bmatrix} \\
\end{align}$
Find 3 points 
$
\begin{align}
\text{Sub t=0,} \\
(x,y) &= (1,2) \\ \\
\text{Sub t=1,} \\
(x,y) &= (1,2)+\begin{bmatrix}2\\5\end{bmatrix} \\
&=(3,7)\\
\text{Sub t=2,}  \\
(x,y) &= (1,2)+2\begin{bmatrix}2\\5\end{bmatrix} \\
&=(5,12)
\end{align}
$

Find vector equation using 2 points

$l$ goes through $P=(0,3)$ and $Q=(9,0)$. Find the vector equation.
Steps:

1. Find out the direction vector ($\overrightarrow{PQ}$)
2. $\overrightarrow{x}=P+\overrightarrow{PQ}$

Working:
$\overrightarrow{PQ}=\left[\begin{array}{c}9\\ -3\end{array}\right]$
$\overrightarrow{x}=(0,3)+t\left[\begin{array}{c}9\\ -3\end{array}\right]$

Linear $\to$ Vector equation

Given $3x+4y=5$, what is the vector equation?
Steps:

1. Find out $P$
2. Find out $\overrightarrow{d}$

Step 1 (Find $P$):

So $P=(0,\frac{5}{4})$
Step 2 (Find $\overrightarrow{d}$):
A line $ax+by=c$ has the normal vector $\left[\begin{array}{c}a\\ b\end{array}\right]$
So, $\overrightarrow{n}=\left[\begin{array}{c}3\\ 4\end{array}\right]⟹\overrightarrow{d}=\left[\begin{array}{c}-4\\ 3\end{array}\right]$

ANS: $\overrightarrow{x}=(0,\frac{5}{4})+t\left[\begin{array}{c}-4\\ 3\end{array}\right]$

Parametric equation

Notation

Let Line ($l$) be a line through $P=(x_0,y_0)$ with a direction $\overrightarrow{d}=\left[\begin{array}{c}a\\ b\end{array}\right]$

This is basically vector equation expanded out

Finding shortest distance

Point to line

Given $\overrightarrow{x}=P+t\overrightarrow{d}$ and $Q(x,y)$