Let $ \vec{x} = (x,y,z)$ be any point on $\delta$
Let $ \vec{p} = (p_1,p_2,p_3)$ be a special point on $\delta$
Then $ \vec{x} - \vec{p} $ is a vector on $\delta$
Let $ \vec{n} = (a,b,c)$ is the normal vector to $\delta$
$ \vec{n} \cdot ( \vec{x} - \vec{p} ) = 0$
this is the normal form for $\delta$
Componentwise:
$ (a, b, c) \cdot (x, y, z) = (a, b, c) \cdot ( p_1, p_2, p_3 ) $
$ ax + by + cz = ap_1 + bp_2 + cp_3 = d $
This is the general form for $\delta$
A plane $\delta$ is generated by two vectors $ \vec{u}, \vec{v} \in \mathbb{R}^3$ then
$ \vec{x} = \vec{p} + s \vec{u} + t \vec{v}, s, t \in \mathbb{R}$
Componentwise:
$$
\left(\begin{array}{r}
x, \\
y, \\
z
\end{array}\right) =
\left(\begin{array}{r}
p_1, \\
p_2, \\
p_3
\end{array}\right) + s
\left(\begin{array}{r}
u_1, \\
u_2, \\
u_3
\end{array}\right) + t
\left(\begin{array}{r}
v_1, \\
v_2, \\
v_3
\end{array}\right)
$$
This is the Parametric From
Example: find Vector form $\delta$ plane passing through $ P = (1, 1, 1), Q = (4, 0, 2), R = (0, 1, -1) $
Let $\vec{u} = P\vec{Q} = \vec{Q} - \vec{P} = (3, -1, 1) $
Let $\vec{u} = P\vec{R} = \vec{R} - \vec{P} = (-1, 0, -2) $
Let $\vec{u}$ and $\vec{v}$ be not parallel
Vector form for $\delta$ is $ \vec{x} = \vec{p} + s \vec{c} \vec + t \vec{v}$
$ = (1, 1, 1) + s(3, -1, 1) + t(-1, 0, -2), s, t \in \mathbb{R} $
Note: cross product is only defined in $\mathbb{R}^3$
Standard unit vectors:
$e_1 = (1, 0, 0)$
$e_2 = (0, 1, 0)$
$e_3 = (0, 0, 1)$
if $\vec{u} \in \mathbb{R}^3$ then it is a linear combination of unit vectors
Definition: the cross product of $\vec{u}$ and $\vec{v}$ is denoted by $ \vec{u} \times \vec{v}$
$$ \vec{u} \times \vec{v} = \left(\begin{array}{rr} \text{ Let } \left[\begin{array}{rr} u_2, u_3 \\ v_2, v_3 \end{array}\right] - \text{ Let } \left[\begin{array}{rr} u_3, u_1 \\ v_3, v_1 \end{array}\right] - \text{ Let } \left[\begin{array}{rr} u_1, u_2 \\ v_1, v_2 \end{array}\right] \end{array}\right) $$
where $ \left[\begin{array}{rr} a, b \\ c, d \end{array}\right] := ad - bc $
$\vec{u} \times \vec{v} = (u_2 v_3 - u_3 v_2) \vec{e_1} + (u_3 v_1 - u_1 v_3) \vec{e_2} + (u_1 v_2 - u_2 v_1) \vec{e_3}$
Note: $ \vec{u} \times \vec{v} = - \vec{v} \times \vec{u} $
Note: $ \vec{u} \times \vec{v} = 0$ if $\vec{v}$ and $\vec{u}$ are parallel
Proposition: $ ( \vec{u} \times \vec{v} ) \circ \vec{u} = 0 $
Proof: $ ( \vec{u} \times \vec{v}) \circ \vec{u} = (u_2 v_3 - u_3 v_2) \vec{u_1} - (u_3 v_1 - u_1 v_3) \vec{u_2} + (u_1 v_2 - u_2 v_1) \vec{u_3}$
$ \Rightarrow ( \vec{u} \times \vec{v}) \circ \vec{u} = u_1 u_2 v_3 - u_1 u_3 v_2 - u_2 u_3 v_1 + u_1 u_2 v_3 + u_1 u_3 v_2 - u_2 u_3 v_1$
$ \Rightarrow 0 $
$ \therefore ( \vec{u} \times \vec{v} ) \circ \vec{u} = 0 $
then length of $ \vec{u} \times \vec{v} $
$ || \vec{u} \times \vec{v} || = || \vec{u} || || \vec{v} || \sin \theta = $ area of the parallelogram
When you add vectors 2 or 3 dimensional geometry you can put them next to each other head to tail, from either vector, this makes no different but gives use 4 sides of a closed shape. This creates the parallelogram.