A plane can be completely illustrated by denoting two intersecting lines which can be translated into a fixed point A and two nonparallel direction vectors. The position vector $\overrightarrow{r}$ of any general point P on the plane passing through point A and having direction vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ is given by the equation
Vector equation of a plane
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}+µ\overrightarrow{c} \lambda, µ∈R (\overrightarrow{AP}=\lambda \overrightarrow{b}+µ\overrightarrow{c})$ 
Parametric equation of a plane: λ , μ are called a parameters λ,μ ∈ R $ (\begin{matrix} x & \\ y & \\ z & \\ \end{matrix} )=(\begin{matrix} a_{1} & \\ a_{2} & \\ a_{3} & \\ \end{matrix} )+\lambda (\begin{matrix} b_{1} & \\ b_{2} & \\ b_{3} & \\ \end{matrix} )+\mu (\begin{matrix} c_{1} & \\ c_{2} & \\ c_{3} & \\ \end{matrix} ) => \begin{matrix} x=a_{1}+\lambda b_{1}+\mu c_{1} & \\ y=a_{2}+\lambda b_{2}+\mu c_{2} & \\ z=a_{3}+\lambda b_{3}+\mu c_{3} & \\ \end{matrix} $
If N is considered to be normal to a given plane, then all other normals to that plane are considered parallel to N which are resultantly scalar multiples of N., In particular,we can say that there are two normals of length 1:
Normal/Scalar product form of vector equation of a plane
Consider a vector n passing through a point A. Only one plane through A can be is perpendicular to the vector. Now consider R being any point on the plane other than A as shown above. Then we can say that
$\overrightarrow{n}.\overrightarrow{AR}=0$
$\overrightarrow{n}.\overrightarrow{(r}-\overrightarrow{a})=0$
$\overrightarrow{n}.\overrightarrow{r}= \overrightarrow{n}.(\overrightarrow{a}+\lambda \overrightarrow{b}+µ\overrightarrow{c})$⇒$ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} or \overrightarrow{n}.(\overrightarrow{r}-\overrightarrow{a})= 0 $
Cartesian equation of a plane
$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} ⤇ n_{1}x+n_{2}y+n_{3}z=n_{1}a_{1}+n_{2}a_{2}+n_{3}a_{3 }=d$
$n_{1}x+n_{2}y+n_{3}z=d$
$D= \vert \overrightarrow{r}.{n}\vert =\vert \overrightarrow{a}.{n}\vert $
{=}$\frac{\vert \overrightarrow{a}.\overrightarrow{n}\vert }{\sqrt{n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}=\frac{\vert n_{1}a_{1}+n_{2}a_{2}+n_{3}a_{3 }\vert }{\sqrt{n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}$
Therefore, the Cartesian form is
where n1, n2 and n3 are the components of n and where n is called the normal vector.
Example: Find the equation of the plane passing through the three points P1(1,-1,4), P2(2,7,-1), and P3(5,0,-1).
$\overrightarrow{b}=\overrightarrow{P_{1}P_{2}}= (\begin{matrix} 1 & \\ 8 & \\ -5 & \\ \end{matrix})$
$\overrightarrow{c}= \overrightarrow{P_{1}P_{3}} = (\begin{matrix} 4 & \\ 1 & \\ -5 & \\ \end{matrix} )$
$P_{1}=(\begin{matrix} 1 & \\ -1 & \\ 4 & \\ \end{matrix} )$
In vector form:
$ \overrightarrow{r}=(\begin{matrix} 1 & \\ -1 & \\ 4 & \\ \end{matrix} ) +\lambda (\begin{matrix} 1 & \\ 8 & \\ -5 & \\ \end{matrix} )+µ(\begin{matrix} 4 & \\ 1 & \\ -5 & \\ \end{matrix} )$
$\overrightarrow{n}=\vert \begin{matrix} {i} & {j} & {k} & \\ 1 & 8 & -5 & \\ 4 & 1 & -5 & \\ \end{matrix} \vert =(\begin{matrix} -35 & \\ -15 & \\ -31 & \\ \end{matrix} )$
Any non-zero scalar multiples of $\overrightarrow{n}$ is also a normal vector of the plane. Therefore, Multiply by -1.
$\overrightarrow{n}=(\begin{matrix} 35 & \\ 15 & \\ 31 & \\ \end{matrix} )$
$(\begin{matrix} 35 & \\ 15 & \\ 31 & \\ \end{matrix} ).(\begin{matrix} 1 & \\ -1 & \\ 4 & \\ \end{matrix} )=144$
$Cartesian form:$
$35x+15y+31z=144$
Example: Find the equation of the plane with normal vector $ (\begin{matrix} 1 & \\ 3 & \\ 5 & \\ \end{matrix} )$containing the point (-2, 3, 4).
$(\begin{matrix} 1 & \\ 3 & \\ 5 & \\ \end{matrix} ).(-2,3,4)=-2+9+20=27$
x+3y+5z=27
Example: Find the distance of the plane $ \overrightarrow{r}.(\begin{matrix} 3 & \\ 2 & \\ -4 & \\ \end{matrix} )$ = 8 from the origin, and the unit vector perpendicular to the plane.
$\vert (\begin{matrix} 3 & \\ 2 & \\ -4 & \\ \end{matrix} ) \vert =\sqrt{29}$
$\frac{1}{\sqrt{29}}\lbrack \overrightarrow{r}.(\begin{matrix} 3 & \\ 2 & \\ -4 & \\ \end{matrix} )\rbrack = \frac{8}{\sqrt{29}}$
$D= \frac{8}{\sqrt{29}} {n}= \frac{1}{\sqrt{29}}(\begin{matrix} 3 & \\ 2 & \\ -4 & \\ \end{matrix} )$
Example: Find the Cartesian equation of the plane through the point A (1, 1, 1) perpendicular to the vector
Solution:
Example: Show that the following vector is perpendicular to the plane containing the points A(1, 0, 2), B(2, 3, -1) and C(2, 2, -1 ).
Solution:
In conclusion, n is a vector that is perpendicular to 2 vectors in the plane so it is perpendicular to the plane.
