Equation of plane in Parametric, Scalar-product & Cartesian form

Parametric form

The vector equation of plane $p$ in parametric form is given by $$ \boxed{ p : \textbf{r} = \textbf{a} + \lambda \textbf{b}_1 + \mu \textbf{b}_2, \phantom{0} \lambda, \mu \in \mathbb{R} } $$

$\textbf{r}$ is the position vector of a point on plane $p$ (i.e. $\overrightarrow{OR}$)

$\textbf{a}$ is the position vector of a known point on plane $p$ (i.e. $\overrightarrow{OA}$)

$\textbf{b}_1$ is a direction vector on plane $p$

$\textbf{b}_2$ is another direction vector on plane $p$

Scalar-product form

The vector equation of plane $p$ in scalar-product form is given by $$ \boxed{ p : \textbf{r} \cdot \textbf{n} = d } $$

$\textbf{r}$ is the position vector of a point on plane $p$ (i.e. $\overrightarrow{OR}$)

$\textbf{n}$ is the normal vector of the plane. It can be obtained from the vector product of two direction vectors on the plane.

$d$ is a constant which is equals to the value of $\textbf{a} \cdot \textbf{n}$, where $a$ is the position vector a known point on plane $p$ (i.e. $\overrightarrow{OA}$)

Cartesian equation

General form: $$ \boxed{ ax + by + cz = d } $$

Convert from one form to the other

\begin{align} p_1: \textbf{r} & = \left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right) + \lambda \left( \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right) + \mu \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right), \phantom{0} \lambda, \mu \in \mathbb{R} \\ \\ \left( \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right) \times \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right) & = \left( \begin{matrix} 0 - (-1) \\ 0 - 1 \\ 1 - 0 \end{matrix} \right) \\ \textbf{n} & = \left( \begin{matrix} 1 \\ -1 \\ 1\end{matrix} \right) \\ \\ d & = \textbf{a} \cdot \textbf{n} \\ & = \left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \\ 1 \end{matrix} \right) \\ & = 1 + (-2) + 3 \\ & = 2 \\ \\ \text{Scalar-product form,} \phantom{0} p_1: \textbf{r} \cdot \left( \begin{matrix} 1 \\ -1 \\ 1 \end{matrix} \right) & = 2 \\ \\ \\ \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \\ 1 \end{matrix} \right) & = 2 \\ x + (-y) + z & = 2 \\ \text{Cartesian equation: } \phantom{0} x - y + z & = 2 \end{align}

\begin{align} \text{Cartesian equation of } p_2 : \phantom{0} x + 2y + z & = 5 \\ \\ \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right) & = 5 \\ \text{Scalar-product form, } \phantom{0} p_2 : \textbf{r} \cdot \left( \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right) & = 5 \end{align}

By guess-and-check or observation, find three points that lie on $p_2$. With the three points, we can form two direction vectors:

\begin{align} \left( \begin{matrix} 5 \\ 0 \\ 0 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right) & = 5 \\ \left( \begin{matrix} 0 \\ 2.5 \\ 0 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right) & = 5 \\ \left( \begin{matrix} 0 \\ 0 \\ 5 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right) & = 5 \\ \therefore (5, 0, 0), (0, 2.5, 0) \text{ and } & (0, 0, 5) \text{ lie on } p_2 \\ \\ \left( \begin{matrix} 5 \\ 0 \\ 0 \end{matrix} \right) - \left( \begin{matrix} 0 \\ 2.5 \\ 0 \end{matrix} \right) & = \left( \begin{matrix} 5 \\ -2.5 \\ 0 \end{matrix} \right) \\ & = 2.5 \left( \begin{matrix} 2 \\ -1 \\ 0 \end{matrix} \right) \\ \\ \left( \begin{matrix} 5 \\ 0 \\ 0 \end{matrix} \right) - \left( \begin{matrix} 0 \\ 0 \\ 5 \end{matrix} \right) & = \left( \begin{matrix} 5 \\ 0 \\ -5 \end{matrix} \right) \\ & = 5\left( \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right) \\ \\ \text{Parametric form, } p_2 : \textbf{r} & = \left( \begin{matrix} 5 \\ 0 \\ 0 \end{matrix} \right) + \alpha \left( \begin{matrix} 2 \\ -1 \\ 0 \end{matrix} \right) + \beta \left( \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right), \phantom{0} \alpha, \beta \in \mathbb{R} \end{align}