Lines and Planes in 3D - Mohsen Hamedi

Lines: Vector Equation

Only one line can pass through any two distinct points in 3-space. Assume that we have two points, $P_1$ and $P_2$ , and there is a point $P$ that lies in the same line. If $\boldsymbol{r} = \overrightarrow{OP}$ , $\boldsymbol{r}_1 = \overrightarrow{OP_1}$ , and $\boldsymbol{r}_2 = \overrightarrow{OP_2}$ , then a vector equation for the line is

(1) $\begin{equation*}\boldsymbol{r} = \boldsymbol{r}_2 + t ( \boldsymbol{r}_2 - \boldsymbol{r}_1 ) \end{equation*}$

$t$ is called a parameter and the non-zero vector $\boldsymbol{a} = ( \boldsymbol{r}_2 - \boldsymbol{r}_1 )$ is called a direction vector. The components of the ditection vector are known as the direction numbers for the line.
There is an alternative vector equation for the line: $\boldsymbol{r} = \boldsymbol{r}_1 + t ( \boldsymbol{r}_2 - \boldsymbol{r}_1 )$ .

Parametric Equations

Consider the aforementioned vectors having the following components:
$\boldsymbol{r} = < x, y, z >, \quad \boldsymbol{r}_1 = < x_1, y_1, z_1 >, \quad \boldsymbol{r}_2 = < x_2, y_2, z_2 >,$
and we know that $\boldsymbol{r} = \boldsymbol{r}_2 + t ( \boldsymbol{r}_2 - \boldsymbol{r}_1 )$ . So,
$x = x_2 + a_1 t , \quad y = y_2 + a_2 t, \quad z = z_2 + a_3 t .$
These equations are called parametric equations for the line passing through $P_1$ and $P_2$ .

Symmetric Equation

Symmetric equations for the line passing through $P_1$ and $P_2$ are defined as

(2) $\begin{equation*}\frac{x-x_2}{a_1} = \frac{y-y_2}{a_2} = \frac{z-z_2}{a_3}\end{equation*}$

Planes: Vector Equation

There is only one plane $\mathcal{P}$ containing point $P_1$ with a vector $\boldsymbol{n}$ normal to the plane. If $P$ is any point on $\mathcal{P}$ , and $\boldsymbol{r} = \overrightarrow{OP}$ and $\boldsymbol{r}_1 = \overrightarrow{OP_1}$ , then a vector equation of the plane is

(3) $\begin{equation*}\boldsymbol{n} . ( \boldsymbol{r} - \boldsymbol{r}_1 ) = 0.\end{equation*}$

Cartesian Equation

If a plane has the normal vector of $\boldsymbol{n} = a \boldsymbol{i} + b \boldsymbol{j} + c \boldsymbol{k}$ and contains the point $P_1 (x_1,y_1,z_1)$ , then the point-normal form of the equation of the plane is

(4) $\begin{equation*}a(x-x_1) + b(y-y_1) + c(z-z_1) = 0.\end{equation*}$

8-Minute Lecture on Lines and Planes in 3D

Reference

Dennis G. Zill. Advanced Engineering Mathematics, $6^{th}$ edition. Jones $&$ Bartlett Learning. 2016.