## Section1.19Lines and Planes

Two points $A$ and $B$ determine a line. But there are also other ways to describe a line. Rather than specifying two points, we can specify just one ($A$), then give a vector $\vv$ along the line. Figure 1.19.1. The geometry behind the vector description of a line.

So let $\AA$ be the vector from the origin to the point $A\text{,}$ and $\vv$ be a vector from $A$ that points along the line. Then any other point $P$ on the line can be reached by going to $A$ along $\AA\text{,}$ then going along $\vv\text{.}$ Thus, the vector

\begin{equation} \rr(u) = \AA + \vv\,u\tag{1.19.1} \end{equation}

from the origin to $P$ provides a vector parameterization of the line. Equivalently, if

\begin{align*} \AA \amp = a_x\,\xhat + a_y\,\yhat + a_z\,\zhat ,\\ \vv \amp = v_x\,\xhat + v_y\,\yhat + v_z\,\zhat , \end{align*}

then the line is given by the parametric equations

\begin{align*} x \amp = a_x + v_x\,u ,\\ y \amp = a_y + v_y\,u ,\\ z \amp = a_z + v_z\,u , \end{align*}

which gives the coordinates $(x,y,z)$ of the point $P$ in terms of the parameter $u\text{.}$ A given line can have many parametrizations, depending not only on the choice of the point $A$ on the line and vector $\vv$ along the line, but also on the choice of parameter $u\text{.}$

How do two points determine a line? Using vector addition, you can think of $\vv$ as $\BB-\AA\text{,}$ and use the vector description above, as shown in Figure 1.19.1.

It takes three points $A\text{,}$ $B\text{,}$ $C$ to determine a plane, but again there are also other descriptions. The orientation of a line is given by a vector $\vv$ along the line. By contrast, the orientation of a plane is given by a vector $\ww$ perpendicular to the plane. A plane can therefore be specified by giving a point $A$ and a normal vector $\ww$ to the plane at the point $A\text{.}$ Figure 1.19.2. The geometry behind the vector description of a plane.

If $\rr$ is the vector from the origin to $P\text{,}$ then $\rr-\AA$ is a vector in the plane, as shown in Figure 1.19.2. If $\ww$ is perpendicular to the plane, then it must be perpendicular to any vector in the plane. In particular, it must be perpendicular to $\rr-\AA\text{,}$ so that

\begin{equation} (\rr-\AA) \cdot \ww = 0\tag{1.19.2} \end{equation}

or equivalently

\begin{equation} \rr \cdot \ww = \AA\cdot\ww .\tag{1.19.3} \end{equation}

Inserting the components

\begin{align*} \rr \amp = x\,\xhat + y\,\yhat + z\,\zhat ,\\ \ww \amp = w_x\,\xhat + w_y\,\yhat + w_z\,\zhat , \end{align*}

and setting

\begin{equation} d = \AA\cdot\ww = \hbox{constant}\tag{1.19.4} \end{equation}

for the equation of the plane through $A$ with normal direction $\ww\text{.}$ This equation should look familiar! Note that the constant coefficients of this linear equation are precisely the components of the normal vector!
How do three points determine a plane? Using vector addition, you can construct two vectors in the plane, such as $\BB-\AA$ and $\CC-\AA\text{.}$ The cross product of these vectors is perpendicular to the plane! Thus, set $\ww=(\BB-\AA)\times(\CC-\AA)\text{,}$ and use the vector description above, as shown in Figure 1.19.2.