- Why is the absolute value here?
- What would the angle describe if the absolute value were omitted?
Section 2.5 Equations of Lines and Planes in Space
References.
- OpenStax Calculus Volume 3, Section 2.5.
1
openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-space - Calculus, Early Transcendentals by Stewart, Section 12.5.
Introduction.
We have seen that dot and cross products have something to do with parallel lines and planes respectively: we now use these properties to describe lines and planes in three dimensional space. It will be convenient to describe a point \(P\) in \(\mathbb{R}^3\) by giving its position vector \(\vec{p} = \veclong{OP}\text{,}\) and then to describe lines and planes as collections of position vectors.
Topics.
Subsection 2.5.1 Lines
Lines in Vector Terms.
We can characterize a line \(L\) by the property that all vectors between any two points on the line are parallel, and so are multiples of some vector \(\vec{v}\text{,}\) which has a role like the slope of a line in the plane. Starting with any one point \(\vec{r}_0\) on the line, the vector \(\vec{r}-\vec{r}_0\) from it to any point \(\vec{r}\) on the line is a multiple \(t\vec{v}\) of \(\vec{v}\text{,}\) so \(L\) consists of the points
\begin{equation}
\vec{r} = \vec{r}_0 + t \vec{v}, \quad \text{ for any real } t.\tag{2.5.1}
\end{equation}
This can be written out in terms of coordinates as three “scalar” equations: with \(\vec{v} = \vector{a , b , c}\text{,}\) \(\vec{r}_0 = \vector{x_0 , y_0 , z_0}\text{,}\) \(\vec{r} = \vector{x , y , z}\text{,}\)
\begin{equation}
x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct\tag{2.5.2}
\end{equation}
which is the general parametric equation for a line in space.
These forms are not unique, since we could replace \(r_0\) by any other point on the line, and replace \(\vec{v}\) by any non-zero scalar multiple of it. The components \(a\text{,}\) \(b\) and \(c\) together are called the direction numbers of line \(L\text{;}\) any other triple of numbers proportional to these also serve as direction numbers of \(L\text{.}\)
The Symmetric Equations for a Line.
Each equation in (2.5.2) can be solved for \(t\text{,}\) and equating the three results gives the symmetric equations of \(L\text{:}\)
\begin{equation}
\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}.\tag{2.5.3}
\end{equation}
If any of the direction numbers is zero, the equations become even simpler: for example if \(a=0\text{,}\)
\begin{equation}
x = x_0, \quad \frac{y - y_0}{b} = \frac{z - z_0}{c}.\tag{2.5.4}
\end{equation}
(But at least one direction number must be non-zero!)
Subsection 2.5.2 Planes
The directions of motion possible within a plane can be described two ways: by a single normal vector \(\vecn\) perpendicular to all vectors \(\veclong{PQ}\) joining two points in the plane, or by giving two non-parallel vectors \(\veclong{PQ}\) and \(\veclong{PR}\) given by three points \(P\text{,}\) \(Q\) and \(R\) in the plane. Given a normal vector \(\vecn\) and point \(\vec{r}_0\) in the plane, any vector \(\vec{r}-\vec{r}_0\) within the plane is perpendicular to \(\vecn\text{,}\) so the plane is given by all solutions \(\vec{r}\) of
\begin{equation}
\vecn \cdot (\vec{r}-\vec{r}_0) = 0, \quad \mbox{or} \quad \vecn \cdot \vec{r} = \vecn \cdot \vec{r}_0\tag{2.5.5}
\end{equation}
Either of these is called a vector equation of the plane.
The scalar equation for a plane.
With \(\vecn = \vector{a , b , c}\) [careful: we are recycling those three letters!] and as before \(\vec{r} = \vector{x , y , z}\text{,}\)
\begin{equation}
a(x - x_0) + b(y - y_0) + c (z - z_0) = 0.\tag{2.5.6}
\end{equation}
This is the scalar equation of the plane through point \(P(x_0,y_0,z_0)\) with normal vector \(\vecn= \vector{a , b , c}\text{.}\) Expanding gives the alternate form
\begin{equation}
ax + by + c z + d = 0 \text{, where } d = -(a x_0 + b y_0 + c z_0).\tag{2.5.7}
\end{equation}
Any choice of constants \(a\text{,}\) \(b\text{,}\) \(c\) (not all zero) and \(d\) given a plane. Equation (2.5.7) is the most general linear equation in three unknowns \(x\text{,}\) \(y\) and \(z\text{,}\) so we now know that the solution of such an equation is always a plane (not a line as the name might suggest, and as is true for a linear equation in two unknowns.)
Three points (usually) determine a plane.
Given three points \(P\text{,}\) \(Q\text{,}\) and \(R\) that do not line on a line, there is a unique plane containing all of them. The two vectors \(\veclong{PQ}\) and \(\veclong{PR}\) are parallel to the plane, so their cross-product \(\vecn= \veclong{PQ} \times \veclong{PR}\) is normal to the plane. Then using \(\vec{r_0}= \veclong{OR}\) as a point on the plane give the equation \(\vecn\cdot(\vec{r}-\vec{r_0})\) as above.
This works unless \(\vecn=\vec{0}\text{:}\) that is the case where \(\veclong{PQ}\) and \(\veclong{PR}\) are parallel, meaning that the three points are collinear, and then there is not a unique plane through them, but many.
A line and point not on it determine a plane.
Choosing any two point on that line converts this to the above case: See Example 50 in OSC3 Section 2.5.
4
openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-spaceThe angle between two planes, and distance to a plane.
Two planes are parallel (and so either do not intersect or are identical) if their normals \(\vecn_1\) and \(\vecn_2\) are parallel. Otherwise, they intersect along a line, and the angle between them is the acute angle between their normals: the smallest angle \(\theta\) with
\begin{equation}
\cos\theta=\frac{| \vecn_1 \cdot \vecn_2 |}{\|\vecn_1\| \|\vecn_2\|}\text{.}\tag{2.5.8}
\end{equation}
Question 2.5.1.
Also, the shortest distance from a point \(P(x_1,y_1,z_1)\) to a point in the plane \(ax+by+cz+d=0\) through a point \(Q(x_0,y_0,z_0)\) with normal \(\vecn\) is
\begin{equation}
D = \frac{\veclong{PQ} \cdot \vecn}{\| \vecn \|} = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}\tag{2.5.9}
\end{equation}
and for \(Q\text{,}\) the closest point in the plane, \(\veclong{PQ}\) is parallel to the normal vector \(\vecn = \vector{a,b,c}\text{.}\)
Note that if we use unit normal vectors \(\hat{n} = \vector{a, b, c}\) and so on, the angle and distance formulas simplify to
\begin{equation}
\cos\theta = | \hat{n}_1 \cdot \hat{n}_2 |.\tag{2.5.10}
\end{equation}
and
\begin{equation}
D = \veclong{PQ} \cdot \hat{n} = |ax_1+by_1+cz_1+d|\tag{2.5.11}
\end{equation}
Study Guide.
Study Section 2.5 of Calculus Volume 3; in particular
7
openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-space- All the Definitions, Theorems, Examples and Checkpoints.
- One or several exercises from each of the following ranges: 243–246; 251–254, 267–270, 271–274 [parts (a) and (b) are enough], 281–282, 289–290.
