Lines and planes in space

Summary

This lesson carries into space the line given by a point and a direction vector, and describes the plane in vector and parametric form, by a point and two direction vectors, and in Cartesian form, by an equation $ax + by + cz = d$ whose coefficients form a normal vector. The cross product connects both forms, and it is proved that every plane has a Cartesian equation, unique up to a factor. The relative positions of planes and lines are translated into linear systems and classified with the Rouché-Capelli theorem (known in the Spanish-speaking world as the Rouché-Frobenius theorem); the scalar triple product decides whether two lines intersect or are skew. The lesson ends with the pencil of planes and with problems on aircraft routes, the inclination of a piece of land and tetrahedra.

Learning objectives

Write a line in space in vector, parametric and symmetric form, also when some component of the direction vector is zero, and prove that exactly one line passes through two distinct points.
Write a plane in vector, parametric and Cartesian form, pass from one form to another with the cross product and construct the plane through three noncollinear points.
Prove that every plane has a Cartesian equation, unique up to a nonzero factor, and that every linear equation in $x$ , $y$ and $z$ with some nonzero coefficient describes a plane.
Decide the relative position of two or of three planes, of a line and a plane and of two lines by means of the ranks of a linear system and the scalar triple product, also when the data depend on a parameter.
Construct with the pencil of planes the plane that contains a line and satisfies another condition, and model with lines and planes situations from air navigation, construction and design.

The line in space

In the lesson of the course Algebra and Geometry I on vectors in the plane and parametric equations of a line, the line in the plane, known as the graph of a first-degree equation since the lesson on the equation of a line, was described as $P_0 + t \vec{d}$ . In space there is no slope, and a first-degree equation in $x$ , $y$ and $z$ describes, as will be seen, a plane. If this is so, it is convenient to reverse the order: the vector description will be the definition, and the other forms will be deduced from it.

Vectors in space and their dot product were studied in the lesson on vectors in space and the dot product; the cross product and the scalar triple product, in the lesson on the cross product and the scalar triple product. As in the plane, $Q = P + \vec{v}$ means $\vec{PQ} = \vec{v}$ , and two nonzero vectors are parallel when one is a multiple of the other. In the figures, the $y$ -axis points to the right, the $z$ -axis upward and the $x$ -axis toward the viewer.

Let $A$ be a point and $\vec{v} = \left( v_1, v_2, v_3 \right)$ a nonzero vector. The line through $A$ with direction vector $\vec{v}$ is the set of the points $P$ for which there exists a real number $t$ such that $P = A + t \vec{v}$ ; the number $t$ is called the parameter of $P$ .

P = A + t \vec{v}, \quad t \in \mathbb{R}

The parameter of each point is unique: if $A + t \vec{v} = A + s \vec{v}$ , then $\left( t - s \right) \vec{v} = \vec{0}$ , and since $\vec{v}$ is not zero, $t = s$ . For $A = (x_0, y_0, z_0)$ and $P = (x, y, z)$ , the vector equation, written in components, gives the parametric equations of the line:

\begin{cases} x = x_0 + t v_1 \\ y = y_0 + t v_2 \\ z = z_0 + t v_3 \end{cases}

If the parameter is time, these equations describe a uniform motion with velocity $\vec{v}$ , as in the plane. The same line admits many starting points and many direction vectors; the theorem below shows that two of its points determine it.

TheoremThe line through two points

Through two distinct points $A$ and $B$ there passes exactly one line, the one with equation $P = A + t \vec{AB}$ . Consequently, the equations $P = A + t \vec{v}$ and $P = C + s \vec{w}$ describe the same line if and only if $C$ lies on the first and $\vec{w}$ is parallel to $\vec{v}$ .

In the figure, $A = (2, 0, 1)$ and $B = (1, 2, 2)$ , so that $\vec{AB} = \left( -1, 2, 1 \right)$ , and the other line passes through $C = (0, 4, 3)$ with direction vector $\vec{v} = \left( 1, -2, -1 \right)$ .

Proof

$A + \resaltar{0} \cdot \vec{AB} = A, \qquad A + \resaltar{1} \cdot \vec{AB} = B$
Existence: the vector $\vec{AB}$ is not zero, because $A \neq B$ , and the line through $A$ with that direction vector contains $A$ , with parameter $0$ , and $B$ , with parameter $1$ .
$\dato{ab}{A = C + t_1 \vec{v}, \qquad B = C + t_2 \vec{v}}$
Uniqueness: let $r$ be any line, $P = C + t \vec{v}$ , that contains $A$ and $B$ , and let $t_1$ and $t_2$ be their parameters. In the figure, $t_1 = 2$ and $t_2 = 1$ .
$\vec{AB} = t_2 \vec{v} - t_1 \vec{v} = \dato{d}{\resaltar{\left( t_2 - t_1 \right)} \vec{v}}$
By the Chasles relation, $\vec{AB} = \vec{CB} - \vec{CA}$ . The parameters are distinct, because $A \neq B$ ; therefore $\vec{AB}$ is parallel to $\vec{v}$ . In the figure, $\vec{AB} = -\vec{v}$ .
$C + t \vec{v} = A + \left( t - t_1 \right) \vec{v} = A + \resaltar{\frac{t - t_1}{t_2 - t_1}} \vec{AB}$
Every point of $r$ lies on the line $AB$ : we start from $A$ instead of $C$ and replace $\vec{v}$ by its expression in terms of $\vec{AB}$ . In the figure, the point of $r$ with parameter $\frac{1}{2}$ is the point of the line $AB$ with parameter $\frac{3}{2}$ .
$A + s \vec{AB} = A + s \left( t_2 - t_1 \right) \vec{v} = C + \resaltar{\left( t_1 + s \left( t_2 - t_1 \right) \right)} \vec{v}$
Conversely, the point with parameter $s$ on the line $AB$ is the one with parameter $t_1 + s \left( t_2 - t_1 \right)$ on $r$ : the two lines have the same points. For the second claim: if the equations describe the same line, $C$ and $C + \vec{w}$ lie on it, and by the third step $\vec{w}$ is a multiple of $\vec{v}$ ; if $C$ lies on the first and $\vec{w} = k \vec{v}$ , the first contains $C$ and $C + \vec{w}$ , and it is the line through them.

It follows that two distinct lines have at most one common point: if they shared two, both would be the only line through them. Three points are collinear when some line contains them. If $A \neq B$ , the point $C$ is collinear with them if and only if $C = A + t \vec{AB}$ for some $t$ , that is, if and only if $\vec{AC}$ is a multiple of $\vec{AB}$ .

The symmetric equations

Solving for the parameter and equating the results gives equations without a parameter, with care for the zero components of the direction vector.

PropositionSymmetric equations of a line

Let $r$ be the line through $A = (x_0, y_0, z_0)$ with direction vector $\vec{v} = \left( v_1, v_2, v_3 \right)$ . If the three components of $\vec{v}$ are nonzero, a point $(x, y, z)$ lies on $r$ if and only if the equalities below hold. If only $v_2$ is zero, the condition is $\frac{x - x_0}{v_1} = \frac{z - z_0}{v_3}$ and $y = y_0$ ; if $v_2$ and $v_3$ are zero, the condition is $y = y_0$ and $z = z_0$ . The cases in which other components vanish are obtained by interchanging the roles of the coordinates.

\frac{x - x_0}{v_1} = \frac{y - y_0}{v_2} = \frac{z - z_0}{v_3}

In the figure, $A = (1, 2, -1)$ and $\vec{v} = \left( 2, 0, 4 \right)$ , the vector that goes from $A$ to $B = (3, 2, 3)$ .

Proof

$\frac{x - x_0}{v_1} = \resaltar{t}, \quad \frac{y - y_0}{v_2} = \resaltar{t}, \quad \frac{z - z_0}{v_3} = \resaltar{t}$
If the three components are nonzero and $P = A + t \vec{v}$ , we solve each parametric equation for $t$ : the three quotients equal $t$ .
$x = x_0 + \resaltar{t} v_1, \quad y = y_0 + \resaltar{t} v_2, \quad z = z_0 + \resaltar{t} v_3$
Conversely, if the three quotients are equal, we call $t$ their common value; multiplying each equality by its denominator recovers the parametric equations, and $P = A + t \vec{v}$ lies on $r$ .
$\frac{x - 1}{2} = \frac{z + 1}{4}, \quad \resaltar{y = 2}$
If only $v_2$ is zero, the second equation is $y = y_0$ for every $t$ , and it is written separately, because dividing by $0$ makes no sense. In the figure, the line lies in the plane $y = 2$ .
$\resaltar{y = 2}, \quad \resaltar{z = -1}$
If $v_2$ and $v_3$ are zero, the last two equations fix $y$ and $z$ , and for each $x$ the parameter $t = \frac{x - x_0}{v_1}$ gives the point: the line is the parallel to the $x$ -axis drawn through $A$ , which appears in the background of the figure.
$\frac{4 - 1}{2} = \frac{5 + 1}{4} = \frac{3}{2}, \qquad \frac{2 - 1}{2} \neq \frac{2 + 1}{4}$
Check in the figure: $C = (4, 2, 5)$ lies on the line, with parameter $\frac{3}{2}$ , and $D = (2, 2, 2)$ does not, although it satisfies $y = 2$ .

Each equality of the symmetric equations is a first-degree equation: the line in the figure is the solution set of $2x - z = 3$ and $y = 2$ . Once it is seen that every such equation describes a plane, this will say that the line is the intersection of two planes, its implicit form.

In the rest of this lesson

The plane in space
The vector equation and the parametric equations
The Cartesian equation and the normal vector
Parallelism and perpendicularity
Relative positions as linear systems
Two planes
Three planes
A line and a plane
Two lines
The pencil of planes
Solved problems
The plane that contains a line and is parallel to another
Three planes that depend on a parameter
The inclination of a piece of land for building
The routes of three aircraft
The bimedians of a tetrahedron
Proposed problems
A line and a plane with two parameters
The projection of a line onto a plane
The line that meets two others
A cutting plane through a cube
The cross sections of a tetrahedron parallel to two opposite edges