2.5 Equations of Lines and Planes in Space

Big idea

In the plane, a line is described by one equation in two variables, and the familiar slope-intercept form hides an assumption: it depends on the line being confined to two dimensions. In space that trick fails. A single linear equation in $x$, $y$ and $z$ describes a plane, not a line, so a line needs a different kind of description.

The description that works is dynamic. Name one point the line passes through and one vector pointing along it, then sweep out the line by travelling from that point along that direction for every possible amount of time. That is the vector equation of a line, and reading it one coordinate at a time gives the parametric equations. Eliminating the parameter gives the symmetric equations.

Planes go the other way. A plane is pinned down not by a direction lying inside it but by a direction perpendicular to it: the normal vector. A point lies on the plane exactly when the vector to it from a fixed point of the plane is orthogonal to the normal, and orthogonality is a dot product set to zero. Expanding that dot product produces the linear equation $ax + by + cz = d$, whose coefficients are the components of the normal, so a plane’s orientation can be read straight off its equation.

With both objects available, the natural questions become computations. Do two lines meet, or pass by each other at different heights? How far is a point from a plane? The answers reduce to the dot product, the cross product and projection, which is why this section arrives where it does.

Decoder

A line in space is the set of points reached from a fixed point by adding all scalar multiples of a fixed direction vector, while a plane is the set of points whose displacement from a fixed point is orthogonal to a fixed normal vector.

Notice the contrast. For a line you are given a direction to move along; for a plane you are given a direction to avoid. One is described by inclusion, the other by exclusion, which is why a line needs a parameter and a plane does not.

That difference explains a fact students often find surprising: the equation of a plane is simpler than the equations of a line, even though a plane is the bigger object. The plane comes from one scalar condition, so one equation suffices. The line requires two independent conditions, which is why the symmetric form is a chain of two equalities, and why a line can also be presented as the intersection of two planes.

The classic mistake is treating the parameter as meaningful. The same line has infinitely many parametrizations, since any point on it serves as the base point and any nonzero multiple of the direction serves as the direction. So when you compare two lines, compare directions and points, never the formulas as written, and when you test whether two lines intersect, give them different parameter names.

Definitions and results

Vector equation of a line. The line through $P_0$ with position vector $\mathbf{r}_0$ and direction $\mathbf{v}$ is

$$ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}, \qquad t \in \mathbf{R} $$

Restricting $t$ to $[0,1]$ gives the segment from $P_0$ to the tip of $\mathbf{r}_0 + \mathbf{v}$.

Parametric equations. Reading the vector equation componentwise with $P_0 = (x_0, y_0, z_0)$ and $\mathbf{v} = \langle a,b,c \rangle$ gives $x = x_0 + at$, $y = y_0 + bt$, $z = z_0 + ct$.

Symmetric equations. If all of $a$, $b$, $c$ are nonzero, solving each parametric equation for $t$ and equating gives

$$ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} $$

If one component of the direction is zero, that variable is constant and is stated separately rather than placed over a zero denominator.

Relative position of two lines. Two lines are parallel when their directions are scalar multiples of each other. Otherwise they either intersect, which happens when the parametric equations have a common solution, or they are skew: not parallel and never meeting.

Normal vector and the point-normal equation. A nonzero $\mathbf{n} = \langle a,b,c \rangle$ perpendicular to a plane through $P_0(x_0,y_0,z_0)$ determines it by

$$ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 $$

Expanding gives the general form $ax + by + cz = d$, where the coefficients are the components of a normal vector.

A plane through three points. Given three non-collinear points, form two vectors between them and take their cross product to get a normal, then use any of the three points in the point-normal equation.

Angles between planes. Two planes are parallel when their normals are parallel and perpendicular when their normals are orthogonal; in general the angle between them is the angle between their normals. Their line of intersection has direction $\mathbf{n}_1 \times \mathbf{n}_2$.

Distance from a point to a plane. The distance from $Q$ to the plane through $P$ with normal $\mathbf{n}$ is the magnitude of the projection of $\overrightarrow{PQ}$ onto $\mathbf{n}$:

$$ d = \frac{\left| \overrightarrow{PQ} \cdot \mathbf{n} \right|}{\|\mathbf{n}\|} $$

Distance from a point to a line. For the line through $P$ with direction $\mathbf{v}$, the distance to $Q$ is

$$ d = \frac{\left\| \overrightarrow{PQ} \times \mathbf{v} \right\|}{\|\mathbf{v}\|} $$

which is the area of the parallelogram divided by its base. The same formula with $\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2$ substituted into the point-to-plane formula gives the distance between two skew lines, using any point on each.

Worked examples

A line through two points

Find equations for the line through $P(1,-2,4)$ and $Q(3,1,0)$.

The direction is $\overrightarrow{PQ} = \langle 2, 3, -4 \rangle$. Using $P$ as the base point,

$$ x = 1 + 2t, \qquad y = -2 + 3t, \qquad z = 4 - 4t $$

$$ \frac{x-1}{2} = \frac{y+2}{3} = \frac{z-4}{-4} $$

Check the endpoints: $t = 0$ gives $(1,-2,4)$ and $t = 1$ gives $(3, 1, 0)$, so both points are on the line. Using $Q$ as base point and $\langle -2,-3,4 \rangle$ as direction would give different-looking equations for the same line.

A plane through three points

Find the plane through $A(1,0,0)$, $B(0,2,0)$ and $C(0,0,4)$.

Two vectors in the plane are $\overrightarrow{AB} = \langle -1, 2, 0 \rangle$ and $\overrightarrow{AC} = \langle -1, 0, 4 \rangle$. A normal is their cross product:

$$ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{i}(8 - 0) - \mathbf{j}(-4 - 0) + \mathbf{k}(0 + 2) = \langle 8, 4, 2 \rangle $$

Any nonzero multiple is also normal, so use $\mathbf{n} = \langle 4, 2, 1 \rangle$. With the point $A$,

$$ 4(x - 1) + 2(y - 0) + 1(z - 0) = 0 \quad \Longrightarrow \quad 4x + 2y + z = 4 $$

Check all three points: each gives $4$ on the left side. The intercepts $1$, $2$ and $4$ read straight off the original points, as they should.

Two distances

First, the distance from $Q(4,1,3)$ to the plane $2x - y + 2z = 5$. A point on the plane is $P(0,-5,0)$, since $0 + 5 + 0 = 5$, and $\overrightarrow{PQ} = \langle 4, 6, 3 \rangle$. With $\mathbf{n} = \langle 2,-1,2 \rangle$ and $\|\mathbf{n}\| = 3$,

$$ d = \frac{|8 - 6 + 6|}{3} = \frac{8}{3} $$

Second, the distance from $Q(3,1,-1)$ to the line through the origin with direction $\mathbf{v} = \langle 1,2,2 \rangle$. Here $\overrightarrow{PQ} = \langle 3,1,-1 \rangle$ and

$$ \overrightarrow{PQ} \times \mathbf{v} = \mathbf{i}(2 + 2) - \mathbf{j}(6 + 1) + \mathbf{k}(6 - 1) = \langle 4, -7, 5 \rangle $$

whose magnitude is $\sqrt{16 + 49 + 25} = \sqrt{90} = 3\sqrt{10}$. Dividing by $\|\mathbf{v}\| = 3$ gives $d = \sqrt{10}$.

Check the second answer by projection instead. The scalar projection of $\overrightarrow{PQ}$ onto $\mathbf{v}$ is $(3 + 2 - 2)/3 = 1$, and $\|\overrightarrow{PQ}\|^2 = 11$, so the perpendicular piece has squared length $11 - 1 = 10$. The two methods agree.

Two lines that never meet

Let $L_1$ be $(1+t, \; 2-t, \; 3+2t)$ and $L_2$ be $(2+s, \; 1+s, \; -1+3s)$.

Their directions $\langle 1,-1,2 \rangle$ and $\langle 1,1,3 \rangle$ are not multiples of each other, so the lines are not parallel. Matching the first two coordinates gives $t - s = 1$ and $-t - s = -1$, so $t = 1$ and $s = 0$. Those values must also satisfy the third coordinate, but $3 + 2(1) = 5$ while $-1 + 3(0) = -1$, so the lines are skew.

For the distance between them, a vector perpendicular to both is

$$ \mathbf{n} = \langle 1,-1,2 \rangle \times \langle 1,1,3 \rangle = \mathbf{i}(-3-2) - \mathbf{j}(3-2) + \mathbf{k}(1+1) = \langle -5, -1, 2 \rangle $$

with $\|\mathbf{n}\| = \sqrt{30}$. Taking $P_1(1,2,3)$ on the first line and $P_2(2,1,-1)$ on the second, $\overrightarrow{P_1P_2} = \langle 1,-1,-4 \rangle$ and

$$ d = \frac{|-5 + 1 - 8|}{\sqrt{30}} = \frac{12}{\sqrt{30}} = \frac{2\sqrt{30}}{5} \approx 2.19 $$

A positive answer is exactly what skewness predicts; intersecting lines would have produced zero.

Practice

Start with lines. These ask you to build vector, parametric and symmetric equations, and to read a direction back out of given equations.

Next, planes. You will find normal vectors, write point-normal and general forms, pass a plane through three points, and test two planes for parallelism.

Finally, the measurements: distance from a point to a plane, distance from a point to a line, and classifying two lines as parallel, intersecting or skew.

Quiz

Six items across lines, planes, and distances in space.