Math Core

In the plane, a single equation $y = mx + b$ describes a line. In space that no longer works: one linear equation in $x$ , $y$ , $z$ describes a plane, and a line needs a different description. Vectors give clean descriptions of both, and the dot and cross products turn questions about angles, intersections and distances into short computations.

Lines

A line in space is determined by a point $P_0(x_0, y_0, z_0)$ on it and a nonzero direction vector $\mathbf{v} = \langle a, b, c \rangle$ parallel to it. A point $P$ lies on the line exactly when $\overrightarrow{P_0P}$ is a multiple of $\mathbf{v}$ , say $t\mathbf{v}$ . Writing $\mathbf{r}_0$ and $\mathbf{r}$ for the position vectors of $P_0$ and $P$ :

Equations of a line

The line through $P_0(x_0, y_0, z_0)$ with direction $\mathbf{v} = \langle a, b, c \rangle$ has

vector equation $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ ,
parametric equations $x = x_0 + at,\ \ y = y_0 + bt,\ \ z = z_0 + ct$ ,
symmetric equations (when $a, b, c \ne 0$ ) $\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c}$ .

Each real value of the parameter $t$ gives one point on the line.

Think of $t$ as time: at $t = 0$ you are at $P_0$ , and each unit of time moves you by $\mathbf{v}$ . A line has infinitely many such descriptions, since any point on it and any nonzero multiple of $\mathbf{v}$ work equally well.

Worked example: The line through two points

Find parametric equations for the line through $P(1, -2, 4)$ and $Q(3, 1, 0)$ , and find where it crosses the $xz$ -plane.

Solution. A direction vector 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.

The $xz$ -plane is $y = 0$ , so $-2 + 3t = 0$ and $t = \dfrac23$ . Then $x = 1 + \dfrac43 = \dfrac73$ and $z = 4 - \dfrac83 = \dfrac43$ . The crossing point is $\left(\dfrac73, 0, \dfrac43\right)$ .

How two lines can meet

Two lines in the plane either intersect or are parallel. In space there is a third option: skew lines are not parallel and never meet, like a road and an overpass crossing it.

If the direction vectors are parallel, the lines are parallel (or identical, if they share a point).
Otherwise, set the coordinates equal using different parameters $t$ and $s$ . Solve two of the three equations. If the solution satisfies the third, the lines intersect; if not, they are skew.

Worked example: Intersecting or skew?

Classify $L_1$ : $x = 1 + t,\ y = 2 - t,\ z = 3t$ and $L_2$ : $x = 2s,\ y = 3 + s,\ z = 1 + 4s$ .

Solution. The directions $\langle 1, -1, 3 \rangle$ and $\langle 2, 1, 4 \rangle$ are not multiples of each other, so the lines are not parallel. Equate $x$ and $y$ : $1 + t = 2s$ and $2 - t = 3 + s$ . Adding the equations gives $3 = 3 + 3s$ , so $s = 0$ and then $t = -1$ . Check $z$ : $L_1$ gives $3t = -3$ , but $L_2$ gives $1 + 4s = 1$ . These disagree, so the lines never meet. They are skew.

Planes

A plane is determined by a point $P_0$ in it and a nonzero normal vector $\mathbf{n} = \langle a, b, c \rangle$ perpendicular to it. A point $P$ lies in the plane exactly when $\overrightarrow{P_0P}$ is orthogonal to $\mathbf{n}$ .

Equation of a plane

The plane through $P_0(x_0, y_0, z_0)$ with normal vector $\mathbf{n} = \langle a, b, c \rangle$ is

\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0, \qquad\text{that is,}\qquad a(x - x_0) + b(y - y_0) + c(z - z_0) = 0.

Expanding gives the linear equation $ax + by + cz = d$ . Conversely, every such equation with $a, b, c$ not all zero is a plane with normal $\langle a, b, c \rangle$ .

So you can read a normal vector straight off the coefficients: $2x - y + 5z = 7$ has normal $\langle 2, -1, 5 \rangle$ . To sketch a plane, find its intercepts and connect them. The plane $2x + 3y + 4z = 12$ meets the axes at $(6, 0, 0)$ , $(0, 4, 0)$ and $(0, 0, 3)$ ; the triangle joining them is the part of the plane in the first octant.

The first-octant piece of the plane 2x + 3y + 4z = 12, drawn by joining its three intercepts.

For a plane through three points $P$ , $Q$ , $R$ , the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ lie in the plane, so their cross product is a normal vector.

Worked example: Plane through three points

Find an equation of the plane through $P(1, 0, 2)$ , $Q(2, -1, 3)$ and $R(0, 2, 1)$ .

Solution. $\overrightarrow{PQ} = \langle 1, -1, 1 \rangle$ and $\overrightarrow{PR} = \langle -1, 2, -1 \rangle$ , so

\mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \langle (-1)(-1) - (1)(2),\ (1)(-1) - (1)(-1),\ (1)(2) - (-1)(-1) \rangle = \langle -1, 0, 1 \rangle.

Using $P$ : $-(x - 1) + 0(y - 0) + (z - 2) = 0$ , which simplifies to $x - z = -1$ . Check: $Q$ gives $2 - 3 = -1$ and $R$ gives $0 - 1 = -1$ .

Angles, intersections and distances

Two planes are parallel if their normals are parallel. Otherwise they meet in a line, and the angle between the planes is the acute angle between their normals.

Worked example: Two intersecting planes

For the planes $x + y + z = 1$ and $x - 2y + 3z = 1$ , find the angle between them and parametric equations for their line of intersection.

Solution. With $\mathbf{n}_1 = \langle 1, 1, 1 \rangle$ and $\mathbf{n}_2 = \langle 1, -2, 3 \rangle$ ,

\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1|\,|\mathbf{n}_2|} = \frac{2}{\sqrt3\sqrt{14}} = \frac{2}{\sqrt{42}}, \qquad \theta \approx 72.0^\circ.

The line of intersection lies in both planes, so it is perpendicular to both normals. Its direction is $\mathbf{n}_1 \times \mathbf{n}_2 = \langle 5, -2, -3 \rangle$ . For a point, set $z = 0$ : then $x + y = 1$ and $x - 2y = 1$ , giving $y = 0$ , $x = 1$ . The line is $x = 1 + 5t,\ y = -2t,\ z = -3t$ .

To find the distance from a point $P_1$ to a plane, pick any point $P_0$ in the plane. The distance is the length of the projection of $\overrightarrow{P_0P_1}$ onto the normal: $D = \dfrac{|\mathbf{n} \cdot \overrightarrow{P_0P_1}|}{|\mathbf{n}|}$ . Writing this out in coordinates gives a formula that doesn't need $P_0$ at all.

Distance from a point to a plane

The distance from $P_1(x_1, y_1, z_1)$ to the plane $ax + by + cz = d$ is

D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}.

For example, the distance from $(1, -2, 4)$ to $3x + 2y + 6z = 5$ is $\dfrac{|3 - 4 + 24 - 5|}{\sqrt{9 + 4 + 36}} = \dfrac{18}{7}$ .

Common mistake

Before using the distance formula for two parallel planes, make their left sides identical, not just proportional. The planes $x + 2y - 2z = 3$ and $2x + 4y - 4z = 18$ are parallel, but you must first divide the second by $2$ to get $x + 2y - 2z = 9$ . Only then is the distance $\dfrac{|9 - 3|}{3} = 2$ .

Practice

Practice 1

A line passes through $(2, -1, 5)$ with direction vector $\langle 3, 0, -2 \rangle$ . Using the parametrization $\mathbf{r}(t) = \langle 2, -1, 5 \rangle + t\langle 3, 0, -2 \rangle$ , find the point with $t = 2$ .

Enter a point like (2, -3)

Practice 2

The plane through $(4, -2, 1)$ with normal vector $\langle 2, -1, 3 \rangle$ has equation $2x - y + 3z = d$ . Find $d$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find the point where the line $x = 1 + 2t,\ y = -3 + t,\ z = 4 - t$ meets the plane $x + 2y + 3z = 11$ .

Enter a point like (2, -3)

Practice 4

Find the distance from the point $(2, 1, -3)$ to the plane $2x - 2y + z = 4$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Which is an equation of the plane through $(1, 2, 3)$ , $(2, 0, 1)$ and $(0, 1, 5)$ ?

Practice 6

Classify the lines $L_1$ : $\mathbf{r} = \langle 1, 2, 0 \rangle + t\langle 2, -1, 1 \rangle$ and $L_2$ : $\mathbf{r} = \langle 3, 0, 4 \rangle + s\langle -4, 2, -2 \rangle$ .

Practice 7

Find the angle, in degrees, between the planes $2x - y + z = 3$ and $x + y + 2z = 1$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Find the distance between the parallel planes $x + 2y - 2z = 3$ and $3x + 6y - 6z = 27$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.