Lines and Planes in Space

Quick answer

In space a line cannot be written as y = mx + b. It is described by a point and a direction vector: r(t) = r₀ + tv, or in components x = x₀ + at, y = y₀ + bt, z = z₀ + ct. A plane is described by a point and a normal vector n perpendicular to it, which gives the equation a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. Cross products supply normals, dot products supply angles, and the formula |ax₀ + by₀ + cz₀ − d|/|n| gives the distance from a point to a plane.

What you'll learn

Write parametric equations of a line in space
Write the equation of a plane from a point and a normal vector
Find the plane through three points with a cross product
Find intersections, angles and distances involving planes

Lines: a point and a direction

In the plane a single equation $y = mx + b$ describes a line. In space one equation in $x$ , $y$ and $z$ describes a surface, so a line needs a different description. Start at a point $\mathbf{r}_0 = (x_0, y_0, z_0)$ and move along a direction vector $\mathbf{v} = (a, b, c)$ .

\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} \qquad\text{or}\qquad x = x_0 + at,\quad y = y_0 + bt,\quad z = z_0 + ct

As $t$ runs over the real numbers the point traces the whole line, the same way parametric equations trace curves in the plane.

Planes: a point and a normal

A plane is decided by one point $P_0$ on it and one vector $\mathbf{n} = (a, b, c)$ perpendicular to it, the normal vector. A point $P$ lies on the plane exactly when the arrow from $P_0$ to $P$ runs along the plane, which means it is perpendicular to $\mathbf{n}$ :

\mathbf{n} \cdot \overrightarrow{P_0P} = 0 \quad\Longleftrightarrow\quad a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

Expanding gives the standard form $ax + by + cz = d$ , and the coefficients of $x$ , $y$ and $z$ are the normal vector, which can be read off at a glance.

Why normals do all the work

Two planes are parallel when their normals are parallel, and the angle between two planes is the angle between their normals, found with a dot product. A normal also measures distance. From a point $Q$ off the plane, the shortest path to the plane runs along $\mathbf{n}$ , so the distance is the length of the projection of $\overrightarrow{P_0Q}$ onto $\mathbf{n}$ . Written out,

\text{distance} = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}

for $Q = (x_0, y_0, z_0)$ and the plane $ax + by + cz = d$ . Everything about a plane’s orientation is carried by its normal vector, so questions about planes become questions about one vector.

Worked examples

Common mistakes

Practice problems

Write the line through $(2, -1, 0)$ with direction $(1, 3, -2)$ , and find its point at $t = 2$ .

Answer

$x = 2 + t$ , $y = -1 + 3t$ , $z = -2t$ ; the point $(4, 5, -4)$

Full solution

Add $t$ times the direction to the starting point, then substitute $t = 2$ .
Write the plane through $(0, 0, 0)$ with normal $(3, -2, 5)$ .

Answer

$3x - 2y + 5z = 0$

Full solution

The coefficients are the normal, and the origin makes $d = 0$ .
Find a normal vector to the plane $4x - y + 7z = 2$ .

Answer

$(4, -1, 7)$

Full solution

Read the coefficients of $x$ , $y$ and $z$ .
Find the plane through $(0, 0, 0)$ , $(1, 2, 0)$ and $(0, 1, 1)$ .

Answer

$2x - y + z = 0$

Full solution

$(1, 2, 0) \times (0, 1, 1) = (2, -1, 1)$ is a normal, and the plane passes through the origin. Check $(1, 2, 0)$ : $2 - 2 + 0 = 0$ ✓.
Find the distance from the origin to the plane $x + 2y + 2z = 9$ .

Answer

$3$

Full solution

$\tfrac{|0 - 9|}{\sqrt{1 + 4 + 4}} = \tfrac{9}{3}$ .
Are the planes $2x - 4y + 6z = 1$ and $x - 2y + 3z = 5$ parallel?

Answer

Yes

Full solution

The first normal $(2, -4, 6)$ is twice the second, $(1, -2, 3)$ . The planes are parallel and different, since $x - 2y + 3z$ equals $\tfrac{1}{2}$ on one and $5$ on the other.
Are the planes $x + y + z = 1$ and $x - y = 0$ perpendicular?

Answer

Yes

Full solution

Their normals $(1, 1, 1)$ and $(1, -1, 0)$ have dot product $0$ .
Find the direction of the line where $x + y + z = 6$ and $x - y + z = 2$ meet, and a point on it.

Answer

Direction $(2, 0, -2)$ ; the point $(2, 2, 2)$

Full solution

The line lies in both planes, so it is perpendicular to both normals: $(1, 1, 1) \times (1, -1, 1) = (2, 0, -2)$ . Subtracting the equations gives $y = 2$ , and then $x + z = 4$ is satisfied by $(2, 2, 2)$ .
Does the line $x = t$ , $y = 1 + t$ , $z = 2t$ meet the plane $x + y - z = 5$ ?

Answer

No; it runs parallel to the plane.

Full solution

Substituting gives $t + 1 + t - 2t = 1$ , so the equation becomes $1 = 5$ , false for every $t$ . The direction $(1, 1, 2)$ is perpendicular to the normal $(1, 1, -1)$ , so the line never reaches the plane.
A student writes the plane through $(1, 2, 3)$ with normal $(2, -1, 1)$ as $x = 1 + 2t$ , $y = 2 - t$ , $z = 3 + t$ . What went wrong?

Hint

How many parameters does a plane need?

Answer

That is the line through the point along the normal, perpendicular to the plane. The plane is $2x - y + z = 3$ .

Full solution

One parameter traces a line. A plane can be written with one equation, $2(x - 1) - (y - 2) + (z - 3) = 0$ , and the student’s line crosses it at a right angle at $(1, 2, 3)$ .

Frequently asked questions

How do you write the equation of a line in space?

Use a point (x₀, y₀, z₀) on it and a direction vector (a, b, c): x = x₀ + at, y = y₀ + bt, z = z₀ + ct.

How do you write the equation of a plane?

Use a point (x₀, y₀, z₀) on it and a normal vector (a, b, c): a(x − x₀) + b(y − y₀) + c(z − z₀) = 0, which expands to ax + by + cz = d.

How do you find the plane through three points?

Form two edge vectors from one point to the other two. Their cross product is a normal vector, and any of the points completes the equation.

How do you find the distance from a point to a plane?

For the plane ax + by + cz = d and the point (x₀, y₀, z₀), the distance is |ax₀ + by₀ + cz₀ − d| divided by √(a² + b² + c²).

When are two planes parallel?

When their normal vectors are parallel, that is, one is a multiple of the other.

What to learn next

Functions of Several Variables and Contour MapsFunctions with two or three inputs: their domains, their graphs as surfaces, and level curves, which draw a surface on flat paper the way a topographic map does.

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