Theory

Lines and planes in space

Equations of planes and lines in space, relative positions, angles, distances, orthogonal projections, pencils of planes, and the line of minimum distance between two skew lines.

On this page

The plane: Cartesian equation and normal vector
The line: parametric and Cartesian equations

The plane: Cartesian equation and normal vector

1. Plane through a point orthogonal to a vector

Let $P_{0}$ be a point in space and $n$ a nonzero vector. The plane through $P_{0}$ orthogonal to $n$ is the set of points $P$ for which the vector $P_{0} P$ is orthogonal to $n$ :

π = {P : n \cdot P_{0} P = 0} .

The vector $n$ is called a normal vector to the plane. A plane is therefore determined by a point and an orthogonal direction; since the condition $n \cdot P_{0} P = 0$ does not change when $n$ is multiplied by a nonzero number, the normal vector is defined up to a factor. For the dot product and orthogonality, see the topic on vectors.

Section by the plane

y = 0

of the plane

π : x + z = 2

, with normal vector

n = (1, 0, 1)

: the vector

P_{0} P

with

P \in π

is orthogonal to

n

, while for the point

Q

off the plane

n \cdot P_{0} Q \neq = 0

2. Cartesian equation of the plane

Every plane in space is the solution set of an equation

a x + b y + cz + d = 0 with (a, b, c) \neq = (0, 0, 0),

and the vector $(a, b, c)$ is normal to the plane. Conversely, every equation of this kind represents a plane with normal vector $(a, b, c)$ . The components of a normal vector can therefore be read off the equation, and a point and a normal vector are enough to write the equation. The plane through $P_{0} = (1, 2, - 1)$ with $n = (2, - 1, 3)$ has equation $2 (x - 1) - (y - 2) + 3 (z + 1) = 0$ , that is, $2 x - y + 3 z + 3 = 0$ .

Proof

Let $π$ be the plane through $P_{0} = (x_{0}, y_{0}, z_{0})$ orthogonal to $n = (a, b, c)$ . Writing the dot product in components, the condition $n \cdot P_{0} P = 0$ becomes

a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0,

that is, $a x + b y + cz + d = 0$ with $d = - (a x_{0} + b y_{0} + c z_{0})$ . Conversely, let the equation be given with $(a, b, c) \neq = (0, 0, 0)$ and suppose, for example, $a \neq = 0$ . The point $P_{0} = (- \frac{d}{a}, 0, 0)$ satisfies the equation, hence $d = - (a x_{0} + b y_{0} + c z_{0})$ with $(x_{0}, y_{0}, z_{0}) = P_{0}$ . Substituting this expression for $d$ , the equation becomes $a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$ , which is the condition $n \cdot P_{0} P = 0$ ; therefore the solutions form the plane through $P_{0}$ orthogonal to $(a, b, c)$ . If $a = 0$ , the same argument works with whichever of $b$ and $c$ is nonzero.

Plane through three points

Three non-collinear points $A$ , $B$ , $C$ determine a unique plane. The vectors $A B$ and $A C$ are parallel to the plane and not parallel to each other, so their cross product is a normal vector:

n = A B \times A C .

Equivalently, the point $P = (x, y, z)$ lies in the plane if and only if the vectors $A P$ , $A B$ , $A C$ are coplanar, that is, if and only if their scalar triple product is zero:

x - x_{A} x_{B} - x_{A} x_{C} - x_{A} y - y_{A} y_{B} - y_{A} y_{C} - y_{A} z - z_{A} z_{B} - z_{A} z_{C} - z_{A} = 0.

For $A = (1, 0, 0)$ , $B = (0, 2, 0)$ , $C = (0, 0, 3)$ we have $A B \times A C = (- 1, 2, 0) \times (- 1, 0, 3) = (6, 3, 2)$ , so the plane is $6 x + 3 y + 2 z - 6 = 0$ . Dividing by $6$ gives $\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1$ , a form in which the intercepts of the plane with the three axes can be read off.

3. Parametric equations of the plane

Let $P_{0}$ be a point and $u$ , $w$ two non-parallel vectors. The plane through $P_{0}$ parallel to $u$ and to $w$ is described by the parametric equations

P = P_{0} + s u + t w, s, t \in R .

Passing to Cartesian form requires a normal vector, which is $n = u \times w$ . The reverse passage consists in solving the Cartesian equation, which has two free unknowns. For the plane $x + y - z = 1$ we set $y = s$ , $z = t$ and obtain $x = 1 - s + t$ , hence $P = (1, 0, 0) + s (- 1, 1, 0) + t (1, 0, 1)$ .

The line: parametric and Cartesian equations

1. Parametric equations of the line

Let $P_{0} = (x_{0}, y_{0}, z_{0})$ be a point and $v = (v_{1}, v_{2}, v_{3})$ a nonzero vector. The line through $P_{0}$ parallel to $v$ is the set of points $P = P_{0} + t v$ for $t \in R$ , that is, of the points whose coordinates satisfy

⎩ ⎨ ⎧ x = x_{0} + v_{1} t y = y_{0} + v_{2} t z = z_{0} + v_{3} t t \in R .

The vector $v$ is called a direction vector of the line and its components are called direction numbers; like the normal vector of a plane, the direction vector is defined up to a nonzero factor. The line through two distinct points $A$ and $B$ has direction vector $A B$ .

Example

The line through $A = (1, - 1, 2)$ with direction vector $v = (2, 1, - 3)$ has parametric equations $x = 1 + 2 t$ , $y = - 1 + t$ , $z = 2 - 3 t$ . The point $(5, 1, - 4)$ belongs to the line, because it is obtained for $t = 2$ . The point $(3, 0, 0)$ does not belong to the line, because the first two equations give $t = 1$ while the third gives $t = \frac{2}{3}$ .

2. Cartesian equations of the line

A line is the intersection of two non-parallel planes, so it is described by the system

{a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 a_{2} x + b_{2} y + c_{2} z + d_{2} = 0 with rk (a_{1} a_{2} b_{1} b_{2} c_{1} c_{2}) = 2.

The two equations are called Cartesian equations of the line. The rank condition says that the normal vectors $n_{1}$ and $n_{2}$ are not parallel; by the Rouché-Capelli theorem the system then has $\infty^{1}$ solutions, which form a line. The Cartesian equations of a line are not unique, because any pair of distinct planes containing it represents it; the set of all these planes is the pencil of planes having the line as its axis.

Passing between the two forms

From parametric to Cartesian: solve for $t$ in an equation where its coefficient is nonzero and substitute it into the other two. For the line of the previous example, $t = y + 1$ gives $x = 2 y + 3$ and $z = - 3 y - 1$ , so the Cartesian equations are

{x - 2 y - 3 = 0 3 y + z + 1 = 0

From Cartesian to parametric: solve the system choosing a free unknown as parameter, or compute a point and a direction vector. The line lies in both planes, so its direction vector is orthogonal to $n_{1}$ and to $n_{2}$ , and one can take

v = n_{1} \times n_{2} .

For the line with equations $x + y - z = 1$ and $2 x - y + z = 2$ we obtain $v = (1, 1, - 1) \times (2, - 1, 1) = (0, - 3, - 3)$ , parallel to $(0, 1, 1)$ . Adding the two equations gives $3 x = 3$ , hence $x = 1$ and $y = z$ ; a point of the line is $(1, 0, 0)$ and the parametric equations are $x = 1$ , $y = t$ , $z = t$ .

The theory continues with 7 more sections (relative positions of two planes and of a line and a plane, relative positions of two lines, angles between lines and planes…), plus all worked examples, the solutions and the PDF.

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