IB Analysis and Approaches HL · Geometry in space · Section 4 of 7

Equations of lines and planes in space

3 worked examples · By Álvaro López

Key idea

In space, a line is fixed by a point and a direction vector, and a plane by a point and two direction vectors (or by a point and a normal vector). Knowing how to write them in all their forms and convert between them is the basis for angles, distances and relative positions, and it appears in Paper 1 and in Paper 2.

Equations of a line in space

The idea is the same as in the plane: a line is determined by a point $A$ on it, with position vector $a$ , and by a direction, given by a direction vector $d \neq = 0$ . The only difference is that the vectors now have three components.

Each value of $λ$ gives a point on the line. There is no single equation: any point on the line and any non-zero multiple of $d$ will do.

If you equate the components you get the parametric equations:

$⎩ ⎨ ⎧ x y z = x_{0} + λ d_{1} = y_{0} + λ d_{2} = z_{0} + λ d_{3}$

And if $d_{1}$ , $d_{2}$ and $d_{3}$ are not zero, solving for $λ$ in all three and equating them gives the symmetric equation:

$\frac{x - x _{0}}{d _{1}} = \frac{y - y _{0}}{d _{2}} = \frac{z - z _{0}}{d _{3}}$

To decide whether a point $P$ lies on the line, substitute its coordinates into the parametric equations and check that the same $λ$ satisfies all three equations. If even one fails, $P$ is not on the line.

Equations of a plane

A plane needs more information than a line. There are two common ways to fix one.

A point and two direction vectors that are not parallel to each other, $u$ and $v$ , which lie in the plane: from $A$ you can reach any point on the plane by moving a certain amount along $u$ and a certain amount along $v$ .
A point and a normal vector $n$ , perpendicular to the plane.

If the plane is given by three points $A$ , $B$ and $C$ that are not collinear, the two direction vectors can be $A B$ and $A C$ , and the point, $A$ .

The normal vector

A non-zero vector $n$ is normal (or perpendicular) to the plane when it is perpendicular to all the vectors contained in it. To check it, it is enough that it is perpendicular to two non-parallel direction vectors, and from this we get how to calculate it.

Plane pi with a point A, another point P and the normal vector n perpendicular to the plane at A

The intuition: the vector product (previous section) gives a vector perpendicular to the two vectors being multiplied, and those two span the plane. The second part says that going from $A$ to any point on the plane means moving across it, and that movement is always perpendicular to $n$ .

Cartesian equation

Let $n = a b c$ be a normal vector and $A (x_{0}, y_{0}, z_{0})$ a point on the plane. The condition $A P \cdot n = 0$ for $P (x, y, z)$ is written

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

Expanding and grouping the constants gives the Cartesian equation.

The two forms, $r \cdot n = a \cdot n$ and $a x + b y + cz = d$ , say the same thing: the second is the first with the components written out, and $d$ is the number $a \cdot n$ (the same for all points on the plane).

A useful consequence: two planes whose normal vectors are proportional are parallel, and the planes $a x + b y + cz = d$ and $a x + b y + cz = d^{'}$ are always parallel to each other.

Converting between forms

Worked examples

Example 1 Basic

A line through two points

The points $A (2, - 1, 3)$ and $B (4, 1, 0)$ are given.

a) Find a direction vector of the line $A B$ with integer components that have no common factor.

b) State whether the point $P (8, 5, - 6)$ lies on the line $A B$ .

c) The point $Q (6, m, k)$ lies on the line $A B$ . Find $m$ and $k$ .

Solution

a) $A B = B - A = 22 - 3$ , and its components have no common factor. The equations of the line, using point $A$ , are

$r = 2 - 1 3 + λ 22 - 3, ⎩ ⎨ ⎧ x y z = 2 + 2 λ = - 1 + 2 λ = 3 - 3 λ, \frac{x - 2}{2} = \frac{y + 1}{2} = \frac{z - 3}{- 3}$

b) Substitute $P$ : $2 + 2 λ = 8$ gives $λ = 3$ ; $- 1 + 2 λ = 5$ gives $λ = 3$ ; $3 - 3 λ = - 6$ gives $λ = 3$ . The same value satisfies all three equations, so $P$ lies on the line.

c) From $x = 6$ : $2 + 2 λ = 6$ , so $λ = 2$ . Then $m = - 1 + 2 \cdot 2 = 3$ and $k = 3 - 3 \cdot 2 = - 3$ .

Example 2 Medium

A plane through three points

The points $A (1, 0, 2)$ , $B (3, 1, 1)$ and $C (2, 3, 0)$ are given.

a) Find a normal vector to the plane $π$ that passes through $A$ , $B$ and $C$ .

b) Find the Cartesian equation of $π$ in the form $a x + b y + cz - d = 0$ , with $a, b, c, d$ integers with no common factor and $a > 0$ .

c) The point $D (k, 1, 1)$ lies on $π$ . Find $k$ .

d) Find the point where $π$ cuts the $x$ -axis.

Solution

a) Two direction vectors are $A B = (2, 1, - 1)$ and $A C = (1, 3, - 2)$ . They are not parallel, because they are not proportional. A normal vector is their vector product:

$A B \times A C = 1 \cdot (- 2) - (- 1) \cdot 3 (- 1) \cdot 1 - 2 \cdot (- 2) 2 \cdot 3 - 1 \cdot 1 = 135$

b) The equation is $x + 3 y + 5 z = d$ . With $A$ : $d = 1 + 0 + 10 = 11$ , so

$x + 3 y + 5 z - 11 = 0$

Check with $B$ : $3 + 3 + 5 = 11$ . With $C$ : $2 + 9 + 0 = 11$ . Both satisfy the equation.

c) Substitute $D$ : $k + 3 \cdot 1 + 5 \cdot 1 = 11$ , so $k = 3$ .

d) The $x$ -axis consists of the points $(x, 0, 0)$ . Substituting: $x = 11$ , so the point of intersection is $(11, 0, 0)$ .

Example 3 Hard

From parametric to Cartesian form, a parallel plane and the foot of the perpendicular

The plane $Π_{1}$ has vector equation

$r = 102 + s 120 + t 013, s, t \in R$

a) Find the Cartesian equation of $Π_{1}$ in the form $a x + b y + cz - d = 0$ , with $a, b, c, d$ integers with no common factor and $a > 0$ .

b) The plane $Π_{2}$ is parallel to $Π_{1}$ and passes through $P (8, - 2, 0)$ . Find $d$ in its equation $6 x - 3 y + z = d$ , writing it first in the form $r \cdot n = a \cdot n$ .

c) The line that passes through $P$ and is perpendicular to $Π_{1}$ meets $Π_{1}$ at the point $Q$ . Find the coordinates of $Q$ .

Solution

a) The direction vectors are $u = (1, 2, 0)$ and $v = (0, 1, 3)$ . The normal vector is

$n = u \times v = 2 \cdot 3 - 0 \cdot 1 0 \cdot 0 - 1 \cdot 3 1 \cdot 1 - 2 \cdot 0 = 6 - 3 1$

So $6 x - 3 y + z = d$ , and with the point $(1, 0, 2)$ : $d = 6 + 0 + 2 = 8$ .

$6 x - 3 y + z - 8 = 0$

Check with another point on the plane, $A + u + v = (2, 3, 5)$ : $12 - 9 + 5 = 8$ .

b) Two parallel planes share a normal vector, so $n = (6, - 3, 1)$ also works for $Π_{2}$ . With $a = (8, - 2, 0)$ :

$r \cdot 6 - 3 1 = 8 - 2 0 \cdot 6 - 3 1 = 48 + 6 + 0 = 54$

Therefore $Π_{2} : 6 x - 3 y + z = 54$ and $d = 54$ .

c) The line perpendicular to $Π_{1}$ through $P$ has $n$ as its direction vector:

$r = 8 - 2 0 + λ 6 - 3 1 ⟹ x = 8 + 6 λ, y = - 2 - 3 λ, z = λ$

Substitute into $6 x - 3 y + z = 8$ :

$6 (8 + 6 λ) - 3 (- 2 - 3 λ) + λ = 8 ⟹ 54 + 46 λ = 8 ⟹ λ = - 1$

Then $Q = (8 - 6, - 2 + 3, - 1) = (2, 1, - 1)$ . Check in $Π_{1}$ : $12 - 3 - 1 = 8$ .

In the exam: in a change of form, marks are given for the set-up: the normal vector correctly calculated (vector product, or two dot products equal to zero) and the substitution of the point to find $d$ . In part c) write the line with its parameter and substitute it into the equation of the plane; the result must be checked on the plane.

Common mistakes

Summary

Line in $R^{3}$ : $r = a + λ d$ ; parametric $x = x_{0} + λ d_{1}$ , $y = y_{0} + λ d_{2}$ , $z = z_{0} + λ d_{3}$ ; symmetric $\frac{x - x _{0}}{d _{1}} = \frac{y - y _{0}}{d _{2}} = \frac{z - z _{0}}{d _{3}}$ . Through two points: direction $A B = B - A$ .
A point lies on the line if the same $λ$ satisfies all three components.
Plane: $r = a + λ u + μ v$ with two non-parallel direction vectors (two parameters). Through three points: $A B$ and $A C$ .
Normal vector: $n = u \times v$ ; $P$ lies on the plane if $A P \cdot n = 0$ .
Cartesian: $a x + b y + cz = d$ with $n = (a, b, c)$ and $d = a \cdot n$ ; this is the same as $r \cdot n = a \cdot n$ . Planes with proportional normal vectors are parallel.
A line in space does not have a single Cartesian equation: it is the intersection of two planes.

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