IB Analysis and Approaches HL · Geometry in space · Section 6 of 7

Relative positions and intersections of lines and planes

3 worked examples · By Álvaro López

Key idea

In space, two lines can pass each other without meeting, and three planes can leave a system with no solution even when no two of them are parallel. To decide each case you only need the direction vectors or normal vectors and a small system to solve; the answer is a point, a line, a plane or the empty set. This is standard material for Paper 1 and, with parameters, for the long questions of Paper 2.

Two lines in space

Let $r : r = a_{1} + λ d_{1}$ and $s : r = a_{2} + μ d_{2}$ . Unlike in the plane, there are four possibilities here, because two lines that are not parallel may have no common point at all.

Direction vectors	Common point	Position
proportional	all points	coincident (the same line)
proportional	none	distinct parallel lines
not proportional	one	intersecting (they meet at one point)
not proportional	none	skew (they do not meet)

Two skew lines are neither parallel nor touching: they do not lie in a common plane. This never happens in $R^{2}$ .

The third equation is what makes the difference from the plane: in $R^{2}$ there are always two equations for two unknowns and, if the direction vectors are not proportional, the lines are guaranteed to meet. In $R^{3}$ you have to check it.

Line and plane

A line $r : r = a + λ d$ and a plane $π : a x + b y + cz = e$ have three possible positions: the line intersects the plane at one point, it is parallel to it (no common points), or it is contained in it.

The quickest method is to substitute the parametric coordinates of the line into the equation of the plane. This leaves an equation in $λ$ of the form $p λ = q$ :

The intuition: $p = d \cdot n$ , where $n = (a, b, c)$ is the normal vector of the plane. If $d \cdot n = 0$ , the direction of the line is perpendicular to the normal, that is, parallel to the plane; then the only question left is whether the line starts inside or outside it. In that case, it is enough to check whether the point $a$ satisfies the equation of the plane.

Two planes

Two planes $π_{1} : n_{1} \cdot r = e_{1}$ and $π_{2} : n_{2} \cdot r = e_{2}$ are compared through their normal vectors:

$n_{1}$ and $n_{2}$ not proportional: the planes intersect in a line.
$n_{1}$ and $n_{2}$ proportional: the planes are parallel or coincident. If the proportionality also extends to the constant term ( $e_{2} = k e_{1}$ with the same constant $k$ ), they are coincident; otherwise they are distinct parallel planes.

Two planes cannot meet at a single point: they either share a whole line, a whole plane, or nothing at all.

Line of intersection of two planes

When the planes meet in a line, that line must be written with a point and a direction vector. There are two ways of doing this.

Three planes and linear systems

Three planes $π_{i} : a_{i} x + b_{i} y + c_{i} z = e_{i}$ form a system of three equations in three unknowns, exactly the kind studied in linear equations and systems in topic 7. Solving it by Gaussian elimination gives the relative position: one common point (a unique solution), a common line or plane (infinitely many solutions), or no common point at all (no solution).

Writing $M$ for the coefficient matrix (the normals as rows) and $M^{*}$ for the augmented matrix (with one more column of constant terms), the cases are:

Rank of $M$	Rank of $M^{*}$	System	Position of the three planes
3	3	consistent, unique solution	intersect at one point
2	2	consistent, one free variable	share a line (three distinct planes through one line, or two coincident planes and a third that cuts them)
2	3	inconsistent	no common point: they form a "prism" (each pair meets in parallel lines), or two are parallel and the third cuts them
1	1	consistent, two free variables	the three planes coincide
1	2	inconsistent	parallel planes (or two coincident planes and one parallel to them)

Within the cases with rank 2, which specific situation applies depends on whether any pair of normals is proportional. If no pair of normals is proportional and the system is inconsistent, you have the triangular prism; if one pair is proportional, you have two parallel planes cut by a third.

Worked examples

Example 1 Basic

A line and three planes

Consider the line $r : r = 120 + t 21 - 1$ and the planes

$π_{1} : 2 x - y + 3 z = 0$ , $π_{2} : 2 x - y + 3 z = 5$ , $π_{3} : x + 2 y - z = 7$ .

a) Study the position of $r$ relative to $π_{1}$ (answer "contained", "parallel" or "intersecting").

b) Study the position of $r$ relative to $π_{2}$ (using the same three words).

c) Find the point at which $r$ meets $π_{3}$ .

Solution

The coordinates of a point on the line are $(1 + 2 t, 2 + t, - t)$ .

a) Substituting into $π_{1}$ :

$2 (1 + 2 t) - (2 + t) + 3 (- t) = 0 ⟹ 2 + 4 t - 2 - t - 3 t = 0 ⟹ 0 = 0$

Every $t$ works: the line is contained in $π_{1}$ .

b) On $π_{2}$ we get the same left-hand side, which is $0$ for every $t$ , while the right-hand side is $5$ : we get $0 = 5$ . There is no solution: the line is parallel to $π_{2}$ .

c) Substituting into $π_{3}$ :

$(1 + 2 t) + 2 (2 + t) - (- t) = 7 ⟹ 5 + 5 t = 7 ⟹ t = \frac{2}{5}$

The point is $(1 + \frac{4}{5}, 2 + \frac{2}{5}, - \frac{2}{5}) = (\frac{9}{5}, \frac{12}{5}, - \frac{2}{5})$ .

Example 2 Medium

Lines that intersect or are skew depending on a parameter

Consider the lines

$r : r = 102 + λ 12 - 1$ and $s_{k} : r = 03 k + μ 2 - 1 1$ , where $k \in R$ .

a) Find the value of $k$ for which $r$ and $s_{k}$ intersect.

b) For that value of $k$ , find the point of intersection.

c) For $k = 3$ , state the relative position of the lines (answer "intersecting", "parallel", "coincident" or "skew").

Solution

The direction vectors $(1, 2, - 1)$ and $(2, - 1, 1)$ are not proportional (the ratios of their first and second components are $\frac{1}{2}$ and $- 2$ , which differ), so the lines are never parallel or coincident: they either intersect or are skew.

a) Equate them using different parameters:

$⎩ ⎨ ⎧ 1 + λ 2 λ 2 - λ = 2 μ = 3 - μ = k + μ$

From the first two equations: $λ = 2 μ - 1$ , and substituting into the second, $4 μ - 2 = 3 - μ$ , so $μ = 1$ and $λ = 1$ . The third equation requires $2 - 1 = k + 1$ , that is, $k = 0$ .

b) With $λ = 1$ in $r$ : $(1 + 1, 2, 2 - 1) = (2, 2, 1)$ . Check with $s_{0}$ and $μ = 1$ : $(2, 3 - 1, 0 + 1) = (2, 2, 1)$ .

c) For $k = 3$ , the values $λ = 1$ and $μ = 1$ (which do not depend on $k$ , since they come from the first two equations) give $1 = 3 + 1 = 4$ in the third equation, which is false. There is no common point and the direction vectors are not proportional: the lines are skew.

Example 3 Hard

Three planes with parameters and their common line

Consider the planes

$π_{1} : x + y + z = 1$ , $π_{2} : x + 2 y + 3 z = 2$ , $π_{3} : 2 x + 3 y + a z = b$ , with $a, b \in R$ .

a) Find a parametric equation of the line of intersection of $π_{1}$ and $π_{2}$ , using the parameter $t$ so that the $x$ -coordinate equals $t$ . Express the answer as a tuple $(x, y, z)$ in terms of $t$ .

b) Discuss, according to the values of $a$ and $b$ , the position of the three planes. Answer with the cases "a=4 and b=3", "a=4 and b≠3" and "a≠4", giving as values "line" (the three planes share a line), "no common point" or "point" (they meet at a single point).

c) For $a = 1$ and $b = 0$ , find the common point of the three planes.

Solution

a) Subtracting $π_{1}$ from $π_{2}$ gives $y + 2 z = 1$ , so $y = 1 - 2 z$ . In $π_{1}$ : $x = 1 - y - z = 1 - (1 - 2 z) - z = z$ . The planes do not have proportional normals, $(1, 1, 1)$ and $(1, 2, 3)$ , so they intersect in a line. With $x = t$ we get $z = t$ and $y = 1 - 2 t$ :

$(x, y, z) = (t, 1 - 2 t, t)$

Check with the vector product: $(1, 1, 1) \times (1, 2, 3) = (1, - 2, 1)$ , which is the direction vector obtained.

b) The points common to the three planes are the points of that line which also satisfy $π_{3}$ . Substitute $(t, 1 - 2 t, t)$ into $2 x + 3 y + a z = b$ :

$2 t + 3 (1 - 2 t) + a t = b ⟹ (a - 4) t = b - 3$

If $a \neq = 4$ : there is a single $t$ , and the three planes meet at one point.
If $a = 4$ and $b = 3$ : we get $0 = 0$ , so any $t$ works. The three planes share the line.
If $a = 4$ and $b \neq = 3$ : we get $0 = b - 3 \neq = 0$ , which has no solution. No common point: since the normals are not proportional in pairs, the planes form a prism.

c) With $a = 1$ and $b = 0$ : $- 3 t = - 3$ , so $t = 1$ and the point is $(1, 1 - 2, 1) = (1, - 1, 1)$ . Check in $π_{3}$ : $2 - 3 + 1 = 0$ .

In the exam: both the method (eliminating variables or using the line of intersection) and the separation of the cases with the exact conditions on $a$ and $b$ earn marks. Give a geometric description of each case (point, line, prism) and check the numerical result in the third equation.

Common mistakes

Summary

Two lines: proportional direction vectors (parallel or coincident) or not (intersecting or skew). With direction vectors that are not proportional, equate with $λ$ and $μ$ ; the third equation decides between intersecting and skew.
Line and plane: substitute the line into the plane. A value of $t$ gives the intersection point; $0 = q \neq = 0$ means parallel; $0 = 0$ means contained.
Two planes: normals not proportional = line of intersection (direction $n_{1} \times n_{2}$ ); proportional = parallel or coincident.
Three planes = $3 \times 3$ system: $rg (M) = rg (M^{*}) = 3$ , one point; $rg (M) = rg (M^{*}) = 2$ , a line; $rg (M) < rg (M^{*})$ , no common point (prism, or parallel planes).
With parameters: reduce to triangular form, locate the values where the last coefficient vanishes, study each case, and check the result in the leftover equation.

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