Equations of a line in space and the position of two lines – Points, Straight Lines and Shapes in 3D (Senior 5 Mathematics) | Ganzaa

Points, Straight Lines and Shapes in 3D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 3D

5 lessons

Lesson 2 of 5

Equations of a line in space and the position of two lines

40 min

Lines that never meet, yet are not parallel

Stand in a classroom and look at the edge where the front wall meets the floor, and the vertical edge in the far corner where two walls meet. The two edges never meet, however far you extend them, yet they are clearly not parallel. In the plane this cannot happen: two lines either meet or are parallel. In space there is a third possibility, skew lines, and telling the three cases apart is the main skill of this lesson.

Definition

The line $L$ through the point $A (x_{1}, y_{1}, z_{1})$ with direction vector $d = (d_1, d_2, d_3) \ne 0$ is the set of points $P$ with $\overrightarrow{AP} \parallel d$ . Its equations are:

vector equation: $r = a + t\,d$ , $t \in \mathbb{R}$ ;
parametric equations: $x = x_{1} + t d_{1}$ , $y = y_{1} + t d_{2}$ , $z = z_{1} + t d_{3}$ ;
Cartesian (symmetric) equations, when $d_1, d_2, d_3 \ne 0$ : $\dfrac{x - x_1}{d_1} = \dfrac{y - y_1}{d_2} = \dfrac{z - z_1}{d_3}$ .

If one component is 0, say $d_{3} = 0$ , write $\frac{x - x_1}{d_1} = \frac{y - y_1}{d_2}$ together with $z = z_{1}$ . The line through two points $A$ and $B$ has direction $\overrightarrow{AB}$ .

Worked example 1: three forms of one line

Write the equations of the line through $A (2, - 1, 3)$ and $B (4, 2, 1)$ .

Solution. $d = \overrightarrow{AB} = (2, 3, -2)$ .

r = (2, -1, 3) + t(2, 3, -2); \qquad x = 2 + 2t,\ y = -1 + 3t,\ z = 3 - 2t; \qquad \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z - 3}{-2} .

Check: $t = 1$ gives $(4, 2, 1) = B$ . ✓

Important

Collinear points. $A, B, C$ are collinear if and only if $\overrightarrow{AB} \parallel \overrightarrow{AC}$ , that is $\overrightarrow{AC} = \lambda\overrightarrow{AB}$ for some $\lambda$ , or equivalently $\overrightarrow{AB} \times \overrightarrow{AC} = 0$ .

Two lines $L_1: r = a_1 + t\,d_1$ and $L_2: r = a_2 + s\,d_2$ are, in space, in exactly one of these positions:

Position	Directions	Common points
coincident	parallel	all (a point of $L_{1}$ lies on $L_{2}$ )
strictly parallel	parallel	none
intersecting	not parallel	exactly one
skew	not parallel	none: the lines are not in one plane

For non-parallel lines, solve $a_1 + t\,d_1 = a_2 + s\,d_2$ : three equations in two unknowns. Solve two of them and test the third. If it holds, the lines intersect; if not, they are skew. (Equivalently, they are coplanar exactly when $[\overrightarrow{A_1A_2}, d_1, d_2] = 0$ .)

Worked example 2: collinear points

Are $A (1, 2, 3)$ , $B (3, 5, 4)$ , $C (7, 11, 6)$ collinear?

Solution. $\overrightarrow{AB} = (2, 3, 1)$ and $\overrightarrow{AC} = (6, 9, 3) = 3\overrightarrow{AB}$ . The vectors are parallel and share the point $A$ , so the three points are collinear, with $C$ three times as far from $A$ as $B$ .

Worked example 3: intersecting or skew?

Let $L_{1} : r = (1, 2, 3) + t (1, - 1, 2)$ , $L_{2} : r = (5, 1, 6) + s (2, 1, - 1)$ and $L_{3} : r = (0, 1, 1) + s (1, 1, 0)$ . Decide how $L_{1}$ lies relative to $L_{2}$ and to $L_{3}$ .

$L_{1}$ and $L_{2}$ . The directions are not parallel. Equate coordinates:

1 + t = 5 + 2s, \qquad 2 - t = 1 + s, \qquad 3 + 2t = 6 - s .

Adding the first two: $3 = 6 + 3 s$ , so $s = - 1$ and $t = 2$ . Third equation: $3 + 4 = 7$ and $6 + 1 = 7$ . ✓ The lines intersect at $(3, 0, 7)$ .

$L_{1}$ and $L_{3}$ . $1 + t = s$ , $2 - t = 1 + s$ , $3 + 2 t = 1$ . The third gives $t = - 1$ , the first $s = 0$ ; the second then reads $3 = 1$ , false. The directions are not parallel and there is no common point: $L_{1}$ and $L_{3}$ are skew.

Important

Angle between two lines. The angle $\theta \in [0^\circ, 90^\circ]$ between lines with directions $d_{1}$ and $d_{2}$ (whether or not they meet) is given by

\cos\theta = \frac{|d_1 \cdot d_2|}{|d_1|\,|d_2|} .

The absolute value picks the acute angle, because a line has no preferred direction. The lines are perpendicular when $d_1 \cdot d_2 = 0$ , even if they are skew.

Worked example 4: the angle between $L_{1}$ and $L_{2}$

For $d_{1} = (1, - 1, 2)$ and $d_{2} = (2, 1, - 1)$ : $d_1 \cdot d_2 = 2 - 1 - 2 = -1$ and $|d_1| = |d_2| = \sqrt{6}$ . So $\cos\theta = \frac{1}{6}$ and $\theta \approx 80.4^\circ$ .

Warning

Use different parameters ( $t$ and $s$ ) for two lines. Using $t$ for both forces the two moving points to arrive at the same time, and you may wrongly conclude "skew".
Test all three equations. Two equations in $t, s$ always have a solution when the directions are not parallel; only the third decides.
Parallel directions need not mean the same line. Check whether one point of $L_{1}$ lies on $L_{2}$ .

Tip

Reading a Cartesian equation: the numbers under the fractions are the direction; the numbers subtracted on top, with their signs changed, are a point. $\frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z - 3}{-2}$ passes through $(2, - 1, 3)$ .

MIT OpenCourseWare: Lecture 5 of 18.02 Multivariable Calculus, Parametric equations for lines and curves, with Denis Auroux (about 51 min). The first part writes lines in space with a parameter t and finds where a line meets a plane.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Write the vector, parametric and Cartesian equations of the line through $(3, - 2, 1)$ parallel to $(2, 1, - 4)$ .
Find the equations of the line through $A (1, 0, 4)$ and $B (3, - 2, 5)$ . Does $C (7, - 6, 7)$ lie on it?
Are $P (2, 1, - 1)$ , $Q (4, 4, 1)$ , $R (8, 10, 6)$ collinear?
Show that $L_{1} : r = (2, 0, 1) + t (1, 1, - 1)$ and $L_{2} : r = (3, 3, - 2) + s (0, 1, - 1)$ intersect, and find the point.
Show that $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{-1}$ and $\frac{x - 3}{4} = \frac{y - 2}{6} = \frac{z + 1}{-2}$ are the same line.
Find the angle between lines with directions $(1, 1, 0)$ and $(3, 0, 4)$ , and the value of $k$ for which lines with directions $(k, 2, - 1)$ and $(2, k, 4)$ are perpendicular.
Explain, with an example from a cuboid, why "two lines that never meet are parallel" is false in space.

Show answersHide answers

$r = (3, - 2, 1) + t (2, 1, - 4)$ ; $x = 3 + 2 t$ , $y = - 2 + t$ , $z = 1 - 4 t$ ; $\frac{x - 3}{2} = \frac{y + 2}{1} = \frac{z - 1}{-4}$ .
$r = (1, 0, 4) + t (2, - 2, 1)$ . Yes: $\overrightarrow{AC} = (6, -6, 3) = 3\overrightarrow{AB}$ , so $t = 3$ .
$\overrightarrow{PQ} = (2, 3, 2)$ , $\overrightarrow{PR} = (6, 9, 7)$ : $6 = 3 \times 2$ and $9 = 3 \times 3$ but $7 \ne 3 \times 2$ . Not collinear.
$2 + t = 3$ gives $t = 1$ ; $t = 3 + s$ gives $s = - 2$ ; third: $1 - 1 = 0$ and $- 2 + 2 = 0$ . ✓ They meet at $(3, 1, 0)$ .
The directions $(2, 3, - 1)$ and $(4, 6, - 2)$ are parallel, and $(3, 2, - 1)$ satisfies the first equation: $\frac{2}{2} = \frac{3}{3} = \frac{-1}{-1} = 1$ . So the lines coincide.
$\cos\theta = \frac{3}{\sqrt{2} \times 5} \approx 0.424$ , $\theta \approx 64.9^\circ$ . Perpendicular: $2 k + 2 k - 4 = 0$ , $k = 1$ .
In the cuboid of the figure, $A B$ and $D H$ never meet (every point of $A B$ has $x = 3$ , every point of $D H$ has $x = 0$ ), but their directions $(0, 1, 0)$ and $(0, 0, 1)$ are not parallel. They are skew.

Sources and credits

Written by Ganzaa: the explanations, the cuboid diagram (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every intersection and angle was checked by computer.
Video: Lec 5: Parametric equations for lines and curves, MIT 18.02 Multivariable Calculus, Fall 2007, Denis Auroux, MIT OpenCourseWare, embedded from YouTube.

1Lines that never meet, yet are not parallel

Lines that never meet, yet are not parallel