The equation of a line in space

A drone takes off from the point $M_0(1, 2, 0)$ in a school yard (unit: metre). It flies in a straight line, moving by the vector $\vec{u}=(2, 1, 2)$ every second. High above, a checkpoint ring is placed at $N(9, 6, 8)$ . Does the drone fly through the ring, and if so, after how many seconds?

As in the plane, a line in space is fixed by one point and one direction. The only difference is that there are now three coordinates.

Parametric equations of a line

Exploring: the drone's positions

After $t$ seconds the drone is at the point $M$ with $\overrightarrow{M_0M}=t\vec{u}$ . We record a few positions:

$t$ (seconds)	$0$	$1$	$2$	$3$
position	$(1, 2, 0)$	$(3, 3, 2)$	$(5, 4, 4)$	$(7, 5, 6)$

The position after $t$ seconds is $(1+2t, 2+t, 2t)$ . To reach $N(9, 6, 8)$ we need $1+2t=9$ , $2+t=6$ , $2t=8$ . All three give $t=4$ , so the drone passes through the ring after $4$ seconds. The distance flown is $4|\vec{u}|=4\cdot3=12$ m.

The line through

M_0

with direction vector

\vec{u}

Definition Direction vector, parametric equations

A vector $\vec{u}\neq\vec{0}$ is a direction vector of the line $\Delta$ if its line is parallel to $\Delta$ or coincides with it. Every vector $k\vec{u}$ ( $k\neq0$ ) is also a direction vector.
The line $\Delta$ through $M_0(x_0, y_0, z_0)$ with direction vector $\vec{u}=(a, b, c)$ has parametric equations $\begin{cases}x=x_0+at\\ y=y_0+bt\\ z=z_0+ct\end{cases}\quad(t\in\mathbb{R}).$
If $a$ , $b$ , $c$ are all non-zero, eliminating $t$ gives the Cartesian (symmetric) equations $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$

Why is it true? A point $M(x, y, z)$ lies on $\Delta$ exactly when $\overrightarrow{M_0M}$ is parallel to $\vec{u}$ , that is, when there is a number $t$ with $\overrightarrow{M_0M}=t\vec{u}$ . Coordinate by coordinate: $x-x_0=at$ , $y-y_0=bt$ , $z-z_0=ct$ . When $abc\neq0$ , all three fractions equal $t$ .

Example A point and a direction vector

The line $\Delta$ passes through $M(1, -2, 3)$ and has direction vector $\vec{u}=(2, 0, -1)$ . Write parametric equations of $\Delta$ , find the point for $t=2$ , and decide whether $A(-3, -2, 5)$ lies on $\Delta$ .

Solution. Substitute the point and the direction vector into the formula:

$\Delta: \begin{cases}x=1+2t\\ y=-2\\ z=3-t.\end{cases}$

For $t=2$ we get the point $(5, -2, 1)$ . For $A$ , the equation $1+2t=-3$ gives $t=-2$ ; then $z=3+2=5$ and $y=-2$ both match.

So $A$ lies on $\Delta$ (for $t=-2$ ). This line has no Cartesian equations, because the second coordinate of $\vec{u}$ is $0$ .

Given two points, we take the vector joining them as a direction vector. When a line is perpendicular to a plane, the normal vector of the plane is a direction vector of the line.

Example Through two points; perpendicular to a plane

a) Write parametric and Cartesian equations of the line $AB$ with $A(1, 2, -1)$ , $B(3, -1, 1)$ .

b) Write parametric equations of the line $d$ through $M(2, 0, -1)$ perpendicular to $(P): x-2y+2z+3=0$ .

Solution. a) Take $\vec{u}=\overrightarrow{AB}=(2, -3, 2)$ and the point $A$ :

$AB: \begin{cases}x=1+2t\\ y=2-3t\\ z=-1+2t\end{cases}\qquad\text{and}\qquad \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{2}.$

b) Since $d\perp(P)$ , the line $d$ has direction vector $\vec{n}_P=(1, -2, 2)$ :

$d: \begin{cases}x=2+t\\ y=-2t\\ z=-1+2t.\end{cases}$

So the equations are as above.

Two lines in space

In space, two lines have four possibilities. Unlike in the plane, two lines can fail to meet without being parallel: such lines are called skew.

Example Intersecting or skew

Let $d_1: \begin{cases}x=1+t\\ y=2t\\ z=2-t\end{cases}$ , $d_2: \begin{cases}x=1+s\\ y=3-s\\ z=-1+2s\end{cases}$ and $d_3: \begin{cases}x=1+s\\ y=3-s\\ z=2s.\end{cases}$ Decide the position of $d_1$ and $d_2$ , and of $d_1$ and $d_3$ .

Solution. The direction vectors $\vec{u}_1=(1, 2, -1)$ and $(1, -1, 2)$ are not parallel (the first coordinates are equal but the second are not). We use two different parameters $t$ , $s$ and solve. For $d_1$ , $d_2$ :

$\begin{cases}1+t=1+s\\ 2t=3-s\\ 2-t=-1+2s\end{cases}\iff\begin{cases}t=s\\ 3t=3\\ 3=3s\end{cases}\iff t=s=1.$

The system has a solution, so $d_1$ meets $d_2$ at the point for $t=1$ , that is $(2, 2, 1)$ .

For $d_1$ , $d_3$ : the first two equations still give $t=s=1$ , but the third needs $2-1=2\cdot1$ , that is $1=2$ , which is impossible.

So $d_1$ meets $d_2$ at $(2, 2, 1)$ , while $d_1$ and $d_3$ are skew.

A line and a plane

To decide how a line $d$ meets a plane $(P)$ , we substitute the parametric equations of $d$ into the equation of $(P)$ and get a linear equation in $t$ .

Example One line and three planes

Let $d: \begin{cases}x=1+2t\\ y=-1+t\\ z=2-t.\end{cases}$ Decide the position of $d$ relative to $(P): x+y+z-6=0$ , $(Q): x-y+z-4=0$ and $(R): x-y+z+1=0$ .

Solution. Substitute $d$ into each equation and simplify:

$\begin{aligned} (P)&: (1+2t)+(-1+t)+(2-t)-6=2t-4=0\iff t=2;\\ (Q)&: (1+2t)-(-1+t)+(2-t)-4=0\ \text{for every}\ t;\\ (R)&: (1+2t)-(-1+t)+(2-t)+1=5=0\ \text{(impossible)}. \end{aligned}$

So $d$ meets $(P)$ at the point for $t=2$ , that is $(5, 1, 0)$ ; $d$ lies in $(Q)$ ; and $d\parallel(R)$ .

Orthogonal projection of a point

The projection $H$ of a point $M$ on a plane $(P)$ is where $(P)$ meets the line through $M$ perpendicular to $(P)$ . The projection $H$ of $M$ on a line $d$ is the point $H\in d$ with $\overrightarrow{MH}\perp\vec{u}$ .

The projection

H

M

on the plane

(P)

Example The point of the hillside nearest the drone

A hillside is the plane $(P): x+2y+2z-5=0$ (unit: metre), and a drone is at $M(4, 5, 9)$ . Find the point $H$ of the hillside nearest the drone and the distance $MH$ .

Solution. The nearest point is the projection of $M$ on $(P)$ . The line through $M$ perpendicular to $(P)$ has direction vector $\vec{n}=(1, 2, 2)$ : $x=4+t$ , $y=5+2t$ , $z=9+2t$ . Substitute into $(P)$ :

$(4+t)+2(5+2t)+2(9+2t)-5=27+9t=0\iff t=-3.$

Then $H(1, -1, 3)$ and $\overrightarrow{HM}=(3, 6, 6)$ , so $MH=\sqrt{9+36+36}=9$ .

So the point nearest the drone is $H(1, -1, 3)$ , $9$ m away.

Check. The distance formula: $\dfrac{|4+10+18-5|}{3}=9$ .

The last example has more steps, so we follow the four stages.

Example The shortest rope to a pipe

A straight water pipe lies along the line $d: x=1+t$ , $y=-1+2t$ , $z=2t$ (unit: metre). A rope as short as possible must run from a hook at $M(0, 0, 4)$ to the pipe. Find the point $H$ of the pipe where the rope is tied and the length of the rope.

Solution.

Understand. The shortest rope is the perpendicular from $M$ to $d$ , so $H$ is the projection of $M$ on $d$ .

Plan. Write $H$ in terms of $t$ , then use the condition $\overrightarrow{MH}\cdot\vec{u}=0$ with $\vec{u}=(1, 2, 2)$ .

Solve. $H(1+t, -1+2t, 2t)$ , so $\overrightarrow{MH}=(1+t, -1+2t, 2t-4)$ . Then:

$\overrightarrow{MH}\cdot\vec{u}=(1+t)+2(-1+2t)+2(2t-4)=9t-9=0\iff t=1.$

So $H(2, 1, 2)$ , $\overrightarrow{MH}=(2, 1, -2)$ and $MH=\sqrt{4+1+4}=3$ . The rope is $3$ m long.

Look back. A single perpendicularity condition is enough to find $t$ . The length $MH$ is the distance from $M$ to the line $d$ .

Watch out Common mistakes

Using one parameter for both lines. Wrong: solving for $d_1$ , $d_2$ with the system $1+t=1+t$ , $2t=3-t$ , ... Why it is wrong: each line has its own parameter; a common point may correspond to different values of $t$ and $s$ . Right: use $t$ for $d_1$ and $s$ for $d_2$ .
Concluding "parallel" just because the lines do not meet. Wrong: $d_1$ and $d_3$ have no common point, so they are parallel. Why it is wrong: in space they can also be skew. Right: parallel lines also need $\vec{u}_1$ , $\vec{u}_2$ to be parallel.
Writing Cartesian equations when a coordinate is $0$ . Wrong: $\Delta$ from Example 1 written as $\dfrac{x-1}{2}=\dfrac{y+2}{0}=\dfrac{z-3}{-1}$ . Why it is wrong: we cannot divide by $0$ . Right: use the parametric equations.
Using the normal vector as the direction of a line parallel to a plane. Wrong: a line parallel to $(P)$ has direction vector $\vec{n}_P$ . Why it is wrong: $\vec{n}_P$ is perpendicular to $(P)$ . Right: $\vec{n}_P$ is a direction vector of a line perpendicular to $(P)$ .

Key terms

English	Tiếng Việt
line	đường thẳng
direction vector	vectơ chỉ phương
parametric equations	phương trình tham số
parameter	tham số
Cartesian (symmetric) equations	phương trình chính tắc
intersecting lines	hai đường thẳng cắt nhau
skew lines	hai đường thẳng chéo nhau
point of intersection	giao điểm
orthogonal projection	hình chiếu vuông góc

Summary

The line through $M_0(x_0, y_0, z_0)$ with direction vector $\vec{u}=(a, b, c)$ : $x=x_0+at$ , $y=y_0+bt$ , $z=z_0+ct$ .
When $abc\neq0$ : $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$ .
Through two points $A$ , $B$ : $\vec{u}=\overrightarrow{AB}$ . Perpendicular to $(P)$ : $\vec{u}=\vec{n}_P$ .
Two lines: if $\vec{u}_1$ , $\vec{u}_2$ are parallel, the lines are parallel or coincide; if not, solve for $t$ , $s$ : a solution means they intersect, no solution means they are skew.
A line and a plane: substituting gives an equation in $t$ ; one solution means they meet, no solution means parallel, every $t$ means the line lies in the plane.
Projection on $(P)$ : where $(P)$ meets the line through $M$ with direction $\vec{n}$ . Projection on $d$ : $H\in d$ with $\overrightarrow{MH}\cdot\vec{u}=0$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Write parametric equations of the line through $A(2, -1, 3)$ with direction vector $\vec{u}=(1, 4, -2)$ . Find the point for $t=-1$ and check whether $B(4, 7, -1)$ lies on the line.

What you need and how to start

What you need: $x=x_0+at$ , $y=y_0+bt$ , $z=z_0+ct$ .

Steps:

Write the parametric equations.
Substitute $t=-1$ .
Find $t$ from the x-coordinate of $B$ , then check the other two coordinates.

Show solution

$x=2+t$ , $y=-1+4t$ , $z=3-2t$ . For $t=-1$ : $(1, -5, 5)$ . For $B$ : $2+t=4$ gives $t=2$ ; then $y=7$ and $z=-1$ , which match.

Answer. $(1, -5, 5)$ ; $B$ lies on the line (for $t=2$ ).

Exercise 2

Write parametric and Cartesian equations of the line through $A(1, 0, -2)$ and $B(3, 2, -1)$ .

What you need and how to start

What you need: $\vec{u}=\overrightarrow{AB}$ .

Steps:

Find $\overrightarrow{AB}$ .
Write both forms using the point $A$ .

Show solution

$\overrightarrow{AB}=(2, 2, 1)$ . $\begin{cases}x=1+2t\\ y=2t\\ z=-2+t\end{cases}\qquad\frac{x-1}{2}=\frac{y}{2}=\frac{z+2}{1}.$

Answer. As above.

Exercise 3

Write the equation of the line $d$ through $M(1, -1, 2)$ perpendicular to $(P): 2x-y+3z+5=0$ . Find the point where $d$ meets $(P)$ .

What you need and how to start

What you need: if $d\perp(P)$ then $\vec{u}_d=\vec{n}_P$ ; for the intersection, substitute the parametric equations into $(P)$ .

Steps:

Write $d$ with $\vec{u}=(2, -1, 3)$ .
Substitute into $(P)$ and solve for $t$ .

Show solution

$d: x=1+2t$ , $y=-1-t$ , $z=2+3t$ . Substituting into $(P)$ : $2(1+2t)-(-1-t)+3(2+3t)+5=14+14t=0\iff t=-1.$ The point of intersection is $(-1, 0, -1)$ .

Answer. $(-1, 0, -1)$ .

Exercise 4

Write parametric and Cartesian equations of the line through $M(0, 1, -1)$ parallel to $\Delta: \dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{3}$ .

What you need and how to start

What you need: parallel lines share a direction vector; read $\vec{u}$ from the denominators.

Steps:

Read off $\vec{u}=(2, -1, 3)$ .
Write the equations through $M$ .

Show solution

$\begin{cases}x=2t\\ y=1-t\\ z=-1+3t\end{cases}\qquad\frac{x}{2}=\frac{y-1}{-1}=\frac{z+1}{3}.$ $M$ does not lie on $\Delta$ (because $\dfrac{0-1}{2}\neq\dfrac{1+3}{-1}$ ), so this line is parallel to $\Delta$ , not the same line.

Answer. As above.

Exercise 5

Let $d_1: \begin{cases}x=1+2t\\ y=-t\\ z=3+t\end{cases}$ and $d_2: \dfrac{x-3}{-4}=\dfrac{y+1}{2}=\dfrac{z-4}{-2}$ . Is each claim true or false?

a) $\vec{u}_1=(2, -1, 1)$ is a direction vector of $d_1$ .

b) The point $M(3, -1, 4)$ lies on $d_1$ .

c) $d_1$ is parallel to $d_2$ .

d) $d_1$ is perpendicular to the plane $(P): 2x-y+z=0$ .

What you need and how to start

What you need: if the direction vectors are parallel, the lines are parallel or coincide; check one more point.

Steps:

Read off $\vec{u}_1$ , $\vec{u}_2$ .
Substitute $M$ into $d_1$ (find $t$ ).
Compare $\vec{u}_1$ with $\vec{n}_P$ .

Show solution

a) True. The coefficients of $t$ are $2$ , $-1$ , $1$ .

b) True. $t=1$ gives $(3, -1, 4)$ .

c) False. $\vec{u}_2=(-4, 2, -2)=-2\vec{u}_1$ , and $M(3, -1, 4)$ also lies on $d_2$ , so the two lines coincide.

d) True. $\vec{n}_P=(2, -1, 1)=\vec{u}_1$ .

Answer. a) True; b) True; c) False; d) True.

Exercise 6

An aeroplane takes off from the point $O(0, 0, 0)$ and flies in a straight line, moving by the vector $\vec{v}=(80, 60, 10)$ every second (unit: metre). When it reaches a height of $1500$ m, how many kilometres from $O$ is the point on the ground directly below it?

What you need and how to start

What you need: the position after $t$ seconds is $(80t, 60t, 10t)$ .

Steps:

Solve $10t=1500$ .
The point on the ground is $(80t, 60t, 0)$ ; find its distance from $O$ .

Show solution

$t=150$ seconds. The point on the ground is $(12000, 9000, 0)$ , at distance $\sqrt{12000^2+9000^2}=15000$ m from $O$ .

Answer. $15$ .

Exercise 7

Let $d: x=2+t$ , $y=1-t$ , $z=3+2t$ . Decide the position of $d$ relative to $(P_1): x+y-1=0$ , $(P_2): x+3y+z-8=0$ , $(P_3): x-y+z-6=0$ .

What you need and how to start

What you need: substitute the parametric equations into the equation of the plane.

Steps:

Substitute into each plane and simplify in $t$ .
Count the solutions.

Show solution

$(P_1)$ : $(2+t)+(1-t)-1=2=0$ , impossible, so $d\parallel(P_1)$ . $(P_2)$ : $(2+t)+3(1-t)+(3+2t)-8=0$ for every $t$ , so $d$ lies in $(P_2)$ . $(P_3)$ : $(2+t)-(1-t)+(3+2t)-6=4t-2=0$ , $t=\dfrac12$ , and the point of intersection is $\left(\dfrac52, \dfrac12, 4\right)$ .

Answer. Parallel to $(P_1)$ ; lies in $(P_2)$ ; meets $(P_3)$ at $\left(\dfrac52, \dfrac12, 4\right)$ .

Exercise 8

Decide the position of the lines $d_1: x=1+t$ , $y=1-t$ , $z=2t$ and $d_2: x=1+2s$ , $y=-2+s$ , $z=5-s$ . If they intersect, find the point of intersection.

What you need and how to start

What you need: if the direction vectors are not parallel, solve the system in $t$ , $s$ .

Steps:

Compare $(1, -1, 2)$ and $(2, 1, -1)$ .
Solve the first two equations, then test the third.

Show solution

The vectors are not parallel. From $1+t=1+2s$ and $1-t=-2+s$ : $t=2s$ , $1-2s=-2+s$ , so $s=1$ , $t=2$ . The third equation: $2\cdot2=5-1$ , which holds.

Answer. They intersect at $(3, -1, 4)$ .

Exercise 9

(Revisiting Operations on vectors in coordinates.) In the previous lesson we proved that $A(1, 2, -1)$ , $B(3, 1, 1)$ , $C(7, -1, 5)$ are collinear using parallel vectors. Write parametric equations of the line $AB$ and give the value of $t$ for the point $C$ .

What you need and how to start

What you need: $\vec{u}=\overrightarrow{AB}$ ; $C$ lies on $AB$ when one value of $t$ works for all three coordinates.

Steps:

$\overrightarrow{AB}=(2, -1, 2)$ .
Solve $1+2t=7$ and check.

Show solution

$AB: x=1+2t$ , $y=2-t$ , $z=-1+2t$ . From $1+2t=7$ we get $t=3$ ; then $y=-1$ and $z=5$ , which match $C$ .

Answer. $t=3$ , which agrees with $\overrightarrow{AC}=3\overrightarrow{AB}$ from the previous lesson.

Exercise 10 Challenge

A laser beam starts at $A(1, 2, 3)$ , reflects off a flat mirror lying in the plane $(P): x+2y+2z-2=0$ , and then reaches $B(3, 2, 2)$ . Find the point $M$ of the mirror where it reflects. (Hint: the path $A\to M\to B$ is shortest when $M$ lies on the segment $A'B$ , where $A'$ is the reflection of $A$ in $(P)$ .)

What you need and how to start

What you need: the projection $H$ of $A$ on $(P)$ ; $H$ is the midpoint of $AA'$ .

Steps:

Find $H$ , then $A'=2H-A$ (in coordinates).
Use the line $A'B$ and find where it meets $(P)$ .

Show solution

The line through $A$ perpendicular to $(P)$ : $x=1+t$ , $y=2+2t$ , $z=3+2t$ . Substituting into $(P)$ : $9+9t=0$ , $t=-1$ , so $H(0, 0, 1)$ and $A'(-1, -2, -1)$ . Substituting $A'$ and $B$ into the left-hand side of $(P)$ gives $-9$ and $9$ , so $A'$ and $B$ are on opposite sides at equal distances from $(P)$ ; therefore $M$ is the midpoint of $A'B$ : $M\left(1, 0, \tfrac12\right),\qquad 1+0+2\cdot\tfrac12-2=0.$ The beam travels a distance $A'B=\sqrt{16+16+9}=\sqrt{41}\approx6.40$ .

Answer. $M\left(1, 0, \dfrac12\right)$ .

Parametric equations of a line

Exploring: the drone's positions

Two lines in space

A line and a plane

Orthogonal projection of a point

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $\vec{u}=(1, -1, 2)$ is a direction vector of $d$ .	True	False
b) $d$ is parallel to $(P)$ .	True	False
c) $d$ meets $(P)$ at the point $(2, 1, 1)$ .	True	False
d) $d$ is perpendicular to the plane $(Q): x-y+2z+7=0$ .	True	False