Coordinate system in space

A classroom is $8$ m long, $6$ m wide and $3$ m high. A lamp hangs $2$ m from the front wall, $3$ m from the left wall and $2.5$ m above the floor. A smoke sensor sits in the opposite floor corner, $8$ m and $6$ m from those two walls. How far is the lamp from the sensor?

In the plane, two numbers are enough to fix the position of a point. In space we need three, and those three numbers let us find distances without drawing anything.

The coordinate system $Oxyz$

Exploring: three numbers for one position

Take the floor corner between the front wall and the left wall as the origin $O$ . From $O$ , the three lines where the floor and the two walls meet give three pairwise perpendicular directions: along the left wall, along the front wall, and straight up. To reach the lamp we go $2$ m in the first direction, $3$ m in the second, then $2.5$ m up. The triple $(2, 3, 2.5)$ fixes exactly one position in the room, and the sensor corresponds to $(8, 6, 0)$ .

If $\vec{i}$ , $\vec{j}$ , $\vec{k}$ are the vectors of length $1$ m in those three directions, then by the parallelepiped rule (previous lesson) the vector from $O$ to the lamp is $2\vec{i}+3\vec{j}+2.5\vec{k}$ .

Coordinates of a point and of a vector

So the coordinates of the point $M$ are exactly the coordinates of the vector $\overrightarrow{OM}$ . In particular $\vec{i}=(1, 0, 0)$ , $\vec{j}=(0, 1, 0)$ , $\vec{k}=(0, 0, 1)$ and $\vec{0}=(0, 0, 0)$ .

The cuboid

OABC.O'A'B'C'

in the system

Oxyz

The first example reads coordinates from the cuboid above, whose edges run along the three axes.

Example Coordinates of the vertices of a cuboid

The cuboid $OABC.O'A'B'C'$ has $A$ on the ray $Ox$ , $C$ on the ray $Oy$ , $O'$ on the ray $Oz$ , with $OA=3$ , $OC=4$ , $OO'=2$ . Find the coordinates of the vertices and of $\overrightarrow{OB'}$ .

Solution. Each vertex is reached by moving along edges parallel to the axes. Vertex $B$ needs $3$ along $Ox$ and $4$ along $Oy$ ; the primed vertices are on the top face, $2$ higher:

$\begin{aligned} &O(0, 0, 0),\quad A(3, 0, 0),\quad B(3, 4, 0),\quad C(0, 4, 0),\\ &O'(0, 0, 2),\quad A'(3, 0, 2),\quad B'(3, 4, 2),\quad C'(0, 4, 2). \end{aligned}$

The coordinates of $\overrightarrow{OB'}$ are those of the point $B'$ : $\overrightarrow{OB'}=3\vec{i}+4\vec{j}+2\vec{k}=(3, 4, 2)$ .

Points on the axes and the coordinate planes

From the definition, we can read the position of a point from the coordinates that are $0$ :

Position of $M$	Condition
on the axis $Ox$ / $Oy$ / $Oz$	$y=z=0$ / $z=x=0$ / $x=y=0$
in the plane $(Oxy)$ / $(Oyz)$ / $(Ozx)$	$z=0$ / $x=0$ / $y=0$

The orthogonal projection of $M(x, y, z)$ onto a coordinate plane is found by setting the remaining coordinate to $0$ ; onto an axis, keep only that axis's coordinate. The distance from $M$ to the plane $(Oxy)$ is $|z|$ , because $M$ is exactly $|z|$ away from its projection $(x, y, 0)$ .

Example Projections and distances to coordinate planes

Let $M(2, -3, 5)$ . Find the projections of $M$ onto $(Oxy)$ , $(Oyz)$ , $(Ozx)$ and onto the axis $Oz$ . Find the distance from $M$ to the plane $(Oxy)$ and to the axis $Oz$ .

Solution. For each plane we replace the "missing" coordinate by $0$ ; for the axis $Oz$ we keep only the z-coordinate:

$(Oxy):\ (2, -3, 0),\qquad (Oyz):\ (0, -3, 5),\qquad (Ozx):\ (2, 0, 5),\qquad Oz:\ (0, 0, 5).$

The distance to $(Oxy)$ is $|5|=5$ . Let $H(0, 0, 5)$ be the projection onto $Oz$ ; $MH$ is horizontal and is the diagonal of a rectangle with sides $2$ and $3$ :

$MH=\sqrt{2^2+(-3)^2}=\sqrt{13}.$

So $M$ is $5$ from $(Oxy)$ and $\sqrt{13}\approx3.61$ from the axis $Oz$ .

Coordinates of the vector $\overrightarrow{AB}$

Since $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ , we subtract component by component along $\vec{i}$ , $\vec{j}$ , $\vec{k}$ . For $A(x_A, y_A, z_A)$ and $B(x_B, y_B, z_B)$ :

$\overrightarrow{AB}=(x_B-x_A,\ y_B-y_A,\ z_B-z_A).$

The familiar rule again: end point minus start point. Two vectors are equal if and only if their three corresponding coordinates are equal.

Example Two equal vectors

Let $A(1, 2, -1)$ , $B(3, -1, 2)$ , $C(-2, 0, 4)$ , $D(0, -3, 7)$ . Prove that $\overrightarrow{AB}=\overrightarrow{CD}$ and say which quadrilateral is a parallelogram.

Solution. Subtract start from end:

$\overrightarrow{AB}=(3-1,\ -1-2,\ 2+1)=(2, -3, 3),\qquad \overrightarrow{CD}=(0+2,\ -3-0,\ 7-4)=(2, -3, 3).$

The three coordinates agree, so $\overrightarrow{AB}=\overrightarrow{CD}$ . Also $\overrightarrow{AC}=(-3, -2, 5)$ is not parallel to $\overrightarrow{AB}$ (the ratios $\frac{-3}{2}\neq\frac{-2}{-3}$ ), so $C$ is not on the line $AB$ .

So $ABDC$ is a parallelogram (note the order of the vertices: $\overrightarrow{AB}=\overrightarrow{CD}$ makes $AB$ and $CD$ opposite sides pointing the same way).

Distance, midpoint and centroid

Write $\overrightarrow{AB}=a\vec{i}+b\vec{j}+c\vec{k}$ with $a=x_B-x_A$ , $b=y_B-y_A$ , $c=z_B-z_A$ . When we square, the scalar products $\vec{i}\cdot\vec{j}$ , $\vec{j}\cdot\vec{k}$ , $\vec{k}\cdot\vec{i}$ are $0$ , so $AB^2=a^2+b^2+c^2$ . This is also the diagonal of a cuboid with edges $|a|$ , $|b|$ , $|c|$ .

The last two formulas come from $\overrightarrow{OM}=\frac12(\overrightarrow{OA}+\overrightarrow{OB})$ and $\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$ : we take the mean of each coordinate, just as in the plane.

Example Midpoint, length, centroid

Let $A(1, 0, 2)$ , $B(3, 4, -2)$ , $C(-1, 2, 3)$ . Find the midpoint $M$ of $AB$ , the length $AB$ and the centroid $G$ of triangle $ABC$ .

Solution. Take the mean of each coordinate for $M$ and $G$ :

$M\left(\frac{1+3}{2},\ \frac{0+4}{2},\ \frac{2-2}{2}\right)=M(2, 2, 0),\qquad G\left(\frac{1+3-1}{3},\ \frac{0+4+2}{3},\ \frac{2-2+3}{3}\right)=G(1, 2, 1).$

For the length, $\overrightarrow{AB}=(2, 4, -4)$ :

$AB=\sqrt{4+16+16}=\sqrt{36}=6.$

So $M(2, 2, 0)$ , $AB=6$ and $G(1, 2, 1)$ .

Now we answer the opening question.

Watch out Common mistakes

Wrong coordinates in a projection. Wrong: the projection of $M(2, -3, 5)$ onto $(Oxy)$ is $(0, 0, 5)$ . Why it is wrong: that is the projection onto the axis $Oz$ . Right: onto $(Oxy)$ , keep $x$ , $y$ and set $z=0$ : $(2, -3, 0)$ .
Distance to an axis confused with distance to a plane. Wrong: the distance from $M(2, -3, 5)$ to the axis $Oz$ is $5$ . Right: $5$ is the distance to $(Oxy)$ ; the distance to the axis $Oz$ is $\sqrt{2^2+3^2}=\sqrt{13}$ .
Subtracting the wrong way round. Wrong: $\overrightarrow{AB}=(x_A-x_B,\ \dots)$ . Right: end point minus start point.
Forgetting a coordinate in a distance. Wrong: $AB=\sqrt{2^2+4^2}$ for $\overrightarrow{AB}=(2, 4, -4)$ . Right: add all three squares, $AB=6$ .

Key terms

English	Tiếng Việt
Oxyz coordinate system	hệ trục toạ độ Oxyz
origin	gốc toạ độ
x-axis, y-axis, z-axis	trục hoành, trục tung, trục cao
coordinate plane	mặt phẳng toạ độ
unit vector	vectơ đơn vị
coordinates of a point	toạ độ của điểm
x-, y- and z-coordinate	hoành độ, tung độ, cao độ
orthogonal projection	hình chiếu vuông góc
midpoint	trung điểm
centroid	trọng tâm

Summary

$Oxyz$ : three pairwise perpendicular axes with unit vectors $\vec{i}$ , $\vec{j}$ , $\vec{k}$ of length $1$ , pairwise perpendicular.
$M(x, y, z)\iff\overrightarrow{OM}=x\vec{i}+y\vec{j}+z\vec{k}$ ; $\vec{a}=(x, y, z)\iff\vec{a}=x\vec{i}+y\vec{j}+z\vec{k}$ .
In $(Oxy)$ : $z=0$ ; on $Ox$ : $y=z=0$ . The projection of $(x, y, z)$ onto $(Oxy)$ is $(x, y, 0)$ .
$\overrightarrow{AB}=(x_B-x_A,\ y_B-y_A,\ z_B-z_A)$ ; $AB$ is the square root of the sum of three squares.
Midpoint, centroid: the mean of each coordinate.

Practice

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

Exercise 1

a) Find the coordinates of the point $M$ given $\overrightarrow{OM}=2\vec{i}-3\vec{j}+\vec{k}$ .

b) Find the coordinates of the vector $\vec{a}=-\vec{i}+4\vec{k}$ .

What you need and how to start

What you need: $\overrightarrow{OM}=x\vec{i}+y\vec{j}+z\vec{k}\iff M(x, y, z)$ ; a missing component has coefficient $0$ .

Steps:

Read off the coefficients of $\vec{i}$ , $\vec{j}$ , $\vec{k}$ in turn.
b) The coefficient of $\vec{j}$ is $0$ .

Show solution

a) The coefficients are $2$ , $-3$ , $1$ , so $M(2, -3, 1)$ .

b) $\vec{a}=-1\cdot\vec{i}+0\cdot\vec{j}+4\vec{k}$ , so $\vec{a}=(-1, 0, 4)$ .

Answer. a) $M(2, -3, 1)$ ; b) $\vec{a}=(-1, 0, 4)$ .

Exercise 2

Let $N(4, -1, 3)$ . Find the projections of $N$ onto the planes $(Oxy)$ , $(Oyz)$ , $(Ozx)$ and onto the axis $Oy$ . Find the distance from $N$ to the plane $(Ozx)$ .

What you need and how to start

What you need: projection onto a coordinate plane: set the remaining coordinate to $0$ ; the plane $(Ozx)$ is $y=0$ .

Steps:

Write the three projections onto the three planes.
Onto the axis $Oy$ : keep only the y-coordinate.
The distance to $(Ozx)$ is $|y|$ .

Show solution

$(Oxy)$ : $(4, -1, 0)$ ; $(Oyz)$ : $(0, -1, 3)$ ; $(Ozx)$ : $(4, 0, 3)$ ; axis $Oy$ : $(0, -1, 0)$ . The distance to $(Ozx)$ is $|-1|=1$ .

Answer. $(4, -1, 0)$ , $(0, -1, 3)$ , $(4, 0, 3)$ , $(0, -1, 0)$ ; the distance is $1$ .

Exercise 3

Let $A(2, -1, 3)$ and $B(-1, 3, 3)$ . Find the coordinates of $\overrightarrow{AB}$ and the length $AB$ .

What you need and how to start

What you need: $\overrightarrow{AB}=(x_B-x_A,\ y_B-y_A,\ z_B-z_A)$ ; $AB=\sqrt{a^2+b^2+c^2}$ .

Steps:

Subtract coordinate by coordinate.
Add the three squares and take the square root.

Show solution

$\overrightarrow{AB}=(-3, 4, 0),\qquad AB=\sqrt{9+16+0}=5.$

Answer. $\overrightarrow{AB}=(-3, 4, 0)$ ; $AB=5$ .

Exercise 4

Let $A(2, 1, -3)$ , $B(0, 5, 1)$ , $C(4, -3, 5)$ . Find the midpoint $M$ of $AB$ and the centroid $G$ of triangle $ABC$ .

What you need and how to start

What you need: midpoint and centroid: the mean of each coordinate.

Steps:

Add each coordinate of $A$ and $B$ , then divide by $2$ .
Add each coordinate of $A$ , $B$ , $C$ , then divide by $3$ .

Show solution

$M\left(\frac{2+0}{2},\ \frac{1+5}{2},\ \frac{-3+1}{2}\right)=M(1, 3, -1),\qquad G\left(\frac{6}{3},\ \frac{3}{3},\ \frac{3}{3}\right)=G(2, 1, 1).$

Answer. $M(1, 3, -1)$ ; $G(2, 1, 1)$ .

Exercise 5

Let $A(1, -2, 3)$ and $M(2, 0, 1)$ . Find the point $N$ such that $M$ is the midpoint of $AN$ .

What you need and how to start

What you need: $M$ is the midpoint of $AN$ $\iff$ $x_M=\dfrac{x_A+x_N}{2}$ (and likewise), that is $x_N=2x_M-x_A$ .

Steps:

Double the coordinates of $M$ .
Subtract the coordinates of $A$ .

Show solution

$N(2\cdot2-1,\ 2\cdot0+2,\ 2\cdot1-3)=N(3, 2, -1).$ Check. The midpoint of $A(1, -2, 3)$ and $N(3, 2, -1)$ is $(2, 0, 1)=M$ .

Answer. $N(3, 2, -1)$ .

Exercise 6

A control station is at the origin $O$ of a system $Oxyz$ (unit: metre; $Ox$ points east, $Oy$ north, $Oz$ up). A drone is flying at the point $P(120, -50, 60)$ . How many metres is the drone from the station? (Round to the nearest whole number.)

What you need and how to start

What you need: $OP=\sqrt{x^2+y^2+z^2}$ .

Steps:

Square the three coordinates and add them.
Take the square root and round.

Show solution

$OP=\sqrt{120^2+(-50)^2+60^2}=\sqrt{14400+2500+3600}=\sqrt{20500}\approx143.18.$

Answer. $143$ .

Exercise 7

Let $M(3, -4, 12)$ . Is each claim true or false?

a) The projection of $M$ onto the plane $(Oxy)$ is $(3, -4, 0)$ .

b) The distance from $M$ to the plane $(Oyz)$ is $4$ .

c) $OM=13$ .

d) The distance from $M$ to the axis $Ox$ is $4$ .

What you need and how to start

What you need: the distance to $(Oyz)$ is $|x|$ ; the distance to the axis $Ox$ is $\sqrt{y^2+z^2}$ .

Steps:

Write the projection onto $(Oxy)$ .
Find $OM$ from the three coordinates.
Decide which coordinates each distance uses.

Show solution

a) True. Keep $x$ , $y$ and set $z=0$ .

b) False. The distance to $(Oyz)$ is $|x|=3$ .

c) True. $OM=\sqrt{9+16+144}=\sqrt{169}=13$ .

d) False. The projection onto $Ox$ is $H(3, 0, 0)$ , so $MH=\sqrt{16+144}=\sqrt{160}=4\sqrt{10}$ .

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

Exercise 8

The cube $ABCD.A'B'C'D'$ with edge $4$ is placed in $Oxyz$ so that $A$ is at $O$ , $B(4, 0, 0)$ , $D(0, 4, 0)$ , $A'(0, 0, 4)$ . Find the coordinates of $C'$ , of the centre $I$ of the face $A'B'C'D'$ , and the length $IB$ .

What you need and how to start

What you need: the parallelepiped rule $\overrightarrow{AC'}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}$ ; the centre of a square is the midpoint of a diagonal.

Steps:

Add the three edge vectors to get $C'$ .
$I$ is the midpoint of $A'C'$ .
Find $IB$ with the distance formula.

Show solution

$\overrightarrow{AC'}=(4, 0, 0)+(0, 4, 0)+(0, 0, 4)=(4, 4, 4)$ , so $C'(4, 4, 4)$ . The midpoint of $A'(0, 0, 4)$ and $C'$ is $I(2, 2, 4)$ . Then $IB=\sqrt{(4-2)^2+(0-2)^2+(0-4)^2}=\sqrt{24}=2\sqrt6.$

Answer. $C'(4, 4, 4)$ , $I(2, 2, 4)$ , $IB=2\sqrt6\approx4.90$ .

Exercise 9

(Revisiting Vectors in the coordinate plane.) In the plane $Oxy$ , let $A(1, 2)$ and $B(4, 6)$ .

a) Find $AB$ with the grade-10 formula.

b) Regard $A$ , $B$ as the points $A(1, 2, 0)$ , $B(4, 6, 0)$ of the plane $(Oxy)$ in space. Lift $B$ straight up to $B'(4, 6, 12)$ . Find $AB'$ .

What you need and how to start

What you need: $AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$ in the plane; in space there is an extra $(z_B-z_A)^2$ .

Steps:

a) Find the two coordinate differences.
b) Add the height difference $12$ and notice that $ABB'$ has a right angle at $B$ .

Show solution

a) $AB=\sqrt{3^2+4^2}=5$ .

b) $\overrightarrow{AB'}=(3, 4, 12)$ , so $AB'=\sqrt{9+16+144}=13$ . This is exactly Pythagoras' theorem in triangle $ABB'$ with the right angle at $B$ : $5^2+12^2=13^2$ .

Answer. a) $AB=5$ ; b) $AB'=13$ .

Exercise 10 Challenge

Find the point $M$ on the axis $Oz$ that is equidistant from $A(1, 2, 3)$ and $B(-1, 0, 5)$ .

What you need and how to start

What you need: a point on $Oz$ has the form $M(0, 0, t)$ ; $MA=MB\iff MA^2=MB^2$ .

Steps:

Write $MA^2$ and $MB^2$ in terms of $t$ .
Set them equal; the $t^2$ terms cancel.

Show solution

$MA^2=1+4+(t-3)^2=t^2-6t+14,\qquad MB^2=1+0+(t-5)^2=t^2-10t+26.$ $t^2-6t+14=t^2-10t+26\iff 4t=12\iff t=3.$ Check. $MA=\sqrt{1+4+0}=\sqrt5$ and $MB=\sqrt{1+0+4}=\sqrt5$ .

Answer. $M(0, 0, 3)$ .

The coordinate system $Oxyz$

Exploring: three numbers for one position

Coordinates of a point and of a vector

Points on the axes and the coordinate planes

Coordinates of the vector $\overrightarrow{AB}$

Distance, midpoint and centroid

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $\overrightarrow{AB}=(2, -4, 4)$ .	True	False
b) $AB=6$ .	True	False
c) The centroid of triangle $ABC$ is $G(3, 3, 3)$ .	True	False
d) The point $D$ that makes $ABCD$ a parallelogram is $D(-3, 7, -3)$ .	True	False

Coordinate system in space

The coordinate system OxyzOxyzOxyz

Exploring: three numbers for one position

Coordinates of a point and of a vector

Points on the axes and the coordinate planes

Coordinates of the vector AB→\overrightarrow{AB}AB

Distance, midpoint and centroid

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

The coordinate system $Oxyz$

Coordinates of the vector $\overrightarrow{AB}$