267 1 3 d coordinate system-n

3D Coordinate Systems
To set up a 3D coordinate system, we add a z-axis
perpendicularly to both the x&y axes.

perpendicularly to both the x&y axes. There are two
ways to add the z-axis.

x
y
z+
Right-hand system

x
y
z+
z+
Right-hand system Left-hand system

x
y
z+
x
y
z+

x
y
z+
x
y
z+
http://www.scientificamerican
.com/article.cfm?id=why-do-
some-chemicals-hav

We use the right-hand system in math/physical
science. The left hand system is used in computer
graphics.
x
y
z+
x
y
z+
some-chemicals-hav

We use the right-hand system in math/physical
science. The left hand system is used in computer
graphics. The real line which is 1D is abbreviated as R.
We write R2 for the 2D xy-plane and R3 for the 3D
rectangular xyz–space.
x
y
z+
x
y
z+
some-chemicals-hav

Every point in space may be addressed by
three numbers, an ordered triple, (x, y, z).
y
z+
x

Given (x, y, z), to find the
location it pinpoints:
y
z+
x

1. find (x, y) in the x&y
coordinate plane,
y
z+
x

coordinate plane,
2. then the z, which specifies
the location above or below
(x, y). Specifically,
z > 0  above
z < 0  below
y
z+
x

coordinate plane,
z > 0  above
z < 0  below
Example A. Draw A(2, 0 , 0),
B(1, 3, 4), C(–2, 1, –3)
y
z+
x

y
z+
B(1, 3, 4), C(–2, 1, –3)
xA(2, 0, 0)
coordinate plane,
z > 0  above
z < 0  below

y
z+
B(1, 3, 4), C(–2, 1, –3)
xA(2, 0, 0)
(1, 3, 0)
coordinate plane,
z > 0  above
z < 0  below

y
z+
B(1, 3, 4), C(–2, 1, –3)
xA(2, 0, 0)
(1, 3, 0)
B(1, 3, 4)
4
coordinate plane,
z > 0  above
z < 0  below

y
z+
B(1, 3, 4), C(–2, 1, –3)
xA(2, 0, 0)
(–2, 1, 0)
4
coordinate plane,
z > 0  above
z < 0  below
(1, 3, 0)
B(1, 3, 4)

y
z+
B(1, 3, 4), C(–2, 1, –3)
xA(2, 0, 0)
(–2, 1, 0)
4
C(–2, 1, –3)
coordinate plane,
z > 0  above
z < 0  below
(1, 3, 0)
B(1, 3, 4)

3D systems may be drawn from different eye
positions:

positions:
x
y
z+
Eye:(1, –1, 1)
Object appears different in its direction and its
orientation depending on the eye position.

positions:
x
y
z+
x
y
z+
Eye:(1, –1, 1) Eye:(1, 1, 1)

positions:
x
y
z+
x
y
z+
x
y
z+
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)

positions:
x
y
z+
x
y
z+
x
y
z+
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)
x
y
z
There are three coordinate planes:

positions:
x
y
z+
x
y
z+
x
y
z+
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)
the xy-plane = {(x, y, 0)}
x
y
z
xy-plane

positions:
x
y
z+
x
y
z+
x
y
z+
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)
the yz-plane = {(0, y, z)}
x
y
z
xy-plane
yz-plane

positions:
x
y
z+
x
y
z+
x
y
z+
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)
the yz-plane = {(0, y, z)}
the xz-plane = {(x, 0, z)} x
y
z
xy-plane
xz-plane
yz-plane

Basic 3D Graphs

Basic 3D Graphs
In general, the graph of an equation in the variables
x, y, and z in R3 is a surface is 3D space.

Basic 3D Graphs
The Constant Equations

Basic 3D Graphs
The graphs of the equations
x = k, y = k, or z = k are planes
that are parallel to the coordinate
planes.

x
z+
a. x = 4 is a plane that’s parallel to the yz-plane.
x = 4
y
Basic 3D Graphs
planes.
Example B.

x
z+
x = 4
b. y = 4 is a plane that’s parallel to the xz-plane.
yy = 4
Basic 3D Graphs
planes.
Example B.

x
z+
x = 4
yy = 4
c. z = 4 is a plane that’s parallel to the xy-plane.
z = 4
Basic 3D Graphs
planes.
Example B.

x
z+
x = 4
yy = 4
c. z = 4 is a plane that’s parallel to the xy-plane.
z = 4
Basic 3D Graphs
planes.
Example B.
d. The graph of (x – 4)(y – 4)(z – 4) = 0 is all three
planes.

The Graphs of Linear Equations

The graphs of linear equations ax + by + cz = d are
planes.

planes. We may use the intercepts to graph the plane.
a, b, c and d are all nonzero:

planes. We may use the intercepts to graph the plane
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.

The three intercepts position the
plane in the coordinate system.

Example C. Sketch 2x – y + 2z = 4.

x
z+
Example C. Sketch 2x – y + 2z = 4
Set x = y = 0  z = 2, we’ve (0, 0, 2),
y

x
z+
Set x = y = 0  z = 2, we’ve (0, 0, 2),
set x = z = 0  y = –4, we’ve (0, –4, 0),
set y = z = 0  x = 2, we’ve (2, 0, 0).
y

x
z+
Set x = y = 0  z = 2, we’ve (0, 0, 2),
set x = z = 0  y = –4, we’ve (0, –4, 0),
set y = z = 0  x = 2, we’ve (2, 0, 0). Plot these
intercepts and the plane containing them is the graph.
y
(0,–4, 0)
(2, 0, 0)
(0, 0, 2)

x
z+
Set x = y = 0  z = 2, we’ve (0, 0, 2),
set x = z = 0  y = –4, we’ve (0, –4, 0),
set y = z = 0  x = 2, we’ve (2, 0, 0). Plot these
intercepts and the plane containing them is the graph.
y
(0, 0, 2)
(0,–4, 0)
(2, 0, 0)

Example D. Sketch y – 2x = 4 in R3.

Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).

Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
We can’t set y = x = 0, so there is no z intercept.

Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
Plot the x and y intercepts, the line y – 2x = 4 in the
xy-plane is part of the graph.

x
z+
Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
The plane y – 2x = 4
y

x
z+
Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
y
(0, 4, 0)
(–2, 0, 0)

x
z+
Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
Because the equation doesn't
have the variable z, therefore the
z coordinate can assume any
value.
y
(0, 4, 0)
(–2, 0, 0)

x
z+
Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
z coordinate assume any value.
Hence all points directly above
and below the line y – 2x = 4,
which form a plane parallel to the
z-axis, is the graph.
y
(0, 4, 0)
(–2, 0, 0)

x
z+
Set x = z = 0  y = 4, we’ve (0, 4, 0),
set y = z = 0  x = –2, we’ve (–2, 0, 0).
z coordinate assume any value.
Hence all points directly above
and below the line y – 2x = 4
which form a plane parallel to the
x-axis, is the graph.
(–2, 0, 0)
(0, 4, 0)
y

General Cylinders (Equation with a missing variable)

In general, if an equation has a missing variable then
the missing variable may assume any value.

Example E. Sketch z = x2

To draw the graph of the equation, draw the 2D
curve in the corresponding coordinate plane first,

Draw the parabola z = x2 in the xz-plane.
z+
x
y

then move the curve in space
parallel to the axis of the missing
variable.
Draw the parabola z = x2 in the xz-plane.
z+
x
y

variable.
Draw the parabola z = x2 in the xz-plane. Extending
this parabola in the y (the missing variable) direction
z+
x
y

variable.
Draw the parabola z = x2 in the xz-plane. Extend this
parabola in the y (the missing variable) direction we
get the (parabolic) cylinder-surface as shown.
x
z+
y

variable. The surface formed is a
general cylinder.
Draw the parabola z = x2 in the xz-plane. Extend this
parabola in the y (the missing variable) direction we
get the (parabolic) cylinder-surface as shown.
x
z+
y

x
z+Equations of Spheres
yThe equation of the sphere with
radius r, centered at (a, b ,c) is
(x – a)2 + (y – b)2 + (z – c)2 = r2.
In particular x2 + y2 + z2 = r2 is the
sphere centered at (0, 0, 0) with
radius r.
(a, b, c)
r
r
(x–a)2+(y–b)2+(z–c)2=r2
x2+y2+z2=r2

x
z+Equations of Spheres
yThe equation of the sphere with
radius r, centered at (a, b ,c) is
(x – a)2 + (y – b)2 + (z – c)2 = r2.
In particular x2 + y2 + z2 = r2 is the
sphere centered at (0, 0, 0) with
radius r.
(a, b, c)
r
r
(x–a)2+(y–b)2+(z–c)2=r2
x2+y2+z2=r2
Equations of Ellipsoids
The graph of the equation
is the ellipsoid
centered at (a, b ,c), with
x–radius = r, y–radius = s, z–radius = t
(x – a)2 (y – b)2 (z – c)2
r2 s2 t2
= 1++
x
z+
y
(a, b, c)
r
s
t

The distance D between (x1, y1, z1), (x2, y2, z2) is
Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
The Distance and the Mid–point Formula in R3

Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
Hence the distance between (2, –1, 1) and (1, –1, 3),
with Δx = 1, Δy = 0, Δz = –2, is D = 1+ 0 + 4 = 5

Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
with Δx = 1, Δy = 0, Δz = –2, is D = 1+ 0 + 4 = 5
The mid–point of
(x1, y1, z1), (x2, y2, z2) is ( x1+ x2 y1+ y2 z1+ z2
2 2 2, ,
).

Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
with Δx = 1, Δy = 0, Δz = –2, is D = 1+ 0 + 4 = 5
The mid–point of
(x1, y1, z1), (x2, y2, z2) is ( x1+ x2 y1+ y2 z1+ z2
2 2 2, ,
).
Example F. Find the equation of the sphere which has
A(2, 1, 3) and B(4, 3, –5) as a diameter.

Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
with Δx = 1, Δy = 0, Δz = –2, is D = 1+ 0 + 4 = 5
The mid–point of
(x1, y1, z1), (x2, y2, z2) is ( x1+ x2 y1+ y2 z1+ z2
2 2 2, ,
).
The center C of the sphere is the mid–point of A and B
so C = (3, 2, –1).

Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
with Δx = 1, Δy = 0, Δz = –2, is D = 1+ 0 + 4 = 5
The mid–point of
(x1, y1, z1), (x2, y2, z2) is ( x1+ x2 y1+ y2 z1+ z2
2 2 2, ,
).
and C = (3, 2, –1). The radius is the distance from
A to C which is 1 + 1 + 16 = 32.

Δx2 + Δy2 + Δz2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
with Δx = 1, Δy = 0, Δz = –2, is D = 1+ 0 + 4 = 5
The mid–point of
(x1, y1, z1), (x2, y2, z2) is ( x1+ x2 y1+ y2 z1+ z2
2 2 2, ,
).
and C = (3, 2, –1). The radius is the distance from
A to C which is 1 + 1 + 16 = 32. So the equation is
(x – 3)2 + (y – 2)2 + (z + 1)2 = 18.

Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

Example G. a. Plot the point (3,120o, 4)
in cylindrical coordinates. Convert it to
rectangular coordinates.

in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
x
y
z

3
120o
4
x
y
z
(3, 120o, 4)

3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 33/2
Hence the point is (–3/2, 33/2, 4)

3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 33/2
b. Convert (3, –3, 1) into
cylindrical coordinates.

3
120o
4
x
y
z
(3, 120o, 4)
 = 315o, r = (9 + 9) = 18
Hence the point is (18, 315o, 1)
in cylindrical coordinates.
x = 3cos(120o) = –3/2
y = 3sin(120o) = 33/2

3
120o
4
x
y
z
(3, 120o, 4)
(18, 315o, 0)
x
y
z
x = 3cos(120o) = –3/2
y = 3sin(120o) = 33/2
 = 315o, r = (9 + 9) = 18

3
120o
4
x
y
z
(3, 120o, 4)
(18, 315o, 0)
x
y
z
(18, 315o, 1) = (3, –3, 1)
1
x = 3cos(120o) = –3/2
y = 3sin(120o) = 33/2
 = 315o, r = (9 + 9) = 18

The constant equations
r = k describe cylinders of
radius k, thus the name
"cylindrical coordinate".

Example H. a. Sketch r = 2
2
x
y
z

x
y
2
 = k describe the vertical
planes through the origin,
at the angle k with x-axis.
b. Sketch θ =3π/4
3π/4
z
x
y
z

x
y
2
 = k describe the vertical
plane through the origin,
at the angle k with x-axis.
b. Sketch θ =3π/4
3π/4
z
x
y
z

267 1 3 d coordinate system-n

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