6 3 d coordinate systems

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.

z+
y

x

Right-hand system

z+
y

x
z+

Right-hand system Left-hand system

ways to add the z-axis. y
z+
y x

x
z+

Right-hand system Left-hand system

z+
y x

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

z+
y x

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

z+
y x

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

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

z+
y

x

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

x

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

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

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

z+
coordinate plane,
2. then the z, which specifies A(2, 0, 0) x
z > 0  above
z < 0  below
B(1, 3, 4), C(–2, 1, –3)

z+
coordinate plane, (1, 3, 0)

z > 0  above
z < 0  below
B(1, 3, 4), C(–2, 1, –3)

three numbers, an ordered triple, (x, y, z). B(1, 3, 4)

z+
location it pinpoints: 4
y
coordinate plane, (1, 3, 0)

z > 0  above
z < 0  below
B(1, 3, 4), C(–2, 1, –3)


z+
y
coordinate plane, (–2, 1, 0)
(1, 3, 0)

z > 0  above
z < 0  below
B(1, 3, 4), C(–2, 1, –3)


z+
y
coordinate plane, (–2, 1, 0)
(1, 3, 0)

z > 0  above C(–2, 1, –3)

z < 0  below
B(1, 3, 4), C(–2, 1, –3)

3D systems may be drawn from different eye
positions:

positions:
z+
y

x

Eye:(1, –1, 1)

Objects appears differently in its direction and its
orientation depending on the eye position.

positions:
z+ z+
y

x y

x
Eye:(1, –1, 1) Eye:(1, 1, 1)


positions:
z+ z+ z+
y x

x y y

x
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)


positions:
z+ z+ z+
y x

x y y

x
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)

There are three coordinate planes: z

y

x

positions:
z+ z+ z+
y x

x y y

x
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)

the xy-plane = {(x, y, 0)}
y

x

xy-plane

positions:
z+ z+ z+
y x

x y y

x
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)

yz-plane
y
the yz-plane = {(0, y, z)}
x

xy-plane

positions:
z+ z+ z+
y x

x y y

x
Eye:(1, –1, 1) Eye:(1, 1, 1) Eye:(1, –1, –1)

yz-plane
There are three coordinate planes: xz-plane z

y
the yz-plane = {(0, y, z)}
the xz-plane = {(x, 0, z)} x

xy-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.

Basic 3D Graphs
x, y, and z in R3 is a surface is 3D space. +
z

that are parallel to the coordinate x = 4 y

planes.
Example B. x

a. x = 4 is a plane that’s parallel to the yz-plane.

Basic 3D Graphs
x, y, and z in R3 is a surface is 3D space. z
+

that are parallel to the coordinate x = 4 y=4 y

planes.
Example B. x

b. y = 4 is a plane that’s parallel to the xz-plane.

Basic 3D Graphs
x, y, and z in R3 is a surface is 3D space. z
+

z=4

planes.
Example B. x

c. z = 4 is a plane that’s parallel to the xy-plane.

Basic 3D Graphs
x, y, and z in R3 is a surface is 3D space. +
z

z=4

planes.
Example B. x

c. z = 4 is a plane that’s parallel to the xy-plane.
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:

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.

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

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

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).

set x = z = 0 to obtain the y intercept, +
z

plane in the coordinate system. y
Set x = y = 0  z = 2, we’ve (0, 0, 2), x
x
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.

set x = z = 0 to obtain the y intercept, z +

(0, 0, 2)
The three intercepts position the (0,–4, 0)
plane in the coordinate system. y
Example C. Sketch 2x – y + 2z = 4 (2, 0, 0)

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

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.

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

y
x

The plane y – 2x = 4

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

(–2, 0, 0)
(0, 4, 0)
y
x


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


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

Because the equation doesn't
have the variable z, therefore the
z coordinate assume any value. (–2, 0, 0)
(0, 4, 0)
Hence all points directly above x y

and below the line y – 2x = 4
which form a plane parallel to the
x-axis, is the graph. The plane y – 2x = 4

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,


z+

y

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

then move the curve in space +
z
parallel to the axis of the missing
variable.
y

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

z
variable.
y

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

z
variable.
y

x
parabola in the y (the missing variable) direction we
get the (parabolic) cylinder-surface as shown.

z
variable. The surface formed is a
general cylinder. y

x
parabola in the y (the missing variable) direction we
get the (parabolic) cylinder-surface as shown.

Equations of Spheres z+
(x–a)2+(y–b)2+(z–c)2=r2
The equation of the sphere with y
radius r, centered at (a, b ,c) is (a, b, c)
r

(x – a)2 + (y – b)2 + (z – c)2 = r2.
In particular x2 + y2 + z2 = r2 is the r x

sphere centered at (0, 0, 0) with x2+y2+z2=r2

radius r.

Equations of Spheres z +
(x–a) +(y–b) +(z–c) =r
2 2 2 2

The equation of the sphere with y
radius r, centered at (a, b ,c) is (a, b, c)
r

(x – a)2 + (y – b)2 + (z – c)2 = r2.
In particular x2 + y2 + z2 = r2 is the r x

sphere centered at (0, 0, 0) with x +y +z =r 2 2 2 2

radius r.
z
Equations of Ellipsoids
+

y
The graph of the equation t
r
(a, b, c)
(x – a)2 (y – b)2 (z – c)2 = 1 s

r2 + s2 + t2
x
is the ellipsoid
centered at (a, b ,c), with
x–radius = r, y–radius = s, z–radius = t

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

√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
and 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 coordinate. Convert it to
rectangular coordinate.

Example G. a. Plot the point (3,120o, 4) z
rectangular coordinate.
3 y
120o
x

(3, 120 , 4)
o

Example G. a. Plot the point (3,120 , 4)
o z

rectangular coordinate. 4

3 y
120o
x

(3, 120 , 4)
o

o z


x = 3cos(120o) = –3/2 3
y = 3sin(120 ) = √3
o y
120 o

Hence the point is (–3/2, √3, 4) x

(3, 120 , 4)
o

o z


x = 3cos(120o) = –3/2 3
y = 3sin(120 ) = √3
o y
120 o


b. Convert (3, –3, 1) into to
cylindrical coordinate.

(3, 120 , 4)
o

o z


x = 3cos(120o) = –3/2 3
y = 3sin(120 ) = √3
o y
120 o


b. Convert (3, –3, 1) into to
θ = 315o, r = √9 + 9 = √18
Hence the point is (√18, 315o,
1) the cylindrical coordinate.

(3, 120 , 4)
o

o z


x = 3cos(120o) = –3/2 3
y = 3sin(120 ) = √3
o y
120 o


b. Convert (3, –3, 1) into to z
θ = 315o, r = √9 + 9 = √18 y

(√18, 315 , 0) o

1) the cylindrical coordinate. x

(3, 120 , 4)
o

Example G. a. Plot the point (3,120 , 4) o z


x = 3cos(120o) = –3/2 3
y = 3sin(120 ) = √3
o y
120 o


b. Convert (3, –3, 1) into to z
cylindrical coordinate. (√18, 315 , 1) = (3, –3, 1)
o

θ = 315o, r = √9 + 9 = √18 1
y

(√18, 315 , 0) o

1) the cylindrical coordinate. x

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

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

Example H. a. Sketch r = 2
z

2
y

x

The constant equations The constant equations
r = k describes cylinders θ = k describes the
of radius k, thus the name vertical plane through the
"cylindrical coordinate". origin, at the angle k with
x-axis.
Example H. a. Sketch r = 2 b. Sketch θ =3π/4
z
z

2 y
y 3π/4

x
x

6 3 d coordinate systems

Recommended

Recommended

More Related Content

What's hot

What's hot (19)

Similar to 6 3 d coordinate systems

Similar to 6 3 d coordinate systems (20)

6 3 d coordinate systems