# The Three-Dimensional Coordinate System 11.1 ```The Three-Dimensional Coordinate
System
11.1
JMerrill, 2010
Solid Analytic Geometry
• The Cartesian plane (rectangular coordinate system) is
determined by 2 perpendicular number line (x- and yaxis) and their point of intersection (the origin).
• To identify a point in space, we need a third dimension.
The geometry of this three-dimensional model is called
solid analytic geometry.
• The 3-D coordinate system is formed by passing a z-axis
perpendicular to both the x- and y-axes at the origin.
Coordinate Planes
Notice we draw the x- and y-axes
in the opposite direction
(x,y,z)
X = directed distance from yzplane to some point P
Y= directed distance from xzplane to some point P
Z= directed distance from xyplane to some point P
So, to plot points you go out, over, up/down
Octants
• The 3-D system can have either a right-handed
or a left-handed orientation.
• We’re only using the right-handed orientation
meaning that the octants (quadrants) are
numbered by rotating counterclockwise around
the positive z-axis.
• There are 8 octants.
Octants
Plotting Points in Space
•
•
•
•
•
Plot the points:
(2,-3,3)
(-2,6,2)
(1,4,0)
(2,2,-3)
Draw a sideways x, then put
a perpendicular line through
the origin.
Formulas
• You can use many of the same formulas that you
already know because right triangles are still formed.
The Distance Formula
• It looks the same in space as it did before
except with a third coordinate:
d 
x 2
 x1    y 2  y1   z 2  z1 
2
2
2
Example
• Find the distance between (1, 0, 2) and
(2, 4,-3)
d 
d 

x 2
 x1    y 2  y1   z 2  z1 
2
2
2
 2  1   4  0     3  2 
2
1  16  25 
2
42
2
Midpoint Formula
• The midpoint formula is
 x1  x 2 y1  y 2 z1  z 2 
,
,


2
2
2


• What is the midpoint if you make a 100 on a test and an
80 on a test?
• So the midpoint is just the average of the x’s, y’s, and
z’s.
Midpoint You Do
• Find the midpoint of the line segment joining
(5, -2, 3) and (0, 4, 4)
7
5
 , 1, 
2
2
Equation of a Sphere
• The equation of a circle is x2 + y2 = r2
• If the center is not at the origin, then the equation is
(x-h)2 + (y-k)2 = r2
• The equation of a sphere whose center is at (h,k,j) with
radius r is (x-h)2 + (y-k)2 + (z–j)2= r2
Finding the Equation of a Sphere
• Find the standard equation of a sphere with center
(2,4,3) and radius 3
• (x-h)2 + (y-k)2 + (z–j)2= r2
• (x-2)2 + (y-4)2 + (z–3)2 = 32
• Does the sphere intersect the plane?
• Yes. The center of the sphere is 3 units above the y-axis
and has a radius of 3. It intersects at (2,4,0).
Finding the Center and Radius of a Sphere
• Find the center and radius of the sphere
given by x2 + y2 + z2 – 2x + 4y – 6z +8 = 0
• This works the same way as it did in 2-D
space. In order to find the center, we must
put the equation into standard form, which
means completing the square.
Finding the Center and Radius of a Sphere
• x2 + y2 + z2 – 2x + 4y – 6z +8 = 0
• (x-1)2 + (y+2)2 + (z-3)2 = 6
• The center is (1,-2,3) and the radius is √6.
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